Expert on Wick's Theorem needed

[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:
<br /> 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.

 

[A. Neumaier]

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

 

[mupsi]

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

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

 

[A. Neumaier]

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

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!

 

[mupsi]

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!

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

 

[mupsi]

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!

And besides. There are papers that mention a reference state which doesn't necessarily have to be the vacuum 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.

 

[A. Neumaier]

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

It is the effective vacuum state.

There is a WT for finite temperatures as well (when H is bilinear).

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.

 

[mupsi]

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.

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.

 

[A. Neumaier]

The Bogoliubov transform seems only to work for the diagonal case.

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.

 

[mupsi]

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.

Thank you. I will have a look at it.

 

[mupsi]

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.

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?

 

[mupsi]

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.

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?

 

[A. Neumaier]

the scattering Matrix S is a transformation between the "new" creators and anhilators and the "old ones" and is quasi-unitary. I believe that equation 3 wouldn't be relevant to my problem? (Correct me if I'm wrong).

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 $S$.

 

[mupsi]

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 $S$.

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 vacuum state exists (such that for for all j $c_{j}|vac>=0$?

 

[A. Neumaier]

if the Hamiltonian is diagonal in the basis of the new creators and annihilators is there a way to proof that an effective vacuum state exists (such that for for all j $c_{j}|vac>=0$?

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