2
$\begingroup$

My notes define:

$$ L_{\pm} = L_{x} \pm i L_{y} $$

and states: $$ [L_{z},L_{\pm}] = \pm \hbar L_{\pm} $$

I'm fine with this as it's easy to show the result with some ugly algebra.

It then says:

Since each component of the angular momentum commutes with $ L^2 $ we can deduce that the action of $ L_± $ on $ |a, b> $ cannot affect the value of a relating to the magnitude of the angular momentum.

I'm happy to plug things in to prove it but I want to see how to deduce it if possible $$ $$ I understand that $ L^2$ commutes with $L_\pm$ because $L^2$ commutes with the individual $L_x$ and $L_y$ that make up $L_\pm$, but I don't see why that means it can't affect the magnitude.

Two operators which commute can be simultaneously known but that doesn't help because the ladder operator isn't Hermitian so isn't an observable.

Any help apreciated!

$$ $$ $$ $$ $$ $$

EDIT: Got it.

$$ [L^2 , L_i] = 0 $$ so $$ L^2 ( L_\pm | a, b > ) = L_\pm (L^2 |a, b>) = a L_\pm |a, b> $$

Forgot the fundamental property that commuting operators... commute.

|cite|improve this question
$\endgroup$
  • $\begingroup$ What is the meaning of the state a, b? Is it the eigenstate of L^2? If so, then this is true because the raising and lowering operators commute with L^2. $\endgroup$ – eqb Sep 15 '14 at 19:55
  • $\begingroup$ Oh I just refreshed after commenting and noticed you found your answer. Good. $\endgroup$ – eqb Sep 15 '14 at 19:55
  • $\begingroup$ |a,b> is the state with eigenvalue of L^2 of a and projection of b. $\endgroup$ – PFC99991 Sep 15 '14 at 19:56
  • 1
    $\begingroup$ Hi PFC99991: Welcome to Phys.SE. Your answer should be posted as an answer, not as an edit to the question. $\endgroup$ – Qmechanic♦ Sep 15 '14 at 20:00
1
$\begingroup$

It doesn't matter that the ladder operators are Hermitian or not. If they commute with some other operator, then you can have simultaneous eigenvalues. The relation between simultaneous eigenstates and commutation relations is a result of mathematics, not physics.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ The ladder operator here do not have eigenvalues or eigenstates (they are nilpotent). Because they commute with $J^2$ their action does not change the eigenvalue of $J^2$ but simply takes one eigenstate of $J^2$ into another with the same eigenvalue. $\endgroup$ – ZeroTheHero Dec 29 '16 at 6:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.