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QM operators

  1. Mar 31, 2008 #1
    ok, i know that if we have operators A and B and a function f(x) that ABf(x) means that first B acts on f(x) and the A acts on the resulting function. But what it my operator looks like [(d^2 /dx^2) - x^2]. How do i apply this to my function. Do i just distribute my function to each part of the operator?
     
  2. jcsd
  3. Mar 31, 2008 #2

    G01

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    Yes that is how you would apply that operator.
     
  4. Mar 31, 2008 #3
    so [(d^2 /dx^2) - x^2]f(x) would just be (d^2 /dx^2)f(x) - x^2 f(x) ?
     
  5. Mar 31, 2008 #4

    G01

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    Yes. That's correct.
     
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