# The significance of commutation

1. Jun 23, 2009

### 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 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. Jun 23, 2009

### 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.

3. Jun 24, 2009

### snoopies622

I think I follow..

i = Eiψi implies ψi*H = ψi*Ei

and this is true because H is Hermitian?

4. Jun 24, 2009

### xepma

Yes. The reason is that:

$$H|\Psi^i\rangle = E^i|\Psi^i\rangle$$
$$(H|\Psi^i\rangle)^\dagger = \langle \Psi^i|H$$
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 ;).

5. Jun 24, 2009

### snoopies622

Got it. Thank you both.