# Built on Facts

## Down the Well

#### July 25th, 2008 · 7 Comments

Disclaimer for casual readers: I write posts which vary wildly in technical difficulty, this one is a little more mathematical than most. Don’t let it scare you off! Even if you’re a little lost, it’s good to have seen it.

The various worked problems I’ve been doing recently have mostly been on the intro undergraduate level. This is mainly because we’re between semesters, and I don’t really have the time to write up long and difficult problems from scratch. Once the fall semester starts up I’ll start writing up solutions to my homework problems.

To assuage my guilt over mainly doing easy problems, how about at least something that’s undergraduate physics major difficulty?

Consider a particle in the ground state of a 1-d infinite square well with sides at positions 0 and L. Suddenly the right-hand wall is moved outward so the walls are now at new positions 0 and 2L. What is the probability of observing the particle in the ground state of this new potential?

The main thing to understand with this problem is the idea that for any potential, there’s only certain quantum states that are allowed. If a particle is in an allowed state of one potential and suddenly the potential changes to something else, the particle’s new state must be expressible in terms of the new allowed states. But a wavefunction can’t just snap instantly from being one thing to being another, so there’s a problem.

Or there would be a problem if there weren’t the superposition principle. Just like you can expand a function in terms of the polynomials when you use Taylor series or in terms of trig functions when you use Fourier series, you can expand the old state in terms of the new allowed states no matter how different the new states are. It seems like magic, but it’s mathematically sound. The allowed eigenfunctions for a given potential form a complete set, so you can expand just about any square-integrable function in terms of the allowed functions over the appropriate interval.

Shall we try it here? The old ground state wavefunction was

The new allowed functions are

So the old ground state has to be expressible in terms of those new allowed functions. In other words,

With the constants Cn to be determined. How can we find those constants? Skipping the derivation (which is in any intro quantum textbook), we use the fact that

Why L for the upper limit of integration? Remember that the original ground state wavefunction is 0 outside of the box. If we forget that and instead calculate as though the sine function keeps going, we’ll get a very wrong answer. The star represents complex conjugation which doesn’t matter here, but it’s good to be accurate.

So plug in the functions with n = 1 (since we’re going from one ground state to another) and integrate. I get

Almost done. Remember it’s not the wavefunction which represents probability, it’s the square of the absolute value of the wavefunction. So the final answer is

It’s actually somewhat more probable that the particle will end up in the n = 2 state. The probability for that happens to be 0.5 exactly. The n = 3 state has a probability of about .1296, the n = 4 state has probability of about 0.00735, and for higher states the probability continues to fall like a stone. Of course as a check you can add up these values to find that it’s very nearly equal to 1, as expected.

Clear as crystal? Probably not. It took me a while to understand this stuff. But don’t worry, there’s tons and tons of good practice problems out there and we’ll end up doing a lot of them.

### 7 responses so far ↓

• 1 Uncle Al // Jul 25, 2008 at 10:23 am

A diverse view: Given the ground state 2L box, adiabatically squeeze the wavefunction until it is L. Work is done on the system by compressing its contents, force through a distance. Flashing ground state L into 2L without doing work leaves the excess energy in there to redistribute among allowed modes.

“An excited wavefunction is forced through a Ranque-Hilsch Wirbelrohr…” “8^>)

• 2 CCPhysicist // Jul 25, 2008 at 3:35 pm

Excellent observation, Uncle Al. Connects it rather nicely to quantum stat mech.

So, Matt, given an ensemble of identical initial states in box L, what is the average energy (temperature) of the ensemble of final states you end up with in box 2L? Two answers are expected: the obvious one based on the intuitive argument Uncle Al gave, and a mathematical one where you sum the infinite series to get that obvious answer.

Matt replies: It’s a royal pain. You have to evaluate

But as expected it ends up just being the energy of the ground state in the previous box of length L.

I was going to comment on your last paragraph. Teaching QM, which is intrinsically a matrix theory, using wave mechanics runs into those conceptual problems because of that odd thing about the vector having an infinite number of basis states. If you start with spin (e.g. polarization of light) as your model, where the vectors have finite dimensions, projecting onto a new basis is no big deal. You can even see it happen. Then, if your math background is appropriate, the problem you just did is more obviously that of a special type of Fourier expansion.

• 3 Ron // Jul 25, 2008 at 6:55 pm

I graphed the functions to picture the superposition here.

Matt replies: Thanks, that’s pretty cool. If you’re interested, all the coefficients for each term in the series is given exactly by

The only tricky case is n = 2 where the denominator becomes 0, but the limit works out to be 1/sqrt(2) as noted previously.

• 4 Ron // Jul 25, 2008 at 7:41 pm

Thanks! Here’s a picture with the general term.

• 5 Carl Brannen // Jul 26, 2008 at 12:33 am

I love this calculation. I think that this is the heart and spirit of quantum mechanics. My feeling is that these sorts of transition probability calculations should also lie at the foundations of elementary particle theory (which is why I do things the way I do).

This reminds me of a practical calculation, one which was my favorite from my QM student days 30+ years ago. It is more practical than this, but otherwise similar.

Suppose you have an atom with nucleus with charge +Z and a single electron in the 1S state. Suppose that the nucleus undergoes a beta decay, what is the probability that the single electron remains in the 1S state?

In beta decay, a neutron converts itself into a proton, an electron, and an anti-neutrino. The anti-neutrino escapes. The electron ends up with extremely high energy and also escapes. The proton remains in the nucleus (which might be in an excited state but this doesn’t matter much to the single electron).

So the system acts like a sudden change in the Z to Z+1. The calculation is similar to the one you’ve shown.

The stuff I do in my physics “research” is to compute the transition probabilities between different spinors. So I’m working in qubit space, that is, there is no spatial dependency so there isn’t any wave functions at least as a function of position. This is fun because it reduces QM to its bare necessitities, but (unlike the examples we’re discussing here) with spin. And it’s easier to do with density operators.

Along this line, there are two philosophies as far as teaching QFT. The usual way is to begin with scalar particles and later add spin as an additional complication. The one I would prefer is to begin with spin-1/2 qubits, and then add momentum as an additional complication. For an intro to the unusual way of learning QFT (which I think is much easier on the student in that it avoids the nasty computations resulting from Fourier transforms of momentum into position), is Quantum Electrodynamics of Qubits, Physical Review A 76, 06106 (2007).

If one wished to rewrite these two example QM problems so they were from a spin-1/2 point of view instead of a wave function point of view, you’d ask for the transition probabilities between two spinor states. But then again, that would be too easy: (1+cos(theta))/2.

• 6 CCPhysicist // Jul 27, 2008 at 7:46 pm

Nice work, Matt!

You make your profs proud. Now go make some beer bets with your fellow grad students on this problem … and apply Uncle Al’s technique to Carl’s homework (which was also in my homework back in the day).

• 7 Rob // Mar 2, 2009 at 9:37 am

Just stumbled accross this, good explanation! But when i looked through the math, surely the function will have a discontinuity at n=2? From what i see, Cn will give an equation containing two sine terms both giving a zero at n=2, and infront of one we get a 1/0. Are we taking here 0/0 to be 1 in order to give a probability of 0.5? Cheers - i hope my explanation of my problem made sense! :)