The Hamiltonian of a free particle is $\hat H = \frac{\hat p^2}{2m}$, in position representation $$ \hat H = -\frac{\hbar^2}{2m} \Delta \;. $$ Now consider two wave functions $\psi_1(x)$ and $\psi_2(x)$ which are smooth enough (say $C^\infty$), have compact support, and their support doesn't intersect. Obviously, $\langle \psi_1 | \psi_2 \rangle = 0$.

Is the matrix element $ \langle \psi_1 | \hat H | \psi_2 \rangle $ zero?

  • On the one hand, the answer should be "obviously yes", since $$ \langle \psi_1 | \hat H | \psi_2 \rangle = -\frac{\hbar^2}{2m} \int \overline{\psi_1(x)}\, \psi^{\prime\prime}_2(x) \,dx = 0 \;. $$

  • On the other hand, it is common knowledge that wave functions spread, and after $dt$ of time their support will be infinite. Therefore I would expect* $$ \langle \psi_1 | \psi_2(dt) \rangle = \langle \psi_1 | \psi_2 \rangle -\frac i \hbar \langle \psi_1 | \hat H | \psi_2 \rangle\, dt + \mathcal O(dt^2) \neq 0 . $$

* To keep it simple, I am only evolving one of the wave functions in time. Otherwise the first order in $dt$ would be zero, but I could ask the same question about the matrix element of $\hat H^2$ appearing at the second order.

  • $\begingroup$ I don't know if this is relevant, but I have a question. Is it possible for two $C^\infty$ functions to have disjoint support? $\endgroup$ – garyp Nov 25 '18 at 13:45
  • $\begingroup$ Just for clarity, are $\psi_{1}(x)$ and $\psi_{2}(x)$ arbitrary wavefunctions or are they specifically energy eigenfunctions? I assume you mean that they are arbitrary solutions to the free-particle Schrodinger equation $\endgroup$ – N. Steinle Nov 25 '18 at 13:52
  • $\begingroup$ @garyp yes: $e^{-1/x^2} x>0$ and $e^{1/(-x)^2} x<0$ $\endgroup$ – Bruce Greetham Nov 25 '18 at 13:53
  • $\begingroup$ @garyp en.wikipedia.org/wiki/Bump_function $\endgroup$ – Noiralef Nov 25 '18 at 14:13
  • 1
    $\begingroup$ @N.Steinle They are arbitrary wave functions. In particular, they are not energy eigenfunctions -- those have infinite support. $\endgroup$ – Noiralef Nov 25 '18 at 14:15

This is an interesting question. It is reminicent of the popular (but fallacious) "proof" that $$ \exp\{ia\hat p\}\psi(x) \equiv \exp\{a\partial_x\}\psi(x)=\psi(x+a) $$ that claims that applying the exponential of the derivative operator to $\psi$ gives the Taylor expanion of $\psi(x+a)$ about $x$. The problem is that if $\psi(x)$ is $C^\infty$ and of compact support, then each term of the Taylor series is always exactly zero outside the support of $\psi(x)$ and so $\psi(x)$ can never become non-zero outside its original region of support. Of course $C^\infty$ functions of compact support do not have Taylor series that converge to the function, and the resolution of this paradox is to realise that the appropriate definition of $\exp\{ia \hat p\}$ comes from its spectral decomposition. In other words we should Fourier expand $\psi(x)= \langle x|\psi\rangle$ to get $\psi(p)\equiv \langle p|\psi\rangle $, multiply by $e^{iap}$ and then invert the Fourier expansion. Then we obtain $\psi(x+a)$.

The same situation applies here. The literal definition of $H$ as a second derivative operator is not sufficiently precise. We must choose a domain for $\hat H$ such that it is truly self-adjoint and so possesses a complete set of eigenfunctions. The action of $\hat H$ on any function in its domain is then defined in terms of the eigenfunction expansion.

| cite | improve this answer | |
  • $\begingroup$ Thanks for the answer! I've never seen that fallacious "proof", but you are right, this seems to be very much related. Just to clarify: Are you saying that a function with compact support is not in the domain of $\hat H$, or that it is in the domain but the action of $\hat H$ can not be calculated as a defivative? $\endgroup$ – Noiralef Nov 25 '18 at 14:37
  • $\begingroup$ @Noiralef It's the "proof" that is usually trotted out in most quantum books/ courses for physicists. My colleagues are always surprised when I tell them that it is fallacious. For your second point. Yes the compactly supported $\psi$ is in any reasonable definition of the domain of $\hat H$, but I'm pretty sure that the action of the unitary evolution operator $\exp\{-it \hat H\}$ (this is what you really want) on it is not given by the derivatives. $\endgroup$ – mike stone Nov 25 '18 at 15:01
  • $\begingroup$ 1. I should have said: I have seen that "proof" many times, and I have probably shown it to students myself, but I had never seen it pointed out as fallacious. 2. Okay, thanks! But sorry - I am still unsure what the answer to the original question is. Would such a matrix element (of $\hat H$) be zero? $\endgroup$ – Noiralef Nov 25 '18 at 15:35
  • $\begingroup$ @Noiralef I suspect that the matrix element is zero. It's just that the power series in $dt$ for $<\psi|\psi(t)>$ that you derive will not converge to $<\psi|\psi(t)>$. $\endgroup$ – mike stone Nov 25 '18 at 16:41
  • $\begingroup$ The "fallacious proof" applies to $C_{0}^{\infty}$ only? I believe it is correct for Schwartz test functions of $\mathcal{S}(\mathbb{R})$. $\endgroup$ – DanielC Nov 26 '18 at 16:26

The core of the issue is that, for unbounded operators $\hat A$, the operator exponential is not defined in terms of the power series $\exp(\hat A) = \sum_{k=0}^\infty \frac{\hat A^n}{n!}$. And it can not be defined that way, as we don't have a guarantee that this series converges. Instead, we use the spectral theorem to define $$ \exp(\hat A) = \int \mathrm e^a\, |a \rangle\!\langle a| \, \mathrm da \;, \tag{1} $$ where $|a \rangle\!\langle a| \, \mathrm da$ is the physicist's notation for the projection-valued measure $\mathrm dP_a$. Crucially, this is the definition used in Stone's theorem on strongly continuous unitary groups.

This means in particular that the time evolution of $|\psi_2\rangle$ is not $|\psi_2(\mathrm dt)\rangle = |\psi_2\rangle - \frac{\mathrm i\, \mathrm dt}{\hbar}\hat H |\psi_2\rangle + \mathcal O(\mathrm dt^2)$ as suggested in the question. Hence it is not a contradiction that $$\langle \psi_1 | \hat H | \psi_2 \rangle = 0 \;. $$

Side note: As explained in [Reed, Simon (1981), VIII.3], definition (1) agrees with the power series for the case of bounded $\hat A$. Further, for all $|\psi\rangle$ that can be written as $|\psi\rangle = \int_{-M}^M |a \rangle\!\langle a|\varphi\rangle \, \mathrm da$ for some $M \in \mathbb R$ and some $|\varphi\rangle$, the power series $\sum_{k=0}^\infty \frac{\hat A^n}{n!} |\psi\rangle$ converges to $\exp(\hat A)|\psi\rangle$ [Reed, Simon (1981), VIII.5].

As mentioned in the answer by mike stone, there is a simpler example demonstrating the same problem. Let $D(\alpha) = \exp(\mathrm i \alpha \hat p)$ be the translation operator ($\hbar=1$). Using definition (1), we immediately see that $$ \langle x | D(\alpha) | \psi \rangle = \int \mathrm e^{\mathrm i \alpha p} \langle x | p \rangle \langle p | \psi \rangle \,\mathrm dp = \langle x+\alpha | \psi \rangle \;. $$ If $\psi$ has compact support, this is obviously different from $$ \sum_{k=0}^\infty \frac{ \langle x | (\mathrm i \alpha \hat p)^n | \psi \rangle }{n!} = \sum_{k=0}^\infty \frac{ (\alpha \partial_x)^n }{n!} \langle x | \psi \rangle = \sum_{k=0}^\infty \frac{(\mathrm i\alpha)^n}{n!} \int p^n \langle x | p \rangle \langle p | \psi \rangle \,\mathrm dp \;. $$ The latter expression is only correct if we can exchange the order of the integral and the series, as explained also in [Holstein, Swift (1972)].

| cite | improve this answer | |
  • 1
    $\begingroup$ I already accepted the answer given by mike stone, but it made me take out my copy of Reed&Simon again and read a bit more. Posting my understanding here in case more people are curious. $\endgroup$ – Noiralef Nov 26 '18 at 15:45
  • $\begingroup$ Can you give the full reference of Holstein & Swift? $\endgroup$ – DanielC Nov 26 '18 at 21:13
  • 1
    $\begingroup$ @DanielC doi.org/10.1119/1.1986678 $\endgroup$ – Noiralef Nov 26 '18 at 22:19

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.