Is the Force Operator in Quantum Mechanics Always Time-Independent?

[Niles]

Homework Statement

Hi

In QM we define the force operator F as (in the Heisenberg picture)
<br /> F = \frac{1}{i\hbar}[p, H] + (d_t F)(t)<br />
What I can't understand is that usually (actually, always) we write
<br /> F = \frac{1}{i\hbar}[p, H]<br />
and neglegt the last time derivative. How can we be so certain that the force is time-independent?Niles.

 

[vela]

Homework Statement

Hi

In QM we define the force operator F as (in the Heisenberg picture)
<br /> F = \frac{1}{i\hbar}[p, H] + (d_t F)(t)<br />

Shouldn't the second term be the derivative of p, not F?

What I can't understand is that usually (actually, always) we write
<br /> F = \frac{1}{i\hbar}[p, H]<br />
and neglegt the last time derivative. How can we be so certain that the force is time-independent?


Niles.

 

[Niles]

You are right, it is the derivative of p. But the velocity is not necessarily time-independent?

 

[vela]

You're looking about the derivative of the operator itself, not the derivative of the momentum of the particle. Second, ∂p/∂t ≠ 0 only if the operator has an explicit time dependence.

 

[Niles]

You are right, thanks for that. In that case it is obvious that the last term is zero.Niles.