Monday, October 17, 2011

More on Perturbation Theory and Ladder Operators

One of the big surprises for me since I started following my blog statistics is the number of hits I've gotten for my musings on ladder operators. It's a topic I've never really understood properly, so when I started writing about it last April my thoughts were pretty speculative. Reading over it today however it seems to me I got off to a pretty good start and it might be worth going a little farther down that road. Hence today's article.
What really annoyed me in university was the day the prof said, right out of the blue, "let's define an operator Q such that Q operating on energy eignestate |n> returns state |n+1>". How does it do that? The professor couldn't care less. "We can define an operator to do anything we want". I have never understood what the hell that even means.
I can't talk about an operator unless I have some idea what it actually looks like. That's why in my last article I started off by trying to construct physical interpretations of the ladder operator. I thought I was doing a pretty good job, and in the end, I said that you take your hydrogen atom, put it between the plates of the capacitor, and the differential change when you add charge to the capacitor in effect creates a differential change in the ground state of the hydrogen atom; and that differential change is exactly proportional to the first excited state. So the operator which changes the ground state to the excited state is simply an electric field in the z direction.
Of course, no one who talks about ladder operators bothers to work them out for the hydrogen atom. They only care what they do to the harmonic oscillator. Interestingly enough, my definition works for the first transition of the harmonic oscillator, because when you multiply the wave function exp(-z^2) by z, you get essentially the first excited state. So in other words, so far so good.
The problem with this definition is that it doesn't exactly work on the next level. What operator takes the pz state and changes it to the d state? It seems that multiplication by z does not quite do the trick. It certainly doesn't do the trick for the harmonic oscillator, where you have to multiply by z and then add in some of the original ground state.
You can actually see that it can't work, because if it did the second excited state would be equal to the ground state multiplied by z^2. This can't be right because this new state isn't even orthogonal to the original ground state, as it must be if it is to be an eigenstate.
There's another very different way you can see that it doesn't work, and that is from the perspective of perturbation theory. It is true that perturbation theory lets us find the new ground state of an atom in a constant field by mixing in a bit of the pz state to the s state. But are all the new eigenstates found so easily? In other words, do we find the modified pz state simply by mixing in a bit of the d state? The answer turns out to be no.
The answer has to be yes for the perturbed s state because adding a bit of pz is the simplest possible perturbation you can make. And the energy change must be linear in the perturbation, because it is simply Hooke's Law applied to an atomic spring.
The perturbation of the second energy level looks a bit the same, because when you add a field in the z direction, you apply it to a charge distribution which is centered on the origin, so you must expect the charge distribution to be displaced along the z axis. The obvious way to do this is to mix in a bit of the d state; and this certainly works: but don't forget, you also get a displacement of the charge cloud when you add in some s state! In fact, it seems the true optimization is to mix in a bit of both the s and the d states. In fact, it's not even totally clear that the net energy of the perturbed state is greater or less than the unperturbed state. I'm thinking it's actually less.
Now I'm going to recall something I said in my April 6th post when I was fishing around for possible candidates for the ladder operator. I said that applying a field in the z direction was promising. When you apply this to an eigenstate, you basically multiply the state by z. But I also said that differentiating the state with respect to z was promising, because that was the first term in a Taylor expansion which basically displaces the state by a fixed amount.
Interestingly enough, when applied to the ground state, these two operations, differentiation and multiplication by x, have the same effect. Why not just add them together?
You can verify that this gives you the correct sequence of eigenstates for the harmonic oscillator. This patch turns out, it seems, to be just what is needed to fix the ladder operator altogether! It seems that when you apply both operations to higher levels, the effect of the second one is just to fix up the error left over by the first one. I'm not sure why or how it works, but it certainly seems to do the trick on the harmonic oscillator, and it just might work on the hydrogen atom as well.
It's all very confusing, and also quite speculative, but it seems to be a promising way to try and understand what is happening. I think I'm going to quit here and maybe six months from now I'll have some more to add.

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