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Having trouble with commutators

  1. Mar 8, 2008 #1
    I'm having trouble with commutators. I have to solve them 2 ways. First, using [tex][x,p]=i\hbar[/tex] and other identities/formulas, and the the second method the "direct way".

    1.) [tex]x,\hat{H}[/tex]

    My work:

    [tex][x,\hat{H}]\psi &= x\hat{H}\psi - \hat{H}x\psi[/tex]
    [tex]= x \left ( \frac{p^2}{2m} + V(x) \right )\psi - \left ( \frac{p^2}{2m} + V(x) \right )x\psi[/tex]
    [tex]= \frac{xp^2\psi}{2m} + x V(x)\psi - \frac{p^2x\psi}{2m} - V(x) x\psi[/tex]

    2.) [tex][\hat{p}, \hat{H} + x][/tex]

    [tex][\hat{p}, \hat{H} + x]\psi &= \hat{p}(\hat{H}+x)\psi + (\hat{H}+x)\hat{p}\psi[/tex]
    [tex]= i\hbar\frac{\partial}{\partial p} \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)\psi + \left(\left( \frac{p^2}{2m} + V(x) \right ) + x\right)i\hbar\frac{\partial}{\partial p}\psi[/tex]

  2. jcsd
  3. Mar 8, 2008 #2


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    What is the original question? i.e the problem stated.

    Use the [x,p] and other commutator idendites, you have not done that,
    for example:

    [x,H] = [x,pp + V] = {i skip the constant factors} = [x,pp] + [x,V] = ? continue
  4. Mar 8, 2008 #3
    The orig question was [x,h] for # 1, I just forgot to put in the brackets...

    Yes, that's the whole point, I'm not sure how. I showed my work.
  5. Mar 8, 2008 #4


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    do you know the commutator relation for

    [A,B*C] = ??

    find that, and use it.
  6. Mar 8, 2008 #5
    Ah yes! That was quite helpful. How does this look:

    [tex][x,\hat{H}]\psi = [x,\frac{p^2}{2m}]\psi[/tex]
    [tex]= [x,p\cdot\frac{p}{2m}]\psi[/tex]
    [tex]= [x,p]\frac{p}{2m}\psi + p[x,\frac{p}{2m}]\psi[/tex]
    [tex]= i\hbar\frac{p}{2m}\psi + p[x,p\cdot\frac{1}{2m}]\psi[/tex]
    [tex]= i\hbar\frac{p}{2m}\psi + p[x,p]\frac{1}{2m}\psi + p[x,\frac{1}{2m}]\psi[/tex]
    [tex]= i\hbar\frac{p}{2m}\psi + \frac{p}{2m}i\hbar + p\frac{x}{2m}\psi + \frac{x}{2m}\psi[/tex]
    [tex]= \boxed{2i\hbar\frac{p}{2m}\psi + 2\frac{px}{2m}\psi}[/tex]

    Not sure if it's right. The direct way, which I tried earlier, is still causing probs. I'm guessing for the second one, I should use [A,B+C] = [A,B] + [A,C]?
  7. Mar 8, 2008 #6


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    So far so good but you don't need to stick in a wavefunction when you do it this way. One has to put a test function only when one uses explicit representations in terms of differential operators.

    :eek: No!
    [tex] [x,\frac{p}{2m}] = \frac{1}{2m} [x,p] [/tex]
    simply (the 1/2m is an overall constant which youcan pull out of the commutator.
  8. Mar 8, 2008 #7


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    For the opening poster... technically, you weren't wrong; you were merely encumbered. ([itex][x, a] = 0[/itex] for any constant a)
  9. Mar 8, 2008 #8


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    Yes, that is true. My apologies if I gave the wrong impression.

    Technically, it is correct to write

    [tex] [x,P/2m] = \frac{1}{2m} [x,P] + [x, \frac{1}{2m}] P [/tex]

    Nobody woul dnormally write it this way but it is indeed correct.
    The next step is to use the fact that [x,1/2m] = 0.

    However, the next line in the demonstration of the OP is incorrect.
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