diff --git a/doc/pub/week39/html/week39-bs.html b/doc/pub/week39/html/week39-bs.html
index fc85670c..673e744e 100644
--- a/doc/pub/week39/html/week39-bs.html
+++ b/doc/pub/week39/html/week39-bs.html
@@ -377,7 +377,27 @@
("Analysis of Hartree-Fock equations and Koopman's theorem",
2,
None,
- 'analysis-of-hartree-fock-equations-and-koopman-s-theorem')]}
+ 'analysis-of-hartree-fock-equations-and-koopman-s-theorem'),
+ ('Hartree-Fock in second quantization and stability of HF '
+ 'solution',
+ 2,
+ None,
+ 'hartree-fock-in-second-quantization-and-stability-of-hf-solution'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ('New operators', 2, None, 'new-operators'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Showing that $|\\tilde{c}\\rangle= |c'\\rangle$",
+ 2,
+ None,
+ 'showing-that-tilde-c-rangle-c-rangle'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem')]}
end of tocinfo -->
@@ -501,6 +521,19 @@
Analysis of Hartree-Fock equations and Koopman's theorem
Analysis of Hartree-Fock equations and Koopman's theorem
Analysis of Hartree-Fock equations and Koopman's theorem
+ Hartree-Fock in second quantization and stability of HF solution
+ Thouless' theorem
+ Thouless' theorem
+ Thouless' theorem
+ Thouless' theorem
+ Thouless' theorem
+ New operators
+ Thouless' theorem
+ Showing that \( |\tilde{c}\rangle= |c'\rangle \)
+ Thouless' theorem
+ Thouless' theorem
+ Thouless' theorem
+ Thouless' theorem
@@ -554,9 +587,9 @@ Week 39, September 23-27, 2
Friday:
Lecture Material: These slides, handwritten notes
Sixth exercise set at https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/Exercises/2024/ExercisesWeek39.pdf
@@ -2148,6 +2181,242 @@
+
+Hartree-Fock in second quantization and stability of HF solution
+
+
We wish now to derive the Hartree-Fock equations using our second-quantized formalism and study the stability of the equations.
+Our ansatz for the ground state of the system is approximated as (this is our representation of a Slater determinant in second quantization)
+
+$$
+|\Phi_0\rangle = |c\rangle = a^{\dagger}_i a^{\dagger}_j \dots a^{\dagger}_l|0\rangle.
+$$
+
+We wish to determine \( \hat{u}^{HF} \) so that
+\( E_0^{HF}= \langle c|\hat{H}| c\rangle \) becomes a local minimum.
+
+
+In our analysis here we will need Thouless' theorem, which states that
+an arbitrary Slater determinant \( |c'\rangle \) which is not orthogonal to a determinant
+\( | c\rangle ={\displaystyle\prod_{i=1}^{n}}
+a_{\alpha_{i}}^{\dagger}|0\rangle \), can be written as
+
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle
+$$
+
+
+
+Thouless' theorem
+
+Let us give a simple proof of Thouless' theorem. The theorem states that we can make a linear combination av particle-hole excitations with respect to a given reference state \( \vert c\rangle \). With this linear combination, we can make a new Slater determinant \( \vert c'\rangle \) which is not orthogonal to
+\( \vert c\rangle \), that is
+
+$$
+\langle c|c'\rangle \ne 0.
+$$
+
+To show this we need some intermediate steps. The exponential product of two operators \( \exp{\hat{A}}\times\exp{\hat{B}} \) is equal to \( \exp{(\hat{A}+\hat{B})} \) only if the two operators commute, that is
+$$
+[\hat{A},\hat{B}] = 0.
+$$
+
+
+
+Thouless' theorem
+
+If the operators do not commute, we need to resort to the Baker-Campbell-Hauersdorf. This relation states that
+$$
+\exp{\hat{C}}=\exp{\hat{A}}\exp{\hat{B}},
+$$
+
+with
+$$
+\hat{C}=\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]+\frac{1}{12}[[\hat{A},\hat{B}],\hat{B}]-\frac{1}{12}[[\hat{A},\hat{B}],\hat{A}]+\dots
+$$
+
+
+
+Thouless' theorem
+
+From these relations, we note that
+in our expression for \( |c'\rangle \) we have commutators of the type
+
+$$
+[a_{a}^{\dagger}a_{i},a_{b}^{\dagger}a_{j}],
+$$
+
+and it is easy to convince oneself that these commutators, or higher powers thereof, are all zero. This means that we can write out our new representation of a Slater determinant as
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}+\left(\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right)^2+\dots\right\}| c\rangle
+$$
+
+
+
+Thouless' theorem
+
+We note that
+$$
+\prod_{i}\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\sum_{b>F}C_{bi}a_{b}^{\dagger}a_{i}| c\rangle =0,
+$$
+
+and all higher-order powers of these combinations of creation and annihilation operators disappear
+due to the fact that \( (a_i)^n| c\rangle =0 \) when \( n > 1 \). This allows us to rewrite the expression for \( |c'\rangle \) as
+
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle,
+$$
+
+which we can rewrite as
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle.
+$$
+
+
+
+Thouless' theorem
+
+The last equation can be written as
+$$
+\begin{align}
+|c'\rangle&=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle=\left(1+\sum_{a>F}C_{ai_1}a_{a}^{\dagger}a_{i_1}\right)a^{\dagger}_{i_1}
+\label{_auto3}\\
+& \times\left(1+\sum_{a>F}C_{ai_2}a_{a}^{\dagger}a_{i_2}\right)a^{\dagger}_{i_2} \dots |0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle.
+\label{_auto4}
+\end{align}
+$$
+
+
+
+New operators
+
+If we define a new creation operator
+$$
+\begin{equation}
+b^{\dagger}_{i}=a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}, \label{eq:newb}
+\end{equation}
+$$
+
+we have
+$$
+|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle,
+$$
+
+meaning that the new representation of the Slater determinant in second quantization, \( |c'\rangle \), looks like our previous ones. However, this representation is not general enough since we have a restriction on the sum over single-particle states in Eq. \eqref{eq:newb}. The single-particle states have all to be above the Fermi level.
+
+
+Thouless' theorem
+
+The question then is whether we can construct a general representation of a Slater determinant with a creation operator
+$$
+\tilde{b}^{\dagger}_{i}=\sum_{p}f_{ip}a_{p}^{\dagger},
+$$
+
+where \( f_{ip} \) is a matrix element of a unitary matrix which transforms our creation and annihilation operators
+\( a^{\dagger} \) and \( a \) to \( \tilde{b}^{\dagger} \) and \( \tilde{b} \). These new operators define a new representation of a Slater determinant as
+
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle.
+$$
+
+
+
+Showing that \( |\tilde{c}\rangle= |c'\rangle \)
+
+We need to show that \( |\tilde{c}\rangle= |c'\rangle \). We need also to assume that the new state
+is not orthogonal to \( |c\rangle \), that is \( \langle c| \tilde{c}\rangle \ne 0 \). From this it follows that
+
+$$
+\langle c| \tilde{c}\rangle=\langle 0| a_{i_n}\dots a_{i_1}\left(\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\dagger} \right)\left(\sum_{q=i_1}^{i_n}f_{i_2q}a_{q}^{\dagger} \right)\dots \left(\sum_{t=i_1}^{i_n}f_{i_nt}a_{t}^{\dagger} \right)|0\rangle,
+$$
+
+which is nothing but the determinant \( det(f_{ip}) \) which we can, using the intermediate normalization condition,
+normalize to one, that is
+
+$$
+det(f_{ip})=1,
+$$
+
+meaning that \( f \) has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations)
+$$
+\sum_{k} f_{ik}f^{-1}_{kj} = \delta_{ij},
+$$
+
+and
+$$
+\sum_{j} f^{-1}_{ij}f_{jk} = \delta_{ik}.
+$$
+
+
+
+Thouless' theorem
+
+Using these relations we can then define the linear combination of creation (and annihilation as well)
+operators as
+
+$$
+\sum_{i}f^{-1}_{ki}\tilde{b}^{\dagger}_{i}=\sum_{i}f^{-1}_{ki}\sum_{p=i_1}^{\infty}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}.
+$$
+
+Defining
+$$
+c_{kp}=\sum_{i \le F}f^{-1}_{ki}f_{ip},
+$$
+
+we can redefine
+$$
+a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}=b_k^{\dagger},
+$$
+
+our starting point.
+
+
+Thouless' theorem
+
+We have shown that our general representation of a Slater determinant
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle=|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle,
+$$
+
+with
+$$
+b_k^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}.
+$$
+
+
+
+Thouless' theorem
+
+This means that we can actually write an ansatz for the ground state of the system as a linear combination of
+terms which contain the ansatz itself \( |c\rangle \) with an admixture from an infinity of one-particle-one-hole states. The latter has important consequences when we wish to interpret the Hartree-Fock equations and their stability. We can rewrite the new representation as
+
+$$
+|c'\rangle = |c\rangle+|\delta c\rangle,
+$$
+
+where \( |\delta c\rangle \) can now be interpreted as a small variation. If we approximate this term with
+contributions from one-particle-one-hole (1p-1h) states only, we arrive at
+
+$$
+|c'\rangle = \left(1+\sum_{ai}\delta C_{ai}a_{a}^{\dagger}a_i\right)|c\rangle.
+$$
+
+
+
+Thouless' theorem
+
+In our derivation of the Hartree-Fock equations we have shown that
+$$
+\langle \delta c| \hat{H} | c\rangle =0,
+$$
+
+which means that we have to satisfy
+$$
+\langle c|\sum_{ai}\delta C_{ai}\left\{a_{a}^{\dagger}a_i\right\} \hat{H} | c\rangle =0.
+$$
+
+With this as a background, we are now ready to study the stability of the Hartree-Fock equations.
+This is the topic for week 40.
+
diff --git a/doc/pub/week39/html/week39-reveal.html b/doc/pub/week39/html/week39-reveal.html
index 7d370438..76c5e3f7 100644
--- a/doc/pub/week39/html/week39-reveal.html
+++ b/doc/pub/week39/html/week39-reveal.html
@@ -221,9 +221,11 @@ Week 39, September 23-27, 2024
Friday:
Lecture Material: These slides, handwritten notes
@@ -2117,6 +2119,304 @@ Analysis of Ha
+
+Hartree-Fock in second quantization and stability of HF solution
+
+We wish now to derive the Hartree-Fock equations using our second-quantized formalism and study the stability of the equations.
+Our ansatz for the ground state of the system is approximated as (this is our representation of a Slater determinant in second quantization)
+
+
+$$
+|\Phi_0\rangle = |c\rangle = a^{\dagger}_i a^{\dagger}_j \dots a^{\dagger}_l|0\rangle.
+$$
+
+
+
We wish to determine \( \hat{u}^{HF} \) so that
+\( E_0^{HF}= \langle c|\hat{H}| c\rangle \) becomes a local minimum.
+
+
+In our analysis here we will need Thouless' theorem, which states that
+an arbitrary Slater determinant \( |c'\rangle \) which is not orthogonal to a determinant
+\( | c\rangle ={\displaystyle\prod_{i=1}^{n}}
+a_{\alpha_{i}}^{\dagger}|0\rangle \), can be written as
+
+
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle
+$$
+
+
+
+
+Thouless' theorem
+
+Let us give a simple proof of Thouless' theorem. The theorem states that we can make a linear combination av particle-hole excitations with respect to a given reference state \( \vert c\rangle \). With this linear combination, we can make a new Slater determinant \( \vert c'\rangle \) which is not orthogonal to
+\( \vert c\rangle \), that is
+
+
+$$
+\langle c|c'\rangle \ne 0.
+$$
+
+
+
To show this we need some intermediate steps. The exponential product of two operators \( \exp{\hat{A}}\times\exp{\hat{B}} \) is equal to \( \exp{(\hat{A}+\hat{B})} \) only if the two operators commute, that is
+
+$$
+[\hat{A},\hat{B}] = 0.
+$$
+
+
+
+
+Thouless' theorem
+
+If the operators do not commute, we need to resort to the Baker-Campbell-Hauersdorf. This relation states that
+
+$$
+\exp{\hat{C}}=\exp{\hat{A}}\exp{\hat{B}},
+$$
+
+
+
with
+
+$$
+\hat{C}=\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]+\frac{1}{12}[[\hat{A},\hat{B}],\hat{B}]-\frac{1}{12}[[\hat{A},\hat{B}],\hat{A}]+\dots
+$$
+
+
+
+
+Thouless' theorem
+
+From these relations, we note that
+in our expression for \( |c'\rangle \) we have commutators of the type
+
+
+$$
+[a_{a}^{\dagger}a_{i},a_{b}^{\dagger}a_{j}],
+$$
+
+
+
and it is easy to convince oneself that these commutators, or higher powers thereof, are all zero. This means that we can write out our new representation of a Slater determinant as
+
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}+\left(\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right)^2+\dots\right\}| c\rangle
+$$
+
+
+
+
+Thouless' theorem
+
+We note that
+
+$$
+\prod_{i}\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\sum_{b>F}C_{bi}a_{b}^{\dagger}a_{i}| c\rangle =0,
+$$
+
+
+
and all higher-order powers of these combinations of creation and annihilation operators disappear
+due to the fact that \( (a_i)^n| c\rangle =0 \) when \( n > 1 \). This allows us to rewrite the expression for \( |c'\rangle \) as
+
+
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle,
+$$
+
+
+
which we can rewrite as
+
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle.
+$$
+
+
+
+
+Thouless' theorem
+
+The last equation can be written as
+
+$$
+\begin{align}
+|c'\rangle&=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle=\left(1+\sum_{a>F}C_{ai_1}a_{a}^{\dagger}a_{i_1}\right)a^{\dagger}_{i_1}
+\tag{10}\\
+& \times\left(1+\sum_{a>F}C_{ai_2}a_{a}^{\dagger}a_{i_2}\right)a^{\dagger}_{i_2} \dots |0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle.
+\tag{11}
+\end{align}
+$$
+
+
+
+
+New operators
+
+If we define a new creation operator
+
+$$
+\begin{equation}
+b^{\dagger}_{i}=a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}, \tag{12}
+\end{equation}
+$$
+
+
+
we have
+
+$$
+|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle,
+$$
+
+
+
meaning that the new representation of the Slater determinant in second quantization, \( |c'\rangle \), looks like our previous ones. However, this representation is not general enough since we have a restriction on the sum over single-particle states in Eq. (12). The single-particle states have all to be above the Fermi level.
+
+
+
+Thouless' theorem
+
+The question then is whether we can construct a general representation of a Slater determinant with a creation operator
+
+$$
+\tilde{b}^{\dagger}_{i}=\sum_{p}f_{ip}a_{p}^{\dagger},
+$$
+
+
+
where \( f_{ip} \) is a matrix element of a unitary matrix which transforms our creation and annihilation operators
+\( a^{\dagger} \) and \( a \) to \( \tilde{b}^{\dagger} \) and \( \tilde{b} \). These new operators define a new representation of a Slater determinant as
+
+
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle.
+$$
+
+
+
+
+Showing that \( |\tilde{c}\rangle= |c'\rangle \)
+
+We need to show that \( |\tilde{c}\rangle= |c'\rangle \). We need also to assume that the new state
+is not orthogonal to \( |c\rangle \), that is \( \langle c| \tilde{c}\rangle \ne 0 \). From this it follows that
+
+
+$$
+\langle c| \tilde{c}\rangle=\langle 0| a_{i_n}\dots a_{i_1}\left(\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\dagger} \right)\left(\sum_{q=i_1}^{i_n}f_{i_2q}a_{q}^{\dagger} \right)\dots \left(\sum_{t=i_1}^{i_n}f_{i_nt}a_{t}^{\dagger} \right)|0\rangle,
+$$
+
+
+
which is nothing but the determinant \( det(f_{ip}) \) which we can, using the intermediate normalization condition,
+normalize to one, that is
+
+
+$$
+det(f_{ip})=1,
+$$
+
+
+
meaning that \( f \) has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations)
+
+$$
+\sum_{k} f_{ik}f^{-1}_{kj} = \delta_{ij},
+$$
+
+
+
and
+
+$$
+\sum_{j} f^{-1}_{ij}f_{jk} = \delta_{ik}.
+$$
+
+
+
+
+Thouless' theorem
+
+Using these relations we can then define the linear combination of creation (and annihilation as well)
+operators as
+
+
+$$
+\sum_{i}f^{-1}_{ki}\tilde{b}^{\dagger}_{i}=\sum_{i}f^{-1}_{ki}\sum_{p=i_1}^{\infty}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}.
+$$
+
+
+
Defining
+
+$$
+c_{kp}=\sum_{i \le F}f^{-1}_{ki}f_{ip},
+$$
+
+
+
we can redefine
+
+$$
+a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}=b_k^{\dagger},
+$$
+
+
+
our starting point.
+
+
+
+Thouless' theorem
+
+We have shown that our general representation of a Slater determinant
+
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle=|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle,
+$$
+
+
+
with
+
+$$
+b_k^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}.
+$$
+
+
+
+
+Thouless' theorem
+
+This means that we can actually write an ansatz for the ground state of the system as a linear combination of
+terms which contain the ansatz itself \( |c\rangle \) with an admixture from an infinity of one-particle-one-hole states. The latter has important consequences when we wish to interpret the Hartree-Fock equations and their stability. We can rewrite the new representation as
+
+
+$$
+|c'\rangle = |c\rangle+|\delta c\rangle,
+$$
+
+
+
where \( |\delta c\rangle \) can now be interpreted as a small variation. If we approximate this term with
+contributions from one-particle-one-hole (1p-1h) states only, we arrive at
+
+
+$$
+|c'\rangle = \left(1+\sum_{ai}\delta C_{ai}a_{a}^{\dagger}a_i\right)|c\rangle.
+$$
+
+
+
+
+Thouless' theorem
+
+In our derivation of the Hartree-Fock equations we have shown that
+
+$$
+\langle \delta c| \hat{H} | c\rangle =0,
+$$
+
+
+
which means that we have to satisfy
+
+$$
+\langle c|\sum_{ai}\delta C_{ai}\left\{a_{a}^{\dagger}a_i\right\} \hat{H} | c\rangle =0.
+$$
+
+
+
With this as a background, we are now ready to study the stability of the Hartree-Fock equations.
+This is the topic for week 40.
+
+
+
diff --git a/doc/pub/week39/html/week39-solarized.html b/doc/pub/week39/html/week39-solarized.html
index f59dd4de..8da60c25 100644
--- a/doc/pub/week39/html/week39-solarized.html
+++ b/doc/pub/week39/html/week39-solarized.html
@@ -404,7 +404,27 @@
("Analysis of Hartree-Fock equations and Koopman's theorem",
2,
None,
- 'analysis-of-hartree-fock-equations-and-koopman-s-theorem')]}
+ 'analysis-of-hartree-fock-equations-and-koopman-s-theorem'),
+ ('Hartree-Fock in second quantization and stability of HF '
+ 'solution',
+ 2,
+ None,
+ 'hartree-fock-in-second-quantization-and-stability-of-hf-solution'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ('New operators', 2, None, 'new-operators'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Showing that $|\\tilde{c}\\rangle= |c'\\rangle$",
+ 2,
+ None,
+ 'showing-that-tilde-c-rangle-c-rangle'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem')]}
end of tocinfo -->
@@ -464,9 +484,9 @@ Week 39, September 23-27, 2024
Friday:
Lecture Material: These slides, handwritten notes
Sixth exercise set at https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/Exercises/2024/ExercisesWeek39.pdf
@@ -2039,6 +2059,242 @@ Analysis of Ha
+
+Hartree-Fock in second quantization and stability of HF solution
+
+
We wish now to derive the Hartree-Fock equations using our second-quantized formalism and study the stability of the equations.
+Our ansatz for the ground state of the system is approximated as (this is our representation of a Slater determinant in second quantization)
+
+$$
+|\Phi_0\rangle = |c\rangle = a^{\dagger}_i a^{\dagger}_j \dots a^{\dagger}_l|0\rangle.
+$$
+
+We wish to determine \( \hat{u}^{HF} \) so that
+\( E_0^{HF}= \langle c|\hat{H}| c\rangle \) becomes a local minimum.
+
+
+In our analysis here we will need Thouless' theorem, which states that
+an arbitrary Slater determinant \( |c'\rangle \) which is not orthogonal to a determinant
+\( | c\rangle ={\displaystyle\prod_{i=1}^{n}}
+a_{\alpha_{i}}^{\dagger}|0\rangle \), can be written as
+
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle
+$$
+
+
+
+Thouless' theorem
+
+Let us give a simple proof of Thouless' theorem. The theorem states that we can make a linear combination av particle-hole excitations with respect to a given reference state \( \vert c\rangle \). With this linear combination, we can make a new Slater determinant \( \vert c'\rangle \) which is not orthogonal to
+\( \vert c\rangle \), that is
+
+$$
+\langle c|c'\rangle \ne 0.
+$$
+
+To show this we need some intermediate steps. The exponential product of two operators \( \exp{\hat{A}}\times\exp{\hat{B}} \) is equal to \( \exp{(\hat{A}+\hat{B})} \) only if the two operators commute, that is
+$$
+[\hat{A},\hat{B}] = 0.
+$$
+
+
+
+Thouless' theorem
+
+If the operators do not commute, we need to resort to the Baker-Campbell-Hauersdorf. This relation states that
+$$
+\exp{\hat{C}}=\exp{\hat{A}}\exp{\hat{B}},
+$$
+
+with
+$$
+\hat{C}=\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]+\frac{1}{12}[[\hat{A},\hat{B}],\hat{B}]-\frac{1}{12}[[\hat{A},\hat{B}],\hat{A}]+\dots
+$$
+
+
+
+Thouless' theorem
+
+From these relations, we note that
+in our expression for \( |c'\rangle \) we have commutators of the type
+
+$$
+[a_{a}^{\dagger}a_{i},a_{b}^{\dagger}a_{j}],
+$$
+
+and it is easy to convince oneself that these commutators, or higher powers thereof, are all zero. This means that we can write out our new representation of a Slater determinant as
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}+\left(\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right)^2+\dots\right\}| c\rangle
+$$
+
+
+
+Thouless' theorem
+
+We note that
+$$
+\prod_{i}\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\sum_{b>F}C_{bi}a_{b}^{\dagger}a_{i}| c\rangle =0,
+$$
+
+and all higher-order powers of these combinations of creation and annihilation operators disappear
+due to the fact that \( (a_i)^n| c\rangle =0 \) when \( n > 1 \). This allows us to rewrite the expression for \( |c'\rangle \) as
+
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle,
+$$
+
+which we can rewrite as
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle.
+$$
+
+
+
+Thouless' theorem
+
+The last equation can be written as
+$$
+\begin{align}
+|c'\rangle&=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle=\left(1+\sum_{a>F}C_{ai_1}a_{a}^{\dagger}a_{i_1}\right)a^{\dagger}_{i_1}
+\label{_auto3}\\
+& \times\left(1+\sum_{a>F}C_{ai_2}a_{a}^{\dagger}a_{i_2}\right)a^{\dagger}_{i_2} \dots |0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle.
+\label{_auto4}
+\end{align}
+$$
+
+
+
+New operators
+
+If we define a new creation operator
+$$
+\begin{equation}
+b^{\dagger}_{i}=a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}, \label{eq:newb}
+\end{equation}
+$$
+
+we have
+$$
+|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle,
+$$
+
+meaning that the new representation of the Slater determinant in second quantization, \( |c'\rangle \), looks like our previous ones. However, this representation is not general enough since we have a restriction on the sum over single-particle states in Eq. \eqref{eq:newb}. The single-particle states have all to be above the Fermi level.
+
+
+Thouless' theorem
+
+The question then is whether we can construct a general representation of a Slater determinant with a creation operator
+$$
+\tilde{b}^{\dagger}_{i}=\sum_{p}f_{ip}a_{p}^{\dagger},
+$$
+
+where \( f_{ip} \) is a matrix element of a unitary matrix which transforms our creation and annihilation operators
+\( a^{\dagger} \) and \( a \) to \( \tilde{b}^{\dagger} \) and \( \tilde{b} \). These new operators define a new representation of a Slater determinant as
+
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle.
+$$
+
+
+
+Showing that \( |\tilde{c}\rangle= |c'\rangle \)
+
+We need to show that \( |\tilde{c}\rangle= |c'\rangle \). We need also to assume that the new state
+is not orthogonal to \( |c\rangle \), that is \( \langle c| \tilde{c}\rangle \ne 0 \). From this it follows that
+
+$$
+\langle c| \tilde{c}\rangle=\langle 0| a_{i_n}\dots a_{i_1}\left(\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\dagger} \right)\left(\sum_{q=i_1}^{i_n}f_{i_2q}a_{q}^{\dagger} \right)\dots \left(\sum_{t=i_1}^{i_n}f_{i_nt}a_{t}^{\dagger} \right)|0\rangle,
+$$
+
+which is nothing but the determinant \( det(f_{ip}) \) which we can, using the intermediate normalization condition,
+normalize to one, that is
+
+$$
+det(f_{ip})=1,
+$$
+
+meaning that \( f \) has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations)
+$$
+\sum_{k} f_{ik}f^{-1}_{kj} = \delta_{ij},
+$$
+
+and
+$$
+\sum_{j} f^{-1}_{ij}f_{jk} = \delta_{ik}.
+$$
+
+
+
+Thouless' theorem
+
+Using these relations we can then define the linear combination of creation (and annihilation as well)
+operators as
+
+$$
+\sum_{i}f^{-1}_{ki}\tilde{b}^{\dagger}_{i}=\sum_{i}f^{-1}_{ki}\sum_{p=i_1}^{\infty}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}.
+$$
+
+Defining
+$$
+c_{kp}=\sum_{i \le F}f^{-1}_{ki}f_{ip},
+$$
+
+we can redefine
+$$
+a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}=b_k^{\dagger},
+$$
+
+our starting point.
+
+
+Thouless' theorem
+
+We have shown that our general representation of a Slater determinant
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle=|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle,
+$$
+
+with
+$$
+b_k^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}.
+$$
+
+
+
+Thouless' theorem
+
+This means that we can actually write an ansatz for the ground state of the system as a linear combination of
+terms which contain the ansatz itself \( |c\rangle \) with an admixture from an infinity of one-particle-one-hole states. The latter has important consequences when we wish to interpret the Hartree-Fock equations and their stability. We can rewrite the new representation as
+
+$$
+|c'\rangle = |c\rangle+|\delta c\rangle,
+$$
+
+where \( |\delta c\rangle \) can now be interpreted as a small variation. If we approximate this term with
+contributions from one-particle-one-hole (1p-1h) states only, we arrive at
+
+$$
+|c'\rangle = \left(1+\sum_{ai}\delta C_{ai}a_{a}^{\dagger}a_i\right)|c\rangle.
+$$
+
+
+
+Thouless' theorem
+
+In our derivation of the Hartree-Fock equations we have shown that
+$$
+\langle \delta c| \hat{H} | c\rangle =0,
+$$
+
+which means that we have to satisfy
+$$
+\langle c|\sum_{ai}\delta C_{ai}\left\{a_{a}^{\dagger}a_i\right\} \hat{H} | c\rangle =0.
+$$
+
+With this as a background, we are now ready to study the stability of the Hartree-Fock equations.
+This is the topic for week 40.
+
© 1999-2024, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license
diff --git a/doc/pub/week39/html/week39.html b/doc/pub/week39/html/week39.html
index 347eafab..1e4848f9 100644
--- a/doc/pub/week39/html/week39.html
+++ b/doc/pub/week39/html/week39.html
@@ -481,7 +481,27 @@
("Analysis of Hartree-Fock equations and Koopman's theorem",
2,
None,
- 'analysis-of-hartree-fock-equations-and-koopman-s-theorem')]}
+ 'analysis-of-hartree-fock-equations-and-koopman-s-theorem'),
+ ('Hartree-Fock in second quantization and stability of HF '
+ 'solution',
+ 2,
+ None,
+ 'hartree-fock-in-second-quantization-and-stability-of-hf-solution'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ('New operators', 2, None, 'new-operators'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Showing that $|\\tilde{c}\\rangle= |c'\\rangle$",
+ 2,
+ None,
+ 'showing-that-tilde-c-rangle-c-rangle'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem'),
+ ("Thouless' theorem", 2, None, 'thouless-theorem')]}
end of tocinfo -->
@@ -541,9 +561,9 @@ Week 39, September 23-27, 2024
Friday:
Lecture Material: These slides, handwritten notes
Sixth exercise set at https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/Exercises/2024/ExercisesWeek39.pdf
@@ -2116,6 +2136,242 @@ Analysis of Ha
+
+Hartree-Fock in second quantization and stability of HF solution
+
+
We wish now to derive the Hartree-Fock equations using our second-quantized formalism and study the stability of the equations.
+Our ansatz for the ground state of the system is approximated as (this is our representation of a Slater determinant in second quantization)
+
+$$
+|\Phi_0\rangle = |c\rangle = a^{\dagger}_i a^{\dagger}_j \dots a^{\dagger}_l|0\rangle.
+$$
+
+We wish to determine \( \hat{u}^{HF} \) so that
+\( E_0^{HF}= \langle c|\hat{H}| c\rangle \) becomes a local minimum.
+
+
+In our analysis here we will need Thouless' theorem, which states that
+an arbitrary Slater determinant \( |c'\rangle \) which is not orthogonal to a determinant
+\( | c\rangle ={\displaystyle\prod_{i=1}^{n}}
+a_{\alpha_{i}}^{\dagger}|0\rangle \), can be written as
+
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle
+$$
+
+
+
+Thouless' theorem
+
+Let us give a simple proof of Thouless' theorem. The theorem states that we can make a linear combination av particle-hole excitations with respect to a given reference state \( \vert c\rangle \). With this linear combination, we can make a new Slater determinant \( \vert c'\rangle \) which is not orthogonal to
+\( \vert c\rangle \), that is
+
+$$
+\langle c|c'\rangle \ne 0.
+$$
+
+To show this we need some intermediate steps. The exponential product of two operators \( \exp{\hat{A}}\times\exp{\hat{B}} \) is equal to \( \exp{(\hat{A}+\hat{B})} \) only if the two operators commute, that is
+$$
+[\hat{A},\hat{B}] = 0.
+$$
+
+
+
+Thouless' theorem
+
+If the operators do not commute, we need to resort to the Baker-Campbell-Hauersdorf. This relation states that
+$$
+\exp{\hat{C}}=\exp{\hat{A}}\exp{\hat{B}},
+$$
+
+with
+$$
+\hat{C}=\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]+\frac{1}{12}[[\hat{A},\hat{B}],\hat{B}]-\frac{1}{12}[[\hat{A},\hat{B}],\hat{A}]+\dots
+$$
+
+
+
+Thouless' theorem
+
+From these relations, we note that
+in our expression for \( |c'\rangle \) we have commutators of the type
+
+$$
+[a_{a}^{\dagger}a_{i},a_{b}^{\dagger}a_{j}],
+$$
+
+and it is easy to convince oneself that these commutators, or higher powers thereof, are all zero. This means that we can write out our new representation of a Slater determinant as
+$$
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}+\left(\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right)^2+\dots\right\}| c\rangle
+$$
+
+
+
+Thouless' theorem
+
+We note that
+$$
+\prod_{i}\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\sum_{b>F}C_{bi}a_{b}^{\dagger}a_{i}| c\rangle =0,
+$$
+
+and all higher-order powers of these combinations of creation and annihilation operators disappear
+due to the fact that \( (a_i)^n| c\rangle =0 \) when \( n > 1 \). This allows us to rewrite the expression for \( |c'\rangle \) as
+
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle,
+$$
+
+which we can rewrite as
+$$
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle.
+$$
+
+
+
+Thouless' theorem
+
+The last equation can be written as
+$$
+\begin{align}
+|c'\rangle&=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle=\left(1+\sum_{a>F}C_{ai_1}a_{a}^{\dagger}a_{i_1}\right)a^{\dagger}_{i_1}
+\label{_auto3}\\
+& \times\left(1+\sum_{a>F}C_{ai_2}a_{a}^{\dagger}a_{i_2}\right)a^{\dagger}_{i_2} \dots |0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle.
+\label{_auto4}
+\end{align}
+$$
+
+
+
+New operators
+
+If we define a new creation operator
+$$
+\begin{equation}
+b^{\dagger}_{i}=a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}, \label{eq:newb}
+\end{equation}
+$$
+
+we have
+$$
+|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle,
+$$
+
+meaning that the new representation of the Slater determinant in second quantization, \( |c'\rangle \), looks like our previous ones. However, this representation is not general enough since we have a restriction on the sum over single-particle states in Eq. \eqref{eq:newb}. The single-particle states have all to be above the Fermi level.
+
+
+Thouless' theorem
+
+The question then is whether we can construct a general representation of a Slater determinant with a creation operator
+$$
+\tilde{b}^{\dagger}_{i}=\sum_{p}f_{ip}a_{p}^{\dagger},
+$$
+
+where \( f_{ip} \) is a matrix element of a unitary matrix which transforms our creation and annihilation operators
+\( a^{\dagger} \) and \( a \) to \( \tilde{b}^{\dagger} \) and \( \tilde{b} \). These new operators define a new representation of a Slater determinant as
+
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle.
+$$
+
+
+
+Showing that \( |\tilde{c}\rangle= |c'\rangle \)
+
+We need to show that \( |\tilde{c}\rangle= |c'\rangle \). We need also to assume that the new state
+is not orthogonal to \( |c\rangle \), that is \( \langle c| \tilde{c}\rangle \ne 0 \). From this it follows that
+
+$$
+\langle c| \tilde{c}\rangle=\langle 0| a_{i_n}\dots a_{i_1}\left(\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\dagger} \right)\left(\sum_{q=i_1}^{i_n}f_{i_2q}a_{q}^{\dagger} \right)\dots \left(\sum_{t=i_1}^{i_n}f_{i_nt}a_{t}^{\dagger} \right)|0\rangle,
+$$
+
+which is nothing but the determinant \( det(f_{ip}) \) which we can, using the intermediate normalization condition,
+normalize to one, that is
+
+$$
+det(f_{ip})=1,
+$$
+
+meaning that \( f \) has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations)
+$$
+\sum_{k} f_{ik}f^{-1}_{kj} = \delta_{ij},
+$$
+
+and
+$$
+\sum_{j} f^{-1}_{ij}f_{jk} = \delta_{ik}.
+$$
+
+
+
+Thouless' theorem
+
+Using these relations we can then define the linear combination of creation (and annihilation as well)
+operators as
+
+$$
+\sum_{i}f^{-1}_{ki}\tilde{b}^{\dagger}_{i}=\sum_{i}f^{-1}_{ki}\sum_{p=i_1}^{\infty}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}.
+$$
+
+Defining
+$$
+c_{kp}=\sum_{i \le F}f^{-1}_{ki}f_{ip},
+$$
+
+we can redefine
+$$
+a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}=b_k^{\dagger},
+$$
+
+our starting point.
+
+
+Thouless' theorem
+
+We have shown that our general representation of a Slater determinant
+$$
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle=|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle,
+$$
+
+with
+$$
+b_k^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}.
+$$
+
+
+
+Thouless' theorem
+
+This means that we can actually write an ansatz for the ground state of the system as a linear combination of
+terms which contain the ansatz itself \( |c\rangle \) with an admixture from an infinity of one-particle-one-hole states. The latter has important consequences when we wish to interpret the Hartree-Fock equations and their stability. We can rewrite the new representation as
+
+$$
+|c'\rangle = |c\rangle+|\delta c\rangle,
+$$
+
+where \( |\delta c\rangle \) can now be interpreted as a small variation. If we approximate this term with
+contributions from one-particle-one-hole (1p-1h) states only, we arrive at
+
+$$
+|c'\rangle = \left(1+\sum_{ai}\delta C_{ai}a_{a}^{\dagger}a_i\right)|c\rangle.
+$$
+
+
+
+Thouless' theorem
+
+In our derivation of the Hartree-Fock equations we have shown that
+$$
+\langle \delta c| \hat{H} | c\rangle =0,
+$$
+
+which means that we have to satisfy
+$$
+\langle c|\sum_{ai}\delta C_{ai}\left\{a_{a}^{\dagger}a_i\right\} \hat{H} | c\rangle =0.
+$$
+
+With this as a background, we are now ready to study the stability of the Hartree-Fock equations.
+This is the topic for week 40.
+
© 1999-2024, Morten Hjorth-Jensen. Released under CC Attribution-NonCommercial 4.0 license
diff --git a/doc/pub/week39/ipynb/ipynb-week39-src.tar.gz b/doc/pub/week39/ipynb/ipynb-week39-src.tar.gz
index 718945e4..75656866 100644
Binary files a/doc/pub/week39/ipynb/ipynb-week39-src.tar.gz and b/doc/pub/week39/ipynb/ipynb-week39-src.tar.gz differ
diff --git a/doc/pub/week39/ipynb/week39.ipynb b/doc/pub/week39/ipynb/week39.ipynb
index f0f6c1bd..4adbcf0b 100644
--- a/doc/pub/week39/ipynb/week39.ipynb
+++ b/doc/pub/week39/ipynb/week39.ipynb
@@ -2,7 +2,7 @@
"cells": [
{
"cell_type": "markdown",
- "id": "c928c76a",
+ "id": "8f006195",
"metadata": {
"editable": true
},
@@ -14,7 +14,7 @@
},
{
"cell_type": "markdown",
- "id": "20201ff0",
+ "id": "9d2c4704",
"metadata": {
"editable": true
},
@@ -27,7 +27,7 @@
},
{
"cell_type": "markdown",
- "id": "8e80f893",
+ "id": "114a1939",
"metadata": {
"editable": true
},
@@ -53,8 +53,10 @@
"2. Friday: \n",
"\n",
" * Hartree-Fock theory and mean field theories\n",
- "\n",
- "\n",
+ "\n",
+ " * [Video of lecture](https://youtu.be/)\n",
+ "\n",
+ " * [Whiteboard notes](https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/HandwrittenNotes/2024/NotesSeptember27.pdf)\n",
"\n",
"* Lecture Material: These slides, handwritten notes\n",
"\n",
@@ -63,7 +65,7 @@
},
{
"cell_type": "markdown",
- "id": "6a3183c7",
+ "id": "ec58e0fd",
"metadata": {
"editable": true
},
@@ -75,7 +77,7 @@
},
{
"cell_type": "markdown",
- "id": "9793f395",
+ "id": "40da8b3e",
"metadata": {
"editable": true
},
@@ -87,7 +89,7 @@
},
{
"cell_type": "markdown",
- "id": "ced9e5f0",
+ "id": "93bad7cf",
"metadata": {
"editable": true
},
@@ -97,7 +99,7 @@
},
{
"cell_type": "markdown",
- "id": "37d6b566",
+ "id": "2775fc77",
"metadata": {
"editable": true
},
@@ -109,7 +111,7 @@
},
{
"cell_type": "markdown",
- "id": "c122416f",
+ "id": "ada9791e",
"metadata": {
"editable": true
},
@@ -121,7 +123,7 @@
},
{
"cell_type": "markdown",
- "id": "6e70dcfa",
+ "id": "f45448d4",
"metadata": {
"editable": true
},
@@ -131,7 +133,7 @@
},
{
"cell_type": "markdown",
- "id": "ca4d30cc",
+ "id": "caa9d822",
"metadata": {
"editable": true
},
@@ -143,7 +145,7 @@
},
{
"cell_type": "markdown",
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@@ -153,7 +155,7 @@
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- "id": "f9882d06",
+ "id": "1fdae7ec",
"metadata": {
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@@ -3044,7 +3046,7 @@
},
{
"cell_type": "markdown",
- "id": "e5c98a15",
+ "id": "fe00dbaa",
"metadata": {
"editable": true
},
@@ -3054,7 +3056,7 @@
},
{
"cell_type": "markdown",
- "id": "46255f45",
+ "id": "f804d9b2",
"metadata": {
"editable": true
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@@ -3066,7 +3068,7 @@
},
{
"cell_type": "markdown",
- "id": "6c005ff5",
+ "id": "3df98d7c",
"metadata": {
"editable": true
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@@ -3079,7 +3081,7 @@
},
{
"cell_type": "markdown",
- "id": "bf5ef400",
+ "id": "9dfe067a",
"metadata": {
"editable": true
},
@@ -3089,7 +3091,7 @@
},
{
"cell_type": "markdown",
- "id": "23fc329a",
+ "id": "6d157727",
"metadata": {
"editable": true
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@@ -3106,7 +3108,7 @@
},
{
"cell_type": "markdown",
- "id": "d05173d7",
+ "id": "5b90a385",
"metadata": {
"editable": true
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@@ -3118,7 +3120,7 @@
},
{
"cell_type": "markdown",
- "id": "ee8d3e7c",
+ "id": "8f64d462",
"metadata": {
"editable": true
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@@ -3130,7 +3132,7 @@
},
{
"cell_type": "markdown",
- "id": "01226122",
+ "id": "04aa6709",
"metadata": {
"editable": true
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@@ -3143,7 +3145,7 @@
},
{
"cell_type": "markdown",
- "id": "3d4ce626",
+ "id": "3c914f8f",
"metadata": {
"editable": true
},
@@ -3154,7 +3156,7 @@
},
{
"cell_type": "markdown",
- "id": "a573723a",
+ "id": "2610257e",
"metadata": {
"editable": true
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@@ -3167,7 +3169,7 @@
},
{
"cell_type": "markdown",
- "id": "c131f166",
+ "id": "e99f983e",
"metadata": {
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@@ -3180,7 +3182,7 @@
},
{
"cell_type": "markdown",
- "id": "69e909d6",
+ "id": "5159c497",
"metadata": {
"editable": true
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@@ -3190,7 +3192,7 @@
},
{
"cell_type": "markdown",
- "id": "119ae768",
+ "id": "d08e2502",
"metadata": {
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@@ -3202,7 +3204,7 @@
},
{
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- "id": "67c53390",
+ "id": "ccdef2e3",
"metadata": {
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@@ -3212,7 +3214,7 @@
},
{
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+ "id": "0e416aec",
"metadata": {
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@@ -3224,7 +3226,7 @@
},
{
"cell_type": "markdown",
- "id": "18fe1079",
+ "id": "19ddf277",
"metadata": {
"editable": true
},
@@ -3244,7 +3246,7 @@
},
{
"cell_type": "markdown",
- "id": "8c42d663",
+ "id": "2b09f31b",
"metadata": {
"editable": true
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@@ -3257,7 +3259,7 @@
},
{
"cell_type": "markdown",
- "id": "39fba9e8",
+ "id": "81017739",
"metadata": {
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@@ -3271,7 +3273,7 @@
},
{
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+ "id": "f811cba2",
"metadata": {
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@@ -3283,7 +3285,7 @@
},
{
"cell_type": "markdown",
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+ "id": "6e8b5a4f",
"metadata": {
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@@ -3294,7 +3296,7 @@
},
{
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+ "id": "9b2835a5",
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@@ -3306,7 +3308,7 @@
},
{
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+ "id": "281ce39d",
"metadata": {
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@@ -3317,7 +3319,7 @@
},
{
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+ "id": "ceba8cdf",
"metadata": {
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@@ -3330,7 +3332,7 @@
},
{
"cell_type": "markdown",
- "id": "4c51a4be",
+ "id": "bad761cb",
"metadata": {
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@@ -3348,7 +3350,7 @@
},
{
"cell_type": "markdown",
- "id": "f5d9b0ae",
+ "id": "9ff690ff",
"metadata": {
"editable": true
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@@ -3358,7 +3360,7 @@
},
{
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+ "id": "98443e66",
"metadata": {
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@@ -3371,7 +3373,7 @@
},
{
"cell_type": "markdown",
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+ "id": "2f9cf056",
"metadata": {
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@@ -3381,7 +3383,7 @@
},
{
"cell_type": "markdown",
- "id": "cda351d1",
+ "id": "4174c0fc",
"metadata": {
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@@ -3392,7 +3394,7 @@
},
{
"cell_type": "markdown",
- "id": "7b0bab43",
+ "id": "dfc7fb10",
"metadata": {
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@@ -3405,7 +3407,7 @@
},
{
"cell_type": "markdown",
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+ "id": "16a4f91a",
"metadata": {
"editable": true
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@@ -3415,7 +3417,7 @@
},
{
"cell_type": "markdown",
- "id": "af99959f",
+ "id": "87f47a07",
"metadata": {
"editable": true
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@@ -3428,7 +3430,7 @@
},
{
"cell_type": "markdown",
- "id": "6ef086d2",
+ "id": "4a02ab43",
"metadata": {
"editable": true
},
@@ -3438,7 +3440,7 @@
},
{
"cell_type": "markdown",
- "id": "45c48dc4",
+ "id": "5e599cb2",
"metadata": {
"editable": true
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@@ -3449,7 +3451,7 @@
},
{
"cell_type": "markdown",
- "id": "0cc89185",
+ "id": "ff66b053",
"metadata": {
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@@ -3462,7 +3464,7 @@
},
{
"cell_type": "markdown",
- "id": "a93c04ae",
+ "id": "5ce389b6",
"metadata": {
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@@ -3472,7 +3474,7 @@
},
{
"cell_type": "markdown",
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+ "id": "6dac3cd2",
"metadata": {
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@@ -3485,7 +3487,7 @@
},
{
"cell_type": "markdown",
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+ "id": "516a7a5d",
"metadata": {
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@@ -3499,7 +3501,7 @@
},
{
"cell_type": "markdown",
- "id": "382c5735",
+ "id": "d3c17cf9",
"metadata": {
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@@ -3511,7 +3513,7 @@
},
{
"cell_type": "markdown",
- "id": "3d07279c",
+ "id": "c6692a94",
"metadata": {
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@@ -3525,7 +3527,7 @@
},
{
"cell_type": "markdown",
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+ "id": "69526a3d",
"metadata": {
"editable": true
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@@ -3535,7 +3537,7 @@
},
{
"cell_type": "markdown",
- "id": "252fc372",
+ "id": "43a6c743",
"metadata": {
"editable": true
},
@@ -3546,7 +3548,7 @@
},
{
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- "id": "5cf513d5",
+ "id": "6016fcbd",
"metadata": {
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@@ -3559,7 +3561,7 @@
},
{
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- "id": "634cb97f",
+ "id": "84a575df",
"metadata": {
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@@ -3569,7 +3571,7 @@
},
{
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+ "id": "a6df55bf",
"metadata": {
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@@ -3582,7 +3584,7 @@
},
{
"cell_type": "markdown",
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+ "id": "11b99892",
"metadata": {
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@@ -3592,7 +3594,7 @@
},
{
"cell_type": "markdown",
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+ "id": "1b8a3e6f",
"metadata": {
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@@ -3604,7 +3606,7 @@
},
{
"cell_type": "markdown",
- "id": "b68dd9f1",
+ "id": "6611adae",
"metadata": {
"editable": true
},
@@ -3615,7 +3617,7 @@
},
{
"cell_type": "markdown",
- "id": "a2ed4e54",
+ "id": "4e7ea340",
"metadata": {
"editable": true
},
@@ -3627,7 +3629,7 @@
},
{
"cell_type": "markdown",
- "id": "b982761a",
+ "id": "59e9e07a",
"metadata": {
"editable": true
},
@@ -3638,7 +3640,7 @@
},
{
"cell_type": "markdown",
- "id": "e1c9894b",
+ "id": "c6882206",
"metadata": {
"editable": true
},
@@ -3657,7 +3659,7 @@
},
{
"cell_type": "markdown",
- "id": "4bc48513",
+ "id": "6ccfc708",
"metadata": {
"editable": true
},
@@ -3669,7 +3671,7 @@
},
{
"cell_type": "markdown",
- "id": "6080bf1a",
+ "id": "a59a7d75",
"metadata": {
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@@ -3683,7 +3685,7 @@
},
{
"cell_type": "markdown",
- "id": "513bb06b",
+ "id": "1e5df9f3",
"metadata": {
"editable": true
},
@@ -3694,7 +3696,7 @@
},
{
"cell_type": "markdown",
- "id": "2989b303",
+ "id": "42bcad87",
"metadata": {
"editable": true
},
@@ -3708,7 +3710,7 @@
},
{
"cell_type": "markdown",
- "id": "14b96b69",
+ "id": "6abe5360",
"metadata": {
"editable": true
},
@@ -3718,7 +3720,7 @@
},
{
"cell_type": "markdown",
- "id": "aa2f4c76",
+ "id": "9417709e",
"metadata": {
"editable": true
},
@@ -3729,7 +3731,7 @@
},
{
"cell_type": "markdown",
- "id": "22c6eb4e",
+ "id": "f1a41836",
"metadata": {
"editable": true
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@@ -3743,7 +3745,7 @@
},
{
"cell_type": "markdown",
- "id": "5be5b818",
+ "id": "c86e929b",
"metadata": {
"editable": true
},
@@ -3753,7 +3755,7 @@
},
{
"cell_type": "markdown",
- "id": "65e34782",
+ "id": "c4ce6b7b",
"metadata": {
"editable": true
},
@@ -3767,7 +3769,7 @@
},
{
"cell_type": "markdown",
- "id": "004e63f6",
+ "id": "2526862d",
"metadata": {
"editable": true
},
@@ -3777,7 +3779,7 @@
},
{
"cell_type": "markdown",
- "id": "486fc9cc",
+ "id": "1e38d109",
"metadata": {
"editable": true
},
@@ -3790,7 +3792,7 @@
},
{
"cell_type": "markdown",
- "id": "c6c7b4c5",
+ "id": "f6fb86f8",
"metadata": {
"editable": true
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@@ -3801,7 +3803,7 @@
},
{
"cell_type": "markdown",
- "id": "7835f119",
+ "id": "a946e785",
"metadata": {
"editable": true
},
@@ -3813,7 +3815,7 @@
},
{
"cell_type": "markdown",
- "id": "23a5a6d2",
+ "id": "13ca5c1b",
"metadata": {
"editable": true
},
@@ -3823,7 +3825,7 @@
},
{
"cell_type": "markdown",
- "id": "5bf2ad18",
+ "id": "494c5584",
"metadata": {
"editable": true
},
@@ -3835,7 +3837,7 @@
},
{
"cell_type": "markdown",
- "id": "cfa4890e",
+ "id": "aed08977",
"metadata": {
"editable": true
},
@@ -3846,7 +3848,7 @@
},
{
"cell_type": "markdown",
- "id": "c69f7825",
+ "id": "7a421098",
"metadata": {
"editable": true
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@@ -3857,7 +3859,7 @@
},
{
"cell_type": "markdown",
- "id": "355afdb0",
+ "id": "dee1b6a1",
"metadata": {
"editable": true
},
@@ -3870,7 +3872,7 @@
},
{
"cell_type": "markdown",
- "id": "1668b012",
+ "id": "98bd7b4d",
"metadata": {
"editable": true
},
@@ -3880,7 +3882,7 @@
},
{
"cell_type": "markdown",
- "id": "16d84fcd",
+ "id": "8494541c",
"metadata": {
"editable": true
},
@@ -3892,7 +3894,7 @@
},
{
"cell_type": "markdown",
- "id": "333837ee",
+ "id": "879a78b9",
"metadata": {
"editable": true
},
@@ -3902,7 +3904,7 @@
},
{
"cell_type": "markdown",
- "id": "b8573745",
+ "id": "8cf546fa",
"metadata": {
"editable": true
},
@@ -3914,7 +3916,7 @@
},
{
"cell_type": "markdown",
- "id": "8efb14d4",
+ "id": "31e82c51",
"metadata": {
"editable": true
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@@ -3924,7 +3926,7 @@
},
{
"cell_type": "markdown",
- "id": "3c841e3e",
+ "id": "232e98e4",
"metadata": {
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@@ -3936,7 +3938,7 @@
},
{
"cell_type": "markdown",
- "id": "aef3e4e2",
+ "id": "b0497fd8",
"metadata": {
"editable": true
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@@ -3947,7 +3949,7 @@
},
{
"cell_type": "markdown",
- "id": "8bc68665",
+ "id": "53c481c7",
"metadata": {
"editable": true
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@@ -3959,7 +3961,7 @@
},
{
"cell_type": "markdown",
- "id": "5785cd8a",
+ "id": "5904aa14",
"metadata": {
"editable": true
},
@@ -3969,7 +3971,7 @@
},
{
"cell_type": "markdown",
- "id": "dd2ee53d",
+ "id": "577d4c62",
"metadata": {
"editable": true
},
@@ -3981,7 +3983,7 @@
},
{
"cell_type": "markdown",
- "id": "6c1fc8ad",
+ "id": "2010454f",
"metadata": {
"editable": true
},
@@ -3993,7 +3995,7 @@
},
{
"cell_type": "markdown",
- "id": "91fd6e0d",
+ "id": "a1e8754d",
"metadata": {
"editable": true
},
@@ -4006,7 +4008,7 @@
},
{
"cell_type": "markdown",
- "id": "baae057c",
+ "id": "807de295",
"metadata": {
"editable": true
},
@@ -4018,7 +4020,7 @@
},
{
"cell_type": "markdown",
- "id": "f59a80d4",
+ "id": "4166b206",
"metadata": {
"editable": true
},
@@ -4028,7 +4030,7 @@
},
{
"cell_type": "markdown",
- "id": "45c7e930",
+ "id": "b50f65ff",
"metadata": {
"editable": true
},
@@ -4040,7 +4042,7 @@
},
{
"cell_type": "markdown",
- "id": "0d4f3c53",
+ "id": "fe87e34c",
"metadata": {
"editable": true
},
@@ -4050,7 +4052,7 @@
},
{
"cell_type": "markdown",
- "id": "a7d08543",
+ "id": "98d23df9",
"metadata": {
"editable": true
},
@@ -4062,7 +4064,7 @@
},
{
"cell_type": "markdown",
- "id": "e4861c06",
+ "id": "79049949",
"metadata": {
"editable": true
},
@@ -4072,7 +4074,7 @@
},
{
"cell_type": "markdown",
- "id": "35b6739e",
+ "id": "ec91f51e",
"metadata": {
"editable": true
},
@@ -4081,6 +4083,746 @@
"\\Delta S_{\\pi}=\\epsilon_{0d^{\\pi}_{5/2}}^{\\mathrm{HF}}-\\epsilon_{0p^{\\pi}_{1/2}}^{\\mathrm{HF}}.\n",
"$$"
]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "12f5b3f8",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Hartree-Fock in second quantization and stability of HF solution\n",
+ "\n",
+ "We wish now to derive the Hartree-Fock equations using our second-quantized formalism and study the stability of the equations. \n",
+ "Our ansatz for the ground state of the system is approximated as (this is our representation of a Slater determinant in second quantization)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c94b9928",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|\\Phi_0\\rangle = |c\\rangle = a^{\\dagger}_i a^{\\dagger}_j \\dots a^{\\dagger}_l|0\\rangle.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "bb61c47f",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "We wish to determine $\\hat{u}^{HF}$ so that \n",
+ "$E_0^{HF}= \\langle c|\\hat{H}| c\\rangle$ becomes a local minimum. \n",
+ "\n",
+ "In our analysis here we will need Thouless' theorem, which states that\n",
+ "an arbitrary Slater determinant $|c'\\rangle$ which is not orthogonal to a determinant\n",
+ "$| c\\rangle ={\\displaystyle\\prod_{i=1}^{n}}\n",
+ "a_{\\alpha_{i}}^{\\dagger}|0\\rangle$, can be written as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "5b79c4f4",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|c'\\rangle=exp\\left\\{\\sum_{a>F}\\sum_{i\\le F}C_{ai}a_{a}^{\\dagger}a_{i}\\right\\}| c\\rangle\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "d6b65527",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "Let us give a simple proof of Thouless' theorem. The theorem states that we can make a linear combination av particle-hole excitations with respect to a given reference state $\\vert c\\rangle$. With this linear combination, we can make a new Slater determinant $\\vert c'\\rangle $ which is not orthogonal to \n",
+ "$\\vert c\\rangle$, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "ac4b3531",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\langle c|c'\\rangle \\ne 0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "5b3935a6",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "To show this we need some intermediate steps. The exponential product of two operators $\\exp{\\hat{A}}\\times\\exp{\\hat{B}}$ is equal to $\\exp{(\\hat{A}+\\hat{B})}$ only if the two operators commute, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "359cdced",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "[\\hat{A},\\hat{B}] = 0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "a7248cd3",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "If the operators do not commute, we need to resort to the [Baker-Campbell-Hauersdorf](http://www.encyclopediaofmath.org/index.php/Campbell%E2%80%93Hausdorff_formula). This relation states that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c8c94128",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\exp{\\hat{C}}=\\exp{\\hat{A}}\\exp{\\hat{B}},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "80411227",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "with"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "9f32f8af",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\hat{C}=\\hat{A}+\\hat{B}+\\frac{1}{2}[\\hat{A},\\hat{B}]+\\frac{1}{12}[[\\hat{A},\\hat{B}],\\hat{B}]-\\frac{1}{12}[[\\hat{A},\\hat{B}],\\hat{A}]+\\dots\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "6e320bf5",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "From these relations, we note that \n",
+ "in our expression for $|c'\\rangle$ we have commutators of the type"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "10b7d7de",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "[a_{a}^{\\dagger}a_{i},a_{b}^{\\dagger}a_{j}],\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "9bb71d0b",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "and it is easy to convince oneself that these commutators, or higher powers thereof, are all zero. This means that we can write out our new representation of a Slater determinant as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "2a170a27",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|c'\\rangle=exp\\left\\{\\sum_{a>F}\\sum_{i\\le F}C_{ai}a_{a}^{\\dagger}a_{i}\\right\\}| c\\rangle=\\prod_{i}\\left\\{1+\\sum_{a>F}C_{ai}a_{a}^{\\dagger}a_{i}+\\left(\\sum_{a>F}C_{ai}a_{a}^{\\dagger}a_{i}\\right)^2+\\dots\\right\\}| c\\rangle\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "1b2491cd",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "We note that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "2afaed7f",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\prod_{i}\\sum_{a>F}C_{ai}a_{a}^{\\dagger}a_{i}\\sum_{b>F}C_{bi}a_{b}^{\\dagger}a_{i}| c\\rangle =0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "60eb88ac",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "and all higher-order powers of these combinations of creation and annihilation operators disappear \n",
+ "due to the fact that $(a_i)^n| c\\rangle =0$ when $n > 1$. This allows us to rewrite the expression for $|c'\\rangle $ as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "dc261b95",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|c'\\rangle=\\prod_{i}\\left\\{1+\\sum_{a>F}C_{ai}a_{a}^{\\dagger}a_{i}\\right\\}| c\\rangle,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "2f3b7509",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "which we can rewrite as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "3bc524a1",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|c'\\rangle=\\prod_{i}\\left\\{1+\\sum_{a>F}C_{ai}a_{a}^{\\dagger}a_{i}\\right\\}| a^{\\dagger}_{i_1} a^{\\dagger}_{i_2} \\dots a^{\\dagger}_{i_n}|0\\rangle.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "4aa9bffb",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "The last equation can be written as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "4606510c",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "|c'\\rangle=\\prod_{i}\\left\\{1+\\sum_{a>F}C_{ai}a_{a}^{\\dagger}a_{i}\\right\\}| a^{\\dagger}_{i_1} a^{\\dagger}_{i_2} \\dots a^{\\dagger}_{i_n}|0\\rangle=\\left(1+\\sum_{a>F}C_{ai_1}a_{a}^{\\dagger}a_{i_1}\\right)a^{\\dagger}_{i_1} \n",
+ "\\label{_auto3} \\tag{10}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "afc83bdc",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation} \n",
+ " \\times\\left(1+\\sum_{a>F}C_{ai_2}a_{a}^{\\dagger}a_{i_2}\\right)a^{\\dagger}_{i_2} \\dots |0\\rangle=\\prod_{i}\\left(a^{\\dagger}_{i}+\\sum_{a>F}C_{ai}a_{a}^{\\dagger}\\right)|0\\rangle.\n",
+ "\\label{_auto4} \\tag{11}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "f356c999",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## New operators\n",
+ "\n",
+ "If we define a new creation operator"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "db95d4f1",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "\n",
+ "\n",
+ "\n",
+ "$$\n",
+ "\\begin{equation}\n",
+ "b^{\\dagger}_{i}=a^{\\dagger}_{i}+\\sum_{a>F}C_{ai}a_{a}^{\\dagger}, \\label{eq:newb} \\tag{12}\n",
+ "\\end{equation}\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "87dbcdb3",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "we have"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c75d6a3c",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|c'\\rangle=\\prod_{i}b^{\\dagger}_{i}|0\\rangle=\\prod_{i}\\left(a^{\\dagger}_{i}+\\sum_{a>F}C_{ai}a_{a}^{\\dagger}\\right)|0\\rangle,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "d11b7ede",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "meaning that the new representation of the Slater determinant in second quantization, $|c'\\rangle$, looks like our previous ones. However, this representation is not general enough since we have a restriction on the sum over single-particle states in Eq. ([12](#eq:newb)). The single-particle states have all to be above the Fermi level."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "9a7eadb6",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "The question then is whether we can construct a general representation of a Slater determinant with a creation operator"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "f91e2af1",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\tilde{b}^{\\dagger}_{i}=\\sum_{p}f_{ip}a_{p}^{\\dagger},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "0d7b0388",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "where $f_{ip}$ is a matrix element of a unitary matrix which transforms our creation and annihilation operators\n",
+ "$a^{\\dagger}$ and $a$ to $\\tilde{b}^{\\dagger}$ and $\\tilde{b}$. These new operators define a new representation of a Slater determinant as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "7447b7ad",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|\\tilde{c}\\rangle=\\prod_{i}\\tilde{b}^{\\dagger}_{i}|0\\rangle.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "139c9d8c",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Showing that $|\\tilde{c}\\rangle= |c'\\rangle$\n",
+ "\n",
+ "We need to show that $|\\tilde{c}\\rangle= |c'\\rangle$. We need also to assume that the new state\n",
+ "is not orthogonal to $|c\\rangle$, that is $\\langle c| \\tilde{c}\\rangle \\ne 0$. From this it follows that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "49e9ec54",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\langle c| \\tilde{c}\\rangle=\\langle 0| a_{i_n}\\dots a_{i_1}\\left(\\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\\dagger} \\right)\\left(\\sum_{q=i_1}^{i_n}f_{i_2q}a_{q}^{\\dagger} \\right)\\dots \\left(\\sum_{t=i_1}^{i_n}f_{i_nt}a_{t}^{\\dagger} \\right)|0\\rangle,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "2dec9b5f",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "which is nothing but the determinant $det(f_{ip})$ which we can, using the intermediate normalization condition, \n",
+ "normalize to one, that is"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "866443b8",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "det(f_{ip})=1,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "017413f7",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "meaning that $f$ has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b65103e1",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\sum_{k} f_{ik}f^{-1}_{kj} = \\delta_{ij},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b2bbb524",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "and"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "f60edb9c",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\sum_{j} f^{-1}_{ij}f_{jk} = \\delta_{ik}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "26eacac4",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "Using these relations we can then define the linear combination of creation (and annihilation as well) \n",
+ "operators as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "a98573aa",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\sum_{i}f^{-1}_{ki}\\tilde{b}^{\\dagger}_{i}=\\sum_{i}f^{-1}_{ki}\\sum_{p=i_1}^{\\infty}f_{ip}a_{p}^{\\dagger}=a_{k}^{\\dagger}+\\sum_{i}\\sum_{p=i_{n+1}}^{\\infty}f^{-1}_{ki}f_{ip}a_{p}^{\\dagger}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "5189dcd1",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "Defining"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c3ea6eef",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "c_{kp}=\\sum_{i \\le F}f^{-1}_{ki}f_{ip},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "e25c8d0a",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "we can redefine"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "7a237674",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "a_{k}^{\\dagger}+\\sum_{i}\\sum_{p=i_{n+1}}^{\\infty}f^{-1}_{ki}f_{ip}a_{p}^{\\dagger}=a_{k}^{\\dagger}+\\sum_{p=i_{n+1}}^{\\infty}c_{kp}a_{p}^{\\dagger}=b_k^{\\dagger},\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "4aff5174",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "our starting point."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b1706653",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "We have shown that our general representation of a Slater determinant"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "d88e90f9",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|\\tilde{c}\\rangle=\\prod_{i}\\tilde{b}^{\\dagger}_{i}|0\\rangle=|c'\\rangle=\\prod_{i}b^{\\dagger}_{i}|0\\rangle,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "014787c3",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "with"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "081709df",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "b_k^{\\dagger}=a_{k}^{\\dagger}+\\sum_{p=i_{n+1}}^{\\infty}c_{kp}a_{p}^{\\dagger}.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "e3509df9",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "This means that we can actually write an ansatz for the ground state of the system as a linear combination of\n",
+ "terms which contain the ansatz itself $|c\\rangle$ with an admixture from an infinity of one-particle-one-hole states. The latter has important consequences when we wish to interpret the Hartree-Fock equations and their stability. We can rewrite the new representation as"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "f9fc2849",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|c'\\rangle = |c\\rangle+|\\delta c\\rangle,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "0188c4e2",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "where $|\\delta c\\rangle$ can now be interpreted as a small variation. If we approximate this term with \n",
+ "contributions from one-particle-one-hole (*1p-1h*) states only, we arrive at"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "d66157f1",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "|c'\\rangle = \\left(1+\\sum_{ai}\\delta C_{ai}a_{a}^{\\dagger}a_i\\right)|c\\rangle.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "7c2e62d5",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "## Thouless' theorem\n",
+ "\n",
+ "In our derivation of the Hartree-Fock equations we have shown that"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "99a52616",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\langle \\delta c| \\hat{H} | c\\rangle =0,\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "0687a547",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "which means that we have to satisfy"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "dbb59f31",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "$$\n",
+ "\\langle c|\\sum_{ai}\\delta C_{ai}\\left\\{a_{a}^{\\dagger}a_i\\right\\} \\hat{H} | c\\rangle =0.\n",
+ "$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "926493ec",
+ "metadata": {
+ "editable": true
+ },
+ "source": [
+ "With this as a background, we are now ready to study the stability of the Hartree-Fock equations.\n",
+ "This is the topic for week 40."
+ ]
}
],
"metadata": {},
diff --git a/doc/pub/week39/pdf/week39.pdf b/doc/pub/week39/pdf/week39.pdf
index 7e79400e..521b4c4a 100644
Binary files a/doc/pub/week39/pdf/week39.pdf and b/doc/pub/week39/pdf/week39.pdf differ
diff --git a/doc/src/week39/week39.do.txt b/doc/src/week39/week39.do.txt
index ee1746f5..8f8e8cae 100644
--- a/doc/src/week39/week39.do.txt
+++ b/doc/src/week39/week39.do.txt
@@ -16,8 +16,8 @@ DATE: Week 39, September 223-27
o Friday:
* Hartree-Fock theory and mean field theories
-# * "Video of lecture":"https://youtu.be/fqWAeBiZ_zg"
-# * "Whiteboard notes":"https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/HandwrittenNotes/2024/NotesSeptember27.pdf"
+ * "Video of lecture":"https://youtu.be/"
+ * "Whiteboard notes":"https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/HandwrittenNotes/2024/NotesSeptember27.pdf"
* Lecture Material: These slides, handwritten notes
* Sixth exercise set at URL:"https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/Exercises/2024/ExercisesWeek39.pdf"
@@ -1703,3 +1703,272 @@ and
+
+!split
+===== Hartree-Fock in second quantization and stability of HF solution =====
+
+We wish now to derive the Hartree-Fock equations using our second-quantized formalism and study the stability of the equations.
+Our ansatz for the ground state of the system is approximated as (this is our representation of a Slater determinant in second quantization)
+!bt
+\[
+|\Phi_0\rangle = |c\rangle = a^{\dagger}_i a^{\dagger}_j \dots a^{\dagger}_l|0\rangle.
+\]
+!et
+We wish to determine $\hat{u}^{HF}$ so that
+$E_0^{HF}= \langle c|\hat{H}| c\rangle$ becomes a local minimum.
+
+In our analysis here we will need Thouless' theorem, which states that
+an arbitrary Slater determinant $|c'\rangle$ which is not orthogonal to a determinant
+$| c\rangle ={\displaystyle\prod_{i=1}^{n}}
+a_{\alpha_{i}}^{\dagger}|0\rangle$, can be written as
+!bt
+\[
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle
+\]
+!et
+
+
+
+!split
+===== Thouless' theorem =====
+
+
+Let us give a simple proof of Thouless' theorem. The theorem states that we can make a linear combination av particle-hole excitations with respect to a given reference state $\vert c\rangle$. With this linear combination, we can make a new Slater determinant $\vert c'\rangle $ which is not orthogonal to
+$\vert c\rangle$, that is
+!bt
+\[
+\langle c|c'\rangle \ne 0.
+\]
+!et
+To show this we need some intermediate steps. The exponential product of two operators $\exp{\hat{A}}\times\exp{\hat{B}}$ is equal to $\exp{(\hat{A}+\hat{B})}$ only if the two operators commute, that is
+!bt
+\[
+[\hat{A},\hat{B}] = 0.
+\]
+!et
+
+!split
+===== Thouless' theorem =====
+
+
+If the operators do not commute, we need to resort to the "Baker-Campbell-Hauersdorf":"http://www.encyclopediaofmath.org/index.php/Campbell%E2%80%93Hausdorff_formula". This relation states that
+!bt
+\[
+\exp{\hat{C}}=\exp{\hat{A}}\exp{\hat{B}},
+\]
+!et
+with
+!bt
+\[
+\hat{C}=\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]+\frac{1}{12}[[\hat{A},\hat{B}],\hat{B}]-\frac{1}{12}[[\hat{A},\hat{B}],\hat{A}]+\dots
+\]
+!et
+
+
+!split
+===== Thouless' theorem =====
+
+From these relations, we note that
+in our expression for $|c'\rangle$ we have commutators of the type
+!bt
+\[
+[a_{a}^{\dagger}a_{i},a_{b}^{\dagger}a_{j}],
+\]
+!et
+and it is easy to convince oneself that these commutators, or higher powers thereof, are all zero. This means that we can write out our new representation of a Slater determinant as
+!bt
+\[
+|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}+\left(\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right)^2+\dots\right\}| c\rangle
+\]
+!et
+
+
+
+!split
+===== Thouless' theorem =====
+
+We note that
+!bt
+\[
+\prod_{i}\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\sum_{b>F}C_{bi}a_{b}^{\dagger}a_{i}| c\rangle =0,
+\]
+!et
+and all higher-order powers of these combinations of creation and annihilation operators disappear
+due to the fact that $(a_i)^n| c\rangle =0$ when $n > 1$. This allows us to rewrite the expression for $|c'\rangle $ as
+!bt
+\[
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle,
+\]
+!et
+which we can rewrite as
+!bt
+\[
+|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle.
+\]
+!et
+
+
+!split
+===== Thouless' theorem =====
+
+The last equation can be written as
+!bt
+\begin{align}
+|c'\rangle&=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle=\left(1+\sum_{a>F}C_{ai_1}a_{a}^{\dagger}a_{i_1}\right)a^{\dagger}_{i_1} \\
+& \times\left(1+\sum_{a>F}C_{ai_2}a_{a}^{\dagger}a_{i_2}\right)a^{\dagger}_{i_2} \dots |0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle.
+\end{align}
+!et
+
+
+!split
+===== New operators =====
+
+
+If we define a new creation operator
+!bt
+\begin{equation}
+b^{\dagger}_{i}=a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}, label{eq:newb}
+\end{equation}
+!et
+we have
+!bt
+\[
+|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle,
+\]
+!et
+meaning that the new representation of the Slater determinant in second quantization, $|c'\rangle$, looks like our previous ones. However, this representation is not general enough since we have a restriction on the sum over single-particle states in Eq.~(ref{eq:newb}). The single-particle states have all to be above the Fermi level.
+
+
+!split
+===== Thouless' theorem =====
+
+The question then is whether we can construct a general representation of a Slater determinant with a creation operator
+!bt
+\[
+\tilde{b}^{\dagger}_{i}=\sum_{p}f_{ip}a_{p}^{\dagger},
+\]
+!et
+where $f_{ip}$ is a matrix element of a unitary matrix which transforms our creation and annihilation operators
+$a^{\dagger}$ and $a$ to $\tilde{b}^{\dagger}$ and $\tilde{b}$. These new operators define a new representation of a Slater determinant as
+!bt
+\[
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle.
+\]
+!et
+
+
+
+!split
+===== Showing that $|\tilde{c}\rangle= |c'\rangle$ =====
+
+
+
+We need to show that $|\tilde{c}\rangle= |c'\rangle$. We need also to assume that the new state
+is not orthogonal to $|c\rangle$, that is $\langle c| \tilde{c}\rangle \ne 0$. From this it follows that
+!bt
+\[
+\langle c| \tilde{c}\rangle=\langle 0| a_{i_n}\dots a_{i_1}\left(\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\dagger} \right)\left(\sum_{q=i_1}^{i_n}f_{i_2q}a_{q}^{\dagger} \right)\dots \left(\sum_{t=i_1}^{i_n}f_{i_nt}a_{t}^{\dagger} \right)|0\rangle,
+\]
+!et
+which is nothing but the determinant $det(f_{ip})$ which we can, using the intermediate normalization condition,
+normalize to one, that is
+!bt
+\[
+det(f_{ip})=1,
+\]
+!et
+meaning that $f$ has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations)
+!bt
+\[
+\sum_{k} f_{ik}f^{-1}_{kj} = \delta_{ij},
+\]
+!et
+and
+!bt
+\[
+\sum_{j} f^{-1}_{ij}f_{jk} = \delta_{ik}.
+\]
+!et
+
+
+
+!split
+===== Thouless' theorem =====
+
+Using these relations we can then define the linear combination of creation (and annihilation as well)
+operators as
+!bt
+\[
+\sum_{i}f^{-1}_{ki}\tilde{b}^{\dagger}_{i}=\sum_{i}f^{-1}_{ki}\sum_{p=i_1}^{\infty}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}.
+\]
+!et
+Defining
+!bt
+\[
+c_{kp}=\sum_{i \le F}f^{-1}_{ki}f_{ip},
+\]
+!et
+we can redefine
+!bt
+\[
+a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}=b_k^{\dagger},
+\]
+!et
+our starting point.
+
+
+!split
+===== Thouless' theorem =====
+
+We have shown that our general representation of a Slater determinant
+!bt
+\[
+|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle=|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle,
+\]
+!et
+with
+!bt
+\[
+b_k^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}.
+\]
+!et
+
+
+!split
+===== Thouless' theorem =====
+
+
+This means that we can actually write an ansatz for the ground state of the system as a linear combination of
+terms which contain the ansatz itself $|c\rangle$ with an admixture from an infinity of one-particle-one-hole states. The latter has important consequences when we wish to interpret the Hartree-Fock equations and their stability. We can rewrite the new representation as
+!bt
+\[
+|c'\rangle = |c\rangle+|\delta c\rangle,
+\]
+!et
+where $|\delta c\rangle$ can now be interpreted as a small variation. If we approximate this term with
+contributions from one-particle-one-hole (*1p-1h*) states only, we arrive at
+!bt
+\[
+|c'\rangle = \left(1+\sum_{ai}\delta C_{ai}a_{a}^{\dagger}a_i\right)|c\rangle.
+\]
+!et
+
+
+!split
+===== Thouless' theorem =====
+
+In our derivation of the Hartree-Fock equations we have shown that
+!bt
+\[
+\langle \delta c| \hat{H} | c\rangle =0,
+\]
+!et
+which means that we have to satisfy
+!bt
+\[
+\langle c|\sum_{ai}\delta C_{ai}\left\{a_{a}^{\dagger}a_i\right\} \hat{H} | c\rangle =0.
+\]
+!et
+With this as a background, we are now ready to study the stability of the Hartree-Fock equations.
+This is the topic for week 40.
+
diff --git a/doc/src/week39/week40add.do.txt b/doc/src/week39/week40add.do.txt
index a6e8a4b8..fbe794dc 100644
--- a/doc/src/week39/week40add.do.txt
+++ b/doc/src/week39/week40add.do.txt
@@ -1,1409 +1,3 @@
-TITLE: Week 39: Full configuration interaction theory
-AUTHOR: Morten Hjorth-Jensen {copyright, 1999-present|CC BY-NC} at Department of Physics and Center for Computing in Science Education, University of Oslo, Norway & Department of Physics and Astronomy and Facility for Rare Isotope Beams, Michigan State University, USA
-DATE: Week 39, September 223-27
-
-!split
-===== Week 39, September 23-27, 2024 =====
-
-* Topics to be covered
- o Thursday:
- * Full configuration interaction (FCI) theory
- * Diagrammatic representation
- * Lipkin model as an example of applications of FCI theory
-# * "Video of lecture":"https://youtu.be/AnAbRonMqPc"
-# * "Whiteboard notes":"https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/HandwrittenNotes/2023/LectureSeptember28.pdf"
-
-o Friday:
- * Hartree-Fock theory and mean field theories
-# * "Video of lecture":"https://youtu.be/fqWAeBiZ_zg"
-# * "Whiteboard notes":"https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/HandwrittenNotes/2023/LectureSeptember29.pdf"
-* Lecture Material: These slides, handwritten notes
-* Sixth exercise set at URL:"https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/Exercises/2024/ExercisesWeek39.pdf"
-
-
-
-!split
-===== Full Configuration Interaction Theory =====
-
-We have defined the ansatz for the ground state as
-!bt
-\[
-|\Phi_0\rangle = \left(\prod_{i\le F}\hat{a}_{i}^{\dagger}\right)|0\rangle,
-\]
-!et
-where the index $i$ defines different single-particle states up to the Fermi level. We have assumed that we have $N$ fermions.
-
-!split
-===== One-particle-one-hole state =====
-
-A given one-particle-one-hole ($1p1h$) state can be written as
-!bt
-\[
-|\Phi_i^a\rangle = \hat{a}_{a}^{\dagger}\hat{a}_i|\Phi_0\rangle,
-\]
-!et
-while a $2p2h$ state can be written as
-!bt
-\[
-|\Phi_{ij}^{ab}\rangle = \hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_j\hat{a}_i|\Phi_0\rangle,
-\]
-!et
-and a general $NpNh$ state as
-!bt
-\[
-|\Phi_{ijk\dots}^{abc\dots}\rangle = \hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_{c}^{\dagger}\dots\hat{a}_k\hat{a}_j\hat{a}_i|\Phi_0\rangle.
-\]
-!et
-
-
-!split
-===== Full Configuration Interaction Theory =====
-
-We can then expand our exact state function for the ground state
-as
-!bt
-\[
-|\Psi_0\rangle=C_0|\Phi_0\rangle+\sum_{ai}C_i^a|\Phi_i^a\rangle+\sum_{abij}C_{ij}^{ab}|\Phi_{ij}^{ab}\rangle+\dots
-=(C_0+\hat{C})|\Phi_0\rangle,
-\]
-!et
-where we have introduced the so-called correlation operator
-!bt
-\[
-\hat{C}=\sum_{ai}C_i^a\hat{a}_{a}^{\dagger}\hat{a}_i +\sum_{abij}C_{ij}^{ab}\hat{a}_{a}^{\dagger}\hat{a}_{b}^{\dagger}\hat{a}_j\hat{a}_i+\dots
-\]
-!et
-
-!split
-===== Intermediate normalization =====
-Since the normalization of $\Psi_0$ is at our disposal and since $C_0$ is by hypothesis non-zero, we may arbitrarily set $C_0=1$ with
-corresponding proportional changes in all other coefficients. Using this so-called intermediate normalization we have
-!bt
-\[
-\langle \Psi_0 | \Phi_0 \rangle = \langle \Phi_0 | \Phi_0 \rangle = 1,
-\]
-!et
-resulting in
-!bt
-\[
-|\Psi_0\rangle=(1+\hat{C})|\Phi_0\rangle.
-\]
-!et
-
-
-!split
-===== Full Configuration Interaction Theory =====
-
-We rewrite
-!bt
-\[
-|\Psi_0\rangle=C_0|\Phi_0\rangle+\sum_{ai}C_i^a|\Phi_i^a\rangle+\sum_{abij}C_{ij}^{ab}|\Phi_{ij}^{ab}\rangle+\dots,
-\]
-!et
-in a more compact form as
-!bt
-\[
-|\Psi_0\rangle=\sum_{PH}C_H^P\Phi_H^P=\left(\sum_{PH}C_H^P\hat{A}_H^P\right)|\Phi_0\rangle,
-\]
-!et
-where $H$ stands for $0,1,\dots,n$ hole states and $P$ for $0,1,\dots,n$ particle states.
-
-!split
-===== Compact expression of correlated part =====
-
-We have introduced the operator $\hat{A}_H^P$ which contains an equal number of creation and annihilation operators.
-
-Our requirement of unit normalization gives
-!bt
-\[
-\langle \Psi_0 | \Phi_0 \rangle = \sum_{PH}|C_H^P|^2= 1,
-\]
-!et
-and the energy can be written as
-!bt
-\[
-E= \langle \Psi_0 | \hat{H} |\Psi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}.
-\]
-!et
-
-
-!split
-===== Full Configuration Interaction Theory =====
-
-Normally
-!bt
-\[
-E= \langle \Psi_0 | \hat{H} |\Psi_0 \rangle= \sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'},
-\]
-!et
-is solved by diagonalization setting up the Hamiltonian matrix defined by the basis of all possible Slater determinants. A diagonalization
-# to do: add text about Rayleigh-Ritz
-is equivalent to finding the variational minimum of
-!bt
-\[
- \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle,
-\]
-!et
-where $\lambda$ is a variational multiplier to be identified with the energy of the system.
-
-!split
-===== Minimization =====
-
-The minimization process results in
-!bt
-\[
-\delta\left[ \langle \Psi_0 | \hat{H} |\Psi_0 \rangle-\lambda \langle \Psi_0 |\Psi_0 \rangle\right]=0,
-\]
-!et
-and since the coefficients $\delta[C_H^{*P}]$ and $\delta[C_{H'}^{P'}]$ are complex conjugates it is necessary and sufficient to require the quantities that multiply with $\delta[C_H^{*P}]$ to vanish. Varying the latter coefficients we have then
-!bt
-\[
-\sum_{P'H'}\left\{\delta[C_H^{*P}]\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-
-\lambda( \delta[C_H^{*P}]C_{H'}^{P'}]\right\} = 0.
-\]
-!et
-
-
-
-
-!split
-===== Full Configuration Interaction Theory =====
-This leads to
-!bt
-\[
-\sum_{P'H'}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-\lambda C_H^{P}=0,
-\]
-!et
-for all sets of $P$ and $H$.
-
-If we then multiply by the corresponding $C_H^{*P}$ and sum over $PH$ we obtain
-!bt
-\[
-\sum_{PP'HH'}C_H^{*P}\langle \Phi_H^P | \hat{H} |\Phi_{H'}^{P'} \rangle C_{H'}^{P'}-\lambda\sum_{PH}|C_H^P|^2=0,
-\]
-!et
-leading to the identification $\lambda = E$.
-
-
-!split
-===== Full Configuration Interaction Theory =====
-
-An alternative way to derive the last equation is to start from
-!bt
-\[
-(\hat{H} -E)|\Psi_0\rangle = (\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0,
-\]
-!et
-and if this equation is successively projected against all $\Phi_H^P$ in the expansion of $\Psi$, then the last equation on the previous slide
-results. As stated previously, one solves this equation normally by diagonalization. If we are able to solve this equation exactly (that is
-numerically exactly) in a large Hilbert space (it will be truncated in terms of the number of single-particle states included in the definition
-of Slater determinants), it can then serve as a benchmark for other many-body methods which approximate the correlation operator
-$\hat{C}$.
-
-
-
-
-!split
-===== FCI and the exponential growth =====
-
-Full configuration interaction theory calculations provide in principle, if we can diagonalize numerically, all states of interest. The dimensionality of the problem explodes however quickly.
-
-The total number of Slater determinants which can be built with say $N$ neutrons distributed among $n$ single particle states is
-!bt
-\[
-\left (\begin{array}{c} n \\ N\end{array} \right) =\frac{n!}{(n-N)!N!}.
-\]
-!et
-
-
-For a model space which comprises the first for major shells only $0s$, $0p$, $1s0d$ and $1p0f$ we have $40$ single particle states for neutrons and protons. For the eight neutrons of oxygen-16 we would then have
-!bt
-\[
-\left (\begin{array}{c} 40 \\ 8\end{array} \right) =\frac{40!}{(32)!8!}\sim 10^{9},
-\]
-!et
-and multiplying this with the number of proton Slater determinants we end up with approximately with a dimensionality $d$ of $d\sim 10^{18}$.
-
-
-
-!split
-===== Exponential wall =====
-!bblock
-This number can be reduced if we look at specific symmetries only. However, the dimensionality explodes quickly!
-
-* For Hamiltonian matrices of dimensionalities which are smaller than $d\sim 10^5$, we would use so-called direct methods for diagonalizing the Hamiltonian matrix
-* For larger dimensionalities iterative eigenvalue solvers like Lanczos' method are used. The most efficient codes at present can handle matrices of $d\sim 10^{10}$.
-!eblock
-
-
-!split
-===== A non-practical way of solving the eigenvalue problem =====
-
-To see this, we look at the contributions arising from
-!bt
-\[
-\langle \Phi_H^P | = \langle \Phi_0|,
-\]
-!et
-that is we multiply with $\langle \Phi_0 |$
-from the left in
-!bt
-\[
-(\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0.
-\]
-!et
-
-!split
-===== Using the Condon-Slater rule =====
-If we assume that we have a two-body operator at most, using the Condon-Slater rule gives then an equation for the
-correlation energy in terms of $C_i^a$ and $C_{ij}^{ab}$ only. We get then
-!bt
-\[
-\langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+
-\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0,
-\]
-!et
-or
-!bt
-\[
-E-E_0 =\Delta E=\sum_{ai}\langle \Phi_0 | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+
-\sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab},
-\]
-!et
-where the energy $E_0$ is the reference energy and $\Delta E$ defines the so-called correlation energy.
-The single-particle basis functions could be the results of a Hartree-Fock calculation or just the eigenstates of the non-interacting part of the Hamiltonian.
-
-
-
-!split
-===== A non-practical way of solving the eigenvalue problem =====
-
-To see this, we look at the contributions arising from
-!bt
-\[
-\langle \Phi_H^P | = \langle \Phi_0|,
-\]
-!et
-that is we multiply with $\langle \Phi_0 |$
-from the left in
-!bt
-\[
-(\hat{H} -E)\sum_{P'H'}C_{H'}^{P'}|\Phi_{H'}^{P'} \rangle=0.
-\]
-!et
-
-
-
-!split
-===== A non-practical way of solving the eigenvalue problem =====
-
-If we assume that we have a two-body operator at most, Slater's rule gives then an equation for the
-correlation energy in terms of $C_i^a$ and $C_{ij}^{ab}$ only. We get then
-!bt
-\[
-\langle \Phi_0 | \hat{H} -E| \Phi_0\rangle + \sum_{ai}\langle \Phi_0 | \hat{H} -E|\Phi_{i}^{a} \rangle C_{i}^{a}+
-\sum_{abij}\langle \Phi_0 | \hat{H} -E|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}=0.
-\]
-!et
-
-!split
-===== Slight rewrite =====
-
-Which we can rewrite
-!bt
-\[
-E-E_0 =\Delta E=\sum_{ai}\langle \Phi_0 | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+
-\sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab},
-\]
-!et
-where the energy $E_0$ is the reference energy and $\Delta E$ defines the so-called correlation energy.
-The single-particle basis functions could be the results of a Hartree-Fock calculation or just the eigenstates of the non-interacting part of the Hamiltonian.
-
-
-
-
-!split
-===== Rewriting the FCI equation =====
-!bblock
-In our discussions of the Hartree-Fock method planned for week 39,
-we are going to compute the elements $\langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle $ and $\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab}\rangle$. If we are using a Hartree-Fock basis, then these quantities result in
-$\langle \Phi_0 | \hat{H}|\Phi_{i}^{a}\rangle=0$ and we are left with a *correlation energy* given by
-!bt
-\[
-E-E_0 =\Delta E^{HF}=\sum_{abij}\langle \Phi_0 | \hat{H}|\Phi_{ij}^{ab} \rangle C_{ij}^{ab}.
-\]
-!et
-!eblock
-
-!split
-===== Rewriting the FCI equation =====
-!bblock
-Inserting the various matrix elements we can rewrite the previous equation as
-!bt
-\[
-\Delta E=\sum_{ai}\langle i| \hat{f}|a \rangle C_{i}^{a}+
-\sum_{abij}\langle ij | \hat{v}| ab \rangle C_{ij}^{ab}.
-\]
-!et
-This equation determines the correlation energy but not the coefficients $C$.
-!eblock
-
-!split
-===== Rewriting the FCI equation, does not stop here =====
-
-We need more equations. Our next step is to set up
-!bt
-\[
-\langle \Phi_i^a | \hat{H} -E| \Phi_0\rangle + \sum_{bj}\langle \Phi_i^a | \hat{H} -E|\Phi_{j}^{b} \rangle C_{j}^{b}+
-\sum_{bcjk}\langle \Phi_i^a | \hat{H} -E|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+
-\sum_{bcdjkl}\langle \Phi_i^a | \hat{H} -E|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=0.
-\]
-!et
-
-!split
-===== Finding the coefficients =====
-This equation will allow us to find an expression for the coefficents $C_i^a$ since we can rewrite this equation as
-!bt
-\[
-\langle i | \hat{f}| a\rangle +\langle \Phi_i^a | \hat{H}|\Phi_{i}^{a} \rangle C_{i}^{a}+ \sum_{bj\ne ai}\langle \Phi_i^a | \hat{H}|\Phi_{j}^{b} \rangle C_{j}^{b}+
-\sum_{bcjk}\langle \Phi_i^a | \hat{H}|\Phi_{jk}^{bc} \rangle C_{jk}^{bc}+
-\sum_{bcdjkl}\langle \Phi_i^a | \hat{H}|\Phi_{jkl}^{bcd} \rangle C_{jkl}^{bcd}=EC_i^a.
-\]
-!et
-
-
-!split
-===== Rewriting the FCI equation =====
-!bblock
-We see that on the right-hand side we have the energy $E$. This leads to a non-linear equation in the unknown coefficients.
-These equations are normally solved iteratively ( that is we can start with a guess for the coefficients $C_i^a$). A common choice is to use perturbation theory for the first guess, setting thereby
-!bt
-\[
- C_{i}^{a}=\frac{\langle i | \hat{f}| a\rangle}{\epsilon_i-\epsilon_a}.
-\]
-!et
-!eblock
-
-!split
-===== Rewriting the FCI equation, more to add =====
-!bblock
-The observant reader will however see that we need an equation for $C_{jk}^{bc}$ and $C_{jkl}^{bcd}$ as well.
-To find equations for these coefficients we need then to continue our multiplications from the left with the various
-$\Phi_{H}^P$ terms.
-
-
-For $C_{jk}^{bc}$ we need then
-!bt
-\[
-\langle \Phi_{ij}^{ab} | \hat{H} -E| \Phi_0\rangle + \sum_{kc}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{k}^{c} \rangle C_{k}^{c}+
-\]
-!et
-!bt
-\[
-\sum_{cdkl}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{kl}^{cd} \rangle C_{kl}^{cd}+\sum_{cdeklm}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{klm}^{cde} \rangle C_{klm}^{cde}+\sum_{cdefklmn}\langle \Phi_{ij}^{ab} | \hat{H} -E|\Phi_{klmn}^{cdef} \rangle C_{klmn}^{cdef}=0,
-\]
-!et
-and we can isolate the coefficients $C_{kl}^{cd}$ in a similar way as we did for the coefficients $C_{i}^{a}$.
-!eblock
-
-
-
-!split
-===== Rewriting the FCI equation, more to add =====
-!bblock
-A standard choice for the first iteration is to set
-!bt
-\[
-C_{ij}^{ab} =\frac{\langle ij \vert \hat{v} \vert ab \rangle}{\epsilon_i+\epsilon_j-\epsilon_a-\epsilon_b}.
-\]
-!et
-At the end we can rewrite our solution of the Schroedinger equation in terms of $n$ coupled equations for the coefficients $C_H^P$.
-This is a very cumbersome way of solving the equation. However, by using this iterative scheme we can illustrate how we can compute the
-various terms in the wave operator or correlation operator $\hat{C}$. We will later identify the calculation of the various terms $C_H^P$
-as parts of different many-body approximations to full CI. In particular, we can relate this non-linear scheme with Coupled Cluster theory and
-many-body perturbation theory.
-!eblock
-
-
-!split
-===== Summarizing FCI and bringing in approximative methods =====
-!bblock
-
-If we can diagonalize large matrices, FCI is the method of choice since:
-* It gives all eigenvalues, ground state and excited states
-* The eigenvectors are obtained directly from the coefficients $C_H^P$ which result from the diagonalization
-* We can compute easily expectation values of other operators, as well as transition probabilities
-* Correlations are easy to understand in terms of contributions to a given operator beyond the Hartree-Fock contribution.
-!eblock
-
-!split
-===== Definition of the correlation energy =====
-
-The correlation energy is defined as, with a two-body Hamiltonian,
-!bt
-\[
-\Delta E=\sum_{ai}\langle i| \hat{f}|a \rangle C_{i}^{a}+
-\sum_{abij}\langle ij | \hat{v}| ab \rangle C_{ij}^{ab}.
-\]
-!et
-The coefficients $C$ result from the solution of the eigenvalue problem.
-
-!split
-===== Ground state energy =====
-The energy of say the ground state is then
-!bt
-\[
-E=E_{ref}+\Delta E,
-\]
-!et
-where the so-called reference energy is the energy we obtain from a Hartree-Fock calculation, that is
-!bt
-\[
-E_{ref}=\langle \Phi_0 \vert \hat{H} \vert \Phi_0 \rangle.
-\]
-!et
-
-
-
-
-
-!split
-===== Why Hartree-Fock? Derivation of Hartree-Fock equations in coordinate space =====
-
-Hartree-Fock (HF) theory is an algorithm for finding an approximative expression for the ground state of a given Hamiltonian. The basic ingredients are
- * Define a single-particle basis $\{\psi_{\alpha}\}$ so that
-!bt
-\[
-\hat{h}^{\mathrm{HF}}\psi_{\alpha} = \varepsilon_{\alpha}\psi_{\alpha}
-\]
-!et
-with the Hartree-Fock Hamiltonian defined as
-!bt
-\[
-\hat{h}^{\mathrm{HF}}=\hat{t}+\hat{u}_{\mathrm{ext}}+\hat{u}^{\mathrm{HF}}
-\]
-!et
- * The term $\hat{u}^{\mathrm{HF}}$ is a single-particle potential to be determined by the HF algorithm.
- * The HF algorithm means to choose $\hat{u}^{\mathrm{HF}}$ in order to have
-!bt
-\[ \langle \hat{H} \rangle = E^{\mathrm{HF}}= \langle \Phi_0 | \hat{H}|\Phi_0 \rangle
-\]
-!et
-that is to find a local minimum with a Slater determinant $\Phi_0$ being the ansatz for the ground state.
- * The variational principle ensures that $E^{\mathrm{HF}} \ge E_0$, with $E_0$ the exact ground state energy.
-
-
-!split
-===== Why Hartree-Fock theory =====
-
-We will show that the Hartree-Fock Hamiltonian $\hat{h}^{\mathrm{HF}}$ equals our definition of the operator $\hat{f}$ discussed in connection with the new definition of the normal-ordered Hamiltonian (see later lectures), that is we have, for a specific matrix element
-!bt
-\[
-\langle p |\hat{h}^{\mathrm{HF}}| q \rangle =\langle p |\hat{f}| q \rangle=\langle p|\hat{t}+\hat{u}_{\mathrm{ext}}|q \rangle +\sum_{i\le F} \langle pi | \hat{V} | qi\rangle_{AS},
-\]
-!et
-meaning that
-!bt
-\[
-\langle p|\hat{u}^{\mathrm{HF}}|q\rangle = \sum_{i\le F} \langle pi | \hat{V} | qi\rangle_{AS}.
-\]
-!et
-
-
-!split
-===== Why Hartree-Fock theory =====
-
-The so-called Hartree-Fock potential $\hat{u}^{\mathrm{HF}}$ brings an
-explicit medium dependence due to the summation over all
-single-particle states below the Fermi level $F$. It brings also in an
-explicit dependence on the two-body interaction (in nuclear physics we
-can also have complicated three- or higher-body forces). The two-body
-interaction, with its contribution from the other bystanding fermions,
-creates an effective mean field in which a given fermion moves, in
-addition to the external potential $\hat{u}_{\mathrm{ext}}$ which
-confines the motion of the fermion. For systems like nuclei, there is
-no external confining potential. Nuclei are examples of self-bound
-systems, where the binding arises due to the intrinsic nature of the
-strong force. For nuclear systems thus, there would be no external
-one-body potential in the Hartree-Fock Hamiltonian.
-
-!split
-===== Variational Calculus and Lagrangian Multipliers =====
-
-The calculus of variations involves
-problems where the quantity to be minimized or maximized is an integral.
-
-In the general case we have an integral of the type
-!bt
-\[
-E[\Phi]= \int_a^b f(\Phi(x),\frac{\partial \Phi}{\partial x},x)dx,
-\]
-!et
-where $E$ is the quantity which is sought minimized or maximized.
-
-
-!split
-===== Variational Calculus and Lagrangian Multipliers =====
-
-The problem is that although $f$ is a function of the variables
-$\Phi$, $\partial \Phi/\partial x$ and $x$, the exact dependence of
-$\Phi$ on $x$ is not known. This means again that even though the
-integral has fixed limits $a$ and $b$, the path of integration is not
-known. In our case the unknown quantities are the single-particle wave
-functions and we wish to choose an integration path which makes the
-functional $E[\Phi]$ stationary. This means that we want to find
-minima, or maxima or saddle points. In physics we search normally for
-minima. Our task is therefore to find the minimum of $E[\Phi]$ so
-that its variation $\delta E$ is zero subject to specific
-constraints. In our case the constraints appear as the integral which
-expresses the orthogonality of the single-particle wave functions.
-The constraints can be treated via the technique of Lagrangian
-multipliers
-
-!split
-===== Variational Calculus and Lagrangian Multipliers =====
-
-Let us specialize to the expectation value of the energy for one particle in three-dimensions.
-This expectation value reads
-!bt
-\[
- E=\int dxdydz \psi^*(x,y,z) \hat{H} \psi(x,y,z),
-\]
-!et
-with the constraint
-!bt
-\[
- \int dxdydz \psi^*(x,y,z) \psi(x,y,z)=1,
-\]
-!et
-and a Hamiltonian
-!bt
-\[
-\hat{H}=-\frac{1}{2}\nabla^2+V(x,y,z).
-\]
-!et
-
-
-!split
-===== Variational Calculus and Lagrangian Multipliers =====
-
-We will, for the sake of notational convenience, skip the variables $x,y,z$ below, and write for example $V(x,y,z)=V$.
-
-The integral involving the kinetic energy can be written as, with the function $\psi$ vanishing
-strongly for large values of $x,y,z$ (given here by the limits $a$ and $b$),
-!bt
- \[
- \int_a^b dxdydz \psi^* \left(-\frac{1}{2}\nabla^2\right) \psi dxdydz = \psi^*\nabla\psi|_a^b+\int_a^b dxdydz\frac{1}{2}\nabla\psi^*\nabla\psi.
-\]
-!et
-We will drop the limits $a$ and $b$ in the remaining discussion.
-Inserting this expression into the expectation value for the energy and taking the variational minimum we obtain
-!bt
-\[
-\delta E = \delta \left\{\int dxdydz\left( \frac{1}{2}\nabla\psi^*\nabla\psi+V\psi^*\psi\right)\right\} = 0.
-\]
-!et
-
-!split
-===== Variational Calculus and Lagrangian Multipliers =====
-
-The constraint appears in integral form as
-!bt
-\[
- \int dxdydz \psi^* \psi=\mathrm{constant},
-\]
-!et
-and multiplying with a Lagrangian multiplier $\lambda$ and taking the variational minimum we obtain the final variational equation
-!bt
-\[
-\delta \left\{\int dxdydz\left( \frac{1}{2}\nabla\psi^*\nabla\psi+V\psi^*\psi-\lambda\psi^*\psi\right)\right\} = 0.
-\]
-!et
-
-
-!split
-===== Variational Calculus and Lagrangian Multipliers =====
-
-We introduce the function $f$
-!bt
-\[
- f = \frac{1}{2}\nabla\psi^*\nabla\psi+V\psi^*\psi-\lambda\psi^*\psi=
-\frac{1}{2}(\psi^*_x\psi_x+\psi^*_y\psi_y+\psi^*_z\psi_z)+V\psi^*\psi-\lambda\psi^*\psi,
-\]
-!et
-where we have skipped the dependence on $x,y,z$ and introduced the shorthand $\psi_x$, $\psi_y$ and $\psi_z$ for the various derivatives.
-
-For $\psi^*$ the Euler-Lagrange equations yield
-!bt
-\[
-\frac{\partial f}{\partial \psi^*}- \frac{\partial }{\partial x}\frac{\partial f}{\partial \psi^*_x}-\frac{\partial }{\partial y}\frac{\partial f}{\partial \psi^*_y}-\frac{\partial }{\partial z}\frac{\partial f}{\partial \psi^*_z}=0,
-\]
-!et
-which results in
-!bt
-\[
- -\frac{1}{2}(\psi_{xx}+\psi_{yy}+\psi_{zz})+V\psi=\lambda \psi.
-\]
-!et
-
-
-!split
-===== Variational Calculus and Lagrangian Multipliers =====
-
-We can then identify the Lagrangian multiplier as the energy of the system. The last equation is
-nothing but the standard
-Schroedinger equation and the variational approach discussed here provides
-a powerful method for obtaining approximate solutions of the wave function.
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-Let us denote the ground state energy by $E_0$. According to the
-variational principle we have
-!bt
-\[
- E_0 \le E[\Phi] = \int \Phi^*\hat{H}\Phi d\mathbf{\tau}
-\]
-!et
-where $\Phi$ is a trial function which we assume to be normalized
-!bt
-\[
- \int \Phi^*\Phi d\mathbf{\tau} = 1,
-\]
-!et
-where we have used the shorthand $d\mathbf{\tau}=dx_1dx_2\dots dx_N$.
-
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-In the Hartree-Fock method the trial function is a Slater
-determinant which can be rewritten as
-!bt
-\[
- \Psi(x_1,x_2,\dots,x_N,\alpha,\beta,\dots,\nu) = \frac{1}{\sqrt{N!}}\sum_{P} (-)^PP\psi_{\alpha}(x_1)
- \psi_{\beta}(x_2)\dots\psi_{\nu}(x_N)=\sqrt{N!}\hat{A}\Phi_H,
-\]
-!et
-where we have introduced the anti-symmetrization operator $\hat{A}$ defined by the
-summation over all possible permutations *p* of two fermions.
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-It is defined as
-!bt
-\[
- \hat{A} = \frac{1}{N!}\sum_{p} (-)^p\hat{P},
-\]
-!et
-with the the Hartree-function given by the simple product of all possible single-particle function
-!bt
-\[
- \Phi_H(x_1,x_2,\dots,x_N,\alpha,\beta,\dots,\nu) =
- \psi_{\alpha}(x_1)
- \psi_{\beta}(x_2)\dots\psi_{\nu}(x_N).
-\]
-!et
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-Our functional is written as
-!bt
-\[
- E[\Phi] = \sum_{\mu=1}^N \int \psi_{\mu}^*(x_i)\hat{h}_0(x_i)\psi_{\mu}(x_i) dx_i
- + \frac{1}{2}\sum_{\mu=1}^N\sum_{\nu=1}^N
- \left[ \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)\hat{v}(r_{ij})\psi_{\mu}(x_i)\psi_{\nu}(x_j)dx_idx_j- \int \psi_{\mu}^*(x_i)\psi_{\nu}^*(x_j)
- \hat{v}(r_{ij})\psi_{\nu}(x_i)\psi_{\mu}(x_j)dx_idx_j\right]
-\]
-!et
-The more compact version reads
-!bt
-\[
- E[\Phi]
- = \sum_{\mu}^N \langle \mu | \hat{h}_0 | \mu\rangle+ \frac{1}{2}\sum_{\mu\nu}^N\left[\langle \mu\nu |\hat{v}|\mu\nu\rangle-\langle \nu\mu |\hat{v}|\mu\nu\rangle\right].
-\]
-!et
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-Since the interaction is invariant under the interchange of two particles it means for example that we have
-!bt
-\[
-\langle \mu\nu|\hat{v}|\mu\nu\rangle = \langle \nu\mu|\hat{v}|\nu\mu\rangle,
-\]
-!et
-or in the more general case
-!bt
-\[
-\langle \mu\nu|\hat{v}|\sigma\tau\rangle = \langle \nu\mu|\hat{v}|\tau\sigma\rangle.
-\]
-!et
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-The direct and exchange matrix elements can be brought together if we define the antisymmetrized matrix element
-!bt
-\[
-\langle \mu\nu|\hat{v}|\mu\nu\rangle_{AS}= \langle \mu\nu|\hat{v}|\mu\nu\rangle-\langle \mu\nu|\hat{v}|\nu\mu\rangle,
-\]
-!et
-or for a general matrix element
-!bt
-\[
-\langle \mu\nu|\hat{v}|\sigma\tau\rangle_{AS}= \langle \mu\nu|\hat{v}|\sigma\tau\rangle-\langle \mu\nu|\hat{v}|\tau\sigma\rangle.
-\]
-!et
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-It has the symmetry property
-!bt
-\[
-\langle \mu\nu|\hat{v}|\sigma\tau\rangle_{AS}= -\langle \mu\nu|\hat{v}|\tau\sigma\rangle_{AS}=-\langle \nu\mu|\hat{v}|\sigma\tau\rangle_{AS}.
-\]
-!et
-The antisymmetric matrix element is also hermitian, implying
-!bt
-\[
-\langle \mu\nu|\hat{v}|\sigma\tau\rangle_{AS}= \langle \sigma\tau|\hat{v}|\mu\nu\rangle_{AS}.
-\]
-!et
-
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-With these notations we rewrite the Hartree-Fock functional as
-!bt
-\begin{equation}
- \int \Phi^*\hat{H_I}\Phi d\mathbf{\tau}
- = \frac{1}{2}\sum_{\mu=1}^N\sum_{\nu=1}^N \langle \mu\nu|\hat{v}|\mu\nu\rangle_{AS}. label{H2Expectation2}
-\end{equation}
-!et
-
-Adding the contribution from the one-body operator $\hat{H}_0$ to
-(ref{H2Expectation2}) we obtain the energy functional
-!bt
-\begin{equation}
- E[\Phi]
- = \sum_{\mu=1}^N \langle \mu | h | \mu \rangle +
- \frac{1}{2}\sum_{{\mu}=1}^N\sum_{{\nu}=1}^N \langle \mu\nu|\hat{v}|\mu\nu\rangle_{AS}. label{FunctionalEPhi}
-\end{equation}
-!et
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-In our coordinate space derivations below we will spell out the Hartree-Fock equations in terms of their integrals.
-
-
-
-
-If we generalize the Euler-Lagrange equations to more variables
-and introduce $N^2$ Lagrange multipliers which we denote by
-$\epsilon_{\mu\nu}$, we can write the variational equation for the functional of $E$
-!bt
-\[
- \delta E - \sum_{\mu\nu}^N \epsilon_{\mu\nu} \delta
- \int \psi_{\mu}^* \psi_{\nu} = 0.
-\]
-!et
-For the orthogonal wave functions $\psi_{i}$ this reduces to
-!bt
-\[
- \delta E - \sum_{\mu=1}^N \epsilon_{\mu} \delta
- \int \psi_{\mu}^* \psi_{\mu} = 0.
-\]
-!et
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-
-Variation with respect to the single-particle wave functions $\psi_{\mu}$ yields then
-!bt
-\[
- \sum_{\mu=1}^N \int \delta\psi_{\mu}^*\hat{h_0}(x_i)\psi_{\mu}
- dx_i
- + \frac{1}{2}\sum_{{\mu}=1}^N\sum_{{\nu}=1}^N \left[ \int
- \delta\psi_{\mu}^*\psi_{\nu}^*\hat{v}(r_{ij})\psi_{\mu}\psi_{\nu} dx_idx_j- \int
- \delta\psi_{\mu}^*\psi_{\nu}^*\hat{v}(r_{ij})\psi_{\nu}\psi_{\mu}
- dx_idx_j \right]+
-\]
-!et
-!bt
-\[
-\sum_{\mu=1}^N \int \psi_{\mu}^*\hat{h_0}(x_i)\delta\psi_{\mu}
- dx_i
- + \frac{1}{2}\sum_{{\mu}=1}^N\sum_{{\nu}=1}^N \left[ \int
- \psi_{\mu}^*\psi_{\nu}^*\hat{v}(r_{ij})\delta\psi_{\mu}\psi_{\nu} dx_idx_j- \int
- \psi_{\mu}^*\psi_{\nu}^*\hat{v}(r_{ij})\psi_{\nu}\delta\psi_{\mu}
- dx_idx_j \right]- \sum_{{\mu}=1}^N E_{\mu} \int \delta\psi_{\mu}^*
- \psi_{\mu}dx_i
- - \sum_{{\mu}=1}^N E_{\mu} \int \psi_{\mu}^*
- \delta\psi_{\mu}dx_i = 0.
-\]
-!et
-
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-Although the variations $\delta\psi$ and $\delta\psi^*$ are not
-independent, they may in fact be treated as such, so that the
-terms dependent on either $\delta\psi$ and $\delta\psi^*$ individually
-may be set equal to zero. To see this, simply
-replace the arbitrary variation $\delta\psi$ by $i\delta\psi$, so that
-$\delta\psi^*$ is replaced by $-i\delta\psi^*$, and combine the two
-equations. We thus arrive at the Hartree-Fock equations
-!bt
-\begin{equation}
-\left[ -\frac{1}{2}\nabla_i^2+ \sum_{\nu=1}^N\int \psi_{\nu}^*(x_j)\hat{v}(r_{ij})\psi_{\nu}(x_j)dx_j \right]\psi_{\mu}(x_i) - \left[ \sum_{{\nu}=1}^N \int\psi_{\nu}^*(x_j)\hat{v}(r_{ij})\psi_{\mu}(x_j) dx_j\right] \psi_{\nu}(x_i) = \epsilon_{\mu} \psi_{\mu}(x_i). label{eq:hartreefockcoordinatespace}
-\end{equation}
-!et
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-Notice that the integration $\int dx_j$ implies an
-integration over the spatial coordinates $\mathbf{r_j}$ and a summation
-over the spin-coordinate of fermion $j$. We note that the factor of $1/2$ in front of the sum involving the two-body interaction, has been removed. This is due to the fact that we need to vary both $\delta\psi_{\mu}^*$ and
-$\delta\psi_{\nu}^*$. Using the symmetry properties of the two-body interaction and interchanging $\mu$ and $\nu$
-as summation indices, we obtain two identical terms.
-
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-The two first terms in the last equation are the one-body kinetic
-energy and the electron-nucleus potential. The third or *direct* term
-is the averaged electronic repulsion of the other electrons. As
-written, the term includes the *self-interaction* of electrons when
-$\mu=\nu$. The self-interaction is cancelled in the fourth term, or
-the *exchange* term. The exchange term results from our inclusion of
-the Pauli principle and the assumed determinantal form of the
-wave-function. Equation (ref{eq:hartreefockcoordinatespace}), in
-addition to the kinetic energy and the attraction from the atomic
-nucleus that confines the motion of a single electron, represents now
-the motion of a single-particle modified by the two-body
-interaction. The additional contribution to the Schroedinger equation
-due to the two-body interaction, represents a mean field set up by all
-the other bystanding electrons, the latter given by the sum over all
-single-particle states occupied by $N$ electrons.
-
-
-!split
-===== Derivation of Hartree-Fock equations in coordinate space =====
-
-The Hartree-Fock equation is an example of an integro-differential
-equation. These equations involve repeated calculations of integrals,
-in addition to the solution of a set of coupled differential
-equations. The Hartree-Fock equations can also be rewritten in terms
-of an eigenvalue problem. The solution of an eigenvalue problem
-represents often a more practical algorithm and the solution of
-coupled integro-differential equations. This alternative derivation
-of the Hartree-Fock equations is given below.
-
-
-
-!split
-===== Analysis of Hartree-Fock equations in coordinate space =====
-
- A theoretically convenient form of the
-Hartree-Fock equation is to regard the direct and exchange operator
-defined through
-!bt
-\begin{equation*}
- V_{\mu}^{d}(x_i) = \int \psi_{\mu}^*(x_j)
- \hat{v}(r_{ij})\psi_{\mu}(x_j) dx_j
-\end{equation*}
-!et
-and
-!bt
-\begin{equation*}
- V_{\mu}^{ex}(x_i) g(x_i)
- = \left(\int \psi_{\mu}^*(x_j)
- \hat{v}(r_{ij})g(x_j) dx_j
- \right)\psi_{\mu}(x_i),
-\end{equation*}
-!et
-respectively.
-
-
-
-!split
-===== Analysis of Hartree-Fock equations in coordinate space =====
-
-The function $g(x_i)$ is an arbitrary function,
-and by the substitution $g(x_i) = \psi_{\nu}(x_i)$
-we get
-!bt
-\begin{equation*}
- V_{\mu}^{ex}(x_i) \psi_{\nu}(x_i)
- = \left(\int \psi_{\mu}^*(x_j)
- \hat{v}(r_{ij})\psi_{\nu}(x_j)
- dx_j\right)\psi_{\mu}(x_i).
-\end{equation*}
-!et
-
-
-!split
-===== Analysis of Hartree-Fock equations in coordinate space =====
-
-We may then rewrite the Hartree-Fock equations as
-!bt
-\[
- \hat{h}^{HF}(x_i) \psi_{\nu}(x_i) = \epsilon_{\nu}\psi_{\nu}(x_i),
-\]
-!et
-with
-!bt
-\[
- \hat{h}^{HF}(x_i)= \hat{h}_0(x_i) + \sum_{\mu=1}^NV_{\mu}^{d}(x_i) -
- \sum_{\mu=1}^NV_{\mu}^{ex}(x_i),
-\]
-!et
-and where $\hat{h}_0(i)$ is the one-body part. The latter is normally chosen as a part which yields solutions in closed form. The harmonic oscilltor is a classical problem thereof.
-We normally rewrite the last equation as
-!bt
-\[
- \hat{h}^{HF}(x_i)= \hat{h}_0(x_i) + \hat{u}^{HF}(x_i).
-\]
-!et
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-Another possibility is to expand the single-particle functions in a
-known basis and vary the coefficients, that is, the new
-single-particle wave function is written as a linear expansion in
-terms of a fixed chosen orthogonal basis (for example the well-known
-harmonic oscillator functions or the hydrogen-like functions etc). We
-define our new Hartree-Fock single-particle basis by performing a
-unitary transformation on our previous basis (labelled with greek
-indices) as
-
-
-!bt
-\begin{equation}
-\psi_p^{HF} = \sum_{\lambda} C_{p\lambda}\phi_{\lambda}. label{eq:newbasis}
-\end{equation}
-!et
-In this case we vary the coefficients $C_{p\lambda}$. If the basis has infinitely many solutions, we need
-to truncate the above sum. We assume that the basis $\phi_{\lambda}$ is orthogonal.
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-It is normal to choose a single-particle basis defined as the eigenfunctions
-of parts of the full Hamiltonian. The typical situation consists of the solutions of the one-body part of the Hamiltonian, that is we have
-!bt
-\[
-\hat{h}_0\phi_{\lambda}=\epsilon_{\lambda}\phi_{\lambda}.
-\]
-!et
-The single-particle wave functions $\phi_{\lambda}(\mathbf{r})$, defined by the quantum numbers $\lambda$ and $\mathbf{r}$
-are defined as the overlap
-!bt
-\[
- \phi_{\lambda}(\mathbf{r}) = \langle \mathbf{r} | \lambda \rangle .
-\]
-!et
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-In deriving the Hartree-Fock equations, we will expand the
-single-particle functions in a known basis and vary the coefficients,
-that is, the new single-particle wave function is written as a linear
-expansion in terms of a fixed chosen orthogonal basis (for example the
-well-known harmonic oscillator functions or the hydrogen-like
-functions etc).
-
-We stated that a unitary transformation keeps the orthogonality. To see this consider first a basis of vectors $\mathbf{v}_i$,
-!bt
-\[
-\mathbf{v}_i = \begin{bmatrix} v_{i1} \\ \dots \\ \dots \\v_{in} \end{bmatrix}
-\]
-!et
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-We assume that the basis is orthogonal, that is
-!bt
-\[
-\mathbf{v}_j^T\mathbf{v}_i = \delta_{ij}.
-\]
-!et
-An orthogonal or unitary transformation
-!bt
-\[
-\mathbf{w}_i=\mathbf{U}\mathbf{v}_i,
-\]
-!et
-preserves the dot product and orthogonality since
-!bt
-\[
-\mathbf{w}_j^T\mathbf{w}_i=(\mathbf{U}\mathbf{v}_j)^T\mathbf{U}\mathbf{v}_i=\mathbf{v}_j^T\mathbf{U}^T\mathbf{U}\mathbf{v}_i= \mathbf{v}_j^T\mathbf{v}_i = \delta_{ij}.
-\]
-!et
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-This means that if the coefficients $C_{p\lambda}$ belong to a unitary or orthogonal trasformation (using the Dirac bra-ket notation)
-!bt
-\[
-\vert p\rangle = \sum_{\lambda} C_{p\lambda}\vert\lambda\rangle,
-\]
-!et
-orthogonality is preserved, that is $\langle \alpha \vert \beta\rangle = \delta_{\alpha\beta}$
-and $\langle p \vert q\rangle = \delta_{pq}$.
-
-This propertry is extremely useful when we build up a basis of many-body Stater determinant based states.
-
-_Note also that although a basis $\vert \alpha\rangle$ contains an infinity of states, for practical calculations we have always to make some truncations._
-
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-Before we develop the Hartree-Fock equations, there is another very
-useful property of determinants that we will use both in connection
-with Hartree-Fock calculations. This applies also to our previous
-discussion on full configuration interaction theory.
-
-Consider the following determinant
-!bt
-\[
-\left| \begin{array}{cc} \alpha_1b_{11}+\alpha_2sb_{12}& a_{12}\\
- \alpha_1b_{21}+\alpha_2b_{22}&a_{22}\end{array} \right|=\alpha_1\left|\begin{array}{cc} b_{11}& a_{12}\\
- b_{21}&a_{22}\end{array} \right|+\alpha_2\left| \begin{array}{cc} b_{12}& a_{12}\\b_{22}&a_{22}\end{array} \right|
-\]
-!et
-
-
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-We can generalize this to an $n\times n$ matrix and have
-!bt
-\[
-\left| \begin{array}{cccccc} a_{11}& a_{12} & \dots & \sum_{k=1}^n c_k b_{1k} &\dots & a_{1n}\\
-a_{21}& a_{22} & \dots & \sum_{k=1}^n c_k b_{2k} &\dots & a_{2n}\\
-\dots & \dots & \dots & \dots & \dots & \dots \\
-\dots & \dots & \dots & \dots & \dots & \dots \\
-a_{n1}& a_{n2} & \dots & \sum_{k=1}^n c_k b_{nk} &\dots & a_{nn}\end{array} \right|=
-\sum_{k=1}^n c_k\left| \begin{array}{cccccc} a_{11}& a_{12} & \dots & b_{1k} &\dots & a_{1n}\\
-a_{21}& a_{22} & \dots & b_{2k} &\dots & a_{2n}\\
-\dots & \dots & \dots & \dots & \dots & \dots\\
-\dots & \dots & \dots & \dots & \dots & \dots\\
-a_{n1}& a_{n2} & \dots & b_{nk} &\dots & a_{nn}\end{array} \right| .
-\]
-!et
-This is a property we will use in our Hartree-Fock discussions.
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-We can generalize the previous results, now
-with all elements $a_{ij}$ being given as functions of
-linear combinations of various coefficients $c$ and elements $b_{ij}$,
-!bt
-\[
-\left| \begin{array}{cccccc} \sum_{k=1}^n b_{1k}c_{k1}& \sum_{k=1}^n b_{1k}c_{k2} & \dots & \sum_{k=1}^n b_{1k}c_{kj} &\dots & \sum_{k=1}^n b_{1k}c_{kn}\\
-\sum_{k=1}^n b_{2k}c_{k1}& \sum_{k=1}^n b_{2k}c_{k2} & \dots & \sum_{k=1}^n b_{2k}c_{kj} &\dots & \sum_{k=1}^n b_{2k}c_{kn}\\
-\dots & \dots & \dots & \dots & \dots & \dots \\
-\dots & \dots & \dots & \dots & \dots &\dots \\
-\sum_{k=1}^n b_{nk}c_{k1}& \sum_{k=1}^n b_{nk}c_{k2} & \dots & \sum_{k=1}^n b_{nk}c_{kj} &\dots & \sum_{k=1}^n b_{nk}c_{kn}\end{array} \right|=det(\mathbf{C})det(\mathbf{B}),
-\]
-!et
-where $det(\mathbf{C})$ and $det(\mathbf{B})$ are the determinants of $n\times n$ matrices
-with elements $c_{ij}$ and $b_{ij}$ respectively.
-This is a property we will use in our Hartree-Fock discussions. Convince yourself about the correctness of the above expression by setting $n=2$.
-
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-With our definition of the new basis in terms of an orthogonal basis we have
-!bt
-\[
-\psi_p(x) = \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x).
-\]
-!et
-If the coefficients $C_{p\lambda}$ belong to an orthogonal or unitary matrix, the new basis
-is also orthogonal.
-Our Slater determinant in the new basis $\psi_p(x)$ is written as
-!bt
-\[
-\frac{1}{\sqrt{N!}}
-\left| \begin{array}{ccccc} \psi_{p}(x_1)& \psi_{p}(x_2)& \dots & \dots & \psi_{p}(x_N)\\
- \psi_{q}(x_1)&\psi_{q}(x_2)& \dots & \dots & \psi_{q}(x_N)\\
- \dots & \dots & \dots & \dots & \dots \\
- \dots & \dots & \dots & \dots & \dots \\
- \psi_{t}(x_1)&\psi_{t}(x_2)& \dots & \dots & \psi_{t}(x_N)\end{array} \right|=\frac{1}{\sqrt{N!}}
-\left| \begin{array}{ccccc} \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_1)& \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{p\lambda}\phi_{\lambda}(x_N)\\
- \sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_1)&\sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{q\lambda}\phi_{\lambda}(x_N)\\
- \dots & \dots & \dots & \dots & \dots \\
- \dots & \dots & \dots & \dots & \dots \\
- \sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_1)&\sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_2)& \dots & \dots & \sum_{\lambda} C_{t\lambda}\phi_{\lambda}(x_N)\end{array} \right|,
-\]
-!et
-which is nothing but $det(\mathbf{C})det(\Phi)$, with $det(\Phi)$ being the determinant given by the basis functions $\phi_{\lambda}(x)$.
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-In our discussions hereafter we will use our definitions of single-particle states above and below the Fermi ($F$) level given by the labels
-$ijkl\dots \le F$ for so-called single-hole states and $abcd\dots > F$ for so-called particle states.
-For general single-particle states we employ the labels $pqrs\dots$.
-
-
-
-
-In Eq.~(ref{FunctionalEPhi}), restated here
-!bt
-\[
- E[\Phi]
- = \sum_{\mu=1}^N \langle \mu | h | \mu \rangle +
- \frac{1}{2}\sum_{{\mu}=1}^N\sum_{{\nu}=1}^N \langle \mu\nu|\hat{v}|\mu\nu\rangle_{AS},
-\]
-!et
-we found the expression for the energy functional in terms of the basis function $\phi_{\lambda}(\mathbf{r})$. We then varied the above energy functional with respect to the basis functions $|\mu \rangle$.
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-Now we are interested in defining a new basis defined in terms of
-a chosen basis as defined in Eq.~(ref{eq:newbasis}). We can then rewrite the energy functional as
-!bt
-\begin{equation}
- E[\Phi^{HF}]
- = \sum_{i=1}^N \langle i | h | i \rangle +
- \frac{1}{2}\sum_{ij=1}^N\langle ij|\hat{v}|ij\rangle_{AS}, label{FunctionalEPhi2}
-\end{equation}
-!et
-where $\Phi^{HF}$ is the new Slater determinant defined by the new basis of Eq.~(ref{eq:newbasis}).
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-Using Eq.~(ref{eq:newbasis}) we can rewrite Eq.~(ref{FunctionalEPhi2}) as
-!bt
-\begin{equation}
- E[\Psi]
- = \sum_{i=1}^N \sum_{\alpha\beta} C^*_{i\alpha}C_{i\beta}\langle \alpha | h | \beta \rangle +
- \frac{1}{2}\sum_{ij=1}^N\sum_{{\alpha\beta\gamma\delta}} C^*_{i\alpha}C^*_{j\beta}C_{i\gamma}C_{j\delta}\langle \alpha\beta|\hat{v}|\gamma\delta\rangle_{AS}. label{FunctionalEPhi3}
-\end{equation}
-!et
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-
-We wish now to minimize the above functional. We introduce again a set of Lagrange multipliers, noting that
-since $\langle i | j \rangle = \delta_{i,j}$ and $\langle \alpha | \beta \rangle = \delta_{\alpha,\beta}$,
-the coefficients $C_{i\gamma}$ obey the relation
-!bt
-\[
- \langle i | j \rangle=\delta_{i,j}=\sum_{\alpha\beta} C^*_{i\alpha}C_{i\beta}\langle \alpha | \beta \rangle=
-\sum_{\alpha} C^*_{i\alpha}C_{i\alpha},
-\]
-!et
-which allows us to define a functional to be minimized that reads
-!bt
-\begin{equation}
- F[\Phi^{HF}]=E[\Phi^{HF}] - \sum_{i=1}^N\epsilon_i\sum_{\alpha} C^*_{i\alpha}C_{i\alpha}.
-\end{equation}
-!et
-
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-
-Minimizing with respect to $C^*_{i\alpha}$, remembering that the equations for $C^*_{i\alpha}$ and $C_{i\alpha}$
-can be written as two independent equations, we obtain
-!bt
-\[
-\frac{d}{dC^*_{i\alpha}}\left[ E[\Phi^{HF}] - \sum_{j}\epsilon_j\sum_{\alpha} C^*_{j\alpha}C_{j\alpha}\right]=0,
-\]
-!et
-which yields for every single-particle state $i$ and index $\alpha$ (recalling that the coefficients $C_{i\alpha}$ are matrix elements of a unitary (or orthogonal for a real symmetric matrix) matrix)
-the following Hartree-Fock equations
-!bt
-\[
-\sum_{\beta} C_{i\beta}\langle \alpha | h | \beta \rangle+
-\sum_{j=1}^N\sum_{\beta\gamma\delta} C^*_{j\beta}C_{j\delta}C_{i\gamma}\langle \alpha\beta|\hat{v}|\gamma\delta\rangle_{AS}=\epsilon_i^{HF}C_{i\alpha}.
-\]
-!et
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-We can rewrite this equation as (changing dummy variables)
-!bt
-\[
-\sum_{\beta} \left\{\langle \alpha | h | \beta \rangle+
-\sum_{j}^N\sum_{\gamma\delta} C^*_{j\gamma}C_{j\delta}\langle \alpha\gamma|\hat{v}|\beta\delta\rangle_{AS}\right\}C_{i\beta}=\epsilon_i^{HF}C_{i\alpha}.
-\]
-!et
-Note that the sums over greek indices run over the number of basis set functions (in principle an infinite number).
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-Defining
-!bt
-\[
-h_{\alpha\beta}^{HF}=\langle \alpha | h | \beta \rangle+
-\sum_{j=1}^N\sum_{\gamma\delta} C^*_{j\gamma}C_{j\delta}\langle \alpha\gamma|\hat{v}|\beta\delta\rangle_{AS},
-\]
-!et
-we can rewrite the new equations as
-!bt
-\begin{equation}
-\sum_{\beta}h_{\alpha\beta}^{HF}C_{i\beta}=\epsilon_i^{HF}C_{i\alpha}. label{eq:newhf}
-\end{equation}
-!et
-The latter is nothing but a standard eigenvalue problem. Compared with Eq.~(ref{eq:hartreefockcoordinatespace}),
-we see that we do not need to compute any integrals in an iterative procedure for solving the equations.
-It suffices to tabulate the matrix elements $\langle \alpha | h | \beta \rangle$ and $\langle \alpha\gamma|\hat{v}|\beta\delta\rangle_{AS}$ once and for all. Successive iterations require thus only a look-up in tables over one-body and two-body matrix elements. These details will be discussed below when we solve the Hartree-Fock equations numerical.
-
-
-!split
-===== Hartree-Fock algorithm =====
-
-Our Hartree-Fock matrix is thus
-!bt
-\[
-\hat{h}_{\alpha\beta}^{HF}=\langle \alpha | \hat{h}_0 | \beta \rangle+
-\sum_{j=1}^N\sum_{\gamma\delta} C^*_{j\gamma}C_{j\delta}\langle \alpha\gamma|\hat{v}|\beta\delta\rangle_{AS}.
-\]
-!et
-The Hartree-Fock equations are solved in an iterative waym starting with a guess for the coefficients $C_{j\gamma}=\delta_{j,\gamma}$ and solving the equations by diagonalization till the new single-particle energies
-$\epsilon_i^{\mathrm{HF}}$ do not change anymore by a prefixed quantity.
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-Normally we assume that the single-particle basis $|\beta\rangle$ forms an eigenbasis for the operator
-$\hat{h}_0$, meaning that the Hartree-Fock matrix becomes
-!bt
-\[
-\hat{h}_{\alpha\beta}^{HF}=\epsilon_{\alpha}\delta_{\alpha,\beta}+
-\sum_{j=1}^N\sum_{\gamma\delta} C^*_{j\gamma}C_{j\delta}\langle \alpha\gamma|\hat{v}|\beta\delta\rangle_{AS}.
-\]
-!et
-The Hartree-Fock eigenvalue problem
-!bt
-\[
-\sum_{\beta}\hat{h}_{\alpha\beta}^{HF}C_{i\beta}=\epsilon_i^{\mathrm{HF}}C_{i\alpha},
-\]
-!et
-can be written out in a more compact form as
-!bt
-\[
-\hat{h}^{HF}\hat{C}=\epsilon^{\mathrm{HF}}\hat{C}.
-\]
-!et
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-
-The Hartree-Fock equations are, in their simplest form, solved in an
-iterative way, starting with a guess for the coefficients
-$C_{i\alpha}$. We label the coefficients as $C_{i\alpha}^{(n)}$, where
-the subscript $n$ stands for iteration $n$. To set up the algorithm
-we can proceed as follows:
-
- * We start with a guess $C_{i\alpha}^{(0)}=\delta_{i,\alpha}$. Alternatively, we could have used random starting values as long as the vectors are normalized. Another possibility is to give states below the Fermi level a larger weight.
- * The Hartree-Fock matrix simplifies then to (assuming that the coefficients $C_{i\alpha} $ are real)
-!bt
-\[
-\hat{h}_{\alpha\beta}^{HF}=\epsilon_{\alpha}\delta_{\alpha,\beta}+
-\sum_{j = 1}^N\sum_{\gamma\delta} C_{j\gamma}^{(0)}C_{j\delta}^{(0)}\langle \alpha\gamma|\hat{v}|\beta\delta\rangle_{AS}.
-\]
-!et
-
-
-
-!split
-===== Hartree-Fock by varying the coefficients of a wave function expansion =====
-
-Solving the Hartree-Fock eigenvalue problem yields then new eigenvectors $C_{i\alpha}^{(1)}$ and eigenvalues
-$\epsilon_i^{HF(1)}$.
- * With the new eigenvalues we can set up a new Hartree-Fock potential
-!bt
-\[
-\sum_{j = 1}^N\sum_{\gamma\delta} C_{j\gamma}^{(1)}C_{j\delta}^{(1)}\langle \alpha\gamma|\hat{v}|\beta\delta\rangle_{AS}.
-\]
-!et
-The diagonalization with the new Hartree-Fock potential yields new eigenvectors and eigenvalues.
-This process is continued till for example
-!bt
-\[
-\frac{\sum_{p} |\epsilon_i^{(n)}-\epsilon_i^{(n-1)}|}{m} \le \lambda,
-\]
-!et
-where $\lambda$ is a user prefixed quantity ($\lambda \sim 10^{-8}$ or smaller) and $p$ runs over all calculated single-particle
-energies and $m$ is the number of single-particle states.
-