2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ In Zee's book, it says that "by the operator $e^{-iHT}$. So maybe $\hbar = 1$ is assumed in the book? $\endgroup$ – War Oct 14 '12 at 13:03
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ In Zee's book, it says that "by the operator $e^{−iHT}$. So maybe ℏ=1 is assumed in the book? $\endgroup$ – War Oct 14 '12 at 13:03
  • 1
    $\begingroup$ I guess, it is a common practice. You can check the text before if it is mentioned. $\endgroup$ – Ondřej Černotík Oct 14 '12 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.