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("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 '
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'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'),
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("Showing that $|\\tilde{c}\\rangle= |c'\\rangle$",
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'showing-that-tilde-c-rangle-c-rangle'),
("Thouless' theorem", 2, None, 'thouless-theorem'),
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("Thouless' theorem", 2, None, 'thouless-theorem'),
("Thouless' theorem", 2, None, 'thouless-theorem')]}
end of tocinfo -->

<body>
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<!-- navigation toc: --> <li><a href="#analysis-of-hartree-fock-equations-and-koopman-s-theorem" style="font-size: 80%;">Analysis of Hartree-Fock equations and Koopman's theorem</a></li>
<!-- navigation toc: --> <li><a href="#analysis-of-hartree-fock-equations-and-koopman-s-theorem" style="font-size: 80%;">Analysis of Hartree-Fock equations and Koopman's theorem</a></li>
<!-- navigation toc: --> <li><a href="#analysis-of-hartree-fock-equations-and-koopman-s-theorem" style="font-size: 80%;">Analysis of Hartree-Fock equations and Koopman's theorem</a></li>
<!-- navigation toc: --> <li><a href="#hartree-fock-in-second-quantization-and-stability-of-hf-solution" style="font-size: 80%;">Hartree-Fock in second quantization and stability of HF solution</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#new-operators" style="font-size: 80%;">New operators</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#showing-that-tilde-c-rangle-c-rangle" style="font-size: 80%;">Showing that \( |\tilde{c}\rangle= |c'\rangle \)</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>
<!-- navigation toc: --> <li><a href="#thouless-theorem" style="font-size: 80%;">Thouless' theorem</a></li>

</ul>
</li>
Expand Down Expand Up @@ -554,9 +587,9 @@ <h2 id="week-39-september-23-27-2024" class="anchor">Week 39, September 23-27, 2
</ol>
<li> Friday:</li>
<ul>
<li> Hartree-Fock theory and mean field theories
<!-- * <a href="https://youtu.be/fqWAeBiZ_zg" target="_self">Video of lecture</a> -->
<!-- * <a href="https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/HandwrittenNotes/2024/NotesSeptember27.pdf" target="_self">Whiteboard notes</a> --></li>
<li> Hartree-Fock theory and mean field theories</li>
<li> <a href="https://youtu.be/" target="_self">Video of lecture</a></li>
<li> <a href="https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/HandwrittenNotes/2024/NotesSeptember27.pdf" target="_self">Whiteboard notes</a></li>
</ul>
<li> Lecture Material: These slides, handwritten notes</li>
<li> Sixth exercise set at <a href="https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/Exercises/2024/ExercisesWeek39.pdf" target="_self"><tt>https://github.com/ManyBodyPhysics/FYS4480/blob/master/doc/Exercises/2024/ExercisesWeek39.pdf</tt></a></li>
Expand Down Expand Up @@ -2148,6 +2181,242 @@ <h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem" class="anchor"
</div>


<!-- !split -->
<h2 id="hartree-fock-in-second-quantization-and-stability-of-hf-solution" class="anchor">Hartree-Fock in second quantization and stability of HF solution </h2>

<p>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)
</p>
$$
|\Phi_0\rangle = |c\rangle = a^{\dagger}_i a^{\dagger}_j \dots a^{\dagger}_l|0\rangle.
$$

<p>We wish to determine \( \hat{u}^{HF} \) so that
\( E_0^{HF}= \langle c|\hat{H}| c\rangle \) becomes a local minimum.
</p>

<p>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
</p>
$$
|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>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
</p>
$$
\langle c|c'\rangle \ne 0.
$$

<p>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</p>
$$
[\hat{A},\hat{B}] = 0.
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>If the operators do not commute, we need to resort to the <a href="http://www.encyclopediaofmath.org/index.php/Campbell%E2%80%93Hausdorff_formula" target="_self">Baker-Campbell-Hauersdorf</a>. This relation states that</p>
$$
\exp{\hat{C}}=\exp{\hat{A}}\exp{\hat{B}},
$$

<p>with </p>
$$
\hat{C}=\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]+\frac{1}{12}[[\hat{A},\hat{B}],\hat{B}]-\frac{1}{12}[[\hat{A},\hat{B}],\hat{A}]+\dots
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>From these relations, we note that
in our expression for \( |c'\rangle \) we have commutators of the type
</p>
$$
[a_{a}^{\dagger}a_{i},a_{b}^{\dagger}a_{j}],
$$

<p>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</p>
$$
|c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}+\left(\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right)^2+\dots\right\}| c\rangle
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>We note that</p>
$$
\prod_{i}\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\sum_{b>F}C_{bi}a_{b}^{\dagger}a_{i}| c\rangle =0,
$$

<p>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
</p>
$$
|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle,
$$

<p>which we can rewrite as </p>
$$
|c'\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle.
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>The last equation can be written as</p>
$$
\begin{align}
|c'\rangle&=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| a^{\dagger}_{i_1} a^{\dagger}_{i_2} \dots a^{\dagger}_{i_n}|0\rangle=\left(1+\sum_{a>F}C_{ai_1}a_{a}^{\dagger}a_{i_1}\right)a^{\dagger}_{i_1}
\label{_auto3}\\
& \times\left(1+\sum_{a>F}C_{ai_2}a_{a}^{\dagger}a_{i_2}\right)a^{\dagger}_{i_2} \dots |0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle.
\label{_auto4}
\end{align}
$$


<!-- !split -->
<h2 id="new-operators" class="anchor">New operators </h2>

<p>If we define a new creation operator </p>
$$
\begin{equation}
b^{\dagger}_{i}=a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}, \label{eq:newb}
\end{equation}
$$

<p>we have </p>
$$
|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle=\prod_{i}\left(a^{\dagger}_{i}+\sum_{a>F}C_{ai}a_{a}^{\dagger}\right)|0\rangle,
$$

<p>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.&nbsp;\eqref{eq:newb}. The single-particle states have all to be above the Fermi level.</p>

<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>The question then is whether we can construct a general representation of a Slater determinant with a creation operator </p>
$$
\tilde{b}^{\dagger}_{i}=\sum_{p}f_{ip}a_{p}^{\dagger},
$$

<p>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
</p>
$$
|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle.
$$


<!-- !split -->
<h2 id="showing-that-tilde-c-rangle-c-rangle" class="anchor">Showing that \( |\tilde{c}\rangle= |c'\rangle \) </h2>

<p>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
</p>
$$
\langle c| \tilde{c}\rangle=\langle 0| a_{i_n}\dots a_{i_1}\left(\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\dagger} \right)\left(\sum_{q=i_1}^{i_n}f_{i_2q}a_{q}^{\dagger} \right)\dots \left(\sum_{t=i_1}^{i_n}f_{i_nt}a_{t}^{\dagger} \right)|0\rangle,
$$

<p>which is nothing but the determinant \( det(f_{ip}) \) which we can, using the intermediate normalization condition,
normalize to one, that is
</p>
$$
det(f_{ip})=1,
$$

<p>meaning that \( f \) has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations)</p>
$$
\sum_{k} f_{ik}f^{-1}_{kj} = \delta_{ij},
$$

<p>and </p>
$$
\sum_{j} f^{-1}_{ij}f_{jk} = \delta_{ik}.
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>Using these relations we can then define the linear combination of creation (and annihilation as well)
operators as
</p>
$$
\sum_{i}f^{-1}_{ki}\tilde{b}^{\dagger}_{i}=\sum_{i}f^{-1}_{ki}\sum_{p=i_1}^{\infty}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}.
$$

<p>Defining </p>
$$
c_{kp}=\sum_{i \le F}f^{-1}_{ki}f_{ip},
$$

<p>we can redefine </p>
$$
a_{k}^{\dagger}+\sum_{i}\sum_{p=i_{n+1}}^{\infty}f^{-1}_{ki}f_{ip}a_{p}^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}=b_k^{\dagger},
$$

<p>our starting point.</p>

<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>We have shown that our general representation of a Slater determinant </p>
$$
|\tilde{c}\rangle=\prod_{i}\tilde{b}^{\dagger}_{i}|0\rangle=|c'\rangle=\prod_{i}b^{\dagger}_{i}|0\rangle,
$$

<p>with </p>
$$
b_k^{\dagger}=a_{k}^{\dagger}+\sum_{p=i_{n+1}}^{\infty}c_{kp}a_{p}^{\dagger}.
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>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
</p>
$$
|c'\rangle = |c\rangle+|\delta c\rangle,
$$

<p>where \( |\delta c\rangle \) can now be interpreted as a small variation. If we approximate this term with
contributions from one-particle-one-hole (<em>1p-1h</em>) states only, we arrive at
</p>
$$
|c'\rangle = \left(1+\sum_{ai}\delta C_{ai}a_{a}^{\dagger}a_i\right)|c\rangle.
$$


<!-- !split -->
<h2 id="thouless-theorem" class="anchor">Thouless' theorem </h2>

<p>In our derivation of the Hartree-Fock equations we have shown that </p>
$$
\langle \delta c| \hat{H} | c\rangle =0,
$$

<p>which means that we have to satisfy</p>
$$
\langle c|\sum_{ai}\delta C_{ai}\left\{a_{a}^{\dagger}a_i\right\} \hat{H} | c\rangle =0.
$$

<p>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.
</p>
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