Dispersion of Momentum Operator - Quantum Mechanics

Hi, hope this is in the right thread, just a little stuck on a Quantum mechanics question.

$\displaystyle \Psi(x) = \sqrt{\frac{2}{a}} \sin{\frac{n \pi x}{a}}$ for $\displaystyle 0 < x < a$, zero otherwise.

I'm then asked to calculate the dispersion $\displaystyle \Delta x$ and the dispersion $\displaystyle \Delta p$. For n = 1, show $\displaystyle \Delta x \Delta p > \frac{\hbar}{2}$

I can easily calculate $\displaystyle \Delta x$, but I'm not really sure about $\displaystyle \Delta p$, which I'm taking as the dispersion of the momentum?

I've tried calculating the expectation of the momentum as:

$\displaystyle \int^a_0 -i \hbar \frac{d}{dx} (\sqrt{\frac{2}{a}} \sin{\frac{n \pi x}{a}})^2$ and following it through this way, but then this gave me the dispersion of the momentum as zero, which doesn't fit in with the question.

Any help on this would be greatly appreciated.

Thanks in advance :)

Craig

Re: Dispersion of Momentum Operator - Quantum Mechanics

Hello Craig.

It's been a while since I've posted but I hope I can help you. I'm guessing that if this were a homework problem, the deadline would have long passed so I'll go into more detail than I normally would. I have still left out some of the calculations (long and tedious) for you to fill in.

Firstly, it's best to think about these problems physically as much as possible, this can guide you on the way to your solution. So, for example, before we start cranking out the calculation for the expectation of the position, we should kind of guess the expected position to be the centre of the well (a/2).

Mathematically we find

$\displaystyle \langle x \rangle_\psi = \frac{2}{a} \int_{0}^{a} x \sin^2\left(\frac{n\pi x}{a}\right) \mathrm{d}x = \frac{a}{2}$ ,

as expected.

Furthermore, the uncertainty comes out as

$\displaystyle \Delta_\psi^2 x = \langle x^2 \rangle_\psi - \langle x \rangle_\psi^2 = \frac{a^2}{12}\left(1-\frac{6}{n^2\pi^2}\right) $

Now for the momentum. It's a little harder to think of the momentum's real life expected behaviour. I'm sure someone can explain it better, but I think of it in terms of the particle's energy, which is usually the first thing you calculate in a QM problem.

The energy levels of the particle in this model (infinite well) are standing waves. In a standing wave there is no net flow of energy or momentum in either direction, so you should guess the expected value of momentum to be zero.

With a little bit of grunge..

$\displaystyle \langle p \rangle_\psi = \frac{-2 i \hbar}{a} \int_{0}^{a} \sin \left(\frac{n\pi x}{a}\right)\ \frac{\mathrm{d}}{\mathrm{d}x} \sin \left(\frac{n\pi x}{a}\right) \mathrm{d}x = 0$

Now, it would be easy, lazy, and incorrect to thus assume that the uncertainty of the momentum is also zero. I know for a fact that was the first thing I done when I tried solving this for the first time. So lets plug on..

$\displaystyle \Delta_\psi^2 p = \langle p^2 \rangle_\psi - \langle p \rangle_\psi^2 = \langle p^2 \rangle_\psi - 0 $

$\displaystyle \langle p^2 \rangle_\psi = - \frac{2 \hbar^2}{a} \int_{0}^{a} \sin \left(\frac{n\pi x}{a}\right)\ \frac{\mathrm{d^2}}{\mathrm{d}x^2} \sin \left(\frac{n\pi x}{a}\right) \mathrm{d}x = \frac{\pi^2 \hbar^2 n^2}{a^2}$

Therefore, throwing everything together,

$\displaystyle \Delta_\psi^2 x \Delta_\psi^2 p = \frac{a^2}{12}\left(1-\frac{6}{n^2\pi^2}\right) \left(\frac{\pi^2 \hbar^2 n^2}{a^2}\right) = \frac{\hbar^2}{4} \left(\frac{n^2 \pi^2}{3} - 2 \right) $

Therefore

$\displaystyle \Delta_\psi x \Delta_\psi p = \frac{\hbar}{2} \sqrt{\left(\frac{n^2 \pi^2}{3} - 2 \right)} $

I'm sure you can see how this satisfies Heisenberg's Uncertainty Principle from here ;)