Heisenberg's Uncertainty Principle — Question 7

Question 7

A momentum boost changes the mean without changing uncertainty

A normalized state $\psi_0$ has $\langle p\rangle_0=0$ and finite variances. Form $\psi=e^{ip_0x/\hbar}\psi_0$ with constant real $p_0$. Use $\widehat p=-i\hbar\partial_x$.

1. Derive the new mean momentum and momentum second moment.

2. Find the new position and momentum variances and the change in mean kinetic energy.

Question 7 - Solution

A momentum boost changes the mean without changing uncertainty

(a) Method and calculation

$$\widehat p\psi=e^{ip_0x/\hbar}(p_0+\widehat p)\psi_0\quad\Rightarrow\quad\boxed{\langle p\rangle=p_0}.$$$$\boxed{\langle p^2\rangle=p_0^2+\langle p^2\rangle_0}.$$

(b) Interpretation and further calculation

$$|\psi|^2=|\psi_0|^2,\quad\boxed{\sigma_x=\sigma_{x,0}},\quad\boxed{\sigma_p^2=\langle p^2\rangle-p_0^2=\sigma_{p,0}^2}.$$$$\boxed{\langle K\rangle-\langle K\rangle_0=p_0^2/(2m)}.$$

Physical check

Uncertainty measures spread around the mean. A large central momentum by itself does not imply a large momentum uncertainty.