# Friday Quantum Mechanics

“So, Blake,” I sez to myself. “You’ve been selected for multiple editions of the Skeptic’s Circle. You’ve been linked, twice, from Pharyngula. Clearly, you’re rising to astonishing heights of science-blogebrity. What worlds are left to conquer?”

“Well,” I replied. “There’s going out for a milkshake with Rebecca Watson.”

I shook my head. “Not gonna happen — she’s just too picky counting tentacles. Anything else?”

“Well, you could do what Revere warned you not to do.”

“Ah, yes, write a sixteen-part series on mathematical modeling! But the modeling of antiviral resistance isn’t really my field.”

“True, but didn’t you spend your spring break in Amsterdam a few years ago, writing that paper which was the first article Prof. Rajagopal ever graded with an A-double-plus?”

“Hey, yeah, on supersymmetric quantum mechanics and the Dirac Equation!”

“So,” I suggested to me, “why don’t you break that paper down into several blag posts, interleave it with some Bill Hicks videos so not all your readers wander away, and have yourself a continuing physics series?”

“Could work, I suppose. But that paper was written for third-term quantum mechanics students, so I’d probably have to build up to it, even just a little.”

“Bah,” I said. “At least you’ll have a purpose in life. And you can start by expounding on the canonical commutation relation for position and momentum. That’ll be your warm-up, after which you can do angular momentum and central potentials —”

“Which I do have written up somewhere,” I interposed, “since I discovered I could type LaTeX as fast as my professors could lecture.”

“Weirdo,” I said.

As I mentioned, at our last DIYU seminar we covered the canonical commutation relation specified in quantum mechanics to relate position and momentum operators. (It’s all part of bringing the mathematicians up to speed.) We used this as our starting point to derive the position-space representation of the momentum operator and thence the spatial form of the momentum eigenstates. Since the math people might not have Cohen-Tannoudji on their shelves, I thought I’d recap this in LaTeX form.

$$[X,P] = i\hbar.$$
With this in hand, and knowing some basic properties of commutators, we can calculate how X interacts with higher powers of P. Let’s start with P2:
$$[X,P^2] = [X, PP] = [X,P]P + P[X,P] = 2i\hbar P.$$
Notice that we have an expression in terms of P2, which works out to an expression involving a factor 2 and a P raised to the 1-th power. This suggests that the commutator of X with Pn will be
$$[X, P^n] = i\hbar n P^{n-1}.$$
We can prove this by induction, since we have an initial case already established. (Exercise: go do it!) Now, since we can talk about powers of P, we can also talk about functions of the operator P expanded as Taylor series. Any reasonably well-behaved function F(P) can be expressed as a sum of coefficients fn multiplied by Pn, and thus we deduce:
$$[X, F(P)] = \sum_n [X, f_n P^n] = \sum_n i\hbar n f_n P^{n-1}.$$
Inspecting this result, and pulling the multiplicative constant out in front, we recognize that the final sum gives us just the derivative of F(P) with respect to P.
$$[X, F(P)] = i\hbar \partial_P F(P).$$
By interchanging P and X in the above argument, we find that a similar relation holds for functions of the position operator:
$$[P, G(X)] = -i\hbar \partial_X G(X).$$
So, the momentum operator P “acting on” a function of X gives -i times Planck’s constant times the derivative of the function. This is an interesting result, which we’d like to extend. If this is how P relates to functions of the position operator, how does it act on states specified in terms of position? Such a result would help us tell, for example, what a momentum eigenstate looks like as a function of spatial coordinates.

To answer this question, we’re going to poke around a little and develop some machinery which we’ll use later. I define a new operator, which we’ll call a translation operator, as the exponential of some constants times P:
$$T(\lambda) = \exp\left(-\frac{i\lambda P}{\hbar}\right).$$
Why do we call this a translation? First, let’s use the equation above to calculate how T(λ) commutes with X:
$$[X, T(\lambda)] = i\hbar \left(-\frac{i\lambda}{\hbar}\right) \exp\left(-\frac{i\lambda P}{\hbar}\right) = \lambda T(\lambda).$$
We see that the commutator just gives us back T(λ) times the parameter λ. Written another way,
$$XT(\lambda) = T(\lambda)(X + \lambda).$$
Next, let’s see how the combination XT(λ) acts on a position eigenstate:
$$XT(\lambda) \ket{x} = T(\lambda) (X + \lambda) \ket{x} = (x + \lambda)T(\lambda)\ket{x}.$$
This result merits close inspection. We started with XT(λ) acting on the eigenstate |x>, which we can just as well call X acting on the state T(λ)|x>, and we ended with (x + λ) — which is just a number — times the state T(λ)|x>. Therefore, T(λ)|x> is an eigenstate of the operator X, with eigenvalue (x + λ).

Acting with the operator T(λ) pushed the state |x> over by an amount λ! Hence, T effects translations of the states it acts upon.

After Ben’s explanations of Lie algebras, we’re naturally curious about the infinitesimal generators of the translation transformation. Since we defined T as an exponential, working it out for small parameters is easy. We just get the identity, plus the small parameter times P (and some constants), plus higher-order terms we lump together because we don’t care about them:
$$T(-\epsilon) = e^{i\epsilon P/\hbar} = \mathbb{I} + i\frac{\epsilon}{\hbar} P + \mathcal{O}(\epsilon^2)$$
Momentum, we note, is (up to some constants) the generator of spatial translations. To push on this further, we introduce a general state |ψ>, which isn’t necessarily an eigenstate or anything else special; it’s just a state of the system. We then compute the Dirac bracket of this state with the position eigenstate |x> and T in the middle:
$$\langle x| T(-\epsilon) |\psi\rangle = \psi(x) + i\frac{\epsilon}{\hbar} \langle x| P |\psi\rangle + \mathcal{O}(\epsilon^2).$$
But because
$$\langle x| T(-\epsilon) |\psi\rangle = \langle x + \epsilon | \psi\rangle = \psi(x + \epsilon),$$
(we flipped signs because the operator is acting on the bra vector, not the ket, which requires taking the adjoint), the Dirac bracket becomes
$$\psi(x + \epsilon) = \psi(x) + i\frac{\epsilon}{\hbar} \bra{x} P | \psi\rangle + \mathcal{O}(\epsilon^2).$$
We can rearrange this to put the wavefunction terms all one one side, giving us
$$\langle x|P|\psi\rangle = \frac{\hbar}{i} \lim_{\epsilon \rightarrow 0} \frac{\psi(x + \epsilon) – \psi(x)}{\epsilon} = -i\hbar \partial_x \psi(x).$$
The ket |ψ> was just an arbitrary state, so this relation doesn’t depend upon our choice of ket. It’s in fact a general statement about the operator P as represented in the position basis:
$$\boxed{P = -i\hbar \partial_x.}$$
Again, we see that “acting with P” involves taking a derivative and multiplying by -i and Planck’s constant.

I leave as a quick exercise the calculation of the momentum eigenfunction in position space. It should not be too difficult to see that for an eigenket |p>,
$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} \exp\left(\frac{ipx}{\hbar}\right).$$
Because the position eigenkets form a complete basis, we can recover the momentum eigenstate by summing over them:
$$|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^\infty dx\, \exp\left(\frac{ipx}{\hbar}\right)|x\rangle.$$
A general state |ψ> can of course be projected onto <p| just as well as <x|, giving a wavefunction in momentum space, thus:
$$\bar{\psi}(p) = \langle p|\psi\rangle.$$
Exercise: what does this say about the Fourier transforms of the position and momentum wavefunctions?

## 5 thoughts on “Friday Quantum Mechanics”

1. I had significant trouble understanding lolcats, and now you go and post this? I got a C in calculus, in summer school!
Quick question(s): If I can not tell if you are being snarky, is it safe to assume that you are? Or is there some kind of snark-o-meter that I am not yet familiar with?

2. You think this is rough? They cover this kind of thing in the first or second term of undergraduate quantum mechanics — junior year, at the latest. You should try reading The n-Category CafÃ©, where the average discussion requires five years of study with ancient and poorly dubbed sages in a secret temple high in the Himalayas before you can even understand the vocabulary.

“Quick, brave Sherpa, what is that over there?”

“That is a span of groupoids, sir.”

“Is that the same as a double coset?”

“Only if the groupoids are connected and form a commutative diagram whose arrows are faithful functors, sir.”

“I see. Oh, brave Sherpa, what is that over there?”

“That’s a Lie derivative, sir. . .”

Ah, I’m killing me. . . but seriously, folks, a human being can only take so much quantum woo before they want to talk about the real thing. Not that I expect everybody to understand the real thing, or even care about the details, but I feel I’d be seriously underusing this blag if I didn’t discuss science at all the levels of which I am capable. (And, of course, riding roughshod into my extensive ignorance as frequently as possible.)

As for the snarkiness issue, I’m afraid you’ll have to use your own proper judgment in that regard.

3. That’s what I figured. Neat coin flip game, did not believe it until I actually sat there flipping for 7 hours. Thanks.

4. Oh dear god I was not expecting that below the cut. I think I kind of miss quantum, though, and it’s too bad blogger doesn’t have any nice LaTeX whatsit yet like WordPress does.

5. manigen says:

I don’t think I really appreciated until now just how much I’d forgotten. This is a great idea though – keep going and I’ll follow for as long as I’m able.