The significance of commutation

706
8
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?
 

Pengwuino

Gold Member
4,854
14
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.
 
706
8
I think I follow..

i = Eiψi implies ψi*H = ψi*Ei

and this is true because H is Hermitian?
 
525
6
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 ;).
 
706
8
Got it. :smile: Thank you both.
 

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