Expectation of Momentum Operator

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SUMMARY

The discussion focuses on proving that the expectation value of the momentum operator, <\hat{p}>, remains constant over time for a free particle in one dimension. The professor advises using the integral <\hat{p}> = \int{p |\phi(k,t)|^2 \, dk} instead of the direct calculation method <\hat{p}> = \int{\Psi^* \hat{p} \Psi \, dx}. The initial state is expressed as \Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\phi(k)e^{ikx} \, dk}, leading to the Fourier transform relationship \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\Psi(x,0)e^{-ikx} \, dx}. The discussion emphasizes the need to derive \phi(k,t) to demonstrate that d

/dt = 0.

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cepheid
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I need help getting started on this problem:

A free particle moving in one dimension is in the initial state [itex]\Psi(x,0)[/itex]. Prove that [itex]<\hat{p}>[/itex] is constant in time by direct calculation (i.e. without recourse to the commutator theorem regarding constants of the motion).


Our professor strongly advised against doing it this way:

[tex]<\hat{p}> = \int{\Psi^* \hat{p} \Psi \, dx}[/tex]

instead saying that we should try this:

[tex]<\hat{p}> = \int{p |\phi(k,t)|^2 \, dk}[/tex]

Since we know that:

[tex]\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\phi(k)e^{ikx} \, dk}[/tex]

and therefore

[tex]\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\Psi(x,0)e^{-ikx} \, dx}[/tex]

Yay Fourier transforms

My questions are:

1. Practically speaking, how does one go about expressing phi(k) using this method?

2. Even if I get a reasonable expression for phi(k), what about the time dependence? How do I then get phi(k,t)? I'm going to need that in order to find <p> and show that d<p>/dt = 0.
 
Last edited:
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Since the particle is free, then the whole time dependence of the state vector (it's actually a functional) can be factorized and, by virtue of being expressed in terms of a complex exponential, vanishes when you consider this baby

[tex]\langle \Psi (t) |\hat{p}| \Psi (t) \rangle[/tex]

Daniel.
 

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