# Matrix elements of the free particle Hamiltonian

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.

• 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? – garyp Nov 25 '18 at 13:45
• 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 – N. Steinle Nov 25 '18 at 13:52
• @garyp yes: $e^{-1/x^2} x>0$ and $e^{1/(-x)^2} x<0$ – Bruce Greetham Nov 25 '18 at 13:53
• – Noiralef Nov 25 '18 at 14:13
• @N.Steinle They are arbitrary wave functions. In particular, they are not energy eigenfunctions -- those have infinite support. – 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.

• 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? – Noiralef Nov 25 '18 at 14:37
• @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. – mike stone Nov 25 '18 at 15:01
• 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? – Noiralef Nov 25 '18 at 15:35
• @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 $<\psi|\psi(t)>$. – mike stone Nov 25 '18 at 16:41
• The "fallacious proof" applies to $C_{0}^{\infty}$ only? I believe it is correct for Schwartz test functions of $\mathcal{S}(\mathbb{R})$. – 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)].

• 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. – Noiralef Nov 26 '18 at 15:45
• Can you give the full reference of Holstein & Swift? – DanielC Nov 26 '18 at 21:13
• @DanielC doi.org/10.1119/1.1986678 – Noiralef Nov 26 '18 at 22:19