8
$\begingroup$

It is the root of quantum mechanics that Heisenberg picture and Schrödinger picture are equivalent?

In most textbooks and wikipedia, the equivalence is proved with a time-independent Hamiltonian. However, some literature uses Heisenberg equation with time-dependent Hamiltonian.

$$i\hbar \frac{dA}{dt}~=~[A(t),H(t)]+i\hbar \frac{\partial A}{\partial t}.$$

So, does Heisenberg equation work with time-dependent Hamiltonian? If so, any proof?

$\endgroup$
1

2 Answers 2

1
$\begingroup$

The short answer is that the time evolution operator for a time-dependent Hamiltonian has two times, the initial and final $U (t,s) $.

Therefore defining $A (t,s)=U (s,t)A U (t,s) $, the Heisenberg equation is obtained differentiating with respect to $t $. Schroedinger equation is obtained differentiating $U (t,s) \psi (s)$ instead. The two are equivalent in the usual sense, i.e. they both give the same time-evolved transition amplitudes.

$\endgroup$
0
0
$\begingroup$

OK, I figured it out myself. Heisenberg equation still holds for time-dependent Hamiltonians:

$$i\hbar \frac{dA}{dt}~=~[A(t),H(t)]+i\hbar \frac{\partial A}{\partial t}.$$

However, now $H(t)$ is defined as $U^\dagger(t)H_S(t)U(t)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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