# QM probability to go from a point to another (Zee)

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?

• Are you sure you reproduced the formulas from Zee correctly ? For instance there is a square missing at the $\hat{p}$. Jan 29, 2019 at 17:25
• @FredericThomas Yes a 2 was missing, then copy pasted for a second typo. The rest should be correct. It is now corrected. Jan 29, 2019 at 17:27

A good question. The answer is that the states $$|q_i>$$ are not normalized to unity, instead they have $$= \delta(q_i-q_j)$$. 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.