I am not so familiar with functional analysis. But in my impression, the Hamiltonian $H$ is often not defined everywhere on the Hilbert space. On the other hand, the time evolution operator $U(t)$, which is formally defined as $U(t)= \exp(-iH t)$, as a unitary operator, can be (and is) defined everywhere on the Hilbert space.

How to reconcile these two facts? We can recover the Hamiltonian from the time evolution operator, right? Then how does the domain shrink in this process?


The domain does not shrink, simply the function $t\mapsto U(t)\psi$ is continuous for all $\psi\in \mathscr{H}$ but it is in general differentiable only when $\psi$ belongs to a dense set (the domain of the Hamiltonian). $H$ is then obtained taking the derivative in $t=0$, and looking at the action on any point of the domain of differentiability.

If and only if the Hamiltonian is bounded, then the map is differentiable everywhere, and in particular $U(t)$ is also differentiable in norm.

Conversely, from the Hamiltonian it is possible to define $U(t)$ rigorously as the complex exponential by the so-called functional calculus of self-adjoint operators. The resulting operator is defined everywhere and unitary by construction.


You recover the Hamiltonian from $U(t)$ in this way: $$ H|\psi\rangle = \lim_{t\to 0} \frac {\mathrm i} t ( U(t) - I) |\psi\rangle $$ This limit does not exist for all $|\psi\rangle$. Let $D$ be the set of all $|\psi\rangle$ where the limit does exist, then $H$ is self-adjoint on the domain $D$.

This is Stone's Theorem.


Here's an explicit example to illustrate what was already stated in the other answers, using the momentum operator ( = the generator of translations in space) instead of the Hamiltonian ( = the generator of translations in time).

Consider the familiar construction in which each vector $|\psi\rangle$ in the Hilbert space is represented by some square-integrable function $\psi(x)$ of a single variable. Define an operator $U(a)$ by $$ U(a)\psi(x)=\psi(x+a). $$ This operator is clearly well-defined on the whole Hilbert space. It is unitary, because \begin{align*} \int dx\ \psi_1^*(x)U(a)\psi_2(x) &= \int dx\ \psi_1^*(x)\psi_2(x+a) \\ &= \int dx\ \psi_1^*(x-a)\psi_2(x) \\ &= \int dx\ \big(U(-a)\psi_1(x)\big)^*\psi_2(x) \\ &= \int dx\ \big(U^{-1}(a)\psi_1(x)\big)^*\psi_2(x), \end{align*} so $U^{-1}(a)=U^\dagger(a)$. When $U(a)$ is acting on a smooth function $\psi(x)$, we can write it as $$ U(a)\psi(x) = \exp\left(a\frac{d}{dx}\right)\psi(x). $$ (I'm omitting factors of $i$ that cancel each other.) The generator $i\,d/dx$ is not defined on the whole Hilbert space. For example, it is not defined on $\psi(x)=|x|^{-1/4}\exp(-x^2)$, even though this function represents a legitimate vector in the Hilbert space, because the derivative of this function is not square-integrable (and therefore does not represent any vector in the Hilbert space). Still, the domain of $d/dx$ is dense in the Hilbert space, because any function on which it's not defined can be approximated arbitrarily well by one on which is is defined. For example, the function $\psi(x)=|x|^{-1/4}\exp(-x^2)$, can be arbitrarily well-approximated by one that remains finite as $|x|\rightarrow 0$.

  • $\begingroup$ Technically, as an $L^2$ function, $|x|e^{-x^2}$ is in the domain of the momentum operator. In fact it is an almost everywhere differentiable function, whose derivative (as an almost everywhere defined function) is still square integrable. $\endgroup$ – yuggib Dec 20 '18 at 7:53
  • $\begingroup$ An example would indeed be a square-integrable function that is no longer square integrable when acted with a derivation. For example take the function on $\mathbb{R}$ $\psi(x)=\frac{1}{x^{\frac{1}{4}}}\chi(x)$, where $\chi(x)$ is one in a neighbourhood of zero, smooth and compactly supported. When acting with the momentum operator, you get a function that is no more square integrable (the integral diverges at zero). Also, you could choose a compactly supported $C^0$ function that is not differentiable in a set with Lebesgue measure strictly bigger than zero. $\endgroup$ – yuggib Dec 20 '18 at 15:10
  • $\begingroup$ I think however that your understanding is not completely flawed, you have just to keep in mind that $L^2$ functions are defined almost everywhere, and that the momentum operator should be intended in a weak (distributional) sense. The easiest way of doing that is in my opinion by keeping in mind that in Fourier space it is the multiplication by the Fourier variable. This also allows to define functions of the momentum operator quite easily. $\endgroup$ – yuggib Dec 20 '18 at 15:11
  • $\begingroup$ @yuggib Thanks again for helping me see my mistake. I edited my answer to replace the original faulty example with a real example, obviously inspired by the one you described, but with $e^{-x^2}$ in place of the factor $\chi(x)$ with compact support, to make the example completely explicit. $\endgroup$ – Chiral Anomaly Dec 21 '18 at 1:59

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