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Commutators: solving with jacobian

  1. May 17, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello:) My problem is as follows:
    Determine the following commutators: [px2,x],[pxx2],[px2,x2],[]. The calculation can be done in two ways, either by inserting a test function, and using the explicit expressions for the operators, or by utilizing Jacobi identity and using the fact that [x,px]=ihbarred . You should master both methods.


    2. Relevant equations
    [A,B]=AB-BA
    [x,px]=ihbarred
    [A,BC]=B[A,C]+[A,B]C
    px=-ihbarred*d/dx
    p2=-hbarred2*d2/dx2


    3. The attempt at a solution
    Here's the problem- I just cannot find good paper with proper explanation how to do it and examples with solutions. I am also not familiar with term and meaning of jacobian- in the textbook I am using there is no explanation of jacobian, or how to solve commutators, not even mentioning solving commutators with jacobians.
     
  2. jcsd
  3. May 17, 2013 #2
    The Jacobi identity is not the same thing as the Jacobian. They are both named after the same person, but they are not the same thing. Here is the Jacobi identity.

    [itex][A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0[/itex]

    It applies to the bracket operation for Lie algebras, so it applies to cross products and Poisson brackets as well as commutators.
     
  4. May 17, 2013 #3
    Also I have no idea why you should use the Jacobi identity with the commutators given to you. The unnamed identities [itex][A,BC] = [A,B]C + B[A,C][/itex] and [itex] [AB,C] = A[B,C] + [A,C]B[/itex] seem like a more appropriate things to use.
     
    Last edited: May 17, 2013
  5. May 17, 2013 #4
    I don't know that either, but that is the content of the problem, so I have to do it. I have no idea how to solve commutator if there is a second derivative (px2).
     
  6. May 17, 2013 #5

    vela

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    Use ##\hat{p}_x^2 = \hat{p}_x\hat{p}_x##. You need to show some effort at working the problem out on your own.
     
  7. May 17, 2013 #6
    ok, but what if there is a factor in front of the commutator? for example like [px2,x]ψ?
    Is it supposed to be like this:

    [px2,x]ψ=p[px,x]ψ+[px,x]ψpx?
     
  8. May 17, 2013 #7

    vela

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    I don't see a factor in front of the commutator. Could you elaborate?
     
  9. May 18, 2013 #8
    One of the ways of solving commutators was to put a function after commutator, like this
    [p,x]ψ - and then solve it. But unfortunately no matter if I put the function there or not- I can't get the correct value. I couldn't find anywhere how to solve commutators [p2,x], or [p2,x[/SUP]2[/SUP]] step by step. When I try do solve this on my own, I just keep getting confused by the order of operators in the equation.
     
  10. May 18, 2013 #9

    vela

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    Show what you tried.
     
  11. May 18, 2013 #10
    ok, so I did the calcuation as follows:
    [pxpx,x]=px[px,x]+x[px,x]px
    px(pxx-xpx)+(pxx-xpx)*px
    -ihbarred(∂/∂x)(-ihbarred(∂x/∂x)+ihbarredx(∂/∂x))+(-ihbarred(∂x/∂x)+ihbarredx(∂/∂x))*(-ihbarred(∂/∂x))
    -ihbarred(∂/∂x)(-ihbarred+ihbarredx(∂/∂x))+(-ihbarred+ihbarredx(∂/∂x))*(-ihbarred(∂/∂x))
    -ihbarred^2(∂/∂x)+xhbarred^2(∂/∂x)-hbarred^2(∂/∂x)+xhbarred^2(∂/∂x)(∂/∂x)

    and that is pretty much the result that I obtain every single time I calculate this. Where am I making mistake?
     
  12. May 18, 2013 #11

    vela

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    Once you get to this point, use the fact that ##[x,p_x]=i\hbar##.

     
  13. May 18, 2013 #12
    But that is what I want to do with the jacobi identity, so the demands in the problem are satisfied. I tried to do this the way I can see step by step what is coming from where, that is why I insisted on step by step explanation.
     
  14. May 18, 2013 #13

    vela

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    Your mistake is when dealing with terms like ##p_x x##. It doesn't simplify to 1, like you have. The operator ##p_x## acts on everything to the right of it, so not just ##x## but the test function ##\psi## as well, i.e.,
    $$p_x x \psi = -i\hbar \frac{\partial}{\partial x} (x\psi) = \cdots$$
     
  15. May 18, 2013 #14
    so in case of [px2,x] (if I want to do it step by step) I am supposed to put the test function after the operator p?
     
  16. May 18, 2013 #15

    vela

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    Yes. Why wouldn't it be at the end?
     
  17. May 19, 2013 #16
    It just confuses me so I wasn't sure if it was supposed to be before or after the p operator. I tried with using ψ function, unfortunately it didn't help. first bracket zeroes:
    -ihbarred(∂/∂x)(-ihbarred(∂x/∂x)ψ+ixhbarred(∂ψ/∂x))
    but the second one in no way gives me the value -2ihbarredpx, which is supposed to be an answer for this problem:
    (-ihbarred(∂x/∂x)+ixhbarred(∂/∂x))(-ihbarred(∂ψ/∂x))
    where am I making mistake?
     
  18. May 19, 2013 #17

    vela

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    Using ##[AB,C] = A[B,C] + [A,C]B##, you have ##[p_x^2,x] = p_x[p_x,x] + [p_x,x]p_x##. Now it's basic substitution:
    $$[p_x^2,x]\psi = (p_x[p_x,x] + [p_x,x]p_x)\,\psi.$$ You can't just move ##\psi## past ##p_x##.

    How'd you get 0? Also, you're missing a term. You should have
    $$p_x [p_x, x] \psi = p_x (p_x x - x p_x)\psi = -i\hbar\frac{\partial}{\partial x} \left[-i\hbar\frac{\partial}{\partial x} (x\psi) - x \left(-i\hbar\frac{\partial}{\partial x}\right) \psi\right].$$ Now use the product rule to expand the first term in the brackets.
     
  19. May 19, 2013 #18
    You were right, I confused the order of calculations, hence the zeroing of the first bracket. My result of the first bracket came out to be -2(hbarred)^2(∂ψ/∂x). The second one comes to -(hbarred)^2(∂ψ/∂x)+x(hbarred)^2(∂/∂x)(∂ψ/∂x). unless I miss something similar to what I did in the first bracket.
     
  20. May 19, 2013 #19
    and also- if the Jacobi identity is [A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0, then for [p2x,x] it would be [px,[px,x]]+[px,[x,px]]+[x,[px,px]]=0, right? But how do I do this if there is commutator like [px2,x2]?
     
  21. May 19, 2013 #20
    I tried calculation with Jacobi identity according to the scheme I wrote above, but the commutators just zero themselves, so it seems kind of pointless. What am I doing wrong?
     
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