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?

  • $\begingroup$ Are you sure you reproduced the formulas from Zee correctly ? For instance there is a square missing at the $\hat{p}$. $\endgroup$ Jan 29, 2019 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. $\endgroup$
    – Noé AC
    Jan 29, 2019 at 17:27

1 Answer 1


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.


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