# Why is amplitude of a wavefunction to propagate from $q$ to $q'$ governed by $e^{-\frac{i}{\hbar}HT}$ unitary operator?

In the textbook Quantum Field Theory by A. Zee, it says:

In quantum mechanics, the amplitude to propagate from a point $q_i$ to a point $q_f$ in time $T$ is governed by the unitary operator $e^{−\frac{i}{\hbar}HT}$, where $H$ is the Hamiltonian.

I am having a hard time understanding this. Can anyone explain this in context of Dirac's formulation and relate this to Schrodinger equation?

-
In Zee's book, it says that "by the operator $e^{-iHT}$. So maybe $\hbar = 1$ is assumed in the book? –  War Oct 14 '12 at 13:03

In Dirac notation, the propagation is given by $|q_i\rangle \to |q_f\rangle = e^{-iHT/\hbar}|q_i\rangle$. That this relation obeys the Schrodinger equation can be checked easily: Define $|q(t)\rangle = e^{-iHt/\hbar}|q_i\rangle$, where $0\le t\le T$. Then, $$\frac{\mathrm{d}}{\mathrm{d}t}|q(t)\rangle = -\frac{i}{\hbar}H |q(t)\rangle$$ (in the derivative, you need to take the derivative of the exponential only). Multiplying this gives by $i\hbar$ gives the traditional form of the Schrondinger equation $$i\hbar\frac{\mathrm{d}}{\mathrm{d}t}|q(t)\rangle = H|q(t)\rangle.$$
In Zee's book, it says that "by the operator $e^{−iHT}$. So maybe ℏ=1 is assumed in the book? –  War Oct 14 '12 at 13:03