Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Consider a Hamiltonian of a system coupled to a bath. Let $H_{sys}=\nu c^{\dagger}c$ ; $H_{env}=\Sigma \omega_r a^{\dagger}a$ ; $H_{int}=\Sigma (g_r ac^{\dagger}+g_r^* ca^{\dagger})$. Then it is generally believed that the mean field ansatz for the system in the form of coherent state $|\alpha\rangle$ is a good description of the system steady state. Is there a way to rigorously justify it? I can attempt one such argument(which leads to further question): When you look at the problem in quantum jump formalism for evolution of the density matrix then the system evolution is given by the effective non hermitian Hamiltonian: $H_{eff}=\nu c^{\dagger}c -i \kappa c^{\dagger}c $, this is supplemented by stochastic jumps by the operator $c$ i.e. the system evolves for some random time according to the $H_{eff}$ and then the resulting state is acted upon by $c$; again the state evolves with effective hamiltonian and so this process repeats ad-infinitum. Now the steady state should be eigenstate of jump operators: which is one indication why in principle $|\alpha \rangle$ could describe a steady state. If this reasoning is correct is there a way to see that any ansatz you start with would eventually tend towards a coherent state via successive jumps? What happens when you have interaction in the system hamiltonian, how does our reasoning fail or what is a good ansatz for steady state description of the system then?

share|improve this question
1  
Maybe this reference would be useful to you (Chapitre III) –  Trimok Jun 14 '13 at 7:54
add comment

1 Answer

You can find in a separate post a rigorous prove that the evolution operator associated to your Hamiltonian is (modulo some phase terms) the displacement operator, which generates the coherent states from the vacuum.

The details of the calculation can be found there

http://physics.stackexchange.com/a/46389/16689

and essentially mean that, as soon as you have a linear interaction term $f\hat{a} + f^{\ast} \hat{a}^{\dagger}$, where $f(t)$ is any function of time, then you will end up in a coherent state. You should avoid higher order terms (like those producing squeezing) otherwise the final state is more complicated. All the details and some other references can be found on the post linked.

My ideas are not clear on the quantum jump problem, but in principle a Fock state sounds still a correct approximation. It's even the simplest one, since your Hamiltonian has Fock states as eigenstates.

There are many books on decoherence that also proof this using different methods, see e.g.

Ulrich Weiss Quantum Dissipative Systems World Scientific (2008)

for generic treatment, or

D.F. Walls and G.J. Milburn Quantum Optics Springer (2008)

for some beautiful applications to quantum optics.

share|improve this answer
1  
The proof is very nice indeed. Thanks! Although I must say that what you have shown is a toy problem discussed in many field theory books. It just says well behaved Yuakawa theory with "classical" fermionic source is exactly soluble(see Michael Stone's field theory book, Sidney Coleman's lecture-9 etc) and the solution is Coherent state of bosons. However my problem is that I would like to see the same result in being held true for an open system where you have to worry about the steady state for the density matrix. There you have to include dissipation and quantum jumps. Thanks again! –  Noob Rev B Jun 23 '13 at 1:10
    
@NoobRevB Sorry then, I missed the point of your question. It's now clear for me that your $c$ operators are fermionic ones, am I correct ? I was missing this point. I think an edit of your question should be done. My answer discussed the problem for $a$ and $c$ two bosonic field operators. Sorry for the misunderstanding. –  FraSchelle Jun 23 '13 at 10:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.