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...

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

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

$$= -i\hbar \int^{\infty}_{-\infty} R R' + i R^2 S' dx$$

= ?

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

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:

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

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.