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Proof help - quantum operators

  1. Sep 22, 2009 #1
    If A and B are hermitian operators, then AB is hermitian only if the commutator=0.

    basically i need to prove that, but i don't really know where to start ofther than the general <f|AB|g> = <g|AB|f>*

    obv physics math is not my strong point. thanks :)
     
  2. jcsd
  3. Sep 22, 2009 #2
    i understand the proof of a single operator and the expansion of the relation to integral notation, but i don't understand the significance of the multiplication of the operators and why there is the condition there.
     
  4. Sep 22, 2009 #3
    how's this?

    <f|AB|g>=<g|AB|f>*

    left=([tex]\sum[/tex]<g|A|m><m|B|f>)*
    left=[tex]\sum[/tex]<m|A*|g><f|B*|m>
    left=[tex]\sum[/tex]<f|B*|m><m|A*|g>
    left=<f|B*A*|g>
    since B and A are Hermitian,
    <f|AB|g> = <f|BA|g>
    so AB-BA=0

    good enough?
     
    Last edited: Sep 22, 2009
  5. Sep 23, 2009 #4

    jambaugh

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    Begin with the definition of "Hermitian operator" i.e. it is equal to its adjoint, then apply it to AB and see what is necessary to make it Hermitian.

    I don't think you need to invoke the "bra", "ket" products.
     
  6. Sep 23, 2009 #5
    so,

    AB=(AB)*
    AB=B*A*
    B and A are Hermitian, so,
    AB=BA
    AB-BA=0

    and that's it?

    i haven't had linear algebra (lol) so the (AB)*=B*A* is a little confusing to me, which i kinda figured out based on using the brackets and the completeness relation. so this would be it?
     
  7. Sep 23, 2009 #6

    jambaugh

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    That's all there is to it!
     
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