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Applying Heisenberg Picture Confusion

  1. Aug 26, 2010 #1
    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 :

    [tex]\partial[/tex]q/[tex]\partial[/tex]t = [tex]\partial[/tex]H/[tex]\partial[/tex]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. jcsd
  3. Aug 26, 2010 #2
    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
  4. Aug 27, 2010 #3
    OK that clears it all up. Thanks!
     
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