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Quick Matrix Element Question using Hermitian Operator

  1. Mar 22, 2013 #1
    Hi there,

    This should be very simple...

    If I have a state <1|AB|2> where A and B are Hermitian operators, can I rewrite this as <2|BA|1> ?

    That would be, taking the complex conjugate of the matrix element and saying that A*=A and B*=B.

    Thank you!
  2. jcsd
  3. Mar 22, 2013 #2
    Well, you need to remember to include the complex conjugation:

    ##\langle 1 | A B | 2 \rangle^* = \langle 2 | B^\dagger A^\dagger | 1 \rangle = \langle 2 | B A | 1 \rangle##

    so ##\langle 1 | A B | 2 \rangle = \langle 2 | B A | 1 \rangle^*##
  4. Mar 22, 2013 #3
    Great, thank you for the help
  5. Mar 22, 2013 #4
    not necessarily, you need to also consider that this is only true if A and B are compatible observables, and it is only a matrix element if |1> and |2> are basis vectors of those same compatible observables.
  6. Mar 22, 2013 #5
    So it appears I have a bigger problem!

    |1> and |2> are eigenstates of the hamiltonian, my system is that of a harmonic oscillator. What I'm trying to prove is that <1|P|2> = -imw<1|X|2>, starting with the matrix element [P,H] where H is the hamiltonian. I thought I could do this just by swapping eigenstates so now I'm more stuck :)

    So far I've tried two methods, one involving writing H = T + V for the harmonic oscillator and finding the commutation relation and the other working with H|1> = E1|1> etc.

    Any suggestions?
  7. Mar 22, 2013 #6


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    What The Duck did holds in general, for arbitrary state vectors |1> and |2>.
  8. Mar 22, 2013 #7
    I agree, my post was in reply to the original poster. Guess i was late to the party
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