One of the postulates of quantum mechanics is that, given a quantum state $\psi_{0}$ at time $t=0$, the state of the system at a posterior time $t > 0$ is given by $\psi_{t} = e^{-iHt}\psi_{0}$, where $H$ is the Hamiltonian operator of the system.

The operator-valued function $\mathbb{R}^{+}\ni t \mapsto U(t) = e^{-iHt}$ is an example of what mathematicians call a strongly continuous one-parameter unitary group. As a matter of fact, I have seen in many mathematical physics books the above postulate being restated as follows: given an initial quantum state $\psi_{0}$, the state at a posterior time $t > 0$ is given by $\psi_{t} = U(t)\psi_{0}$, where $U(t)$ is a strongly continuous one-parameter unitary group. It turns out that Stone's Theorem ensures that for each strongly continuous one-parameter unitary group $U(t)$, there exists a self-adjoint operator $H$ such that $U(t) = e^{-iHt}$, so we are now back to the previous formulation by taking $H$ to be the Hamiltonian of the system.

This second approach, however, made me think about the nature of this postulate. Stone's Theorem ensures the existence of some self-adjoint operator $H$, which is not necessarily the Hamiltonian of the system. So, my question is pretty basic and straightfoward: is there any reason or interpretation why the time-evoution is defined in terms of the Hamiltonian $e^{-iHt}$, and not any other self-adjoint operator/observable? Or is it just another postulate of quantum mechanics that we should accept without any apparent reason?

Note that, by setting $U(t) = e^{-iHt}$, the associated Schrödinger equation is then reduced to finding eigenvalues of $H$; of course, if another observable, say $A$, was used instead of $H$, one would find eigenvalues of $A$ instead. Hence, I would expect that this postulate has some sort of "we want to find a basis of eigenvectors of $H$" explanation, but I am not quite sure if this is the reason behind the postulate, since it is not always possible to find such a basis of eigenstates anyways.

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    $\begingroup$ As a motivation: In classical mechanics the Hamiltonian is the generator of time-evolution (too). $\endgroup$ Mar 25 at 19:03
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    $\begingroup$ What do you mean by the Hamiltonian of the system, if not the operator which generates time-evolution? $\endgroup$
    – J. Murray
    Mar 25 at 19:10
  • $\begingroup$ The usual sign convention is $U(t)=e^{-iHt}$ not $e^{+iHt}$ $\endgroup$
    – hft
    Mar 25 at 19:16
  • $\begingroup$ @hft fixed the sign convention. Thanks! $\endgroup$
    – IamWill
    Mar 25 at 19:26
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    $\begingroup$ That's wrong. As @TobiasFünke wrote above, that's also what happens classically. The Hamiltonian is a function on phase space that describes the dynamical evolution of the system through the EoM (Hamilton's equations classically, Schrödinger equation in a nR quantum system). In other words, the Hamiltonian generates time evolution. $\endgroup$ Mar 25 at 19:41

1 Answer 1


You should re-read the statement of Stone's theorem : it doesn't ensure the existence of some self-adjoint operator $H$ associated to a given strongly continuous one-parameter unitary group $U(t)$, but precisely a unique one, and vice versa, hence the unambiguous correspondence between the Hamiltonian and time evolution in the present case.

  • $\begingroup$ I would suggest to modify the sentence that says that "quantum mechanics postulates Schroedinger's equation". What you are trying to express about time evolution in QM is far broader than the (very limited and unphysical) SE. $\endgroup$ Mar 25 at 20:47
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    $\begingroup$ @FlatterMann As you wish, that's done. $\endgroup$
    – Abezhiko
    Mar 26 at 7:44
  • $\begingroup$ The Schroedinger equation is $i \frac{\partial}{\partial t} \left | \psi \right > = H \left | \psi \right >$. Nothing about this is limited, unphysical or different from time evolution. $\endgroup$ Mar 26 at 11:45

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