Jtdylktuy

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
  $$ langle psi_1 | hat H | psi_2 rangle = -frac{hbar^2}{2m} int overline{psi_1(x)}, psi^{primeprime}_2(x) ,dx = 0 ;. $$




• 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.

This is an interesting question. It is reminicent of the popular (but fallacious) "proof" that
$$
exp{iahat p}psi(x) equiv exp{apartial_x}psi(x)=psi(x+a)
$$
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 $psi(x)= langle x|psirangle$ to get $psi(p)equiv langle p|psirangle $, 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 $|psi_2rangle$ is not $|psi_2(mathrm dt)rangle = |psi_2rangle - frac{mathrm i, mathrm dt}{hbar}hat H |psi_2rangle + mathcal O(mathrm dt^2)$ 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 $|psirangle$ that can be written as $|psirangle = int_{-M}^M |a rangle!langle a|varphirangle , mathrm da$ for some $M in mathbb R$ and some $|varphirangle$, the power series $sum_{k=0}^infty frac{hat A^n}{n!} |psirangle$ converges to $exp(hat A)|psirangle$ [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
$$ sum_{k=0}^infty frac{ langle x | (mathrm i alpha hat p)^n | psi rangle }{n!} = sum_{k=0}^infty frac{ (alpha partial_x)^n }{n!} langle x | psi rangle = sum_{k=0}^infty frac{(mathrm ialpha)^n}{n!} int p^n langle x | p rangle langle p | psi rangle ,mathrm dp ;. $$
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)].