My question essentially revolves around multi-electron atoms and spectroscopic terms. I understand the idea that the total wavefunction for Fermions should be antisymmetric. Consider as an example, the $2p^2$ electrons in a partially filled p shell; that is, the outer shell of Carbon. The two electrons both have $l=1$, and hence total orbital angular momentum takes the values:

$L = L1+L2, L1+(L2-1),...,|L1-L2| = 0,1,2$ and $S = 0,1$

I can sort of intuitively see that $L=2$ must refer to a symmetric spatial wavefunction and hence an antisymmetric spin wavefunction. I can handwave and say that for $L=2$, we must have $m_{l1}=m_{l2}=\pm1$ and hence they must have opposing spin to satisfy PEP which gives S=0 - but I'm not sure how to express that in terms of an actual wavefunction and it seems to be a bit of a circular argument. However, I don't see why $L=1$ must have $S=1$ (a triplet) and $L=0$, $S=0$ (another singlet).

Can anyone shed some light on this?


  • 2
    $\begingroup$ It is all a bit more complicated. In general the orbital wave-function is given by a Young tableau and the spin wave-function by the conjugate Young tableau (en.wikipedia.org/wiki/Young_tableau). For two particles, you can figure it out only looking at the symmetry of the wavefunction. $\endgroup$
    – Fabian
    Apr 30, 2011 at 6:49

2 Answers 2


Dear Josh, the wave functions are only perfectly symmetric and/or perfectly antisymmetric - each of the factors is - in the case of two particles. As Fabian correctly says, for more than 2 particles, the wave function isn't perfectly symmetric and isn't perfectly antisymmetric with respect to particular transpositions of the two particles. The general character of the wave function's behavior is given by a Young diagram.

That's why what you want to be proved can only be proved for two particles. In the text below, I will therefore assume that there are just two electrons. What you want to be proved is that if the total $L=L_1+L_2$ of the two electrons is even, the wave function is even under the exchange of the two electrons, and it's odd if $L$ is odd.

It's not hard to see. By rotational symmetry, the states with a given value of $L$ form a multiplet with $2L+1$ components because $L_z$ may go from $-L$ to $+L$ with the spacing one. Without the loss of generality, you may focus on the wave function with the maximum value of $L_3$, namely $L_z=L$.

To get this maximum value, you need $L_z=L_{z,1}+L_{z,2}$ to be composed of two equally maximal, equally oriented terms. Because $L_{z,1}$ and $L_{z,2}$ also go between $-L_1$ and $+L_1$, or similarly between $-L_2$ and $+L_2$, it's clear that the only way to get $L_z=L$ is to have $L_{z,1}=L_1$ and $L_{z,2}=L_2$: both $z$-components must be maximum, too. But if it is so, then the wave function is simply $$ \psi_{L_3=L} = \psi_{L_{z,1}=L_1} \otimes \psi_{L_{z,2}=L_2} $$ For $L_1=L_2$, you can easily see that this wave function has to be symmetric under the permutation of $1$ and $2$. After all, it is the tensor product of the two equal pieces. If you antisymmetrized it, you would get zero. And in fact, the total wave function cannot have any complete symmetry or complete antisymmetry under the exchange of the particles $1,2$ if $L_1\neq L_2$. It's because $L_1^2$ acting on the total wave function gives you $L_1(L_1+1)$ times the wave function, while $L_2^2$ acting on the total wave function gives you $L_2(L_2+1)$. Because the two eigenvalues are not equal for $L_1\neq L_2$, the total wave function can't be symmetric under the exchange of $1,2$.

Again, the question about the symmetry of the orbital wave function only has a sharp answer is there are two particles and if they have the same $L_1=L_2$ - and in that case, the function is symmetric under the permutation.

Similarly, one can prove that the exchange of the two particles with the same half-integer spins $S_1=S_2$ produces a minus sign - assuming that $S_1=S_2$ differs from an integer by $1/2$. To do such things, it's useful to imagine that the components of the multiplet with a given $J$ are organized into a symmetric spintensor.

All the $2J+1$ components of the multiplet with the angular momentum $J$ may be expressed as a completely symmetric "tensor" with 2-valued spinor indices, $$ T_{abc\dots z} $$ Each index is either $0$ or $1$. The number of ones goes from $0$ to the number of indices $N$ - so there are $N+1$ components of this tensor.

Because each index brings $1/2$ to the total angular momentum, it's clear that the number of indices is $N=2J$. Indeed, then we have $N+1=2J+1$ components.

If there were pairs of indices that are antisymmetric with respect to the exchange of the two indices, one could factorize $\epsilon_{ab}$, a totally antisymmetric object. So only the total symmetry is relevant for our simplest case. Now, the spintensor for an angular momentum $J=J_1+J_2$ object may be written as the symmetrization of $$ T_{abc\dots z} = T^{1}_{(abc\dots m} T^{2}_{nop\dots z)} $$ where the parentheses represent the complete symmetrization - because the multiplet is represented by the totally symmetric tensor, as we said. The tensors $T^1$ and $T^2$ correspond to the $J_1$ and $J_2$ pieces.

However, if the total angular momentum $J$ differs by an odd number from its maximum value $J_1+J_2$, like in the case of the singlet where $J_1=J_2=1/2$ but $J=0$, then we must factorize those epsilons. $$ T_{abc\dots z, \,\,{\rm missing\,\,}{mn}} = T^{1}_{(abc\dots l} T^{2}_{op\dots z)} \epsilon^{mn} $$ The epsilon was added to the right hand side to reduce the number of indices by two - i.e. the total spin by one. Because the epsilon is antisymmetric, it changes the symmetry of the whole $T$ under the exchange of the two groups of indices. Each time you reduce the total $J$ by one, the symmetry property changes from symmetry to antisymmetry or vice versa. So the sign obtained from the permutation is given, for $J_1=J_2$, by $(-1)^{J-J_1-J_2}$.

It's also useful to know that the orbital wave functions of a single particle that can be expressed as spherical harmonics $Y_{lm}$ pick the factor of $(-1)^{l}$ under parity i.e. the factor of $(-1)^{l+m}$ under $\theta\to\pi-\theta$.

  • $\begingroup$ Thanks Luboš, I think that may be a little more complicated than I was hoping for, though I realise you clearly spent a decent amount of time writing. The point you made at the end was mentioned in my lecture notes, which I presumed led to a simpler explanation (if you substitute the spherical harmonics corresponding to the various combinations of $l$, $m_l$, etc into the a two-particle symmetric or anti-symmetric wavefunction). $\endgroup$
    – Josh
    Apr 30, 2011 at 17:48

There are several ways of knowing what states should be there. For simple cases such as this, the easiest way is just by counting all the possibilities or micro-states.

Since you have 2 "equivalent" electrons in $p^2$ (equivalent meaning that they share quantum numbers $n$ and $l$, related to the energy of the system) there are $$\left( \begin{array}{l} 6 \\ 2 \end{array} \right) = 6 \cdot 5/2 = 15$$ micro-states (possible ways of assigning $n$, $l$, $m_l$ and $m_s$ to the outer (valence) electrons. Know you have to count all the possible (allowed by Pauli's principle) different arrangements of $m_{l,1}$, $m_{l,2}$, $m_{s,1}$, $m_{s,2}$.

You can find them explicitly in any Physical Chemistry book (e.g. McQuarrie and Simon). Writing $m_{l,1}, m_{l,2}$ and $m_s$ omitted ($m_s = 1/2$) or with an over bar (for spin $m_s = -1/2$), the microstates are: $(1,\bar{1}), (1,0),(1,\bar{0}), (1,-1),(1,-\bar{1})$, $(0,1),(0,\bar{1}), (0,\bar{0}), (0,-1),(0,-\bar{1})$, $(-1,1),(-1,\bar{1}), (-1,0),(-1,\bar{0}), (-1,-\bar{1})$ Now you have to group these states by characterising $L$ and $S$ (since, barring spin-orbit coupling, for a given $L$ and $S$ the $M_L$ and $M_S$ states form a manifold of degenerate states).

There are several ways of doing so, but I'll be sketchy to avoid lengthiness. For instance you can first think in those cases with $m_{s,1} \neq m_{s,2}$ where $m_{l,1}$ can be equal to $m_{l,2}$. Then $M_S = m_{s,1} + m_{s,2} = 0$ ($S$ can still be 1 -triplet or 0 - singlet). $M_L = m_{l,1} + m_{l,2} = 2,1,0$. So you can have states with $L = 2,1,0$. The states with $L = 2$ must be $S = 0$ since, as you realised, $m_{s,1}$ cannot be equal to $m_{s,2}$. Thus you identify 5 micro-states ($M_L = +L, \ldots, 0, \ldots -L$) corresponding to a single level $^1$D.

Of the remaining 10 states you can clearly see from the listing of micro-states that your $L=1$ level is a triplet (you can find microstates with $M_S = 1$ and $M_L = 1$ so the remaining $M_S = 0,-1$ and $M_S = 0, -1$ have to be there too). For a $^3$P level there are $3\times 3 = 9$ microstates. Finally, the remaining micro-state must correspond to a $^1$S state.

Regarding your question of the wave function. Again, think only on your last 2 electrons (it is not difficult to "enlarge" your Slater determinant with the other electrons). Each of your previous micro-states would correspond to a single Slater determinant. For example, for $(1,\bar{0})$ you would have the wave function $$ \frac{1}{\sqrt{2}} \left| \begin{array}{cc} 2p_1(1)\alpha(1) & 2p_0(1)\beta(1) \\ 2p_1(2)\alpha(2) & 2p_0(2)\beta(2) \end{array} \right|$$ where $2p_{m}$ is the wave function in atomic orbital $2p_{m}$ and $\alpha$ and $\beta$ are the spin functions. Some states belonging to the levels with well-defined $L$ and $S$ correspond directly to the micro-states. For instance $(1,\bar{1})$ is exactly $|L=2,S=0,M_L=2,M_S=0\rangle$. However, most states must be written as linear combinations of Slater determinants. For instance, to the state $|L=2,M_L=1,S=0,M_S=0\rangle$ corresponds the wave function $$\frac{1}{2} \left| \begin{array}{cc} 2p_1(1)\alpha(1) & 2p_0(1)\beta(1) \\ 2p_1(2)\alpha(2) & 2p_0(2)\beta(2) \end{array} \right| + \frac{1}{2} \left| \begin{array}{cc} 2p_0(1)\alpha(1) & 2p_1(1)\beta(1) \\ 2p_0(2)\alpha(2) & 2p_1(2)\beta(2) \end{array} \right|$$ whereas for the state $|L=2,M_L=0,S=0,M_S=0\rangle$ the wave function would be $$\frac{1}{2\sqrt{3}} \left| \begin{array}{cc} 2p_1(1)\alpha(1) & 2p_{-1}(1)\beta(1) \\ 2p_1(2)\alpha(2) & 2p_{-1}(2)\beta(2) \end{array} \right| + \frac{1}{\sqrt{3}} \left| \begin{array}{cc} 2p_0(1)\alpha(1) & 2p_0(1)\beta(1) \\ 2p_0(2)\alpha(2) & 2p_0(2)\beta(2) \end{array} \right| + \frac{1}{2\sqrt{3}} \left| \begin{array}{cc} 2p_{-1}(1)\alpha(1) & 2p_{1}(1)\beta(1) \\ 2p_{-1}(2)\alpha(2) & 2p_{1}(2)\beta(2) \end{array} \right|$$

  • $\begingroup$ Thanks Perplexity, that's the sort of answer I was looking for. So am I right in saying you first work out that $L=0,1,2$, then work out the relevant $M_L$ and $M_S$ values and then work out the states that correspond to each $M_L$,$M_S$ state? Again, the case for $L=2$ $(1,\bar 1)$ and $(-1,\bar{-1})$ seem straightforward ($M_L=2,-2$ I would think?), but which states refer to $M_L=\pm1,0$? Apologies if this seems like a simple question - I know I'm overthinking this, but please humour me :) $\endgroup$
    – Josh
    Apr 30, 2011 at 18:25
  • $\begingroup$ You cannot identify a single microstate with states $|L=2,M_L=1,S=0,M_S=0\rangle$ or $|L=2,M_L=0,S=0,M_S=0\rangle$. For the first one the solution (Clebsch-Gordan coefficients) is: $1/\sqrt{2} (1 \bar{0} ) + 1/\sqrt{2} (0 \bar{1})$ (where $(a b)$ is typically written as a Slater determinant). For the second one, $(L=2 M_L=0) = 1/\sqrt{6} (1 -\bar{1}) + \sqrt{2/3} (0, \bar{0} ) + 1/\sqrt{6} (-1 \bar{1} )$ $\endgroup$
    – perplexity
    May 1, 2011 at 11:44
  • $\begingroup$ Ok, thanks, a superposition of the various states makes sense (sneaky edit there!) :) $\endgroup$
    – Josh
    May 3, 2011 at 21:07

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