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Operators and expectation values.

  1. May 5, 2013 #1
    I'm stuck on a question in atkins molecular quantum mechanics 4e (self test 1.9).

    If (Af)* = -Af, show that <A> = 0 for any real function f.

    I think you are expected to use the completeness relation sum,s { |s><s| = 1.

    I'm sure the answer is simple but I'm stumped.
     
    Last edited: May 5, 2013
  2. jcsd
  3. May 5, 2013 #2

    tom.stoer

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    What are A and f?
     
  4. May 5, 2013 #3
    I'm assuming A is any real or complex operator for which the relation holds and f is any real function.
     
  5. May 5, 2013 #4

    tom.stoer

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    And what is <f> ?

    In QM <f> would be the expectation value of an operator f sandwitched between two states. But I guess you have something in mind like ∫ dx f(x)
     
  6. May 5, 2013 #5
    Sorry, my mistake. Need to show that the expectation value of the operator A is zero. <f|A|f> = 0
     
  7. May 5, 2013 #6

    tom.stoer

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    Ok: If

    [tex](Af)^\ast = -Af[/tex]

    show that

    [tex]\langle f|A|f\rangle = 0 [/tex]

    for any real function f.
     
  8. May 5, 2013 #7
  9. May 5, 2013 #8
    Does that make sense?
     
  10. May 5, 2013 #9

    tom.stoer

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    I think something is missing.

    What I can show immediately is

    [tex] a = \int dx \, fAf = \int dx \, f (-Af)^\ast = -a^\ast[/tex]

    and therefore that a must be purely imaginary.

    Note that I ommited the range of integration which must be symmetric [-L,L]
     
  11. May 5, 2013 #10
  12. May 5, 2013 #11

    tom.stoer

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    As I said, something is missing. In the context of Atkins 1.9 only hermitean operators a considered.

    1) so by the singe line above [itex]\text{Re}\,a = 0[/itex]
    2) by hermiticity [itex]\text{Im}\,a = 0[/itex]

    Therefore [itex]a=0[/itex]
     
  13. May 5, 2013 #12

    vela

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    Why don't you try rereading the problem yourself and then state the problem statement here again with all the relevant information? You might notice you've made mistakes or overlooked some important info.
     
  14. May 5, 2013 #13
    Thanks vela. This isn't actually a homework assignment, I'm just working my way through the textbook. I got the same result as Tom, (Af)* = (Af*)* = -Af = -Af* => a = -a* => a = 0 given hermiticity. This result is actually stated in the book in one of the worked-through examples. I was wondering if somebody knew how to solve the problem in the context of the textbook i.e. using matrix notation and the completeness relation. Do you think the textbook intends that the problem be solved that way? I know atkins mqm4 has got a reputation for not being overly clear.
     
    Last edited: May 5, 2013
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