P implicit vs. explicit function of time

  • Thread starter mmwave
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  • #1
mmwave
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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.
 

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  • #2
HallsofIvy
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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.
 
  • #3
mmwave
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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.
 

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