What is the Expectation Value of Momentum for a Wave Function?

[Feldoh]

Homework Statement

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

Homework Equations

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

\hat{p} = -i \hbar \frac{\partial}{\partial x}

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 <p> = \int^{\infty}_{-\infty} \psi^{*} \hat{p} \psi dx = \int^{\infty}_{-\infty} \hat{p} \psi^{*} \psi dx ?

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

So...

&lt;p&gt; = \int^{\infty}_{-\infty} R exp(-i S) \hat{p} R exp(i S) dx

= -i\hbar \int^{\infty}_{-\infty} R exp(-i S) * [R&#039; exp(i S) + i R S&#039; exp(i S)]dx

= -i\hbar \int^{\infty}_{-\infty} R R&#039; + i R^2 S&#039; dx

= ?

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.

 

[CompuChip]

Can I just say that &lt;p&gt; = \int^{\infty}_{-\infty} \psi^{*} \hat{p} \psi dx = \int^{\infty}_{-\infty} \hat{p} \psi^{*} \psi dx ?

No, you cannot.
The first equality is true, but the second is not. You cannot just pull an operator through a function, because
\psi^* \frac{\partial}{\partial x} \psi \neq \frac{\partial}{\partial x} \psi^* \psi
which is ambiguous too, as it could mean either
\frac{\partial}{\partial x} \left( \psi^* \psi \right) = \frac{\partial \psi^*}{\partial x} \psi + \psi^* \frac{\partial\psi}{\partial x}
or
\left( \frac{\partial}{\partial x} \psi^* \right) \psi

Instead you just plug in the wave-function. Start by writing down what is
\hat p \psi(x, t)

 

[Feldoh]

I all ready did apply the momentum operator like that in my derivation:

i\hbar [R&#039; exp(i S) + i R S&#039; exp(i S)]

I just assumed that the second expression was wrong and did it the (semi)correct way.

 

[CompuChip]

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.

 

[Feldoh]

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.