Wave packet in free fall

Let us calculate the two relevant commutators. First, $[\hat H,\hat X]=\frac{1}{2m}[\hat P^2,\hat X]=-\frac{i\hbar}{m}\hat P$, since the potential term commutes with $\hat X$. Second, $[\hat H,\hat P]=mg[\hat X,\hat P]=i\hbar mg$. Ehrenfest's theorem gives

$$\frac{d\langle\hat X\rangle}{dt}=\frac{\langle\hat P\rangle}{m},\qquad\frac{d\langle\hat P\rangle}{dt}=-mg .$$

Integrating the second equation gives $\langle\hat P\rangle(t)=p_0-mgt$, and then the first gives

$$\langle\hat X\rangle(t)=x_0+\frac{p_0}{m}t-\frac12gt^2 .$$

The centre of the packet follows exactly the parabolic trajectory of a classical particle, whatever the shape of the packet. This is the general case for potentials of at most quadratic order, for which the mean force $-\langle V'(\hat X)\rangle$ is equal to the force at the mean position; here, the force is in fact constant.

Let us calculate the evolution of $\langle\hat P^2\rangle$. Only the potential term does not commute with $\hat P^2$:

$$[\hat H,\hat P^2]=mg[\hat X,\hat P^2]=2i\hbar mg\,\hat P, \qquad \frac{d\langle\hat P^2\rangle}{dt}=\frac i\hbar\cdot2i\hbar mg\langle\hat P\rangle=-2mg\langle\hat P\rangle .$$

Furthermore, $\frac{d}{dt}\langle\hat P\rangle^2=2\langle\hat P\rangle\frac{d\langle\hat P\rangle}{dt}=-2mg\langle\hat P\rangle$. Thus

$$\frac{d\sigma_P^2}{dt}=\frac{d\langle\hat P^2\rangle}{dt}-\frac{d\langle\hat P\rangle^2}{dt}=0 .$$

The constant force shifts all momenta by the same amount $-mgt$, without changing their spread.

Let us begin with $\hat X^2$. Only the kinetic energy contributes, and using the product rule for commutators,

$$[\hat P^2,\hat X^2]=\hat X[\hat P^2,\hat X]+[\hat P^2,\hat X]\hat X=-2i\hbar\bigl(\hat X\hat P+\hat P\hat X\bigr),$$

which gives $\frac{d}{dt}\langle\hat X^2\rangle=\frac i\hbar\cdot\frac{1}{2m}\cdot(-2i\hbar)\langle\hat X\hat P+\hat P\hat X\rangle=\frac1m\langle\hat X\hat P+\hat P\hat X\rangle$. For $\hat X\hat P+\hat P\hat X$, the kinetic term gives $[\hat P^2,\hat X\hat P+\hat P\hat X]=[\hat P^2,\hat X]\hat P+\hat P[\hat P^2,\hat X]=-4i\hbar\hat P^2$, and the potential term gives $mg[\hat X,\hat X\hat P+\hat P\hat X]=mg\bigl(\hat X[\hat X,\hat P]+[\hat X,\hat P]\hat X\bigr)=2i\hbar mg\hat X$. Thus

$$\frac{d}{dt}\langle\hat X\hat P+\hat P\hat X\rangle=\frac i\hbar\Bigl(-\frac{4i\hbar}{2m}\langle\hat P^2\rangle+2i\hbar mg\langle\hat X\rangle\Bigr)=\frac2m\langle\hat P^2\rangle-2mg\langle\hat X\rangle .$$

It follows, using the results from the first question, that

$$\frac{d\sigma_X^2}{dt}=\frac{d\langle\hat X^2\rangle}{dt}-2\langle\hat X\rangle\frac{d\langle\hat X\rangle}{dt}=\frac1m\langle\hat X\hat P+\hat P\hat X\rangle-\frac2m\langle\hat X\rangle\langle\hat P\rangle=\frac{2C}{m}, \frac{dC}{dt}=\frac1m\langle\hat P^2\rangle-mg\langle\hat X\rangle-\frac{\langle\hat P\rangle^2}{m}-\langle\hat X\rangle(-mg)=\frac{\langle\hat P^2\rangle-\langle\hat P\rangle^2}{m}=\frac{\sigma_P^2}{m}.$$

Since $\sigma_P$ is constant, the second equation integrates to $C(t)=C(0)+\sigma_P^2t/m$, and the first then gives

$$\sigma_X^2(t)=\sigma_X^2(0)+\frac{2C(0)}{m}\,t+\frac{\sigma_P^2}{m^2}\,t^2 .$$

This expression does not contain $g$: it is identical to that for a free particle. Gravity accelerates the centre of the packet, but changes neither the momentum spread nor the spreading of the packet. This can be understood as follows: in the freely falling reference frame moving with the centre of the packet, gravity disappears, and the packet evolves like a free packet. At long times, the term in $t^2$ always dominates, and $\sigma_X(t)\simeq\sigma_Pt/m$: the packet spreads at a rate equal to the spread of velocities $\sigma_P/m$. For an initially Gaussian packet with no correlation between position and momentum, $C(0)=0$ and $\sigma_P=\hbar/(2\sigma_X(0))$, and we recover the formula from Lesson 5.