So based on my understand $m_l,m_s$ and $m_j$ represent the projection of $L,S$ and $J$ angular momentum vectors to the Z axis. In a lecture my professor said that it that you can not define the $m_j$ quantum number when there is no external magnetic field to define the Z axis since it makes no sense to talk about projections to the Z axis without any Z axis defined. But if this is true, why can we say an electron is in say the $n=2 ,l=1, m_l=0$ state? We appear to still be able to talk about $m_l$ and $m_s$ without any external $B$ field but not $m_j$?

  • $\begingroup$ Can you please give some context as we were not present in the same lecture as you were. $\endgroup$ – Sam Jan 10 at 15:15
  • $\begingroup$ It was used in the context of the questions "For the hydrogen atom, and taking into account the fine structure, how many transitions are allowed from $n = 4$ directly into the ground state". He then began to list the quantum numbers of the relevant states and said we wont use $m_j$ since no Z axis is defined. $\endgroup$ – Vishal Jain Jan 10 at 15:30

A magnetic moment $\vec \mu \propto g_l \vec L + g_s \vec S$ in a B field possesses the energy $E=-\vec \mu \cdot \vec B$. Thus, if the B field vanishes, $B=0T$, the energy levels of the states $|l, m_l; s, m_s\rangle$ are degenerated. Furthermore, since electrons with the same energy are indistinguishable, we are unable to tell which of the six $|l=1; s=1/2\rangle$ electrons is in which $|m_l, m_s\rangle$ state. Therefore, it does not make sense to say an electron is in particular $m_l$ or $m_s$ state, if $B=0T$.

If we include the fine structure of the atom, we add a term to the Hamiltonian which is proportional to $\vec L \cdot \vec S$. However, by using the definition of the total angular momentum $J$ we obtain \begin{align} \vec J^2 &= (\vec L + \vec S)^2 \\ &= \vec L^2 + \vec S^2 + 2 \vec L \cdot \vec S \\ \Rightarrow \vec L \cdot \vec S &= \frac{1}{2} \left[ \vec J^2 - (\vec L^2 + \vec S^2) \right] \end{align} I'm not sure whether the $g$ factors have to be included in the definition. However, they are only constants and therefore do not change the principle of the calculation.

The key to the fine structure energy levels is the above relationship and using the coupled basis vectors $|l, s; j, m_j\rangle$. Using this basis the calculation of the energy levels for the operator $\vec L \cdot \vec S$ becomes simple, because they are eigenstates. The operator $\vec J^2$ has the eigenvalue $\hbar^2 \, J(J+1)$, and analog for $\vec L^2$ and $\vec S^2$. Thus, again states possessing different magnetic moments still have the same energy.

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  • $\begingroup$ @Vishal Jain: I edited my answer. $\endgroup$ – Semoi Jan 10 at 20:38
  • $\begingroup$ Sorry one thing I am not understanding upon review of this answer is for the fine structure energy levels section of your answer, havent you just proved that unless there is an external field, there is no spin orbit splitting of energy levels? This is not observed to be the case though? $\endgroup$ – Vishal Jain Jan 15 at 19:49

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