I encountered the derivation of Schrodinger time dependent equation using expansion of a unitary time propagation operator into power series in a small quantity $\delta t$, working to first order in $\delta t$:

$U(\delta t)|\psi(t)\rangle = (I - i \delta tH)|\psi(t)\rangle = |\psi(t+\delta t)\rangle \rightarrow i\frac{|\psi(t+\delta t) - |\psi(t)\rangle}{\delta t} = i\frac{\partial |\psi(t)\rangle}{\partial t} = H|\psi(t)\rangle$

Two questions to this. Firstly, where did $\hbar$ go? Is it the case that I should actually write $\epsilon$ rather than $\delta t$ and then $\epsilon$ encompasses the missing $\hbar$? It seems reasonable since that would the units match, as H is Hamiltonian.

Secondly, how do we know that H is the Hamiltonian in the first place? If I had no prior knowledge that H is the Hamiltonian, all I know is that H is some sort of a Hermitian operator; there is no $\hbar$ either, since I don't even know I need it to make the units match, since I don't know what H stands for. How do we go from a piece of maths to the actual physical interpretation?

  • $\begingroup$ In some unit systems, $\hbar=1$. $\endgroup$ – Kyle Kanos Sep 24 '15 at 13:15
  • $\begingroup$ $\hbar=1$ in natural units. So - no problem here. Furthermore, how do you know what a Hamiltonian is anyway? I guess the point is that you don't assume Hamiltonians, but you assume a reasonable time evolution operators and work from there... $\endgroup$ – Martin Sep 24 '15 at 13:40
  • $\begingroup$ I'm not sure if I know what you mean, @Martin . So how did we come up with ideas what H could be? How did we realise that H is an operator involving the potential energy and such? $\endgroup$ – Glo Sep 24 '15 at 15:27
  • $\begingroup$ What $H$ looks like comes from experiment. This is like the Newton's second law $F=ma$, which in some sense is just the definition of $F$. Only when one supplies independently the form of $F$ we can say something useful. $\endgroup$ – Meng Cheng Sep 24 '15 at 16:11
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    $\begingroup$ You don't have to guess. H should have the same unit as $1/t$, which is frequency. From the very early days of quantum mechanics, we know that frequency and energy are basically the same thing, and the proportionality constant is $\hbar$ (which you have set to 1 in your equations). $\endgroup$ – Meng Cheng Sep 24 '15 at 20:17

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