# Matrix elements of the free particle Hamiltonian

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.

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)].