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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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.

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  • $\begingroup$ 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
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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)].

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    $\begingroup$ 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
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    $\begingroup$ @DanielC doi.org/10.1119/1.1986678 $\endgroup$ – Noiralef Nov 26 '18 at 22:19

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