In class, my teacher was motivating Schrodinger's Equation, but there is one step that I do not understand or even have intuition for. I'll give the argument leading up to the step I do not understand for context.(adsbygoogle = window.adsbygoogle || []).push({});

Let [itex] \hat{U}(t) [/itex] be the operator that gives the wave function after time t, given the initial wave function. [itex] \langle\Psi(t) |=\hat{U}(t) |\Psi(0) \rangle [/itex]

Then [itex] \langle \Psi(t) | \Psi(t) \rangle = \langle \Psi(0) | \hat{U}^\dagger(t) \hat{U}(t)| \Psi(t)\rangle =1 [/itex]

This implies that [itex] \hat{U} [/itex] is unitary and we can write any unitary operator in the form [itex] e^{i\hat{A}} [/itex], where [itex] \hat{A} [/itex] is Hermitian. Now this is the step I do not understand: In this case, [tex] \hat{A} = \frac{-t}{\hbar} \hat{H} [/tex] where [itex] \hat{H} [/itex] is the Hamiltonian. Once we got past this step, the derivation of Schrodinger's Equation was easy, but I am confused as to how we know what A is.

Now, there are a couple math concepts used that are not completely clear, but they are at least logical. The derivative of a state is defined in the natural way and the ability to write a unitary operator as an exponential makes sense once you realize what Hermitian conjugation does to an exponential.

But how do we know the correct operator is related to the Hamiltonian in this way? Is this an axiom or was the reasoning not presented because it is too complicated? Part of the reason why this is confusing is that the only other time we have seen [itex] \hbar [/itex] used was in the commutator relation [itex] [\hat{x},\hat{p}]=i\hbar [/itex], which does not seem to be very relevant to this problem.

Edit: I said matrix exponential in the title because I was thinking of the Hamiltonian as a diagonal matrix with its energy eigenvalues on the diagonal, but I probably should have said operator.

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# Matrix exponential in QM

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