Raising and Lowering Operators in the Lipkin Model

[Rubiss]

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?

 

[TSny]

$\prod_{p} \prod_{p''} \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)$

The possible values for p are 1, 2, 3, and 4. The possible values of sigma are -1 and 1.

If p' must simultaneously equal p and p'', what can you say about p and p''?

 

[Rubiss]

They must be equal... I know that. I'm trying to be very explicit in all my steps though. So if the kronecker deltas say p=p'=p'', how does that affect the products?

 

[TSny]

$\prod_{p} \prod_{p''} \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)$

If p ≠ p'', then what is the value of $\sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)$

If you let $a_{pp''} = \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)$, then your expression is of the form $\prod_{p} \prod_{p''} a_{pp''}$. This is a product of terms $a_{pp''}$. If anyone of the $a_{pp''}$ is zero, what is the value of the overall expression?

 

[Rubiss]

$\prod_{p} \prod_{p''} \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)$

If p ≠ p'', then what is the value of $\sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)$

0

If you let $a_{pp''} = \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)$, then your expression is of the form $\prod_{p} \prod_{p''} a_{pp''}$. This is a product of terms $a_{pp''}$. If anyone of the $a_{pp''}$ is zero, what is the value of the overall expression?

0. I'm still not seeing how I am going to get the value of 4 I need.

Let me show you what my problem is

\begin{align*}
\prod_{p} \prod_{p''} \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right)
&=\prod_{p} \prod_{p''} \left( \delta_{p1}\delta_{1p''} + \delta_{p2}\delta_{2p''} + \delta_{p3}\delta_{3p''} + \delta_{p4}\delta_{4p''} \right) \\
&= \prod_{p} \left( \delta_{p1}\delta_{11} + \delta_{p2}\delta_{21} + \delta_{p3}\delta_{31} + \delta_{p4}\delta_{41} \right)
\left( \delta_{p1}\delta_{12} + \delta_{p2}\delta_{22} + \delta_{p3}\delta_{32} + \delta_{p4}\delta_{42} \right)
\left( \delta_{p1}\delta_{13} + \delta_{p2}\delta_{23} + \delta_{p3}\delta_{33} + \delta_{p4}\delta_{43} \right)
\left( \delta_{p1}\delta_{14} + \delta_{p2}\delta_{24} + \delta_{p3}\delta_{34} + \delta_{p4}\delta_{44} \right) \\
&= \prod_{p} \left( \delta_{p1} \delta_{11} \right) \left( \delta_{p2} \delta_{22} \right) \left( \delta_{p3} \delta_{33} \right) \left( \delta_{p4} \delta_{44} \right) \\
&= \prod_{p} \left( \delta_{p1} \right) \left( \delta_{p2} \right) \left( \delta_{p3} \right) \left( \delta_{p4} \right) \\
&=0 \neq 4
\end{align*}

Sorry if I am not understanding you or am missing something that is totally obvious.

 

[TSny]

Yes, I agree that $\prod_{p} \prod_{p''} \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right) = 0.$

I don't understand how you justify reordering the products and summation when you wrote

$\langle \Psi | H_{0} | \Psi \rangle =$

$\left[ \langle 0 | \prod_{p} \sum_{\sigma} \left( C_{\alpha\sigma}^{*} a_{p\sigma} \right) \right] \left[ \left( \frac{1}{2} \right) \epsilon \sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) \right] \left[ \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle \right]=$

$=\left( \frac{1}{2} \right) \epsilon \langle 0 | \prod_{p} \prod_{p''} \sum_{p'} \sum_{\sigma} \sum_{\sigma'} \sum_{\sigma''} C_{\alpha\sigma}^{*} C_{\alpha\sigma''} \sigma' a_{p\sigma} a^{\dagger}_{p'\sigma'} a_{p'\sigma'} a_{p''\sigma''}^{\dagger} | 0 \rangle$

For example I think you can easily check that in general

$( \prod_{i=1}^2 \sum_{j=1}^2A_{ij})(\prod_{i\,'=1}^2 \sum_{j\,'=1}^2B_{i\,'j\,'}) \neq \prod_{i=1}^2\prod_{i\,'=1}^2\sum_{j=1}^2\sum_{j\,'=1}^2A_{ij}B_{i\,'j\'}$.

 

[Rubiss]

Yes, I agree that $\prod_{p} \prod_{p''} \sum_{p'} \left( \delta_{pp'} \delta_{p'p''} \right) = 0.$

I don't understand how you justify reordering the products and summation when you wrote



For example I think you can easily check that in general

$( \prod_{i=1}^2 \sum_{j=1}^2A_{ij})(\prod_{i\,'=1}^2 \sum_{j\,'=1}^2B_{i\,'j\,'}) \neq \prod_{i=1}^2\prod_{i\,'=1}^2\sum_{j=1}^2\sum_{j\,'=1}^2A_{ij}B_{i\,'j\'}$.

Ok...if I write out the original order (product, sum, product), I will still get zero

\begin{align*}
\prod_{p}\sum_{p'}\prod_{p''}\delta_{pp'}\delta_{p'p''}
&=\prod_{p}\sum_{p'}\delta_{pp'}\delta_{p'1}\delta_{p'2}\delta_{p'3} \delta_{p'4} \\
&=\prod_{p}\left(\delta_{p1}\delta_{11}\delta_{12}\delta_{13}\delta_{14}+\delta_{p2}\delta_{21}\delta_{22}\delta_{23}\delta_{24}+\delta_{p3} \delta_{31}\delta_{32}\delta_{33}\delta_{34}+\delta_{p4}\delta_{41} \delta_{42}\delta_{43}\delta_{44} \right) \\
&= \prod_{p} \left(0 + 0 + 0 + 0 \right) \\
&= 0 \neq 4
\end{align*}

So something is still wrong...

 

[TSny]

I was able to get the result, but only by writing things out explicitly and assuming $\small |C_{\alpha+}|^{2}+|C_{\alpha-}|^{2} = 1$.

Note that the expression
$\left[ \langle 0 | \prod_{p} \sum_{\sigma} \left( C_{\alpha\sigma}^{*} a_{p\sigma} \right) \right] \left[ \left( \frac{1}{2} \right) \epsilon \sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) \right] \left[ \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle \right]$

has three main factors. The factor on the right is

$\left[ \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle \right] = \left(C_{\alpha+} a_{1+}^{\dagger} + C_{\alpha-} a_{1-}^{\dagger} \right) \left(C_{\alpha+} a_{2+}^{\dagger} + C_{\alpha-} a_{2-}^{\dagger} \right) \left(C_{\alpha+} a_{3+}^{\dagger} + C_{\alpha-} a_{3-}^{\dagger} \right) \left(C_{\alpha+} a_{4+}^{\dagger} + C_{\alpha-} a_{4-}^{\dagger} \right)$

I found it convenient to let

$\left[1 \right] = \left(C_{\alpha+} a_{1+}^{\dagger} + C_{\alpha-} a_{1-}^{\dagger} \right)$ $\;\;\;$ $\left[2 \right] = \left(C_{\alpha+} a_{2+}^{\dagger} + C_{\alpha-} a_{2-}^{\dagger} \right)$ $\;\;\;$ etc.

so that

$\prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle = [1][2][3][4] | 0 \rangle$

The middle main factor of the general expression contains

$\sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) = \left( n_{1+}-n_{1-} \right) + \left( n_{2+}-n_{2-} \right) + \left( n_{3+}-n_{3-} \right) + \left( n_{4+}-n_{4-} \right)$

where $n_{1+} = a_{1+}^{\dagger}a_{1+}\;\;$ etc.

You can show that $\left( n_{1+}-n_{1-} \right) [1][2][3][4]| 0 \rangle = [\overline{1}][2][3][4]| 0 \rangle$, where $[\overline{1}]= \left(C_{\alpha+} a_{1+}^{\dagger} - C_{\alpha-} a_{1-}^{\dagger} \right)$

Then see if you can show

$\left[ \sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) \right] \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle = \left( [\overline{1}][2][3][4] + [1][\overline{2}][3][4] + [1][2][\overline{3}][4] + [1][2][3][\overline{4}] \right) | 0 \rangle$

Finally you can try to see what happens when you apply the final factor

$\langle 0 | \prod_{p} \sum_{\sigma} \left( C_{\alpha\sigma}^{*} a_{p\sigma} \right) = \langle 0 | \{1\}\{2\}\{3\}\{4\}$ where $\{1\} = \left(C_{\alpha+}^{*} a_{1+} + C_{\alpha-}^{*} a_{1-} \right)$ $\;\;\;$ etc.

Note that $\{1\}[1]| 0 \rangle = \left(|C_{\alpha+}|^2 + |C_{\alpha-}|^2 \right) | 0 \rangle = | 0 \rangle$ and $\{1\}[\overline{1}] | 0 \rangle = \left(|C_{\alpha+}|^2 - |C_{\alpha-}|^2 \right) | 0 \rangle$

Sorry, I don't see an elegant, compact way using Kronecker deltas, etc. I get confused if I try to move around product and sum symbols.

 

[Rubiss]

I was able to get the result, but only by writing things out explicitly and assuming $\small |C_{\alpha+}|^{2}+|C_{\alpha-}|^{2} = 1$.

Note that the expression
$\left[ \langle 0 | \prod_{p} \sum_{\sigma} \left( C_{\alpha\sigma}^{*} a_{p\sigma} \right) \right] \left[ \left( \frac{1}{2} \right) \epsilon \sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) \right] \left[ \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle \right]$

has three main factors. The factor on the right is

$\left[ \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle \right] = \left(C_{\alpha+} a_{1+}^{\dagger} + C_{\alpha-} a_{1-}^{\dagger} \right) \left(C_{\alpha+} a_{2+}^{\dagger} + C_{\alpha-} a_{2-}^{\dagger} \right) \left(C_{\alpha+} a_{3+}^{\dagger} + C_{\alpha-} a_{3-}^{\dagger} \right) \left(C_{\alpha+} a_{4+}^{\dagger} + C_{\alpha-} a_{4-}^{\dagger} \right)$

I found it convenient to let

$\left[1 \right] = \left(C_{\alpha+} a_{1+}^{\dagger} + C_{\alpha-} a_{1-}^{\dagger} \right)$ $\;\;\;$ $\left[2 \right] = \left(C_{\alpha+} a_{2+}^{\dagger} + C_{\alpha-} a_{2-}^{\dagger} \right)$ $\;\;\;$ etc.

so that

$\prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle = [1][2][3][4] | 0 \rangle$

The middle main factor of the general expression contains

$\sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) = \left( n_{1+}-n_{1-} \right) + \left( n_{2+}-n_{2-} \right) + \left( n_{3+}-n_{3-} \right) + \left( n_{4+}-n_{4-} \right)$

where $n_{1+} = a_{1+}^{\dagger}a_{1+}\;\;$ etc.

You can show that $\left( n_{1+}-n_{1-} \right) [1][2][3][4]| 0 \rangle = [\overline{1}][2][3][4]| 0 \rangle$, where $[\overline{1}]= \left(C_{\alpha+} a_{1+}^{\dagger} - C_{\alpha-} a_{1-}^{\dagger} \right)$

Then see if you can show

$\left[ \sum_{p'} \sum_{\sigma'} \left( \sigma' a^{\dagger}_{p'\sigma'} a_{p'\sigma'} \right) \right] \prod_{p''} \sum_{\sigma''} \left( C_{\alpha\sigma''} a_{p''\sigma''}^{\dagger} \right) | 0 \rangle = \left( [\overline{1}][2][3][4] + [1][\overline{2}][3][4] + [1][2][\overline{3}][4] + [1][2][3][\overline{4}] \right) | 0 \rangle$

Finally you can try to see what happens when you apply the final factor

$\langle 0 | \prod_{p} \sum_{\sigma} \left( C_{\alpha\sigma}^{*} a_{p\sigma} \right) = \langle 0 | \{1\}\{2\}\{3\}\{4\}$ where $\{1\} = \left(C_{\alpha+}^{*} a_{1+} + C_{\alpha-}^{*} a_{1-} \right)$ $\;\;\;$ etc.

Note that $\{1\}[1]| 0 \rangle = \left(|C_{\alpha+}|^2 + |C_{\alpha-}|^2 \right) | 0 \rangle = | 0 \rangle$ and $\{1\}[\overline{1}] | 0 \rangle = \left(|C_{\alpha+}|^2 - |C_{\alpha-}|^2 \right) | 0 \rangle$

Sorry, I don't see an elegant, compact way using Kronecker deltas, etc. I get confused if I try to move around product and sum symbols.

Thanks for the reply! I will try to work through everything you have written. By the way, your assumption $\small |C_{\alpha+}|^{2}+|C_{\alpha-}|^{2} = 1$ is correct because the operators were constructed from a unitary transformation.