3
$\begingroup$

Quantum mechanics is governed by Schrodinger's equation:

$$\hat{H}\psi=i\hbar\partial_t \psi$$

It seems that Hamiltonian acts on wave functions like a time derivative. Just out of curiosity, is there any Hamiltonian that contains time derivatives, either first order or second order?

$\endgroup$

marked as duplicate by ACuriousMind quantum-mechanics Jun 13 '16 at 10:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2
$\begingroup$

The quick answer is: no.

The Hamiltonian operator is a unitary operator that maps state vectors to other state vectors in a given Hilbert Space, regardless of time. Lubos's answer in this thread discusses this distinction very clearly: Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator?

Another point that you may be interested in is: Hamiltonians can't contain time derivative operators, but they certainly CAN be time dependent.

For any particular interaction, you will have a predetermined Hamiltonian. For example, if a particle is free, then $$\hat H = \hat P^2/2m$$ If a particle is subject to some kind of scalar potential energy V(x), then $$\hat H = \hat P^2/2m + V(x)$$

Most operators you see in introductory quantum mechanics, like the two written above, are time-independent. But in general, operators can be time dependent. For example, you can apply a potential energy that is changing over time (a finite square wall that varies in height maybe) Sometimes, these operators are divided into time-independent and time-dependent parts. See the link below for detailed discussions and examples of important time-dependent Hamiltonians in atomic physics: http://ocw.mit.edu/courses/nuclear-engineering/22-51-quantum-theory-of-radiation-interactions-fall-2012/lecture-notes/MIT22_51F12_Ch5.pdf

$\endgroup$
  • $\begingroup$ Do you mean that time derivatives are not unitary? I think $-i\partial_t$ is unitary. $\endgroup$ – Chong Wang Jun 13 '16 at 10:21
  • $\begingroup$ What I mean is: the time derivative operators are NOT linear operators on the Hilbert space. They are mappings from "Hilbert space + time as a parameter" to Hilbert space. So the time derivative is not of the same nature as the Hamiltonian itself. $\endgroup$ – Zhengyan Shi Jun 13 '16 at 13:07

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