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QM operators

  • Thread starter eit32
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
G01
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Yes that is how you would apply that operator.
 
  • #3
21
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so [(d^2 /dx^2) - x^2]f(x) would just be (d^2 /dx^2)f(x) - x^2 f(x) ?
 
  • #4
G01
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Yes. That's correct.
 

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