Free Particle Mass m: Probability and Wavefunction Solutions

[Domnu]

Problem
A free particle of mass $$m$$ moving in one dimension is known to be in the initial state

$$\psi(x, 0) = \sin(k_0 x)$$

a) What is $$\psi(x, t)$$?
b) What value of $$p$$ will measurement yield at the time $$t$$, and with what probabilities will these values occur?
c) Suppose that $$p$$ is measured at $$t=3 s$$ and the value $$\hbar k_0$$ is found. What is $$\psi(x, t)$$ at $$t > 3 s$$?

Attempt at Solutions
Well one question I have is this: how is this a valid state function for a free particle if it is non-square integrable? Generally, for any free particle, doesn't the wavefunction have to be square-integrable?

 

[Dick]

Exactly the opposite. In general, free particles having wavefunctions like exp(ikx) are NOT square integrable. Since they aren't localized.

 

[Domnu]

Okay, so part a) would then just be

$$\psi(x,t) = e^{-i \omega_0 t} \sin k_0 x$$​

since $$\psi(x,0)$$ is an eigenfunction of the momentum operator. Now, when I try to evaluate part b), I get

$$b(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \sin k_0 x e^{-ikx} dx$$

where $$\int_{a}^b |b(k)|^2 dk$$ represents the probability of a particle having a wavenumber between $$a$$ and $$b$$. How do I actually evaluate the above integral? Also, the question asks for the probability of certain momenta... how is this allowed if momenta aren't discrete but are, instead, continuous?

 

[Dick]

You are making this much harder than it actually is. sin(k0*x) is a linear combination of exactly two momentum eigenfunctions exp(ikx). Which two? Don't think 'integral', think 'deMoivre'.

 

[Domnu]

Aah, I see it now... sin(k0*x) = 1/2 exp(i*k0*x) - 1/2 exp(-i*k0*x). But we need to normalize this... so we finally have

$$\psi(x, t) = e^{-i \omega_0 t} \cdot \left(\frac{\sqrt{2}}{2} \phi_{k_0} - \frac{\sqrt{2}}{2} \phi_{-k_0}\right)$$​

Thus, the particle has a 50% chance of having momenta $$\pm \hbar k_0$$. This solves part b). As for part c), the wavefunction collapses onto

$$\psi(x, 0) = e^{-i \omega_0 t} \cdot \frac{1}{\sqrt{2\pi}} e^{i k_0 x}$$​

 

[Dick]

Actually, it's sin(kx)=(exp(ikx)-exp(-ikx))/(2i). And you don't need to do any particular normalization. The only thing that's important is that the magnitude of the two coefficients are equal.

 

[Domnu]

Heh, yes that's what I meant :P Thanks a bunch