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Commutator problem

  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    a = √(mω/2ħ)x + i√(1/2ħmω), a = √(mω/2ħ)x - i√(1/2ħmω),

    find [a,a]

    the solution is given. it should be 1.
    2. Relevant equations
    [a,b] = ab -ba

    3. The attempt at a solution
    im guessing there is something I'm missing or I'm not doing something somewhere.
    I'm just doing what the formula says.

    (√(mω/2ħ)x + i√(1/2ħmω))(√(mω/2ħ)x - i√(1/2ħmω)) -(√(mω/2ħ)x - i√(1/2ħmω))(√(mω/2ħ)x + i√(1/2ħmω))

    and from that i get zero. what i am forgetting to do?
     
  2. jcsd
  3. Oct 10, 2016 #2

    strangerep

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    I think you mis-typed the definitions of ##a## and ##a^\dagger##. Shouldn't there be a ##p## (momentum) in the 2nd term of each?
     
  4. Oct 10, 2016 #3
    i just noticed that. i copied it straight from the homework so I'm guessing the teacher made a mistake.

    here i am banging my head on the table.
     
  5. Oct 10, 2016 #4

    TSny

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    Yes, the teacher left out the p in the second term of each expression. x and p are operators here. Do they commute?
     
  6. May 15, 2017 #5
    It still seems like zero to me. How can I commute those terms?
     
  7. May 15, 2017 #6

    TSny

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    Hello Sugyned, welcome to PF!

    Please show your work.
     
  8. May 16, 2017 #7
    Shouldnt poisson parenthesis be:

    (dA/dQ)(dA†/dP) - (dA/dP)(dA†/dQ)
    ?
    That would result in:
    -aib-iba = -2iab

    if a=√(mω/2ħ)
    and b = √(1/2ħmω)


    I don't know much about this, maybe i'm completely wrong.
    My university is protesting because the teachers weren't payed some salaries that the government owes to them, so the quantum mechanics teacher isn't giving classes, thus, we have a test coming up next tuesday. :smile:
     
  9. May 16, 2017 #8

    strangerep

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    The original question is not about Poisson brackets -- it's about quantum operators.

    Maybe you should start a new thread, and state what your problem you're trying to answer?
     
  10. May 16, 2017 #9
    It's actualy this exact same problem shown up here. The notes I have are quite confusing, I can't just simply read it and understand it.
     
  11. May 16, 2017 #10

    strangerep

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    Note that the original problem as stated in post #1 is wrong. See my post #2 where I corrected it.

    With this correction,... are you familiar with the canonical commutation relations between ##x## and ##p##?

    From these, you should be able to calculate ##[a, a^\dagger]##, using the linearity property of commutators. It's little more than a 1-liner.
     
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