# Applying Heisenberg Picture Confusion

1. Aug 26, 2010

### nateHI

I was trying to follow http://www.youtube.com/watch?v=dCua1R9VIiQ&p=EFD655A9E0B979B7&playnext=1&index=54" lecture at the 4:15 mark but am having a little difficulty. In particular, why doesn't he have to take the commutator of all four of the terms you get when you square (p-eA).

Is he using :

$$\partial$$q/$$\partial$$t = $$\partial$$H/$$\partial$$p and the chain rule? If so, why doesn't he differentiate what's inside the parantheses of the Hamiltonian?

-Nate

Last edited by a moderator: Apr 25, 2017
2. Aug 26, 2010

### qbert

it's a trick. you can take the commutator of all four terms, and you should
get the same result. the trick is [x, f(p)] = i h-bar f'(p). where prime here
is derivative wrt p.

lets check // (i'll assume x,p 1 dimensional to not worry about the vector stuff
but it's no big deal)

(p - eA)^2 = p^2 -epA -eAp + e^2A^2
so [x,(p - eA)^2] = [x, p^2] -e [x, pA] -e[x, Ap] + [x, e^2A^2 ]

now we can use the rule [A, BC] = [A,B]C + B[A,C].
[x,p^2] = [x,p]p + p[x,p] = 2 i h-bar p
[x,pA] = [x,p]A + p[x,A] = i h-bar A + 0
[x,Ap] = [x,A]p + A[x,p] = 0 + A(i h-bar)
[x,A^2] = 0

so [x,(p-eA)^2] = 2 i h-bar p - e (i h-bar A) - e (i h-bar A) = (2 i h-bar)(p - eA).

now take derivative of f(p)= (p-eA)^2 wrt p => f'(p) = 2(p-eA)
[x,f(p)] = i h-bar f'(p) = i h-bar 2 (p-eA).

it all works.

Last edited by a moderator: Apr 25, 2017
3. Aug 27, 2010

### nateHI

OK that clears it all up. Thanks!