I always like to show the following to those who have just learned quantum theory. The commutation relation between poisition and momentum is [tex]$[x,p]=i hbar$[/tex] or [tex]$xp-px=i hbar$[/tex]. Now act on an eigenstate of the [tex]$x$[/tex] operator [tex]$|x_0rangle$[/tex] and you get [tex]$(xp-px)|x_0rangle=xp|x_0rangle- p x_0 |x_0rangle$[/tex]. Take the inner product of this state with the ket [tex]$langle x_0|$[/tex]: [tex]$langle x_0 | (xp|x_0rangle- p x_0 |x_0rangle)= langle x_0| (x_0 p – x_0 p)|x_0rangle=0$[/tex]. But if we carry out the same procedure on the right hand side of the commutation relation we get [tex]$langle x_0| i hbar |x_0rangle=ihbar$[/tex], which, last time I checked, was not zero. Snicker. It’s so mean to give this to those who’ve just learned quantum theory, but shucks, it’s also pretty fun to watch them squirm and figure out what went wrong.
O brave new quantum world!
You divided out by [tex]langle x_0 |[tex] in the second section of the equality (after applying in the ket) but left [tex]x_0[/tex] one the lefthand side inside the parantheses? Or am I missing something deeper?
Oh, fsck. that was supposed to be “You divided out by
Argh! “divided out by |x_0> in the second equality after applying the ket but left behind a x_0 on the left-hand side of the parantheses.
If this doesn’t work, I give up. 🙂
Taking the trace of both sides is also likely to give rise to hours of enjoyment.
Great! you just proved that h=0.
Yep. And I’ll bet I’m wrong. But where am I wrong? 😉
It is unfortunate that physicists often refer to calculations as “the math” and concepts as “the physics”. In reality mathematicians study the conceptual framework of calculations (and non-calculations); indeed many math papers have fewer calculations than the typical physics paper. (But I suppose that Dave is only using this terminology ironically.)
Anyway, a mathematician would note that p|x_0> is ill-defined. Despite appearances, |x_0> is not a vector in the relevant Hilbert space, L^2(R). It is an idealized eigenvector of x, which is a continuous-spectrum operator without any true eigenvectors.
As someone who has only just finished his first year of quantum (chemistry in my case) I can safely say “aiee!”
My five-second diagnosis is that for your relations to make sense one has to assume that the position and momentum operators already commute (which they don’t hence the crazy). Or in other words the expectation value of momentum (ie: ) is simply undefined. So subtracting an undefined quantity from itself doesn’t really mean anything (and quite possibly not zero).
But then again I’m a chemist, what do I know?
my attempt at math was taken to be html so I shall try again: expectation value== <x_0| p |x_0>
That’s not the problem, because x = x_dagger.
The problem is that |x_0> is, in physics speak, a non-normalizable wave function.
Well maybe afterwards you should tell the poor souls to try to see if the same problem holds for finite operators as well…
you can’t act x to the left in the first matrix element. only x_dagger can act on the bra and give the eigenvalue x_0. or at least that is what i think went wrong.
The technical responses that there are no x-eigenvalues in the Hilbert space, let alone in the domain of p, are all very well, but they are inappropriate to the level at which the question is posed. They also fail to answer the question of what happens with a wavefunction which is very highly peaked in x.
The response I would suggest is that the uncertainty principle tells us that the more x is close to definite, the more p is indefinite. Thus we are looking at something like the difference between two approximations to a very large number. We have no reason to believe that the errors in our approximations are bounded as the large number increases. In the limit we get zero times infinity minus infinity times zero which, as Allan says, is undefined.
Let’s look at the same problem for momentum eigenstates. In particular, let’s evaluate . (Let’s use periodic boundary conditions, with period L, so that the eigenstates are normalizable.)
On the one hand, you can let p act to the left, and reduce it to k .
On the other hand, you can evaluate it by explicit integration (using p = -i d/dx).
= -i/L integral exp(-ikx) d/dx x exp(ikx) dx
Note: d/dx x exp(ikx) = (1 + ikx) exp(ikx).
So we have
= -i
= -i + k
What gives? Well, the justification for d/dx acting to the left is integration by parts:
Integral f(x) d/dx g(x) dx
= f(x) g(x) – Integral (d/dx f(x)) g(x) dx
where the first term is evaluated over the “boundary”. Usually, the assumption is that this term is zero, but in the particular case f(x) = exp(-ikx), g(x) = x exp(ikx), that boundary term is just x, which is nonzero.
However, I think you can say that the problem is that x is not well-defined in a system with periodic boundary conditions, because x=0 and x=L are the same point.
Since I never developed a deep intuition about classical phase spaces (mea culpa), I’ve no physical interpretation for the commutator, but from a formal point of view…
A position eigenstate in position representation’s just a delta function. Apply momentum to it, you get the derivative of a delta function. Maybe Hardy-Littlewood maximal function theory can give this a sensible interpretation, but for my purposes it’s just infinite. Then you integrate against a delta function, which selects the infinity. Subtracting infinity from infinity’s just not well posed.
So in the end, you have to choose how big the difference between the infinities is. If they cancel, you get classical mechanics. If they don’t, you get something else. If they leave something real I believe you get diffusion.