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.


1 Answer 1


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.

  • 3
    $\begingroup$ You might add that the operators must be compatible (commuting) observables when used to label a ket. $\endgroup$
    – DrEntropy
    May 27, 2014 at 14:58
  • $\begingroup$ @DrEntropy good point. Done. $\endgroup$
    – Wildcat
    May 27, 2014 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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