1
$\begingroup$

In Zee's QFT in a Nutshell book at p.10-11 it is piecewise said that : In quantum mechanics, for a Hamiltonian $\hat{H}=\hat{p}^2/2m$, the amplitude to propagate from a point $q_j$ to a point $q_{j+1}$ in time $\delta t$ is : $$ \langle q_{j+1}|e^{-i\delta t(\hat{p}^2/2m)}|q_j\rangle = \left(\frac{-i2\pi m}{\delta t}\right)^{1/2} e^{i\delta t (m/2) [(q_{j+1}-q_j)/\delta t]^2} $$ Because of this, I am lead to think that the probability density to go from a point $q_j$ to a point $q_{j+1}$ in time $\delta t$ is : $$ |\langle q_{j+1}|e^{-i\delta t(\hat{p}^2/2m)}|q_j\rangle|^2 = \frac{2\pi m}{\delta t} $$ which doesn't seem to make any sense because it doesn't depend on $q_j$ and $q_{j+1}$.

What is going on here?

$\endgroup$
  • $\begingroup$ Are you sure you reproduced the formulas from Zee correctly ? For instance there is a square missing at the $\hat{p}$. $\endgroup$ – Frederic Thomas Jan 29 at 17:25
  • $\begingroup$ @FredericThomas Yes a 2 was missing, then copy pasted for a second typo. The rest should be correct. It is now corrected. $\endgroup$ – NAC Jan 29 at 17:27
2
$\begingroup$

A good question. The answer is that the states $|q_i>$ are not normalized to unity, instead they have $<q_i|q_j>= \delta(q_i-q_j)$. As a result you can't get the probability in the usual way. If you start in a position eigenstate $|q_i>$, your momentum uncertainty is infinite, and you can get arbitrarily far away from the initial point in an arbitrarily short time --- this is why the "probability" you get from squaring Tony Zee's formula is independent of how far you go. What you can do with his formula is to compute $$ \psi(q_2,t) =\int dq_1 K(q_2,q_1,t) \psi(q_1,0), $$
(where $K$ is his matrix element) which gives the amplitude to evolve (without any interactions) from a normalized wavepacket $\psi(q,t=0)$ centered about some point to the more spread-out $\psi(q,t)$. The prefactor $\sqrt{-i2\pi m/\delta t}$ is arranged so that the normalization is preserved by the evolution.

$\endgroup$

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.