# A Expert on Wick's Theorem needed

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1. Jan 27, 2017

### mupsi

Hi everyone,

I use Wick's theorem to decompose expectation values of a string of bosonic creation and annihilation operators evaluated at the vacuum state. This can only be done when the time evolution is driven by a Hamiltonian of the form:
$$H=\sum_{i,j}{\epsilon_{i,j} c^{\dagger}_{i}c_{j}}$$
which follows from the functional field integral in the coherent state basis. Now I am being told, that WT can also be applied when the Hamiltonian contains quadratic terms that don't conserve particle number (c dagger, c dagger and c, c terms). Can anyone confirm this? I am still trying to figure out how that is supposed to lead to a gaussian exponential. I'd appreciate it if anyone can provide links or explain me why this is legitimate.

2. Jan 27, 2017

### A. Neumaier

You may apply a Bogoliubov transformation that defines transformed c/a operators in which the Hamiltonian has the desired form.

3. Jan 27, 2017

### mupsi

but wouldn't then Wick's theorem only be valid when evaluated at the BCS ground state?

4. Jan 27, 2017

### A. Neumaier

The vacuum state is always the ground state, and Wick's theorem holds only in the ground state. Even when the Hamiltonian has no double creation/annihilation terms!

5. Jan 27, 2017

### mupsi

I disagree. The "vacuum state" of the new creation and anihilation operators *is* the BCS groundstate. It's not a vaccum state in the sense of the "usual" fock space.

6. Jan 27, 2017

### mupsi

And besides. There are papers that mention a reference state which doesn't necessarily have to be the vaccum state. There is a WT for finite temperatures as well (when H is bilinear). When the you let T->0 you get WT with regard to the ground state of the Hamiltonian.

7. Jan 27, 2017

### A. Neumaier

It is the effective vacuum state.
But then the representation is not a Fock representation at all but a KMS representation. In this representation, the Hamiltonian is unbounded below, and the meaning of the c/a operators is questionable.

8. Feb 2, 2017

### mupsi

Sorry for the late reply. Wick's theorem is indeed vaild for the BCS groundstate, when the Hamiltonian has the form of a BCS Hamiltonian. Now I want to modify the BCS Hamiltonian such that the single particle part (c, c dagger) is not diagonal. Is there a way to do that? I tried the Bogoliubov transform but it didn't work out. All of the coefficients were trivially zero. If it doesn't work for a general single particle Hamiltonian: does it work in certain cases? The Bogoliubov transform seems only to work for the diagonal case.

9. Feb 2, 2017

### A. Neumaier

No. Please have a look at my paper

U. Leonhardt and A. Neumaier, Explicit effective Hamiltonians for general linear quantum-optical networks, J. Optics B: Quantum Semiclass. Opt. 6 (2004), L1-L4. quant-ph/0306123

Even when your context is different, the techniques are the same.

10. Feb 2, 2017

### mupsi

Thank you. I will have a look at it.

11. Feb 5, 2017

### mupsi

I read your paper. Very interesting. A few questions remain however. Here is how understood your paper and what I believe seems to be relevant to my problem: the scattering Matrix S is a transformation between the "new" creators and anhilators and the "old ones" and is quasi-unitary. These transformations have generators that obey the Lie-Algebra. This seems like the more general case of the kind of Bogoliubov transformation one encounters in BCS theory with a diagonal single particle part. I believe that equation 3 wouldn't be relevant to my problem? (Correct me if I'm wrong). I want to choose S such that H will be diagonal in the basis of the new creators and annihilators which means using standard linear algebra and hoping that this Matrix has all the properties of S. I believe that method will fail. Alternatively, it would be sufficient to arrive at a transformed Hamiltonian H' such that c,c and c dagger, c dagger don't mix. What would be the general procedure to do that? Wouldn't S have to be unitary then?

Last edited: Feb 5, 2017
12. Feb 6, 2017

### mupsi

I just realized that it's obvious that S must be unitary because of the anti-commutation relations. So the same arguments wouldn't apply, I assume?

13. Feb 6, 2017

### A. Neumaier

Equation 3 relates the first 1-particle c/a operators and second quantized (multiparticle) c/a operators and is always relevant. $\widehat S$ is unitary. As you can see, it is enough to do everything on the 1-particle level. By general Lie properties one can diagonalize any Hermitian Hamiltonian in a Lie *-algebra by means of group elements of the corresponding group, and the corresponding eigenvector matrix serves as the 1-particle $S$.

14. Feb 6, 2017

### mupsi

Thanks for your help so far. As it turns out a diagonalisation of the Hamiltonian won't help us much. Just out of curiosity: if the Hamiltonian is diagonal in the basis of the new creators and annihilators is there a way to proof that an effective vaccum state exists (such that for for all j $c_{j}|vac>=0$?

15. Feb 7, 2017

### A. Neumaier

Of course. After the B-transform one has an ordinary Fock space with its vacuum state.