Given that the eigenstates of the position operator can be written as $\delta(x-x')$, and suppose we measure a particle in an infinite potential with walls at $x=0$ and $x=L$. I measure the particle to be in the position $x=L/2$, so the particle is in the eigenstate $ |x \rangle = \delta(x-L/2)$. Suppose now that I want to measure the energy of the particle. The eigenstates of the energy operator are given by:

$$ |\psi_n\rangle = \sqrt{\frac{2}{L}}\sin \left( \frac{n\pi x}{L} \right) $$

In order to measure energy I understand that I have to expand the original eigenstate in terms of the new energy eigenstates: $$ |x\rangle = \sum|\psi_n\rangle\langle\psi_n|x\rangle $$ where the probability of collapse into the eigenstate is given by:

$$ P_n = |\langle\psi_n|x\rangle|^2 $$

But now I sort of run into an issue. Sure then, I can say that: $$ \langle\psi_n|x\rangle = \int \sqrt{\frac{2}{L}}\sin \left( \frac{n\pi x}{L} \right)\delta(x-L/2)dx $$ and since

$$ \int \delta(x-x')f(x)dx = f(x') $$ I can say

$$ \langle\psi_n|x\rangle = \sqrt{\frac{2}{L}}\sin \left( \frac{n\pi }{2} \right) $$ and, $$ P_n=|\langle\psi_n|x\rangle|^2 = \frac{2}{L}\sin^2 \left( \frac{n\pi}{2} \right) $$

I know that this means that all odd values of n are equally probably and all even values are not possible, but probability is supposed to be dimensionless, so what's happened here? What rookie error have I made?

  • $\begingroup$ Rookie. $|\psi\rangle$ is dimensionless, but $|x\rangle$ has dimension $L^{-1/2}$, as you may check from $\langle x| x'\rangle=\delta(x-x')$. $\endgroup$ – Cosmas Zachos Mar 23 at 22:05
  • $\begingroup$ Gotcha. So would it be correct to say that $\psi (x) = \langle x|L/2 \rangle$? And $|\psi\rangle = |L/2\rangle$? So then if my energy eigenstates are represented by $|\phi\rangle$ then $P_n = |\langle \phi| L/2 \rangle|^2$? $\endgroup$ – monkeyofscience Mar 23 at 23:01
  • 2
    $\begingroup$ No, normalized w.f.s have dimension $L^{-1/2}$, but what you wrote has dimensions of 1/L, like a delta function. So that is your dimensional mismatch (your "And ...?"). You may try a truly normalized Gaussian w.f. with labile width, and take the width to zero in a limit, but the expressions you write in this comment are visibly inconsistent. $\endgroup$ – Cosmas Zachos Mar 23 at 23:10
  • $\begingroup$ Dang it. So how on earth do I get the position eigenfunctions to a point where I can talk meaningfully about the probability of an energy eigenfunction? $\endgroup$ – monkeyofscience Mar 23 at 23:17
  • $\begingroup$ Yes, basically it says, the eigenfunction of the position operator can be written as $\delta (x-x')$ such that $\hat{x}\delta (x-x') = x'\delta(x-x')$. A particle is in an infinite potential well with walls at $x=0$ and $x=L$. A measurement of its position finds $x=L/2$. I am supposed to show that in a subsequent measurement of the energy, it is equally probable to find the particle in an odd-n energy eigenstate. $\endgroup$ – monkeyofscience Mar 23 at 23:39

The formula $$ P_n = |\langle\psi_n|\psi\rangle|^2 $$ assumes that the pre-measurement state $|\psi\rangle$ and the observable's eigenstates $|\psi_n\rangle$ are both normalized to be unit state-vectors. In other words, the formula for arbitrary non-zero state-vectors is $$ P_n = \frac{|\langle\psi_n|\psi\rangle|^2}{ \langle\psi_n|\psi_n\rangle\,\langle\psi|\psi\rangle}. $$ Notice that this expression for $P_n$ is dimensionless by construction.

The problem with the case described in the OP is that $|x\rangle=\delta(x-L/2)$ is not normalizable: it does not belong to the Hilbert space, so it can't be used for the pre-measurement state $|\psi\rangle$.

That's not a problem in principle, because real position-measurments don't have infinite precision, so the state after a real position-measurement would not be $|x\rangle$. It would be some normalizable state-vector $|\psi\rangle$ whose corresponding wavefunction is sharply concentrated near a particular position, but with a non-zero width that makes it normalizable.

  • 1
    $\begingroup$ How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|\langle x|\psi\rangle|^2$ ? $\endgroup$ – Aaron Stevens Mar 24 at 2:15
  • 1
    $\begingroup$ @AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|\langle x|\psi\rangle|^2/\langle\psi|\psi\rangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density." $\endgroup$ – Chiral Anomaly Mar 24 at 2:21
  • 1
    $\begingroup$ @AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed. $\endgroup$ – Chiral Anomaly Mar 24 at 2:22
  • 1
    $\begingroup$ That all makes sense. It just seems odd to me that one is doable and one is not because $|\langle x|\psi\rangle|^2=\langle \psi|x\rangle\langle x |\psi\rangle=\langle x|\psi\rangle\langle\psi|x\rangle=|\langle\psi|x\rangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that). $\endgroup$ – Aaron Stevens Mar 24 at 2:28
  • 1
    $\begingroup$ @AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric. $\endgroup$ – Chiral Anomaly Mar 24 at 2:31

The problem, as formed, is malformed and meaningless, as it is obviously dimensionally inconsistent, as pointed out in my comments. Nevertheless, there is method in its madness: it has, of course, a good point, and can be redeemed/salvaged by tweaking it into something more meaningful.

First, recall the $\delta$-function is but the vanishing-width limit of the Gaussian, $$ \delta(x)= \lim_{a\to 0}\frac{1}{a\sqrt{\pi}} e^{-x^2/a^2}. $$ Note $a$ has dimensions of length, so the above has dimensions of inverse length: a warning sign. Since the integral of this is 1, but the integral of its square is singular, you'd better not use this as a wavefunction!

If you wanted a wavefunction peaking at 0, you might as well call the above a probability density, instead, and take your wavefunction as its square root, before taking the limit, $$ \psi_a(x)=\langle x|\psi\rangle = \frac{1}{\sqrt{a}~\pi^{1/4}}e^{-x^2/2a^2}, $$ obviously normalized, and with the right dimensions.

In your case, you center it at L/2, so it is $$ \psi_a(x)=\langle x|\psi_a\rangle = \frac{1}{\sqrt{a}~\pi^{1/4}}\exp \left ({-\frac{(x-L/2)^2}{2a^2}}\right ) , $$ so that $$ P_n= |\langle \psi_n|\psi_a\rangle |^2 , $$ with $$ \langle \psi_n| \psi_a\rangle=\frac{\sqrt{2}}{\sqrt{aL}~\pi^{1/4}}\int_0^L dx ~ \sin (n\pi x/L)~e^{-\frac{(x-L/2)^2}{2a^2}} , $$ now dimensionless.

From the symmetry of the integral, you exclude even values for n, as you did, and shift and rescale variables to $$ \langle \psi_n| \psi_a\rangle=\frac{\sqrt{2a}}{\sqrt{L}~\pi^{1/4}}\int_{-L/2a}^{L/2a} dy ~ \sin (n\pi (ya/L +1/2))~e^{-y^2/2} . $$ You note that in the $a\to 0$ limit, this is independent of n, but, of course, the normalization is comeasurately going to 0, as it should, given equipartition to an infinity of equal modes.

  • $\begingroup$ Thanks for all your help. Spoke to lecturer today. She says, yes it's physically unrealistic, but that doesn't matter, because you can still get across the general idea of equal probability for finding the particle in odd-n energy eigenstates, and zero for even-n eigenstates, that was all the question was intending to show. $\endgroup$ – monkeyofscience Mar 29 at 5:14
  • 1
    $\begingroup$ Yes, so this is how one always fixes it... she should know this standard move.... $\endgroup$ – Cosmas Zachos Mar 29 at 11:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.