Recent content by 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 vacuum state exists (such that for for all j...

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?

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...

Thank you. I will have a look at it.

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...

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.

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.

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

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}}...

The derivative of the Green's function is: i \dfrac{dG_{A,B}(t)}{dt} =\delta(t) \left< {[A,B]}\right>+G_{[A,H],B}(t) the Fourier transform is: \omega G_{A,B}(t)=\left< {[A,B]}\right>+G_{[A,H],B}(\omega) but this would require that the Green's function is 0 for t->inf. Why is that the case...

*integral over time

If you'd calculate the transitions of one particle you would only use the golden rule. The probability to find the electron in the continuum with energy close to Ea would then be an integral of the rate -no distribution function needed. When you use many particle systems you have to use...

When calculating the mean lifetime of an electron in a solid under the influence of a perturbation (for example electron-phonon interaction) we often apply Fermi's golden rule but the rate always has to be weighted by appropriate distribution functions (for example fermi functions). These take...

but that approach is fundamentally different. I think that the argument is time complexity. In general it takes more time for the system to relax into the groundstate (as a function of the number of qubits) than if you perform an adiabatic evolution. The computation time is related to the number...

you could argue that entangled states are a particular subgroup of superposition states of many particle systems, namely superpositions that you can not express as a product (generally speaking: tensor product) of single particle wave functions. It is not easy to "quantify" entanglement when it...