[ 2 posts ] 
 Exercise [21.07] 
Author Message

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Exercise [21.07]
\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]<br />= i\hbar\frac{\partial}{\partial t}\frac{\partial}{\partial x^3}\psi<br />= i\hbar\frac{\partial}{\partial x^3}\frac{\partial}{\partial t}\psi<br />= i\hbar\frac{\partial}{\partial x^3}\frac{E}{i\hbar}\psi<br />= P^3(t)\frac{E}{i\hbar}\psi\\ \\<br />= 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


Last edited by deant on 01 Sep 2012, 12:10, edited 4 times in total.

01 Sep 2012, 11:42

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [21.07]
Note: I'm not entirely happy with this solution, since it relies on knowing the fact that commuting operators have simultaneous eigenstates (i.e. functions that are eigenstates of both operators) - something that hasn't really been discussed in the book.

If anyone has a better way of demonstrating that commutation with the \frac{\partial}{\partial t} operator implies conservation, please post it!


01 Sep 2012, 11:58
   [ 2 posts ] 


cron