Expectation value in Heisenberg picture: creation and annihilation

  • #1
Bruno Cardin
6
1
TL;DR Summary
Hi. I posted this in homework, but it isn't really homework. I'm just someone who has spent 2 years in classical general relativity and find myself lost trying to re-do my final exam.
So, I have a hamiltonian for screening effect, written like:

$$ H=\sum_{k}^{}\epsilon_{k}c_{k}^{\dagger}c_{k}+ \frac{1}{\Omega}\sum_{k,q}^{}V(q,t)c_{k+q}^{\dagger}c_{k} $$

And I have to find an equation for the time evolution of the expected value of the operator ##c_{k-Q}^{\dagger}c_{k}##.

I wrote this, initially

$$ i\hbar\frac{d}{dt}c_{k-Q}^{\dagger}(t)c_{k}(t)= [c_{k-Q}^{\dagger}c_{k} , H] $$

as the time evolution equation for the operator in the Heisenberg picture. What I procceed to do is to plug a bra in the left <phi| and a ket in the right |phi> , with phi being an energy eigenstate, and then start raising and lowering energy levels since the operators ##c_{k}## are the anihilation operators (and with the dagger they switch to creation operators). But the result I have to get to, according to the exam's solution is:

$$ i\hbar\frac{d}{dt} < c_{k-Q}^{\dagger}(t)c_{k}(t) > = (\epsilon_{k}-\epsilon_{Q-k})< c_{k-Q}^{\dagger}c_{k}>+\frac{1}{\Omega}\sum_{k}^{}V(q,t)[<c_{k-Q}^{\dagger}c_{k-q} >- <c_{k+q-Q}^{\dagger}c_{k}>]$$

which has the expression of V in it.. this means I have to "open" the hamiltonian. I'm so rusty that that didn't even cross my mind. I don't get it. Could anyone help? Thank you in advance.
 
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
This needs a bit more information. What (anti)commutation relations are satisfied by the ##c_k##'s ? Is your vacuum annihilated by ##c_k##? If not, then how is your vacuum defined?
 
  • Like
Likes vanhees71

Similar threads

Replies
0
Views
840
Replies
3
Views
1K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
15
Views
2K
Back
Top