Does the Expectation of Momentum Vanish for a Real Wavefunction?

[Domnu]

Problem
Consider a free particle moving in one dimension. The state functions for this particle are all elements of $$L^2$$. Show that the expectation of the momentum $$\langle p_x \rangle$$ vanishes in any state that is purely real. Does this property hold for $$\langle H \rangle$$? Does it hold for $$\langle H \rangle$$?

Solution
For $$\langle p_x \rangle$$, we have

$$\langle p_x \rangle = \int_{-\infty}^{\infty} \phi^* \hat{p_x} \phi dx$$
$$= - i \hbar \int_{-\infty}^{\infty} (Ae^{ikx} + Be^{-ikx})(-Aik \cdot e^{-ikx} + Bik \cdot e^{ikx}) dx$$
$$= -\hbar k \int_{-\infty}^{\infty} [A^2 - B^2]$$,

since we need to have $$kx = n\pi$$ to satisfy the condition that the wavefunction must be real. But, the above integral diverges to infinity (assuming that $$A \neq B$$).

I'll post the second and third parts a bit later, but have I correctly shown that the expectation of the momentum vanishes?

 

[Zizy]

I would do this:
First try to get purely real functions that are elements of L2; squaring wavefunction gives you something like (A^2 +B^2)* exp(2ikx) + 2AB -> 0, when x-> inf. Considering purely real functions gets you A=-B... put that in your equation in the end and you get desired result.

 

[Dick]

kx=n*pi has nothing to do with the wavefunction being real. Nor does A=(-B). The condition you want is B=conjugate(A). That must be true for every momentum component of the wave function. exp(ikx) and exp(-ikx) represent opposite direction momentum components. What does that tell you?

 

[Domnu]

Ah, yes, I understand why this must be true now... because we need $$\phi^* = \phi$$... that was a careless error on my part heh. Also, what exactly does it mean for it to 'vanish'? Does this mean go to 0 or go to infinity? Infinity would be nicer...

 

[Dick]

I've never heard of 'vanish' meaning 'go to infinity'. In your example the contribution of the exp(ikx) part is equal and opposite to the contribution of the exp(-ikx) part. They cancel. They are also orthogonal, so there are no cross terms. A general wavefunction is a superposition of such functions where A is a function of time and the wavenumber k, A(k,t). But the conclusion still holds.

 

[Domnu]

Hmm... okay i think I understand now. Thanks =)

 

[nagendra]

what is the uncertainity of a free particle moving in one dimension

 

[borgwal]

Your proof is incomplete: you focus on a very particular type of real wavefunction, namely with a fixed value for $p_x^2$. However, what you have to prove is that for *any* real wavefunction $\langle p_x\rangle =0.$ There is no need to expand your wavefunction in plane waves, although, as Dick said, it gives a simple intuitive picture for why the momentum is zero on average.