Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This is a very simple question. I am learning about angular momentum. In my lecture notes, the symbol $|\lambda,m_l \rangle$ was defined as a eigenfunction of a central potential. Two assumptions are introduced: $L^2 = \lambda \hbar^2$ and $L_z = m_l \hbar$. So, $$ \hat{L}^2 |\lambda,m_l \rangle = \lambda \hbar^2|\lambda,m_l \rangle \\ \hat{L_z}|\lambda,m_l \rangle = m_l \hbar|\lambda,m_l \rangle $$

But what does $|\lambda,m_l \rangle$ mean exactly? I am comfortable with $|\psi \rangle$, but I do not understand what having two variables in the ket means.

share|improve this question
add comment

1 Answer 1

up vote 10 down vote accepted

Quite often everything inside bra or ket is just a label. In this particular case the meaning of $|λ,m_l⟩$ is "a state with the square of the angular momentum being equal to $λ$ (in atomic units, where $\hbar=1$) and with the projection of the angular momentum in some direction ($z$-axis conventionally) being equal to $m_l$".

That is, $|λ,m_l⟩$ state is the simultaneous eigenstate of both $\hat{L}^2$ and $\hat{L}_z$, i.e. it is an eigenstate of $\hat{L}^2$ (with eigenvalue $λ \hbar^2$, or just $λ$ in atomic units) and at the same time it is an eigenstate of $\hat{L}_z$ (with eigenvalue $m_l \hbar$, or just $m_l$ in atomic units). Such simultaneous eigenstates exist because the corresponding operators commute $[\hat{L}^2, \hat{L}_z] = 0$, or, in other words, because the corresponding observables are compatible.

share|improve this answer
3  
You might add that the operators must be compatible (commuting) observables when used to label a ket. –  DrEntropy May 27 at 14:58
    
@DrEntropy good point. Done. –  Wildcat May 27 at 16:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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