1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook