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Expectation Value of Momentum

  • Thread starter Feldoh
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  • #1
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Homework Statement


Consider a wave function [tex]\psi (x,t) = R(x,t) exp(i S(x,t))[/tex] what is the expectation value of momentum?


Homework Equations


[tex] <f(x)> = \int^{\infty}_{-\infty} \psi^* f(x) \psi dx[/tex]

[tex] \hat{p} = -i \hbar \frac{\partial}{\partial x} [/tex]


The Attempt at a Solution


This is for an intro to modern class so I don't really have a formal background with eigenvalues/vectors yet so this is a bit confusing.

Can I just say that [tex] <p> = \int^{\infty}_{-\infty} \psi^{*} \hat{p} \psi dx = \int^{\infty}_{-\infty} \hat{p} \psi^{*} \psi dx[/tex] ?

If so by the normalization condition <p> = -i hbar which I don't think can be the case.

So...

[tex]<p> = \int^{\infty}_{-\infty} R exp(-i S) \hat{p} R exp(i S) dx[/tex]

[tex]= -i\hbar \int^{\infty}_{-\infty} R exp(-i S) * [R' exp(i S) + i R S' exp(i S)]dx [/tex]

[tex]= -i\hbar \int^{\infty}_{-\infty} R R' + i R^2 S' dx [/tex]

= ???

I don't really see anything from there...

I'm tempted to just say 0, but I'm not sure that the function being evaluated is odd.
 
Last edited:

Answers and Replies

  • #2
CompuChip
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Can I just say that [tex] <p> = \int^{\infty}_{-\infty} \psi^{*} \hat{p} \psi dx = \int^{\infty}_{-\infty} \hat{p} \psi^{*} \psi dx[/tex] ?
No, you cannot.
The first equality is true, but the second is not. You cannot just pull an operator through a function, because
[tex]\psi^* \frac{\partial}{\partial x} \psi \neq \frac{\partial}{\partial x} \psi^* \psi[/tex]
which is ambiguous too, as it could mean either
[tex]\frac{\partial}{\partial x} \left( \psi^* \psi \right) = \frac{\partial \psi^*}{\partial x} \psi + \psi^* \frac{\partial\psi}{\partial x} [/tex]
or
[tex]\left( \frac{\partial}{\partial x} \psi^* \right) \psi[/tex]

Instead you just plug in the wave-function. Start by writing down what is
[tex]\hat p \psi(x, t) [/tex]
 
  • #3
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I all ready did apply the momentum operator like that in my derivation:

[tex]i\hbar [R' exp(i S) + i R S' exp(i S)][/tex]

I just assumed that the second expression was wrong and did it the (semi)correct way.
 
  • #4
CompuChip
Science Advisor
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Ah, I see that now.
Yes you did it right.
The result isn't something very beautiful, but I suppose one would not expect that... after all the wave function depends on R and S, both of which are arbitrary functions on spacetime, so one cannot "predict" the expectation value on physical grounds.
 
  • #5
1,341
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Ah, all right I just had in my mind that it would come out to be something the come be simplified a bit more. Thanks.
 

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