Saturday 15 September 2018

quantum mechanics - Heisenberg Picture with a time-dependent Schrödinger Hamiltonian


So when the Hamiltonian is time-independent, we can define the Heisenberg state vectors by evolving the Schrödinger state vectors back in time:


$$ | \psi \rangle_H = \hat{U}^\dagger (t)|\psi(t) \rangle_S=e^{i\hat{H}t} |\psi(t)\rangle_S $$



and we define operators


$$ \hat{A}_H(t) = \hat{U}^\dagger (t) \hat{A}_S \hat{U}(t)$$


which gives us the Heisenberg equation: $$ \frac{d\hat{A}_H(t)}{dt} = -i[\hat{A}_H(t),\hat{H}]. $$


If, in the Schrödinger picture, we have a time-dependent Hamiltonian, the time evolution operator is given by


$$ \hat{U}(t) = T[e^{-i \int_0^t \hat{H}(t')dt'}] $$


If I define the Heisenberg operators in the same way with the time evolution operators and calculate $ dA_H(t)/dt $ I find


$$ \frac{d}{dt} \hat{A}_H(t)= \frac{d\hat{U}^\dagger(t)}{dt} \hat{A}_S\hat{U}(t) + \hat{U}^\dagger(t) \hat{A}_S \frac{d\hat{U}(t)}{dt} \\ = i \hat{U}^\dagger (t) \hat{H(t)} \hat{A}_S \hat{U}(t) - i\hat{U}^\dagger (t)\hat{A}_S\hat{H}(t)\hat{U}(t). $$


At this point, I am not sure how to proceed. I can't commute $\hat{H}(t)$ through $\hat{U}(t) $ because $[\hat{H}(t),\hat{H}(t')] \neq 0$. How do I show derive Heisenberg's equation for a time-dependent Hamiltonian?




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...