Math Infinitum

Mapping the lesson.

Loading the workspace…

Intuition

Where is classical mechanics hiding in all this? In the averages. The average position moves with the average momentum over the mass, and the average momentum changes at the rate of the average force. These are Newton’s laws for expectation values — exactly Newton’s laws when the force is the same across the whole packet, and nearly so whenever the packet is small compared with how fast the force changes.

A flock of birds wheels and spreads, but its centre moves as if it were one bird pushed by the average of the winds on all of them. The expectation value of position is the centre of the quantum flock.

Newton’s laws for averages

Applying the rate formula of the lesson on conserved quantities to $\hat{x}$ and $\hat{p}$ , with the commutators of the last chapter, gives Ehrenfest's theorem. The force appears averaged over the probability density.

\frac{d\langle x\rangle}{dt}=\frac{\langle p\rangle}{m},\qquad \frac{d\langle p\rangle}{dt}=-\Bigl\langle\frac{dV}{dx}\Bigr\rangle

When averages move classically

Combining the two: $m\,\frac{d^{2}\langle x\rangle}{dt^{2}}=-\langle V'(x)\rangle$ .

Ehrenfest’s theorem

\frac{d\langle p\rangle}{dt}=\frac{i}{\hbar}\bigl\langle[\hat{H},\hat{p}]\bigr\rangle=-\langle V'(\hat{x})\rangle

Apply the rate formula to $\hat{x}$ and to $\hat{p}$ . For position only the kinetic energy fails to commute, and $[\hat{p}^{2},\hat{x}]=-2i\hbar\hat{p}$ gives . For momentum only the potential fails to commute, and gives minus the average force.

Proof steps

\frac{d\langle x\rangle}{dt}=\frac{i}{\hbar}\Bigl\langle\Bigl[\frac{\hat{p}^{2}}{2m},\hat{x}\Bigr]\Bigr\rangle=\frac{i}{\hbar}\cdot\frac{-2i\hbar\langle p\rangle}{2m}

Only the kinetic term fails to commute with $\hat{x}$ , and $[\hat{p}^{2},\hat{x}]=-2i\hbar\hat{p}$ .

=\frac{\langle p\rangle}{m}

$i\cdot(-i)=1$ .

\frac{d\langle p\rangle}{dt}=\frac{i}{\hbar}\bigl\langle[V(\hat{x}),\hat{p}]\bigr\rangle

The kinetic term commutes with $\hat{p}$ ; only the potential remains.

[V(\hat{x}),\hat{p}]=i\hbar V'(\hat{x})

The rule for a function of position from the last chapter.

\frac{d\langle p\rangle}{dt}=\frac{i}{\hbar}\cdot i\hbar\langle V'\rangle=-\langle V'(\hat{x})\rangle

$i\cdot i=-1$ : minus the average slope of the potential, the average force.

Applications

Practice

Average Velocity

The rate of change of the average position is the average momentum divided by the mass, exactly as for a classical particle.

\frac{d\langle x\rangle}{dt}=\frac{\langle p\rangle}{m}

Try it

A particle of mass 2 has $\langle p\rangle=6$ at some moment. How fast is $\langle x\rangle$ changing?