- #1
Rubiss
- 21
- 0
Homework Statement
I am trying to calculate the expectation value of an operator in the Lipkin model of nuclear physics. The background isn't important because my problem in really just a math problem.
Homework Equations
The anticommutation relation
\begin{align*}
a_{p\sigma} a_{p'\sigma'}^{\dagger} + a^{\dagger}_{p'\sigma'} a_{p\sigma} = \delta_{pp'}\delta_{\sigma\sigma'}
\end{align*}
Whenever an annihilation operator acts on the vacuum, you get 0.
The Attempt at a Solution
I will be very explicit in what I am doing.
\begin{align*}
\langle \Psi | H_{0} | \Psi \rangle
&= \left[ \langle 0 | \prod_{p} \sum_{\sigma} \left( C_{\alpha\sigma}^{*} a_{p\sigma} \right) \right] \left[ \left( \frac{1}{2} \right) \epsilon \sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) \right] \left[ \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle \right] \\
%
&= \left( \frac{1}{2} \right) \epsilon \langle 0 | \prod_{p} \prod_{p''} \sum_{p'} \sum_{\sigma} \sum_{\sigma'} \sum_{\sigma''} C_{\alpha\sigma}^{*} C_{\alpha\sigma''} \sigma' a_{p\sigma} a^{\dagger}_{p'\sigma'} a_{p'\sigma'} a_{p''\sigma''}^{\dagger} | 0 \rangle \\
%
&= \left( \frac{1}{2} \right) \epsilon \langle 0 | \prod_{p} \prod_{p''} \sum_{p'} \sum_{\sigma} \sum_{\sigma'} \sum_{\sigma''} C_{\alpha\sigma}^{*} C_{\alpha\sigma''} \sigma' \left( \delta_{pp'}\delta_{\sigma\sigma'} - a_{p'\sigma'}^{\dagger} a_{p\sigma} \right) \left( \delta_{p'p''}\delta_{\sigma'\sigma''} - a_{p''\sigma''}^{\dagger} a_{p'\sigma'} \right) | 0 \rangle \\
%
&= \left( \frac{1}{2} \right) \epsilon \langle 0 | \prod_{p} \prod_{p''} \sum_{p'} \sum_{\sigma} \sum_{\sigma'} \sum_{\sigma''} C_{\alpha\sigma}^{*} C_{\alpha\sigma''} \sigma' \left( \delta_{pp'} \delta_{p'p''} \delta_{\sigma\sigma'} \delta_{\sigma'\sigma''} \right) | 0 \rangle \\
&= \left( \frac{1}{2} \right) \epsilon \langle 0 | \prod_{p} \prod_{p''} \sum_{p'} \sum_{\sigma'} \sum_{\sigma} C_{\alpha\sigma}^{*} C_{\alpha\sigma'} \sigma' \left( \delta_{pp'} \delta_{p'p''} \delta_{\sigma\sigma'} \right) | 0 \rangle \\
&= \left( \frac{1}{2} \right) \epsilon \langle 0 | \prod_{p} \prod_{p''} \sum_{p'} \sum_{\sigma'} C_{\alpha\sigma'}^{*} C_{\alpha\sigma'} \sigma' \left( \delta_{pp'} \delta_{p'p''} \right) | 0 \rangle \\
&= \left( \frac{1}{2} \right) \epsilon \langle 0 | \prod_{p} \prod_{p''} \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right) \sum_{\sigma'} \left( C_{\alpha\sigma'}^{*} C_{\alpha\sigma'} \sigma' \right) | 0 \rangle \\
\end{align*}
The possible values for p are 1, 2, 3, and 4. The possible values of sigma are -1 and 1.
The final total answer should be
\begin{align*}
2 \epsilon \left( |C_{\alpha+}|^{2}-|C_{\alpha-}|^{2}\right)
\end{align*}
When I sum over sigma', I will get
\begin{align*}
|C_{\alpha+}|^{2}-|C_{\alpha-}|^{2}
\end{align*}
This means to get the answer I am supposed to get, everything that has to do with p, p', and p'' must equal 4. I have tried numerous times writing out explicitly with p,p',p'' = 1, 2, 3, 4 and end up with something that is very messy and that will not equal 4. I also tried moving the sums and products of the p's around, and that doesn't seem to help either.
Does anyone see what I am doing wrong?