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Orthonormal Energy Eigenstates

  1. Jul 9, 2010 #1
    I'm a self learner with a decent math background trying to follow this lecture.



    I have been able to follow him perfectly up to this point but at 16:00 Professor Balakrishnan states that "<n|a|n> = 0" because |n-1> is perpendicular to |n>. My question is why and how do we know that an energy eigenstate is perpendicular to another energy eigenstate.

    -Nate
     
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Jul 9, 2010 #2
    This is one of the "standard results" of linear algebra.

    For a hermitian operator H, eigenvectors with different eigenvalues
    are always orthogonal, while for any eigenvectors with the same eigenvalue
    we can use Graham-Schmidt to find an orthogonal basis of the subspace.


    Here are some proofs. I don't feel like tex'ing my answers so
    I didn't use dirac notation.


    H is hermitian if (Y,H.X) =(H.Y,X) for all vectors X,Y
    [and (,) is an inner-product s.t. (X,Y) = (Y,X)* {complex conjugate},
    and (X, aY) = a(X,Y) {for a scalar a}, so that together we get
    (aY,X) = (X,aY)* = (a(X,Y))* = a*(X,Y)*=a*(Y,X), and
    i will use H.X to mean H(X) {ie the operator acting on x}]

    Let H be a hermitian operator

    claim 1 (eigenvals are real)
    s'pose X is an eigenvector of H with eigenvalue x so that H.X = xX
    then x is real. Since (X,H.X) = (X,xX) = x(X,X) on one hand
    on the other (X,H.X)=(H.X,X) = (xX,X) = x*(X,X) on the other.


    claim 2 (eigenvectors with different eigenvalues are orthogonal)
    let X,Y be eigenvectors of H with eigenvals x,y respectively
    and suppose they are different eigenvals.
    Then (Y,H.X) = (Y, xX) = x(Y,X)
    and also (Y, H.X) = (H.Y, X) = (yY, X) = y(Y,X)
    subtracting these we get
    (x-y)(Y,X) = 0
    and since x-y isn't 0 we have (Y,X)=0.


    claim 3. (eigenspaces)
    all vectors that have the same eigenvalue of H form a subspace.
    pf. check vector space properties . only interesting one is closure
    let H.X = xX and H.Y = xY be two vectors
    then H.(aX +bY)= aH.X + bH.Y = axX+bxY = x(aX + bY).

    thus we can use Graham Schmidt to orthogonalize any basis of
    the subspace.
     
  4. Jul 9, 2010 #3
    That clears up a lot. Thanks!
     
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