Expectation Value of Momentum for Wavepacket

  • #1
torq123
8
0

Homework Statement



What is the average momentum for a packet corresponding to this normalizable wavefunction?

[itex]\Psi(x) = C \phi(x) exp(ikx) [/itex]

C is a normalization constant and [itex]\phi(x) [/itex] is a real function.

Homework Equations


[itex]\hat{p}\rightarrow -i\hbar\frac{d}{dx}[/itex]

The Attempt at a Solution



[itex]\int\Psi(x)^{*}\Psi(x)dx = \int C^2 \phi(x)^{2}dx= 1 [/itex]

Plugging in the momentum operator and using the chain rule:

[itex]<\hat{p}> = \hbar k \int C^2 \phi(x)^2 dx - i \hbar \int C^2 \phi^{'}\phi dx [/itex]

The second term is always imaginary since [itex]\phi(x)[/itex] is real, so I said the momentum is [itex]\hbar k[/itex] which I think might be right, but for the wrong reasons? I didn't think Hermetian operators could give imaginary expectation values...
 
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  • #2
torq123 said:
[itex]<\hat{p}> = \hbar k \int C^2 \phi(x)^2 dx - i \hbar \int C^2 \phi^{'}\phi dx [/itex]

The second term is always imaginary since [itex]\phi(x)[/itex] is real, so I said the momentum is [itex]\hbar k[/itex] which I think might be right, but for the wrong reasons? I didn't think Hermetian operators could give imaginary expectation values...
Try working on the 2nd term a bit more. Hint: use integration by parts.
 
  • #3
Are you saying that the second term must be zero since [itex]\phi[/itex] vanishes at ±∞ and the integral evaluates to [itex]\phi^2(x)/2[/itex]? That makes sense to me.
 
  • #4
torq123 said:
Are you saying that the second term must be zero since [itex]\phi[/itex] vanishes at ±∞ and the integral evaluates to [itex]\phi^2(x)/2[/itex]?
That's the idea.
 
  • #5
Awesome. Thanks for the help.
 

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