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update week 39
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Expand Up @@ -1762,6 +1762,361 @@ <h2 id="hartree-fock-by-varying-the-coefficients-of-a-wave-function-expansion">H
</p>
</section>

<section>
<h2 id="using-the-density-matrix">Using the density matrix </h2>

<p>The equations are often rewritten in terms of a so-called density matrix,
which is defined as
</p>
<p>&nbsp;<br>
$$
\begin{equation}
\rho_{\gamma\delta}=\sum_{i=1}^{N}\langle\gamma|i\rangle\langle i|\delta\rangle = \sum_{i=1}^{N}C_{i\gamma}C^*_{i\delta}.
\tag{9}
\end{equation}
$$
<p>&nbsp;<br>

<p>It means that we can rewrite the Hartree-Fock Hamiltonian as</p>
<p>&nbsp;<br>
$$
\hat{h}_{\alpha\beta}^{HF}=\epsilon_{\alpha}\delta_{\alpha,\beta}+
\sum_{\gamma\delta} \rho_{\gamma\delta}\langle \alpha\gamma|V|\beta\delta\rangle_{AS}.
$$
<p>&nbsp;<br>

<p>It is convenient to use the density matrix since we can precalculate in every iteration the product of two eigenvector components \( C \). </p>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>We can rewrite the ground state energy by adding and subtracting \( \hat{u}^{HF}(x_i) \) </p>
<p>&nbsp;<br>
$$
E_0^{HF} =\langle \Phi_0 | \hat{H} | \Phi_0\rangle =
\sum_{i\le F}^A \langle i | \hat{h}_0 +\hat{u}^{HF}| j\rangle+ \frac{1}{2}\sum_{i\le F}^A\sum_{j \le F}^A\left[\langle ij |\hat{v}|ij \rangle-\langle ij|\hat{v}|ji\rangle\right]-\sum_{i\le F}^A \langle i |\hat{u}^{HF}| i\rangle,
$$
<p>&nbsp;<br>

<p>which results in</p>
<p>&nbsp;<br>
$$
E_0^{HF}
= \sum_{i\le F}^A \varepsilon_i^{HF} + \frac{1}{2}\sum_{i\le F}^A\sum_{j \le F}^A\left[\langle ij |\hat{v}|ij \rangle-\langle ij|\hat{v}|ji\rangle\right]-\sum_{i\le F}^A \langle i |\hat{u}^{HF}| i\rangle.
$$
<p>&nbsp;<br>

<p>Our single-particle states \( ijk\dots \) are now single-particle states obtained from the solution of the Hartree-Fock equations.</p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>Using our definition of the Hartree-Fock single-particle energies we obtain then the following expression for the total ground-state energy</p>
<p>&nbsp;<br>
$$
E_0^{HF}
= \sum_{i\le F}^A \varepsilon_i - \frac{1}{2}\sum_{i\le F}^A\sum_{j \le F}^A\left[\langle ij |\hat{v}|ij \rangle-\langle ij|\hat{v}|ji\rangle\right].
$$
<p>&nbsp;<br>

<p>This form will be used in our discussion of Koopman's theorem.</p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b>Atomic physics case</b>
<p>
<p>We have </p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(N)]
= \sum_{i=1}^H \langle i | \hat{h}_0 | i \rangle +
\frac{1}{2}\sum_{ij=1}^N\langle ij|\hat{v}|ij\rangle_{AS},
$$
<p>&nbsp;<br>

<p>where \( \Phi^{\mathrm{HF}}(N) \) is the new Slater determinant defined by the new basis of Eq.&nbsp;<a href="#mjx-eqn-4">(4)</a>
for \( N \) electrons (same \( Z \)). If we assume that the single-particle wave functions in the new basis do not change
when we remove one electron or add one electron, we can then define the corresponding energy for the \( N-1 \) systems as
</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(N-1)]
= \sum_{i=1; i\ne k}^N \langle i | \hat{h}_0 | i \rangle +
\frac{1}{2}\sum_{ij=1;i,j\ne k}^N\langle ij|\hat{v}|ij\rangle_{AS},
$$
<p>&nbsp;<br>

<p>where we have removed a single-particle state \( k\le F \), that is a state below the Fermi level. </p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>Calculating the difference </p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(N)]- E[\Phi^{\mathrm{HF}}(N-1)] = \langle k | \hat{h}_0 | k \rangle +
\frac{1}{2}\sum_{i=1;i\ne k}^N\langle ik|\hat{v}|ik\rangle_{AS} \frac{1}{2}\sum_{j=1;j\ne k}^N\langle kj|\hat{v}|kj\rangle_{AS},
$$
<p>&nbsp;<br>

<p>we obtain</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(N)]- E[\Phi^{\mathrm{HF}}(N-1)] = \langle k | \hat{h}_0 | k \rangle +
\frac{1}{2}\sum_{j=1}^N\langle kj|\hat{v}|kj\rangle_{AS}
$$
<p>&nbsp;<br>

<p>which is just our definition of the Hartree-Fock single-particle energy</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(N)]- E[\Phi^{\mathrm{HF}}(N-1)] = \epsilon_k^{\mathrm{HF}}
$$
<p>&nbsp;<br>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>Similarly, we can now compute the difference (we label the single-particle states above the Fermi level as \( abcd > F \))</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(N+1)]- E[\Phi^{\mathrm{HF}}(N)]= \epsilon_a^{\mathrm{HF}}.
$$
<p>&nbsp;<br>

<p>These two equations can thus be used to the electron affinity or ionization energies, respectively.
Koopman's theorem states that for example the ionization energy of a closed-shell system is given by the energy of the highest occupied single-particle state. If we assume that changing the number of electrons from \( N \) to \( N+1 \) does not change the Hartree-Fock single-particle energies and eigenfunctions, then Koopman's theorem simply states that the ionization energy of an atom is given by the single-particle energy of the last bound state. In a similar way, we can also define the electron affinities.
</p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>As an example, consider a simple model for atomic sodium, Na. Neutral sodium has eleven electrons,
with the weakest bound one being confined the \( 3s \) single-particle quantum numbers. The energy needed to remove an electron from neutral sodium is rather small, 5.1391 eV, a feature which pertains to all alkali metals.
Having performed a Hartree-Fock calculation for neutral sodium would then allows us to compute the
ionization energy by using the single-particle energy for the \( 3s \) states, namely \( \epsilon_{3s}^{\mathrm{HF}} \).
</p>

<p>From these considerations, we see that Hartree-Fock theory allows us to make a connection between experimental
observables (here ionization and affinity energies) and the underlying interactions between particles.
In this sense, we are now linking the dynamics and structure of a many-body system with the laws of motion which govern the system. Our approach is a reductionistic one, meaning that we want to understand the laws of motion
in terms of the particles or degrees of freedom which we believe are the fundamental ones. Our Slater determinant, being constructed as the product of various single-particle functions, follows this philosophy.
</p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-koopman-s-theorem">Analysis of Hartree-Fock equations, Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>

<p>With similar arguments as in atomic physics, we can now use Hartree-Fock theory to make a link
between nuclear forces and separation energies. Changing to nuclear system, we define
</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(A)]
= \sum_{i=1}^A \langle i | \hat{h}_0 | i \rangle +
\frac{1}{2}\sum_{ij=1}^A\langle ij|\hat{v}|ij\rangle_{AS},
$$
<p>&nbsp;<br>

<p>where \( \Phi^{\mathrm{HF}}(A) \) is the new Slater determinant defined by the new basis of Eq.&nbsp;<a href="#mjx-eqn-4">(4)</a>
for \( A \) nucleons, where \( A=N+Z \), with \( N \) now being the number of neutrons and \( Z \) th enumber of protons. If we assume again that the single-particle wave functions in the new basis do not change from a nucleus with \( A \) nucleons to a nucleus with \( A-1 \) nucleons, we can then define the corresponding energy for the \( A-1 \) systems as
</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(A-1)]
= \sum_{i=1; i\ne k}^A \langle i | \hat{h}_0 | i \rangle +
\frac{1}{2}\sum_{ij=1;i,j\ne k}^A\langle ij|\hat{v}|ij\rangle_{AS},
$$
<p>&nbsp;<br>

<p>where we have removed a single-particle state \( k\le F \), that is a state below the Fermi level. </p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>Calculating the difference </p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(A)]- E[\Phi^{\mathrm{HF}}(A-1)]
= \langle k | \hat{h}_0 | k \rangle +
\frac{1}{2}\sum_{i=1;i\ne k}^A\langle ik|\hat{v}|ik\rangle_{AS} \frac{1}{2}\sum_{j=1;j\ne k}^A\langle kj|\hat{v}|kj\rangle_{AS},
$$
<p>&nbsp;<br>

<p>which becomes </p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(A)]- E[\Phi^{\mathrm{HF}}(A-1)]
= \langle k | \hat{h}_0 | k \rangle +
\frac{1}{2}\sum_{j=1}^A\langle kj|\hat{v}|kj\rangle_{AS}
$$
<p>&nbsp;<br>

<p>which is just our definition of the Hartree-Fock single-particle energy</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(A)]- E[\Phi^{\mathrm{HF}}(A-1)]
= \epsilon_k^{\mathrm{HF}}
$$
<p>&nbsp;<br>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>Similarly, we can now compute the difference (recall that the single-particle states \( abcd > F \))</p>
<p>&nbsp;<br>
$$
E[\Phi^{\mathrm{HF}}(A+1)]- E[\Phi^{\mathrm{HF}}(A)]= \epsilon_a^{\mathrm{HF}}.
$$
<p>&nbsp;<br>

<p>If we then recall that the binding energy differences </p>
<p>&nbsp;<br>
$$
BE(A)-BE(A-1) \hspace{0.5cm} \mathrm{and} \hspace{0.5cm} BE(A+1)-BE(A),
$$
<p>&nbsp;<br>

<p>define the separation energies, we see that the Hartree-Fock single-particle energies can be used to
define separation energies. We have thus our first link between nuclear forces (included in the potential energy term) and an observable quantity defined by differences in binding energies.
</p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>We have thus the following interpretations (if the single-particle field do not change)</p>
<p>&nbsp;<br>
$$
BE(A)-BE(A-1)\approx E[\Phi^{\mathrm{HF}}(A)]- E[\Phi^{\mathrm{HF}}(A-1)]
= \epsilon_k^{\mathrm{HF}},
$$
<p>&nbsp;<br>

<p>and</p>
<p>&nbsp;<br>
$$
BE(A+1)-BE(A)\approx E[\Phi^{\mathrm{HF}}(A+1)]- E[\Phi^{\mathrm{HF}}(A)] = \epsilon_a^{\mathrm{HF}}.
$$
<p>&nbsp;<br>

<p>If we use \( {}^{16}\mbox{O} \) as our closed-shell nucleus, we could then interpret the separation energy</p>
<p>&nbsp;<br>
$$
BE(^{16}\mathrm{O})-BE(^{15}\mathrm{O})\approx \epsilon_{0p^{\nu}_{1/2}}^{\mathrm{HF}},
$$
<p>&nbsp;<br>

<p>and</p>
<p>&nbsp;<br>
$$
BE(^{16}\mathrm{O})-BE(^{15}\mathrm{N})\approx \epsilon_{0p^{\pi}_{1/2}}^{\mathrm{HF}}.
$$
<p>&nbsp;<br>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>Similalry, we could interpret</p>
<p>&nbsp;<br>
$$
BE(^{17}\mathrm{O})-BE(^{16}\mathrm{O})\approx \epsilon_{0d^{\nu}_{5/2}}^{\mathrm{HF}},
$$
<p>&nbsp;<br>

<p>and </p>
<p>&nbsp;<br>
$$
BE(^{17}\mathrm{F})-BE(^{16}\mathrm{O})\approx\epsilon_{0d^{\pi}_{5/2}}^{\mathrm{HF}}.
$$
<p>&nbsp;<br>

<p>We can continue like this for all \( A\pm 1 \) nuclei where \( A \) is a good closed-shell (or subshell closure)
nucleus. Examples are \( {}^{22}\mbox{O} \), \( {}^{24}\mbox{O} \), \( {}^{40}\mbox{Ca} \), \( {}^{48}\mbox{Ca} \), \( {}^{52}\mbox{Ca} \), \( {}^{54}\mbox{Ca} \), \( {}^{56}\mbox{Ni} \),
\( {}^{68}\mbox{Ni} \), \( {}^{78}\mbox{Ni} \), \( {}^{90}\mbox{Zr} \), \( {}^{88}\mbox{Sr} \), \( {}^{100}\mbox{Sn} \), \( {}^{132}\mbox{Sn} \) and \( {}^{208}\mbox{Pb} \), to mention some possile cases.
</p>
</div>
</section>

<section>
<h2 id="analysis-of-hartree-fock-equations-and-koopman-s-theorem">Analysis of Hartree-Fock equations and Koopman's theorem </h2>
<div class="alert alert-block alert-block alert-text-normal">
<b></b>
<p>
<p>We can thus make our first interpretation of the separation energies in terms of the simplest
possible many-body theory.
If we also recall that the so-called energy gap for neutrons (or protons) is defined as
</p>
<p>&nbsp;<br>
$$
\Delta S_n= 2BE(N,Z)-BE(N-1,Z)-BE(N+1,Z),
$$
<p>&nbsp;<br>

<p>for neutrons and the corresponding gap for protons</p>
<p>&nbsp;<br>
$$
\Delta S_p= 2BE(N,Z)-BE(N,Z-1)-BE(N,Z+1),
$$
<p>&nbsp;<br>

<p>we can define the neutron and proton energy gaps for \( {}^{16}\mbox{O} \) as</p>
<p>&nbsp;<br>
$$
\Delta S_{\nu}=\epsilon_{0d^{\nu}_{5/2}}^{\mathrm{HF}}-\epsilon_{0p^{\nu}_{1/2}}^{\mathrm{HF}},
$$
<p>&nbsp;<br>

<p>and </p>
<p>&nbsp;<br>
$$
\Delta S_{\pi}=\epsilon_{0d^{\pi}_{5/2}}^{\mathrm{HF}}-\epsilon_{0p^{\pi}_{1/2}}^{\mathrm{HF}}.
$$
<p>&nbsp;<br>
</div>
</section>



</div> <!-- class="slides" -->
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