In non-relativistic quantum mechanics:

By definition $$\langle x_1|x_1\rangle=\int_{-\infty}^\infty |x_1(x)|^2dx.$$

On the other hand,

$$\langle x_1|x_2\rangle=\delta(x_2-x_1).$$

Where $x_1$ and $x_2$ are positions and $\delta$ is Dirac delta function.

Take $x_1=x_2$, $$\langle x_1|x_1\rangle=\delta(0)=\int_{-\infty}^\infty |x_1(x)|^2dx~ ?!$$

Could you please correct my false understanding?

  • 5
    $\begingroup$ Ignoring the fact that every term is undefined $\int dx |x_1(x)|^2 = \int dx \delta(x-x_1)\delta(x-x_1) = \delta(x_1-x_1) = \delta(0)$ so this is consistent. What exactly is your question? $\endgroup$ Jun 13 '17 at 17:34
  • $\begingroup$ @BySymmetry You are a true physicist. $\endgroup$
    – High GPA
    Jun 13 '17 at 17:58
  • $\begingroup$ Related: physics.stackexchange.com/q/47934/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 13 '17 at 18:26

It's a general phenomenon that if an eigenstate corresponds to a discrete eigenvalue (like the bound eigenstates of the hydrogen atom or harmonic potential hamiltonians), the state is normalizable, and if it corresponds to a continuous eigenvalue, the state is not normalizable. By "continuous eigenvalue" I mean an eigenvalue that belongs to a continuous part of the spectrum. The position operator $\hat{x}$ has a continuous spectrum, so, it's not really a problem that asking for the norm of $|x\rangle$ doesn't make sense, because we shouldn't expect that state to have a well-defined norm in the first place.

Things look weird in your example because of the delta function, but the weirdness occurs elsewhere too. For example, consider the momentum operator $\hat{p}$. An eigenvector of $\hat{p}$ with eigenvalue $p$ is:

$$\psi_p(x)=e^{i p x}$$

The integral of the norm of this over all of space clearly diverges. This is exactly the same issue as the "$\delta(0)$" problem. So one has to choose its normalization by another condition. Usually this is done via the identity $\int_{-\infty}^\infty e^{ikx}dk=2\pi\delta(x)$, which can be made rigorous other ways. Then we usually define $$\langle x|p\rangle=\psi_p(x)=\frac{1}{\sqrt{2\pi}}e^{-i p x}$$ as the correct normalization, because then

\begin{align*} \langle p | p'\rangle&=\int \langle p | x \rangle\langle x | p'\rangle dx \\ &=\int \frac{1}{\sqrt{2\pi}}e^{-ip x}\frac{1}{\sqrt{2\pi}}e^{ip' x}dx \\ &=\int \frac{1}{2\pi}e^{ix(p'-p)}dx \\ &=\delta(p'-p) \end{align*} This definitely is a distinct normalization condition from $\langle p | p\rangle=1$! Other normalizations are useful for other contexts. If I recall correctly, the "probability 1 for 1 unit cube" normalization (corresponding to the first case with no pi's in it) is more useful when calculating scattering cross sections.

This is generally all possible to make rigorous, but you don't get that much out of it. For the spectrum of the $\hat{x}$ operator, these eigenstates are not technically in the Hilbert space. To make the second set of equations rigorous, you generally act on a test function (so instead of worrying about whether $\int_{-\infty}^\infty e^{ikx}dk=2\pi\delta(x)$, you worry about whether $\int_{-\infty}^\infty\int_{-\infty}^\infty e^{ikx} f(x)dkdx=2\pi f(0)$, for various restrictions on the function $f$ and various limits of the integral. It's useful paying some attention to these technicalities, but you generally don't get much physics out of it.

  • 2
    $\begingroup$ If people do want to dig into those technicalities, this question is a good starting point. $\endgroup$ Jun 13 '17 at 21:37

Assuming everything we're doing is well defined (which is not really true, but last time I checked this wasn't math.stackexchange), we have the following:

$$\begin{align} \langle x_1 | x_1 \rangle &= \int \mathrm{dx}\ |x_1(x)|^2 \\ &= \int \mathrm{dx}\ \delta(x-x_1)^2 \\ &= \int \mathrm{dx}\ \delta(x-x_1)\delta(x-x_1) \\ &= \delta(x_1 - x_1) \\ &= \delta(0) \end{align} $$

so everything works out. The crucial step is the fourth equality, where I've used $\int \mathrm{d}x\ \delta(x-x_1) f(x) = f(x_1)$ with $f(x) = \delta(x-x_1)$.

  • $\begingroup$ I'm sorry, but even with all the non-rigorous concessions I'm willing to make when doing physics, this is not one of them: The square of the delta function does not exist. Strictly speaking, $x_1(x)$ does not exist either, $\delta$ is not a function. And $\lvert x\rangle$ are also ill-defined, and only certain manipulations are allowed with them - this site is riddled with questions where falsely assuming these are well-behaved quantum states leads to contradictions. Instead of supporting that line of thought, I think a better response is to advise caution when using position "eigenstates". $\endgroup$
    – ACuriousMind
    Jun 13 '17 at 18:31
  • $\begingroup$ I am certain I've seen many a professor write or say "We could ask the norm but obviously that's just $\delta(0)$ so [...]". Maybe not the best pedagogy, but it's not exactly blasphemy. $\endgroup$
    – user12029
    Jun 13 '17 at 19:16
  • $\begingroup$ @ACuriousMind Thank you commenting. Could you please elaborate a little bit on how $|x\rangle$ is ill-defined? $\endgroup$
    – High GPA
    Jun 13 '17 at 23:46

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