A fundamental principle of quantum mechanics, as far as I can tell, states that the Hamiltonian generates time evolution. A common result about generators are the following: let $\mathrm T$ be the Hermitian generator of $\mathrm U(\tau)$, then we have

$$i\frac{\partial}{\partial\tau}\mathrm U\vert\psi\rangle = \mathrm T\vert\psi\rangle$$

If $\mathrm H$ generates $\mathrm U(t)$ (specifically time evolution this time) then, writing for the evolved state $\vert\psi'\rangle=\mathrm U\vert\psi\rangle$:

$$i\frac{\partial\vert\psi'\rangle}{\partial t} = \mathrm H\vert\psi\rangle$$ and we've lost an $\hbar$ from the real TDSE. Does this then actually imply that time evolution is generated by $\frac{\mathrm H}{\hbar}$? I ask only because I've heard from multiple places the exact statement "the Hamiltonian generates time evolution".

  • $\begingroup$ Note that $\hbar$ is a number, $\hat{H}$ is an operator. $\endgroup$
    – Kyle Kanos
    Commented Nov 8, 2014 at 15:37
  • 1
    $\begingroup$ But $\frac{\hat{H}}{\hbar}\neq\hat{H}$, so it isn't equivalent. $\endgroup$
    – theage
    Commented Nov 8, 2014 at 15:39
  • $\begingroup$ True, but what is $\hbar|\psi(t)\rangle$ vs $\hat{H}|\psi(t)\rangle$? $\endgroup$
    – Kyle Kanos
    Commented Nov 8, 2014 at 15:40

1 Answer 1


Yes, you need to put in an extra factor of $\hbar$ to get the units to work out. This is easily seen since, if an operator $T$ generates evolution in a parameter $\tau$, the corresponding evolution operator is $e^{iT\tau}$, so $T\tau$ must be unitless. When $\tau$ is time then $T$ has to be $H/\hbar$.

This is a quirk of language usage that you'll have to get used to; when people say "the Hamiltonian generates time evolution" they are using "the Hamiltonian" to mean something proportional to the Hamiltonian, $aH$ for some nonzero constant $a$ (in this case, $a = 1/\hbar$). This is not unreasonable because rescaling the Hamiltonian by a constant doesn't change the solutions to the equations of motion, so you can consider the equivalence class $\{aH\ \forall\ a\neq 0\}$ to be "the Hamiltonian." (Apologies to the mathematicians for my horrible misuse of set notation, but the point should be clear.)

  • $\begingroup$ Cantor is spinning in his grave! $\endgroup$ Commented Jul 2, 2015 at 18:25

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