Exercise [21.07]

deant · 01 Sep 2012, 11:42

$\mathcal{H}$ independent of (say) x^3

implies that:

$\frac{\partial}{\partial x^3}\mathcal{H}\psi = \mathcal{H}\frac{\partial}{\partial x^3}\psi$

but from the Schrödinger equation, $\mathcal{H} = i\hbar\frac{\partial}{\partial t}$ so we have:

$i\hbar\frac{\partial}{\partial x^3}\frac{\partial}{\partial t}\psi = i\hbar\frac{\partial}{\partial t}\frac{\partial}{\partial x^3}\psi$

i.e. $\frac{\partial}{\partial x^3}$ and $\frac{\partial}{\partial t}$ commute.

Now on p.521 it is noted that if quantum operators commute, you can always find wavefunctions that are eigenstates of both operators simultaneously. When $\psi$ is a momentum eigenstate, $i\hbar\frac{\partial}{\partial x^3}\psi = P^3(t)\psi$ , where P^3(t)

is the classical momentum along the x^3

axis - a function of time, not a differential operator. Similarly for an energy eigenstate, $i\hbar\frac{\partial}{\partial t}\psi = E\psi$ , where E is the energy (a constant).

So let $\psi$ be an eigenfunction both of energy and of momentum along x^3

. Then:

$\frac{\partial}{\partial t}\left[P^3(t)\psi\right] = i\hbar\frac{\partial}{\partial t}\frac{\partial}{\partial x^3}\psi = i\hbar\frac{\partial}{\partial x^3}\frac{\partial}{\partial t}\psi = i\hbar\frac{\partial}{\partial x^3}\frac{E}{i\hbar}\psi = P^3(t)\frac{E}{i\hbar}\psi\\ \\ = P^3(t)\frac{\partial}{\partial t}\psi$

But $\frac{\partial}{\partial t}\left[P^3(t)\psi\right] = P^3(t)\frac{\partial}{\partial t}\psi$ only when P^3(t)

is not actually a function of t, i.e. when P^3

is a constant that does not change with time.

So when $\frac{\partial}{\partial x^3}$ commutes with $\frac{\partial}{\partial t}$ , P^3

is conserved.

$\blacksquare$