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The significance of commutation

  1. Jun 23, 2009 #1
    I'm in chapter two of H. S. Green's Matrix Mechanics and at a sticking point. In section 2.2 he gives the following scenario:

    An atom emits a photon with angular velocity ω, it has energy Ei before the emission and Ef after, so Ei - Ef = ħω. (That I can understand.) ψi and ψf are eigenvectors of the energy operator H, while Ei and Ef are their corresponding eigenvalues, respectively. He then gives this equation, where A is "any observable":

    ψf * (AH-HA) ψi = ( Ei - Ef ) ψ f * A ψ i.

    Although he briefly mentions commutation a couple paragraphs before this point, it's not enough to explain where this relationship comes from.

    Can anyone help me out?
     
  2. jcsd
  3. Jun 23, 2009 #2

    Pengwuino

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    Gold Member

    Well, commutation in this case is [A,H] = AH - HA. If it is equal to 0, it is said that the 2 operators, H and A, commute, that is they share eigenfunctions. Otherwise, they do not share eigenfunctions as this problem kinda shows. Getting to the meat of the problem, [tex]\psi ^f (AH - HA)\psi ^i = \psi ^f AH\psi ^i - \psi ^f HA\psi ^i [/tex]. Now the important part is that since H is hermitian, it can act to the left in the second term, that is [tex]\psi ^f HA\psi ^i = \psi ^f E^f A\psi ^i = E^f \psi ^f A\psi ^i [/tex]. You can't do anything with A since you don't know what it is and that is how you get the right hand side.
     
  4. Jun 24, 2009 #3
    I think I follow..

    i = Eiψi implies ψi*H = ψi*Ei

    and this is true because H is Hermitian?
     
  5. Jun 24, 2009 #4
    Yes. The reason is that:

    [tex]H|\Psi^i\rangle = E^i|\Psi^i\rangle[/tex]
    so taking the adjoint gives
    [tex](H|\Psi^i\rangle)^\dagger = \langle \Psi^i|H [/tex]
    on the left hand side, and
    [tex](E^i|\Psi^i\rangle)^\dagger = \langle \Psi^i|(E^i)^* = \langle \Psi^i|E^i [/tex]
    on the right hand side (the fact that all eigenvalues are real is crucial).

    Mathematically, it's a bit sloppy, since the adjoint of a vector doesn't exist. The conclusion is correct, nevertheless ;).
     
  6. Jun 24, 2009 #5
    Got it. :smile: Thank you both.
     
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