The significance of commutation

[snoopies622]

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 E^f after, so Eⁱ - E^f = ħω. (That I can understand.) ψⁱ and ψ^f are eigenvectors of the energy operator H, while Eⁱ and E^f are their corresponding eigenvalues, respectively. He then gives this equation, where A is "any observable":

ψ^f * (AH-HA) ψⁱ = ( Eⁱ - E^f ) 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]

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, \psi ^f (AH - HA)\psi ^i = \psi ^f AH\psi ^i - \psi ^f HA\psi ^i. Now the important part is that since H is hermitian, it can act to the left in the second term, that is \psi ^f HA\psi ^i = \psi ^f E^f A\psi ^i = E^f \psi ^f A\psi ^i. 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.

 

[snoopies622]

I think I follow..

Hψⁱ = Eⁱψⁱ implies ψⁱ*H = ψⁱ*Eⁱ

and this is true because H is Hermitian?

 

[xepma]

Yes. The reason is that:

H|\Psi^i\rangle = E^i|\Psi^i\rangle
so taking the adjoint gives
(H|\Psi^i\rangle)^\dagger = \langle \Psi^i|H
on the left hand side, and
(E^i|\Psi^i\rangle)^\dagger = \langle \Psi^i|(E^i)^* = \langle \Psi^i|E^i
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 ;).

 

[snoopies622]

Got it. Thank you both.