When we shift the system's time from $t=0$ to $t = t$, we can define the following operator $\hat{U}$.

$$\hat{U} = e^{- i \hat{H} t / \hbar} \, .\tag{1}$$

So many (as far as I read, almost all of) documents assume $\hat{H}$ is Hamiltonian and $\hat{H} = \hat{H}^\dagger$ to prove that $\hat{U}$ is unitary.

I don't understand the reason why we can say $\hat{H}$ in (eq.1) is Hamiltonian. I believe $\hat{H}$ in $(1)$ is just an operator at this time and there is no reasonable context to conclude $\hat{H}$ here is nothing else but Hamiltonian we know.

Could anyone please tell me the reason?


1st point of view:

If you accept the Schrödinger equation $$ \mathrm i\hbar\, \partial_t \psi = \hat H \psi $$ with self-adjoint $\hat H$, then your equation 1 follows directly and $\hat U$ is unitary.

2nd point of view:

Time evolution must have the following properties:

  • $\hat U$ must be norm-preserving so that probability is conserved.
  • $\hat U$ should be invertible so that information is conserved.

Those two properties together imply that $\hat U$ is unitary. If you add the fact that $\hat U(t)$ should be a group, your equation 1 follows and it implies Schrödinger's equation with self-adjoint $\hat H$.

  • $\begingroup$ Thank you for your answer. Is the statement "$\hat{U}$ must be norm-preserving" trivial? In some documents, I saw "time-evolution must conserve probability" but I'm not convinced of the claim. I do understand $\int _{-\infty} ^{\infty} |\psi (x, t)|^2 dx = 1$ at all times but don't understand $|\psi (x, t)|$ in itself conserves at all times. $\endgroup$ – ynn May 15 '17 at 15:05
  • $\begingroup$ $|\psi(x, t)| $ is not the norm, that integral expression is the norm. You're correct that $|\psi(x, t)|$ can change when the time evolution operator acts on the wave function, but not the norm. $\endgroup$ – Señor O May 15 '17 at 15:29
  • $\begingroup$ Thank you. Now I understand the meaning of "probability must be conserved." thanks to you. $\endgroup$ – ynn May 15 '17 at 15:42
  • $\begingroup$ Does invertibility need to be assumed separately from norm preservation? Doesn't norm preservation (for the entire Hilbert space) alone guarantee unitarity? $\endgroup$ – tparker Jul 6 '18 at 17:57
  • 1
    $\begingroup$ @tparker Not if the Hilbert space is infinite-dimensional, see math.stackexchange.com/a/900311/224757 . Technically, we only need norm-preservation (isometry) and surjectivity. $\endgroup$ – Noiralef Jul 6 '18 at 20:20

The assumption is that the wave function is a probability amplitude. In particular, it's a vector that is normalized. In Dirac's notation, this is the statement: $$\langle \psi |\psi\rangle = 1.$$ This can be made more concrete with: $$\begin{align} \mathrm{ordinary\ vectors\ } &\sum_{i} \psi^\star_i \psi_i = 1, \\ \mathrm{wave\ functions\ } &\int \psi^\star(x) \psi(x) \operatorname{d}x = 1,\ \mathrm{or} \\ \mathrm{even\ wave\ functionals\ } & \int \left[\mathcal{D}\phi(x)\right] \Psi^\star[\phi(x)] \Psi[\phi(x)] = 1. \end{align}$$ Dont' worry if that last one is cryptic - it's for when you're dealing with quantum field theory.

The important point is that the wave function is confined to exist in only a part of the vector space; like how unit vectors are confined to lie on the surface of a sphere. Transformations that respect this constraint are called unitary. Thus that constraint means that every allowed transformation of $|\psi\rangle$ is unitary. Rotations, spatial translations, reflections, etc, all must respect the requirement that the wave function remains normalized.

The rest follows from the requirement that the time translation operation is a continuous change in $|\psi\rangle$ and that quantum mechanics maps on to classical mechanics on average (see: the correspondence principle). That means that $\hat{H}$, the generator of time translations in quantum mechanics, has to correspond with the generator of time translations in classical mechanics, the Hamiltonian.

There is one exception I know of to the unitarity requirement. That is time reflections. Time refletion is anti-unitary. For details, see the Wikipedia article on $T$-symmetry.

  • $\begingroup$ Thank you for your answer. What you wrote for me is not easy but seems substantive. In particular, the part "has to correspond with the generator of time translations in classical mechanics, the Hamiltonian." would help me a lot. Perhaps that's not rigorous, but such a reasoning is easy to accept to beginners of QM like me. $\endgroup$ – ynn May 15 '17 at 15:28

An intuitive approach would be to notice that the adjoint $U^{\dagger}(t)$ is the same as $U(-t)$.

Thus, if $U(t)|\psi(0) \rangle = |\psi(t)\rangle$

And $U^{\dagger}(t)|\psi(0) \rangle = |\psi(-t) \rangle$

Then $U^{\dagger}(t)U(t) |\psi(0)\rangle = |\psi(0)\rangle$

Meaning $U^{\dagger}U = I$, the requirement for a unitary operator.

  • $\begingroup$ Thank you for your answer. I don't understand the intuitive approach you told me. I believe $A^\dagger = ({}^t A)^*$. I have no idea why the adjoint of time-evolution operator $\hat{U}$ corresponds to time-reversing operator. (Perhaps, I'm on more elementary phase. I'm really a beginner of QM.) $\endgroup$ – ynn May 15 '17 at 15:13
  • $\begingroup$ Oh it's super simple - since $U(t) = e^{-iHt/\hbar}$, then $U^{\dagger}(t) = e^{+iHt/\hbar}$ which is the same as you would get for using -t as the argument for U: $U(-t) = e^{-iH(-t)/\hbar} = e^{+iHt/\hbar} = U^{\dagger}(t)$ $\endgroup$ – Señor O May 15 '17 at 15:16
  • 1
    $\begingroup$ That's what I want to ask at this post. I think you assume $H^\dagger = H$. I don't know the reasonable reason, so I posted this entry. $\endgroup$ – ynn May 15 '17 at 15:20
  • 1
    $\begingroup$ That's true just by the definition of the hamiltonian. If you put a non-hermitian operator in place of H, then you're correct that U would not be unitary. $\endgroup$ – Señor O May 15 '17 at 15:23
  • 1
    $\begingroup$ Thanks to you and other helpers, now I understand what I wanted to know. Thank you again. $\endgroup$ – ynn May 15 '17 at 15:44

protected by Qmechanic May 15 '17 at 17:26

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.