# Particle In a box, momentum, and velocity

On a homework assignment, we are give the width of a well $a$, the mass of the particle $m$, and we want to find the average velocity of the particle at the $n=1$ state. Here is my attempt at the solution:

Using, $\psi (x) = \sqrt{\frac{2}{a}} \sin(\frac{\pi x}{a})$, I was able to integrate $$\langle p^2\rangle=\int_0^adx\psi^*(x) (-i)^2\hbar ^2\frac{d^2}{dx^2} \psi (x)$$ which yielded $$\langle p^2 \rangle = \frac{\hbar ^2\pi ^2}{a^2} \, .$$

I'm stuck on the remaining parts. The problem gives us a hint to use the relation $p^2 = (mv)^2$. I'm not sure how I would go about finding average velocity from $\langle p^2\rangle$, since $\langle v^2\rangle$ and $\langle v\rangle^2$ are not equivalent.

According to the problem, the speed is supposed to come from the kinetic energy, which I don't understand (I derived average KE by using $\langle p^2 \rangle$, so I don't understand why finding KE is necessary).

• I'm stuck on your wavefunction. Is there a typo? What I see is a constant function, and besides it being the wrong wavefunction, its derivatives are zero ... Is it an infinitely deep well? – garyp May 14 '14 at 21:02
• Yeah, sorry about that. I forgot to out in the sine portion. I've corrected it in the question. – user46558 May 14 '14 at 21:04
• Also, I made a typo with $<p^2>$ which has also now been fixed. Sorry for the typos. – user46558 May 14 '14 at 21:12
• I think you are supposed to find ⟨v^2⟩, because if you calculate the expected value of p, it vanishes and if v=p/m, then the expected value of v also vanishes. But now that you have KE, you can say that vˆ2 is (2/m)KE. But I see your point, I'm not sure if what I'm saying is the answer of the question. Maybe it is interested in the dispersion of the velocity, which would be expected value of vˆ2 - 0, since v=0. – gcsantucci May 14 '14 at 23:20
• I haven't followed this line of reasoning, but $v=\partial E / \partial p$. Perhaps that's what the question is looking for with the KE hint. – garyp May 15 '14 at 1:53

The average of $v$, i.e. the expectation value $\langle v \rangle$ must be zero, as $\langle p \rangle = 0$ for the eigenstate $n$ (here $n = 1$). For $\langle v \rangle \neq 0$, the particle couldn't stay inside the box.

But $\langle p^2 \rangle \neq 0$:

$\langle p^2 \rangle = \frac{\hbar ^2\pi ^2}{a^2}$, as noted by the poster.

And with $\langle p^2 \rangle = m^2\langle v^2 \rangle$, $\langle v^2 \rangle$ can then be calculated.

Alternatively, we know the eigenvalue for $n = 1$ is:

$E_1 = \frac{h^2}{8ma^2}$ and since as $U(x) = 0$, so the total energy is equal to the kinetic energy, so:

$E_1 = \frac{m\langle v^2 \rangle}{2}$

From that identity isolate $\langle v^2 \rangle$.