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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?

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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
up vote 2 down vote accepted

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 $$ i\hbar\frac{\mathrm{d}}{\mathrm{d}t}|q(t)\rangle = H|q(t)\rangle. $$

share|cite|improve this answer
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
I guess, it is a common practice. You can check the text before if it is mentioned. – Ondřej Černotík Oct 14 '12 at 14:03

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