Using Vicks Theorem to Calculate Average of Bose Operators

[Petar Mali]

Vick's theorem help us to find average value of product of even number of operators. For example look the case of four Bose operators

$$\langle \hat{b}_1\hat{b}_2\hat{b}_3\hat{b}_4 \rangle =\langle \hat{b}_1\hat{b}_2 \rangle \langle\hat{b}_3\hat{b}_4 \rangle +\langle \hat{b}_1\hat{b}_3 \rangle \langle\hat{b}_2\hat{b}_4 \rangle +\langle \hat{b}_1\hat{b}_4 \rangle \langle\hat{b}_2\hat{b}_3 \rangle$$

In some cases when we have just product of four Bose operators we use decoupling

$$\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j$$

Why not

$$\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_j \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_j +\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j$$?

similarly

$$\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i$$

and no

$$\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_i \rangle \hat{b}_i\hat{b}_j+\langle \hat{b}_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_i$$

Thanks for your answer!

 

[xepma]

The correlator $$\left\langle b^\dag b^\dag\right\rangle$$ is effectively the overlap between a state with N bosons and a state with N+2 bosons. This is because the operator $$b^\dag$$ creates a boson on top of the state.

The overlap between two wavefunctions with different particle numbers is zero. Hence, these terms are omitted in Wick's expansion. Technically, you are correct that you should sum over all possible contractions -- it's just that the omitted contractions are simply zero.

The combination $$\left\langle b_i^\dag b_j\right\rangle$$ removes a boson from the j'th state, and creates one in the i'th state. The particle number before and after this process is the same, so the overlap is usually non-zero.

 

[DrDu]

You can't neglect this terms in general, neither for Bosons nor for fermions. They are important in lasers, superfluids etc.

 

[Petar Mali]

This is some self consistent spin wave theory. You know something about this approximation?

 

[Petar Mali]

@ DrDu
You can't neglect this terms in general, neither for Bosons nor for fermions. They are important in lasers, superfluids etc.

For example if I have antiferromagnet and need to decoupling

$$\hat{b}_j^+\hat{a}_i^+\hat{a}_i^+\hat{a}_i$$?

where j represent sublattice B, and i represent sublattice A

where

$$[\hat{a}_i,\hat{a}_j]=[\hat{b}_i,\hat{b}_j]=[\hat{a}^+_i,\hat{a}^+_j]=[\hat{b}^+_i,\hat{b}^+_j]=[\hat{a}^+_i,\hat{b}_j]=[\hat{a}_i,\hat{b}^+_j]=[\hat{a}^+_i,\hat{b}^+_j]=[\hat{a}_i,\hat{b}_j]=0$$

 

[DrDu]

I am sorry, I am not too familiar with this problem.

 

[shawl]

@ DrDu
You can't neglect this terms in general, neither for Bosons nor for fermions. They are important in lasers, superfluids etc.

The General means the ground states which excludes the condensed systems. In General, these double creation and double destruction terms are not considered.

These terms should be calculated when the system in a new equilibrium state in the dissipative structure, such as Lasers, Super-fluids or Superconductor, etc. In these states, the system is condensed.

PS:
In the cases, double creation (destruction) terms means a positive feedback process in the non-equilibrium statistics.

 

[DrDu]

Petar, maybe you could enlighten us how exactly the boson operators appear in the problem of an anti-ferromagnet, which is primarily about spins, not bosons. There are schemes for bosonization, however, you could explain which one you are using.

 

[Petar Mali]

Approximation is not so important. I decoupling product of for Bose operators. My question

$$\langle \hat{A} \rangle =\sum_n\langle n|\hat{A}e^{-\beta\hat{H}}|n\rangle$$

Can for example

$$\langle \hat{a}^+_i\hat{b}^+_j\rangle \neq 0$$ ,$$\langle \hat{a}_i\hat{b}_j\rangle \neq 0$$, $$\langle \hat{a}^+_i\hat{b}_j\rangle \neq 0$$...

in any case?

 

[Petar Mali]

You have here more about my problem

https://www.physicsforums.com/showthread.php?t=445228

 

[shawl]

In some cases when we have just product of four Bose operators we use decoupling

$$\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j$$

Why not

$$\hat{b}^+_i\hat{b}_i\hat{b}^+_j\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_j\hat{b}_j+\langle \hat{b}^+_j\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_j \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_j +\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_j\hat{b}_i+\langle \hat{b}^+_j\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j$$?

similarly

$$\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i$$

and no

$$\hat{b}^+_i\hat{b}^+_i\hat{b}_i\hat{b}_j=\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_i \rangle \hat{b}^+_i\hat{b}_j+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}_i+\langle \hat{b}^+_i\hat{b}^+_i \rangle \hat{b}_i\hat{b}_j+\langle \hat{b}_i\hat{b}_j \rangle \hat{b}^+_i\hat{b}^+_i$$

Thanks for your answer!

This is a method of mean field theory.

In Hartree-Fock approximation, there is only <a+a>, but no <a+a+> or <aa>. because the latter is zero.

But, for some condensed system, such as Lasers and Superconductors, the double creation and double destruction process is important. In the mean field theory, one can also average the operators, just like <a+a+> and <aa>. In that time, they are not zero.

 

[Petar Mali]

This is a method of mean field theory.

In Hartree-Fock approximation, there is only <a+a>, but no <a+a+> or <aa>. because the latter is zero.

But, for some condensed system, such as Lasers and Superconductors, the double creation and double destruction process is important. In the mean field theory, one can also average the operators, just like <a+a+> and <aa>. In that time, they are not zero.

Is that only in case of non equlibrium statistics?

 

[shawl]

This is a mean field theory.

The mean field theory contains various approximations, such as the Hartree-Fock one I mentioned.

It is not ONLY in the case of nonequilibrium statistics.

 

[Petar Mali]

But if I don't have equlibrium statistics average values

$$\langle \hat{A} \rangle =\sum_n\langle n|\hat{A}e^{-\beta\hat{H}}|n\rangle$$

$$\langle \hat{a}^+_i\hat{b}^+_j\rangle$$, $$\langle \hat{a}_i\hat{b}_j\rangle$$... is going to be zero!

 

[shawl]

You may be right.

I think the equilibrium statistics, generally, do not contain the condensed systems. Especially, the wick's theorem is not suitable for the condensed systems.

In the condensed systems, one may consider the double creation or destruction effect, e.g. <a+b+> or <ab>.

 

[DrDu]

Petar, maybe you find the book by Mattuck, A guide to Feynman Diagrams in many body systems enlightening. He discusses "anomalous propagators" in the chapter on phase transitions.

 

[Petar Mali]

I have question when I'm doing Fourier transformation of some operators

https://www.physicsforums.com/showthread.php?t=452110

Nobody answers me! So... let's try here. Maybe shawl knows?