Expectation Value of Momentum for Wavepacket

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Homework Help Overview

The discussion revolves around calculating the average momentum for a wavepacket represented by a specific wavefunction, which includes a normalization constant and a real function. The problem involves the application of quantum mechanics principles, particularly the use of the momentum operator.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of the expectation value of momentum using the given wavefunction and the momentum operator. There is an exploration of the implications of the imaginary component in the expectation value and whether it can yield a valid physical interpretation.

Discussion Status

Some participants have offered hints and suggestions for refining the approach, particularly regarding the treatment of the second term in the momentum expectation value calculation. There is an acknowledgment of the need to consider boundary conditions and the behavior of the wavefunction at infinity.

Contextual Notes

Participants are navigating the constraints of the problem, including the normalization of the wavefunction and the properties of Hermitian operators in quantum mechanics. The discussion reflects uncertainty about the validity of certain assumptions regarding the expectation value calculations.

torq123
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Homework Statement



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

\Psi(x) = C \phi(x) exp(ikx)

C is a normalization constant and \phi(x) is a real function.

Homework Equations


\hat{p}\rightarrow -i\hbar\frac{d}{dx}

The Attempt at a Solution



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

Plugging in the momentum operator and using the chain rule:

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

The second term is always imaginary since \phi(x) is real, so I said the momentum is \hbar k 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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torq123 said:
<\hat{p}> = \hbar k \int C^2 \phi(x)^2 dx - i \hbar \int C^2 \phi^{'}\phi dx

The second term is always imaginary since \phi(x) is real, so I said the momentum is \hbar k 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.
 
Are you saying that the second term must be zero since \phi vanishes at ±∞ and the integral evaluates to \phi^2(x)/2? That makes sense to me.
 
torq123 said:
Are you saying that the second term must be zero since \phi vanishes at ±∞ and the integral evaluates to \phi^2(x)/2?
That's the idea.
 
Awesome. Thanks for the help.
 

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