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Find Adjoints of certain Operators

  1. May 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the adjoints for x2d/dx and d2/dx2


    2. Relevant equations
    I know that (x2)_dagger=x2 and that (d/dx)_dagger=-(d/dx).


    3. The attempt at a solution
    I solved x2d/dx by doing the following:

    (x2)_dagger * (d/dx)_dagger= (x2) * (-d/dx)

    thus the answer should be -x2d/dx


    For the second adjoint can I do the same step as I did previously?

    Is it right for me to separate (d2/dx2)_dagger into (d/dx)_dagger * (d/dx)_dagger?

    I got d2/dx2 as the final answer.
     
  2. jcsd
  3. May 27, 2013 #2

    TSny

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    In general, it is not true that ##(AB)^\dagger = A^\dagger B^\dagger##. Have you learned the correct way to express ##(AB)^\dagger## in terms of ##A^\dagger## and ##B^\dagger##?
     
  4. May 27, 2013 #3
    Yes I believe I figured it out yesterday!

    ##(AB)^\dagger##=(##B^\dagger##)(##A^\dagger##)

    But for this d2/dx2, it will be the same result right or should I not be able to use that rule?
     
  5. May 27, 2013 #4

    TSny

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    Yes, that's right. The order of the operators is switched.

    No. The operator O1 = (x2 d/dx) is not the same as the operator O2 = (d/dx x2), as you can see if you compare the results of letting O1 and O2 operate on an arbitrary function f(x).
     
  6. May 27, 2013 #5
    So I messed up for the first part:

    (##(x2d/dx)^\dagger##) = (##(d/dx)^\dagger##)(##(x2)^\dagger##)

    Then the result would be:

    (-d/dx)*(x2)= -2x?
     
    Last edited: May 27, 2013
  7. May 27, 2013 #6

    TSny

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    The left side is correct, but the right side isn't. See what you get if you let the operator (-d/dx*x2) act on a function f(x). First you will multiply f(x) by x2 and then take the derivative -d/dx.
     
    Last edited: May 27, 2013
  8. May 27, 2013 #7
    I see so I should've included the function as well.

    So it would be -d(x2f(x))/dx= -2xf(x)-x2df(x)/dx. So what should my operator look like? Is it everything except for the f(x)?
     
  9. May 27, 2013 #8

    TSny

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    Yes, that's right.
     
  10. May 27, 2013 #9
    Thank you very much!

    Is my second part correct where I applied the same concept but this time separate the second derivative?
     
  11. May 27, 2013 #10

    TSny

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    Second part is correct. Good work!
     
  12. May 27, 2013 #11
    I really appreciate the help. Thank you :]
     
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