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Operator acting on an operator

  1. Nov 8, 2013 #1
    Hello everyone I've got these things buzzing in my head and not exactly knowing how to solve them.
    1. The problem statement, all variables and given/known data
    Operator Ahat = (d/dx + x) and Bhat = (d/dx - x)

    a. Chat = AhatAha
    b. Chat = AhatBhat

    What do the position and momentum operator Xhat = x and Phat = -i*hbar*d/dx, give when they combine as Phat^2Xhat^2?

    Thanks a lot
     
  2. jcsd
  3. Nov 8, 2013 #2

    CAF123

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    Gold Member

    Given ##\hat{P}##, do you know how to find ##\hat{P}^2##?
     
  4. Nov 8, 2013 #3
    well I can assume it is -H^2*d^2/dx^2 (without being sure)
     
  5. Nov 8, 2013 #4

    berkeman

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    Staff: Mentor

    Please remember that you must show your attempt at solving the problem before we can be of tutorial help. That's in the PF Rules link at the top of the page under Site Info.
     
  6. Nov 8, 2013 #5

    CAF123

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    It's correct, assuming H = ##\hbar##. What makes you unsure?
     
  7. Nov 8, 2013 #6
    Well what makes me unsure is the way that this acts on x^2. as it goes like d/dx d/dx x x
     
  8. Nov 8, 2013 #7
    And on the Ahat*Ahat I get a constant in an operator, which I don't know if it is right.
     
  9. Nov 8, 2013 #8

    CAF123

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    The important thing to remember about operators is that they act on functions. I think what you have to do is to express the operator ##\hat{P^2} \hat{X^2}## in terms of ##\hat{A}## and ##\hat{B}##.
    As you said, $$\hat{P^2}\hat{X^2} \equiv -\hbar^2 \frac{d^2}{dx^2} x^2$$ Now act on a function f(x). You will need product rule.
     
  10. Nov 9, 2013 #9
    So is it left like that or is it written as -2*hbar^2 (differentiate twice x^2)?
     
  11. Nov 10, 2013 #10

    CAF123

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    What did you get when you computed $$-\hbar^2 \frac{d^2}{dx^2} (x^2 \psi)?$$
    I don't know if ##\hat{A}## or ##\hat{B}## have any relevance to the problem or not.

    It is okay if you have a constant as part of the operator, as long as you are adding it to a dimensionless quantity.
     
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