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I am very confused about the microstates for a specific orbital. Let's suppose I have an electronic configuration $$ \ce{[\dots] 2p^5} $$ Considering the symmetry of the orbitals and the "Electron-Hole" idea, one would get for this state the same Term Symbols for the possible states, i.e.: \begin{align} L &\in \{1\}\\ S &\in \{1/2\}\\ J &\in \{1/2,3/2\} \end{align} With that, one gets the following terms: $$^2P_{1/2} \text{ and } ^2P_{3/2}$$

My confusion starts here. If I simply draw every possible state with the electron-box diagram, I would get 6 possible states:

\begin{array}{|l|c|c|c|}\hline 1:& \uparrow\downarrow & \uparrow\downarrow & \uparrow\\\hline 2:& \uparrow\downarrow & \uparrow\downarrow & \downarrow\\\hline 3:& \uparrow & \uparrow\downarrow & \uparrow\downarrow\\\hline 4:& \downarrow & \uparrow\downarrow & \uparrow\downarrow\\\hline 5:& \uparrow\downarrow & \uparrow & \uparrow\downarrow\\\hline 6:& \uparrow\downarrow & \downarrow & \uparrow\downarrow\\\hline \end{array}

What am I thinking or doing wrong?

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  • $\begingroup$ Do you know Hund's rules ? Most of your possible states are inconsistent $\endgroup$
    – M06-2x
    Commented Jun 24, 2023 at 1:27
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    $\begingroup$ Everything looks correct so far. You can calculate the M_S and M_L values of the configurations, then draw a table with M_S and M_L as row and column and then extract two tables that correspond to your term symbols. One table should be 2x2 (S=1/2, L=1, J=1/2) and the other table 4x4 (S=1/2, L=1, J=3/2). What exactly are you unsure about? $\endgroup$
    – Hans Wurst
    Commented Jun 24, 2023 at 7:21
  • $\begingroup$ Makes sense. But if the energy shift of the $L \cdot S$ coupling for fixed L and S only depends on the values for J : $\Delta E_{FS}=c/2\cdot(j(j+1)-l(l+1)-s(s+1)$, can we say that the ground state, according to the Hund's rules, can actually assume any one of the 2 states of j=1/2, as stated by you? $\endgroup$
    – Kubrik
    Commented Jun 24, 2023 at 14:11

1 Answer 1

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I will explain in details why I stated (in the comments) that some states are inconsistent. The spin-orbit (S.O) coupling correlates the orbital angular momentum $\vec{L}$ and the spin angular momentum $\vec{L}$ in a way that the eigenvalues of these operators are not good quantum numbers. This means naively that the ground-state is not an eigenvector of these operators anymore, but an eigenvector of the total angular momentum $\vec{J}$. This is a non surprising result as the orbit of a charged particle independent of its spin around a heavy charged particle creates has an angular momentum, a magnetic moment and a magnetic field which acts on the spin of the particle : in a more intuitive picture, one has to consider a relative motion of the heavy charge around the light particle, giving in the frame of the latter the superposition of $\vec{L}$ and $\vec{S}$. When someone is used to deal with dipole interactions, there is an intuitive deduction that the orbital magnetic moment derived from $\vec{L}$ and the spin magnetic moment proportional to $\vec{S}$ should be antiparallel in the ground-state : therefore the classical direction of the rotational motion should be chosen depending on the orientation of the spin to the particle. Considering a weak interaction ($^2P$), the S.O. coupling gives the correction :

$$ \Delta H_{S.O.} = \lambda \vec{L}\cdot \vec{S} = \lambda L.S.\cos(\theta)$$

There is no need to use this expression $ \Delta E_{S.O.} = \beta \left[j(j+1)-l(l+1)-s(s+1)\right]$ to derive the third rule : if $\lambda > 0 $, the only stable configuration is $\cos(\theta) < 0$, this means $ \frac{\pi}{2} < \theta < \frac{3\pi}{2}$. Because, $\vec{L}$ is quantized along $z$, $L_z |l,m\rangle = \hbar m_l|l,m\rangle$ with $m_l$ integers, and $S_z |l,m\rangle = \hbar m_s|s,m\rangle$ with $m_s$ half integers, the angle $\theta \approx \frac{9}{10} \pi \neq \pi$ for an antiparallel configuration of $\vec{L}$ and $\vec{S}$ (should be checked). With S.O. coupling, $m_j$ eigenvalues of $\vec{J}$ are good quantum numbers. The previous results shows that $\vec{S}$ and $\vec{L}$ should be antiparallel for a stable configuration. This means that if you consider $\vec{J} = \vec{L} + \vec{S}$ as the total angular momentum, $m_l<0$ should match $m_s>0$ and inversely, the filling is given step-by-step in the table below :

\begin{array}{|l|c|c|c|c|}\hline m_l & +1 & 0 & -1\\\hline & \downarrow & & \\\hline & \downarrow & \downarrow & \\\hline & \downarrow & \downarrow & \downarrow \\\hline & \downarrow & \downarrow & \uparrow\downarrow \\\hline (4) & \downarrow & \uparrow\downarrow & \uparrow\downarrow \\\hline \end{array}

The final states gives $J=|-1+\frac{1}{2} |= 3/2$ and is the ground state, some of the other states do not satisfy $\Delta E_{SO} <0$ especially if $J=1/2$. The state (3) is also correct only if one starts by $m_l=-1$ :

\begin{array}{|l|c|c|c|c|}\hline m_l & -1 & 0 & +1\\\hline (3) & \uparrow & \uparrow\downarrow & \uparrow\downarrow \\\hline \end{array}

However, the state (1) represent the conventional filling picture, from left to right, where the empty states are on the right, this is the reason why the third rule has the term $J=|L-S|$ for less than half filled and $J=L+S$ for more than half filled shell. Obviously, if you fill the shell from left to right until an half filled shell an then right to left, $J\equiv |L+S |$.

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  • $\begingroup$ But your assuption for a stable state is for the ground state. My main question,initially, didn't have to do only with the ground state, but rather with all allowed states, even if they aren't stable. But at the same time, your answer clarified my comment about the possible ground states, i.e. two of them are equally possible, if I understood correctly. $\endgroup$
    – Kubrik
    Commented Jun 27, 2023 at 21:19
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    $\begingroup$ The possibles states are (permutation of 2 spins in 3 levels) and (2 spins -> 3 levels) gives $3!\times 3!$ = 36 possibilities without Pauli exclusion and reduced to 6 by applying the exclusion principle, so you was right. Three possible values of $L$, therefore two states with $L=0$ ($J=1/2$), two states $L=-1$, one with $S=1/2$ ($|J|=1/2$) and two states $L=1$, one ($S=-1/2$) ($|J|=1/2$). To summarize four states $|J|=1/2$ and your are right two of them are the ground state ($|J|=3/2$). My answer should not have been so long and so far from the main question. $\endgroup$
    – M06-2x
    Commented Jun 27, 2023 at 22:14

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