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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.

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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.

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    $\begingroup$ 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

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