# P implicit vs. explicit function of time

## Main Question or Discussion Point

As part of proving that d/dt <p> = <-Del V> you have to use the fact that < dp/dt> = 0 when p is not an explicit function of time.
I'm not clear on what this means. Any insights to share?

From classical mechanics, if there is a potential V = -k/r then there will be a force on a particle and the momentum will evolve over time. In Q-M, this is an implicit function of time and so still <dp/dt > = 0.

An explicit function of time would be V = -kcos(t) / r and now
< dp/dt> is not equal to zero. p still evolves over time.

I can't see why implicit or explicit function makes any difference mathematically.

Related Quantum Physics News on Phys.org
HallsofIvy
Homework Helper
Does your textbook actually use the words "implicit" and "explicit"? Ah, now I see, you are talking about measurable quantities in quantum mechanics.

Notice that <dp/dt> is NOT just the derivative with respect to time. It is, rather, an operator. In order to calculate that derivative, you do "dp/dt" symbolically- that is if p= f(x,v) (x, v are position and velocity) then dp/dt= 0 even though x and v themselves depend on time- you don't use the chain rule here.

Originally posted by HallsofIvy

Notice that <dp/dt> is NOT just the derivative with respect to time. It is, rather, an operator. In order to calculate that derivative, you do "dp/dt" symbolically- that is if p= f(x,v) (x, v are position and velocity) then dp/dt= 0 even though x and v themselves depend on time- you don't use the chain rule here.
There is a subtlety here that escapes me - why being an operator means do not apply the chain rule.

The complete relation used is

d/dt <Qhat> = i/hbar * < [Hhat,Qhat]> + < dQhat/dt>

where [] = commutator, Qhat is any operator and the last derivative is a partial with respect to t. In my example, Qhat = phat the momentum operator hbar/i * d /dt.