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Show that inner product is zero.

  1. Oct 15, 2013 #1
    Let A be a Hermitian operator with n eigenkets: [itex]A|u_i\rangle = a_i |u_i\rangle[/itex] for i=1,2,...,n.

    Suppose B is an operator that commutes with A. How could I show that
    [tex]\langle u_i | B | u_j \rangle = 0 \qquad (a_i \neq a_j)?[/tex]

    I have tried the following but not sure how to proceed:
    [tex]AB - BA=0\\ \implies \langle u_i | AB | u_j \rangle - \langle u_i | BA | u_j \rangle = 0[/tex]
     
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  3. Oct 15, 2013 #2

    WannabeNewton

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    Is ##A## Hermitian or is it self-adjoint? If it is the latter then the result is immediate from what you wrote down because for any self-adjoint operator ##A##, if ##|u_i \rangle## is an eigenket of ##A## with eigenvalue ##a_i## then ##\langle u_i|## is an eigenbra of ##A## with the same eigenvalue i.e. ##\langle u_i|A = a_i\langle u_i |## so you would be left with ##(a_i -a_j) \left \langle u_i|B|u_j \right \rangle = 0##.
     
  4. Oct 15, 2013 #3
    Actually I think I got it:

    [tex]a_i \langle u_i |B|u_j \rangle - \langle u_i |B| u_j \rangle a_j = 0[/tex] using the Hermiticity of A for the first term, and then since [itex]a_i \neq a_j[/itex] we get
    [tex]\langle u_i |B| u_j \rangle = 0[/tex]
     
  5. Oct 15, 2013 #4

    WannabeNewton

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    This is a technical point and I don't know if it matters for your class however it should be noted that while ##\langle u_i |A = a_i \langle u_i |## is certainly true if ##A## is self-adjoint, it won't necessarily be true if ##A## is only Hermitian.
     
  6. Oct 15, 2013 #5
    Thanks for that - I think we both posted at the same time! My question is related to quantum mechanics. I think physicists use the words Hermitian and self-adjoint interchangeably. I know that pure mathematicians will distinguish them but it's not important for my area. Thanks!
     
  7. Oct 15, 2013 #6

    WannabeNewton

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    Alrighty then! Glad you got it worked out :)
     
  8. Oct 15, 2013 #7

    dextercioby

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    Just by using the Dirac bra-ket formalism (one of the most vicious inventions in the history of science) means that you're hiding a lot of mathematical beauty under the carpet. Let's proceed then and leave finesse aside and say that if A and B commute, then they have a common set of eigenvectors. Thus the ui's of A are the ui's of B, so that you need to prove that 2 eigenvectors pertaining to 2 different eigenvalues are orthogonal one on each other.
     
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