# Raising and Lowering Operators in the Lipkin Model

• Rubiss
A_{ij}B_{i\,'j\,'})##I think the misunderstanding is that you are assuming p=p'=p'' as a single unit, but it is not. It is three separate indices. So the way you have written it, you actually have ##\sum_{p} ##terms which are each a product of three delta functions. However, because the summation over p' is independent of the summation over p, you can factor the product of these sums into the product of
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?

Rubiss said:
## \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''?

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?

## \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?

TSny said:
## \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

TSny said:
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.

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\'}##.

TSny said:
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...

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.

TSny said:
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.

## 1. What is the Lipkin model?

The Lipkin model is a theoretical model used in nuclear and quantum physics to study the collective motion of nucleons (protons and neutrons) in a nucleus.

## 2. What are raising and lowering operators?

Raising and lowering operators are mathematical operators that are used to create or destroy particles in a quantum system. In the Lipkin model, these operators are used to describe the collective motion of nucleons.

## 3. How are raising and lowering operators related to angular momentum?

In the Lipkin model, the raising and lowering operators are related to angular momentum because they correspond to the creation or annihilation of spin states, which are related to angular momentum.

## 4. What is the significance of raising and lowering operators in the Lipkin model?

The raising and lowering operators play a crucial role in the Lipkin model as they allow for the calculation of physical observables such as energy levels and transition probabilities.

## 5. How are raising and lowering operators used in the Lipkin model to study nuclear structure?

In the Lipkin model, raising and lowering operators are used to describe the collective motion of nucleons in a nucleus and to calculate properties such as energy levels, transition probabilities, and other observables that are important in understanding nuclear structure.

• Advanced Physics Homework Help
Replies
5
Views
2K
• Advanced Physics Homework Help
Replies
2
Views
749
• Advanced Physics Homework Help
Replies
12
Views
2K
• Advanced Physics Homework Help
Replies
1
Views
2K
• Advanced Physics Homework Help
Replies
1
Views
845
• Advanced Physics Homework Help
Replies
1
Views
1K
• Advanced Physics Homework Help
Replies
1
Views
1K
• Advanced Physics Homework Help
Replies
1
Views
1K
• Advanced Physics Homework Help
Replies
2
Views
1K
• Advanced Physics Homework Help
Replies
1
Views
947