I've learned how to add two 1/2-spins, which you can do with C-G-coefficients. There are 4 states (one singlet, three triplet states). States are symmetric or antisymmetric and the quantum numbers needed are total spin and total z-component.

But how do you add three 1/2-spins? It should yield 8 different eigenstates. Which quantum numbers do you need to characterise the 8 states?

It is not as easy as using C-G-coefficients and the usual quantum numbers as for the total momentum the doubly degenerate 1/2 state and quadruple degenerate 3/2 state can describe only 6 or the 8 states. You will need an additional quantum number for the degeneracy.

So how do you get the result?

(I actually tried out myself with a large 8x8 matrix. The total spin 1/2 is each doubly degenerate. For the additional quantum number I chose the cyclic permutation. Spin 1/2 states are neither symmetric nor antisymmetric. But what is the usual way to derive this?)

EDIT: For reference I'm adding my results for up to 4 spins from some time ago: enter image description here

If you recall the basics of quantum mechanics with matrices it is actually a straightforward matrix diagonalization and requires no specialized knowledge. However, you still need to find an additional operator which breaks degeneracy. I chose the cyclic permutation, which seems to do the job. Please refer to the below answer, since I haven't checked all details.


I looked in Edmonds, which is usually the standard reference, and he doesn't mention any standard approach at breaking the degeneracy.

You need two linearly independent $s=1/2,\,m=1/2$ solutions, and you can get three different solutions by first coupling one of the three different pairs to the singlet $s=0$ state and then adding an up state. This yields the three vectors $\newcommand{\ket}[1]{\left|#1\right\rangle}$ $$\ket{\psi_1}={1\over\sqrt{2}}\left(\ket{\uparrow\uparrow\downarrow}-\ket{\uparrow\downarrow\uparrow}\right),$$ $$\ket{\psi_2}={1\over\sqrt{2}}\left(\ket{\downarrow\uparrow\uparrow}-\ket{\uparrow\uparrow\downarrow}\right),$$ $$\ket{\psi_3}={1\over\sqrt{2}}\left(\ket{\uparrow\downarrow\uparrow}-\ket{\downarrow\uparrow\uparrow}\right),$$ which add to zero so only two are linearly independent.

Edmonds shows, in particular, that there is a unitary transformation linking any of the three representations linked to the three vectors above (which is of course no surprise) and that this unitary transformation is independent of spatial orientation (which is not automatic but by the Wigner-Eckart theorem ought to happen). He then goes on to define appropriate invariant transformation coefficients (the Wigner $6j$ symbols) and spends a good deal of time exploring them, but he doesn't say how to (canonically) break the degeneracy.

If it's a basis you want, then take any two of the three above. If you need (like you should!) an orthonormal basis, then you can take linear combinations like $$\ket{\psi_{23}}={1\over\sqrt{6}}\left(\ket{\uparrow\uparrow\downarrow}-2\ket{\downarrow\uparrow\uparrow}+\ket{\uparrow\downarrow\uparrow}\right)$$ which obeys $\langle\psi_1|\psi_{23}\rangle=0$.

However, I don't think there is any way to treat the problem symmetrically in the three electrons. I had a quick go and I think one can prove there are no linear combinations of the three states that are symmetric or antisymmetric w.r.t. all three electron exchanges.

One way to see this is noting that you have three linearly dependent, unit-norm vectors that span a two-dimensional vector space and sum to zero. This is like having three unit vectors on a plane, symmetrically arranged at $120^\circ$ to each other. (The analogy is precise: the Gram matrices, $G_{ij}=\langle\psi_i|\psi_j\rangle=-\frac12+\frac32\delta_{ij}$, coincide, and these encode all the geometrical information about any set of vectors - see problem 8.5 in these notes by F. Jones at Rice.) There is then no way to choose a basis for the plane that is symmetric in the three "electron" exchanges, i.e. one whose symmetry group is the same as the three original vectors, including all three reflections.

enter image description here

On the other hand, there are two approaches to this problem that do retain some of the exchange symmetry. One is to form an electron-exchange invariant resolution of the identity, of the form $$ \frac{2}{3}\sum_{j=1}^3\ket{\psi_j}\langle\psi_j|=1|_{S={1\over2},m= +{1\over2}} $$ This also holds for the three vectors in the plane and expresses the fact that they form a tight vector space frame for $\mathbb{R}^2$. This is also a consequence of Schur's lemma, as both vector spaces carry irreducible representations of the exchange group of three electrons; the sum above is the Haar integral over the orbit of any one state and commutes with all matrices in the representation.

The other approach is due to the OP, who provided this image (with slight errors), and which I'll write in full here for completeness. An alternative basis for the plane, which does play well with the electron exchange group - though not as symmetric as one might wish - is to use a complex-valued basis (which is of course perfectly all right) and which corresponds to the circular polarization basis if we think of the plane as the Jones vectors for the polarization of an EM wave. In this analogy, the vectors in the image represent polarizations about those directions. Circular polarization is then invariant - up to a phase - under rotations, but individual electron exchange reflections will flip left$\leftrightarrow$right circular polarizations.

To cut the waffle, the trick in the plane is to take as basis vectors $$ \mathbf{e}_L=\begin{pmatrix}1\\i\end{pmatrix} =\frac23\sum_{j=1}^3 e^{\frac{2\pi i}{3}(j-1)}v_j \text{ and } \mathbf{e}_R=\begin{pmatrix}1\\-i\end{pmatrix} =\frac23\sum_{j=1}^3 e^{-\frac{2\pi i}{3}(j-1)}v_j. $$ These are taken to each other, up to a phase, by the reflections, and to themselves up to a phase by the rotations.

Similarly, for the three electrons you can take the combinations $$ |\psi_+\rangle =\frac{1}{\sqrt{3}} \left[\ket{\uparrow\uparrow\downarrow}+e^{2\pi i/3}\ket{\uparrow\downarrow\uparrow}+e^{-2\pi i/3}\ket{\downarrow\uparrow\uparrow}\right] =\frac{\sqrt{2}}{3}e^{-i\pi/6}\sum_{j=1}^3e^{-\frac{2\pi i}{3}(j-1)}|\psi_j\rangle $$ and $$ |\psi_-\rangle =\frac{1}{\sqrt{3}} \left[\ket{\uparrow\uparrow\downarrow}+e^{-2\pi i/3}\ket{\uparrow\downarrow\uparrow}+e^{2\pi i/3}\ket{\downarrow\uparrow\uparrow}\right] =\frac{\sqrt{2}}{3}e^{+i\pi/6}\sum_{j=1}^3e^{+\frac{2\pi i}{3}(j-1)}|\psi_j\rangle $$ which are eigenvectors of the cyclic permutations with eigenvalue $e^{\pm 2\pi i/3}$, and for which the individual exchanges act as $$P_{12}|\psi_+\rangle=|\psi_-\rangle, \ P_{23}|\psi_+\rangle=e^{\frac{2\pi i}{3}}|\psi_-\rangle, \text{ and }P_{31}|\psi_+\rangle=e^{\frac{-2\pi i}{3}}|\psi_-\rangle . $$

So, in conclusion: this method is not perfect, as it does not give a way to lift the degenerate subspace into two distinct subspaces which are invariant under the full electron exchange group, and which therefore carry separate representations of it. However, it does give a basis that's got a definite action under the exchange group. I would be interested to know what the formal analysis of this action is, and how this generalizes to more than three spins. Maybe for another time!

  • 1
    $\begingroup$ Thanks. I tried a brute force approach diagonalizing the 8x8 matrix for total angular momentum. The result was img31.imageshack.us/img31/6228/3spins.png. The cyclic operator turned out to be nice and also works for 4 spins. I need to study your approach - I forgot the basics :) A bit surprising that there is no easy answer. $\endgroup$ – Gerenuk Jun 5 '12 at 15:30
  • $\begingroup$ Yes, it is quite surprising. Your solution is also quite nice, and it does allow a symmetrical treatment of the three electrons. Why not add it as another answer? $\endgroup$ – Emilio Pisanty Jun 5 '12 at 18:04

$\newcommand{\Ket}[1]{\left|#1\right>}$You can build them just from the highest one, $\Ket{\frac{1}{2}\frac{1}{2}\frac{1}{2}}$, using the lowering operator $S_{-}=S_{1,-}+S_{2,-}+S_{3,-}$. Now remember that each operator in this sum acts only on it's respective space of states. Also, it gets messy with the numerical coefficients, but remember, after each step you can check to see if the norm is 1. Lets do one together. $$ \begin{align} S_{-}\Ket{\frac{3}{2},\frac{3}{2}} &= %\sqrt{\left(\frac{3}{2}+\frac{3}{2}\right)\left(\frac{3}{2}-\frac{3}{2}+1\right)} \sqrt3 \hbar\Ket{\frac{3}{2},\frac{1}{2}} \tag A \\ (S_{1,-}+S_{2,-}+S_{3,-})\Ket{\frac{1}{2};\frac{1}{2};\frac{1}{2}} &= %\sqrt{\left(\frac{1}{2}+\frac{1}{2}\right)\left(\frac{1}{2}-\frac{1}{2}+1\right)} \hbar\left(\Ket{{-\frac{1}{2}}\frac{1}{2}\frac{1}{2}}+\Ket{\frac{1}{2}{-\frac{1}{2}}\frac{1}{2}}+\Ket{\frac{1}{2}\frac{1}{2}{-\frac{1}{2}}}\right) \tag B \end{align} $$ The numerical factor out front comes from the lowering operator, $$ S_\pm\Ket{s,m} = \hbar \sqrt{ s(s+1) - m(m\pm1) } \Ket{s, m\pm1} $$ which is what terminates the series if you try to raise or lower beyond $|m|=s$.

You can do this a couple more times to to get the other $3/2$ ones. But after this first one, you can build a $\Ket{\frac{1}{2}\frac{1}{2}}$ one thats perpendicular to the other ones. I think you can just use Gram-Schmidt, or eye-ball it. Then with the $\Ket{\frac{1}{2}\frac{1}{2}}$ you just use the lowering operator some more.

I did this for learning a little while back, I hope this helps/is correct.

  • $\begingroup$ The problem is that there will be two linearly independent $\left|\frac{1}{2}\frac{1}{2}\right\rangle$ states, representing in total four ($=8-(2\times\frac{3}{2}+1)$) Hilbert space dimensions orthogonal to the $s=\frac{3}{2}$ subspace. The problem is choosing, hopefully in a symmetric manner, these two states. $\endgroup$ – Emilio Pisanty Jun 4 '12 at 14:40

I don't understand the answers given, nor the reference to an 8x8 matrix. When combing 3 doublets (aka spin 1/2), the guiding principle is that $$2 \times 2 \times 2 = 4 + 2 + 2.$$ That is, the tensor product can be decomposed into a tensor sum of composite states that are a quartet (spin 3/2) and 2 doublets (spin 1/2).

The quartet is symmetric under interchange and is (up to normalization): \begin{align} |3/2, 3/2⟩ & = |↑↑↑⟩\\ |3/2, 1/2⟩ & = \frac{|↑↑↓⟩+|↑↓↑⟩+|↓↑↑⟩}{\sqrt{3}}\\ |3/2,-1/2⟩ & = \frac{|↓↓↑⟩+|↓↑↓⟩+|↑↓↓⟩}{\sqrt{3}}\\ |3/2,-3/2⟩ & = |↓↓↓⟩ \end{align}

The two doublets are combinations of: \begin{align} |↑⟩(|↑↓⟩-|↓↑⟩) & =|↑↑↓⟩-|↑↓↑⟩,\\ |↓⟩(|↑↓⟩-|↓↑⟩) & =|↓↑↓⟩-|↓↓↑⟩ \end{align} and \begin{align} (|↑↓⟩-|↓↑⟩)|↑⟩ & =|↑↓↑⟩-|↓↑↑⟩,\\ (|↑↓⟩-|↓↑⟩)|↓⟩ & =|↑↓↓⟩-|↓↑↓⟩, \end{align} and appear to have mixed symmetry. For example, one doublet is: \begin{align} |1/2, 1/2⟩^{(1)} & = \frac{|↑↑↓⟩+|↑↓↑⟩-2|↓↑↑⟩}{\sqrt{6}},\\ |1/2,-1/2⟩^{(1)} & = \frac{|↓↓↑⟩+|↑↓↓⟩-2|↑↓↓⟩}{\sqrt{6}} \end{align} and an orthogonal combination is \begin{align} |1/2, 1/2⟩^{(2)} & = \frac{2|↑↑↓⟩-|↑↓↑⟩-|↓↑↑⟩}{\sqrt{6}},\\ |1/2,-1/2⟩^{(2)} & = \frac{2|↓↓↑⟩-|↑↓↓⟩-|↑↓↓⟩}{\sqrt{6}}. \end{align}

Any other combination with a zero spin-3/2 component is a linear combination of these two.

The $8 \times 8$ matrix indicates a misunderstanding of the problem: while we can make product states where we know each particles spin-there by justifying an $8 \times 8$ operator, those states are not eigenstates of total angular momentum, and hence we don't want to consider them.

We consider the combinations that are eigenstates of total angular momentum, and the way to find them is as the first answer stated: pair the first and second spin into a spin 1 triplet and a spin 0 singlet and then, using Clebsch-Gordan coefficients, take their products with a doublet: \begin{align} \text{spin-1 times spin 1/2: } & 3 \times 2 = 4 + 2\\ \text{spin-0 times spin 1/2: } & 1 \times 2 = 2 \end{align} (which is how the doublets broke down explicitly, as shown above).

So recapping, given the product of 3 doublets, break it down pairwise: \begin{align} 2 \times 2 \times 2 & = (2 \times 2) \times 2 \\ (2 \times 2) \times 2 & = (3 + 1) \times 2 \\ (3 + 1) \times 2 & = (3 \times 2) + (1 \times 2) \\ (3 \times 2) + (1 \times 2) & = (4 + 2) + (2) = 4 + 2 + 2 \end{align}

Also: the desire for symmetry is great, but in general only the extremal case $|J,J⟩$ is symmetric, and the other states have mixed symmetry. There may or may not be an antisymmetric case. See the Wikipedia article on Young tableaux for more on that.

  • $\begingroup$ One short comment: the 8$\times$8 matrix is pretty well justified - it is the unitary matrix taking the product basis into the basis of common $J^2,J_z$ eigenstates. Thus, you're diagonalizing $J^2$ and $J_z$ as 8$\times$8 matrices in the product basis, and looking at the resulting unitary. The original poster (not active since January) quite clearly understood that this was a valid but non-ideal method. $\endgroup$ – Emilio Pisanty May 28 '15 at 21:24
  • $\begingroup$ Your final comments are quite interesting. Are you claiming that in the general case of $n$ spins there are provably no decompositions which respect the antisymmetry? What about paired representations which are swapped by single electron exchanges - are those provably not possible in the general case? $\endgroup$ – Emilio Pisanty May 28 '15 at 21:33
  • $\begingroup$ I am working on a python package for real-world Euclidean-space-as-we-known-it vector/tensor processing, and I wanted to understand symmetries beyond the rank-2 case. The symmetries for rank-N tensors look a lot the combination of N spin-1 (vector) particles. One can go brute force with Clebsch-Gordon coefficients, or turn to Young tableaux and the representation theory of the permutation group. That leads to the spectacular "Hook-Length Formula" which allows you to calculate the dimensions of a symmetry Irrep in any dimension. For 3D, rank-4 tensor, fully antisymmetric: it is 0. $\endgroup$ – JEB Aug 13 '15 at 15:33
  • $\begingroup$ The only nontrivial claims in this answer are unjustified, so it should really be considered incorrect unless the concerns in my comment above are addressed by the poster. $\endgroup$ – Emilio Pisanty Jan 4 '16 at 16:42
  • $\begingroup$ @EmilioPisanty When I make an antisymmetric young tableau for N-spins (so a column of N boxes), and use the hook-length formula (en.wikipedia.org/wiki/…) I get zero if $N>2$ and the representation is 2 dimensional. $\endgroup$ – JEB Apr 17 '18 at 13:55

$\newcommand{\rket}[1]{\left|#1\right>}% \renewcommand{\ket}{\rket}% \newcommand{\up}{\uparrow}\newcommand{\dn}{\downarrow}% $This answer is in the same spirit as the answer by kηives ("you figure it out from the ladder operators"), but more explicit about breaking the degeneracy between the two spin-half combinations. The trick is to notice that both of the spin-3/2 states in kηive's answer can be written with the first two spins combined in a spin-one triplet: $$ \begin{alignat}2 \rket{\frac32, +\frac32} &= \rket{\up\up\up} &&= \ket{\up\up}\ket{\up} = \ket{1,1} \ket\up \\ \sqrt3\rket{\frac32, +\frac12} &= \rket{\dn\up\up} +\rket{\up\dn\up} +\rket{\up\up\dn} &&= \sqrt2\left(\frac{ \ket{\up\dn} + \rket{\dn\up} }{\sqrt2}\right)\rket\up + \rket{\up\up}\rket\dn \\&&&= \sqrt2\rket{1,0}\rket\up + \rket{1,1}\rket\dn \end{alignat} $$ These are the same linear combinations as with the ordinary Clebsch-Gordan coefficients for spin-one with spin-half, and suggests that one of the spin-half combinations should be $$ \begin{alignat}3 \sqrt3\rket{\frac12,+\frac12}_\text{triplet} &= \rket{1,0}\rket\up - \sqrt2\ket{1,1}\ket\dn \\&= \left(\frac1{\sqrt2}\ket{\up\dn\up} + \frac1{\sqrt2}\ket{\dn\up\up}\right) -\sqrt2\ket{\up\up\dn} \end{alignat} $$ You can check that this "triplet" combination is orthogonal to the others and gets killed by the raising operator, $$ S_+\ket{s,m} = \hbar\sqrt{s(s+1) - m(m+1)} \ket{s, m+1}, $$ regardless of whether you use the spin-half raising operator on the individual $\ket\dn$ or the spin-one raising operator $S_+\ket{1,0}=\hbar\sqrt2\ket{1,1}$ on the linear combination.

The remaining state that we haven't yet used for the first two particles is the singlet, $$ \begin{align} \rket{\frac12,+\frac12}_\text{singlet} &= \ket{0,0}\ket\up = \frac1{\sqrt2}\ket{\dn\up\up} - \frac1{\sqrt2}\ket{\up\dn\up} \end{align} $$ which gets killed by the raising operator as well. Of course you can construct all the states with negative $m$ using the lowering operator.

The four states with total spin $\hbar/2$ have mixed symmetry under exchange (but definite symmetry under exchange of the first two particles). There isn't a completely antisymmetric state to be constructed out of three two-state particles, but you could construct a completely mixed-symmetry state with definite spin by taking a linear combination of my "singlet" and "triplet" states.


I only show how to obtain the C-G coefficients for the two states $|\frac{1}{2},\frac{1}{2}\rangle_1$ and $|\frac{1}{2},\frac{1}{2}\rangle_2$.

First, it is easy to see that $|\frac{3}{2},\frac{1}{2}\rangle=\frac{1}{\sqrt{3}}(|-++\rangle+|+-+\rangle+|++-\rangle)$.

By writing $|\frac{1}{2},\frac{1}{2}\rangle_i=\alpha_i|-++\rangle+\beta_i|+-+\rangle+\gamma_i|++-\rangle$ ($i=1,2$), then the matrix $U=\left( \begin{array}{ccc} \frac{1}{\sqrt{3}} & \alpha_1 & \alpha_2 \\ \frac{1}{\sqrt{3}} & \beta_1 & \beta_2 \\ \frac{1}{\sqrt{3}} & \gamma_1 & \gamma_2 \\ \end{array} \right)$ should be an orthogonal matrix, due to the normalization of $|\frac{1}{2},\frac{1}{2}\rangle_1$ and $|\frac{1}{2},\frac{1}{2}\rangle_2$, and the fact that the three states with same $m$, $|\frac{3}{2},\frac{1}{2}\rangle$, $|\frac{1}{2},\frac{1}{2}\rangle_1$, and $|\frac{1}{2},\frac{1}{2}\rangle_2$, should be mutually orthogonal.

We thus have \begin{eqnarray} &&\alpha^2_i+\beta^2_i+\gamma^2_i=1,\\ &&\alpha_i+\beta_i+\gamma_i=0,\\ &&\alpha_1\alpha_2+\beta_1\beta_2+\gamma_1\gamma_2=0 \end{eqnarray} By eliminating $\gamma_i$ from the first two equations, we get $ \alpha^2_i+\alpha_i\beta_i+(\beta^2_i-\frac{1}{2})=0$, which gives $\alpha_i=\frac{1}{2}(-\beta_i\pm\sqrt{2-3\beta^2_i })$. So we have \begin{eqnarray} (\alpha_i,\beta_i,\gamma_i)=\left(\frac{1}{2}(-\beta_i\pm\sqrt{2-3\beta^2_i }),\beta_i,\frac{1}{2}(- \beta_i\mp\sqrt{2-3\beta^2_i })\right) \end{eqnarray} Now we can choose \begin{eqnarray} (\alpha_{1/2},\beta_{1/2},\gamma_{1/2})=\left(\frac{1}{2}(-\beta\pm\sqrt{2-3\beta^2 }),\beta,\frac{1}{2}(- \beta\mp\sqrt{2-3\beta^2 })\right) \end{eqnarray} so that the orthogonal condition $\alpha_1\alpha_2+\beta_1\beta_2+\gamma_1\gamma_2=0$ gives \begin{eqnarray} \frac{1}{4}(\beta^2-2+3\beta^2)+\beta^2+\frac{1}{4}(\beta^2-2+3\beta^2)=0\to3\beta^2-1=0\to\beta=\pm\frac{1}{\sqrt{3}} \end{eqnarray} By choosing $\beta=\frac{1}{\sqrt{3}}$, we finally obtain \begin{eqnarray} |\frac{1}{2},\frac{1}{2}\rangle_1&=&\frac{1}{2}(1-\frac{1}{\sqrt{3}})|-++\rangle+\frac{1}{\sqrt{3}}|+-+\rangle-\frac{1}{2}(1+\frac{1}{\sqrt{3}})|++-\rangle,\nonumber\\ |\frac{1}{2},\frac{1}{2}\rangle_2&=&\frac{1}{2}(1+\frac{1}{\sqrt{3}})|-++\rangle-\frac{1}{\sqrt{3}}|+-+\rangle-\frac{1}{2}(1-\frac{1}{\sqrt{3}})|++-\rangle. \end{eqnarray}

Other choices appearing in previous answers do not satisfy the condition $\alpha_1\alpha_2+\beta_1\beta_2+\gamma_1\gamma_2=0$ [for example $(\alpha_{1/2},\beta_{1/2},\gamma_{1/2})=\left(e^{\mp 2\pi i/3},e^{\pm 2\pi i/3},1\right)$ or $(\alpha_{1},\beta_{1},\gamma_{1})=\frac{1}{\sqrt{6}}\left(-2,1,1\right)$ and $(\alpha_{2},\beta_{2},\gamma_{2})=\frac{1}{\sqrt{6}}\left(-1,-1,2\right)$].

Another choice of $(\alpha,\beta,\gamma)$ can be found in the following lecture note (P48): https://public.lanl.gov/mparis/qmp.pdf

There, \begin{eqnarray} |\frac{1}{2},\frac{1}{2}\rangle_1&=&-\frac{1}{\sqrt{2}}|-++\rangle+\frac{1}{\sqrt{2}}|+-+\rangle,\nonumber\\ |\frac{1}{2},\frac{1}{2}\rangle_2&=&-\frac{1}{\sqrt{6}}|-++\rangle-\frac{1}{\sqrt{6}}|+-+\rangle+\frac{2}{\sqrt{6}}|++-\rangle. \end{eqnarray} are chosen. I just noticed that these are already mentioned in Emilio Pisanty's answer above.

This is possible since we have 6 unknowns and only 5 constraints, so the solutions are not unique. As stated in the above reference, the additional quantum number to resolve the degeneracy is just the permutation operator between spin 1 and spin 2, $P_{12}$: $P_{12}|\frac{1}{2},\frac{1}{2}\rangle_{1,2}=\mp |\frac{1}{2},\frac{1}{2}\rangle_{1,2}$.

It is easy to see that $|\frac{1}{2},\frac{1}{2}\rangle_{1,2}$ are also eigenstates of $\vec{S}_1\cdot\vec{S}_2$, with eigenvalues $-\frac{3}{4}$ and $\frac{1}{4}$, respectively.

  • $\begingroup$ Hi. It's been a while since I did physics and it will take me a while to think about that. Is my solution also working? Which quantum number would you suggest to resolve the degenerate states? $\endgroup$ – Gerenuk Dec 18 '18 at 21:05
  • $\begingroup$ Please see that update. $\endgroup$ – Enigma Dec 19 '18 at 6:10
  • $\begingroup$ I have the impression U should be unitary and not orthogonal?! In that case my attempt with complex coefficients satisfies all needed conditions? Can you still you a spin swap operator when you have 4 spins and a triple-degeneracy? $\endgroup$ – Gerenuk Dec 19 '18 at 8:03
  • $\begingroup$ Textbooks tell us the C-G coefficients can always chosen real (at least for addition of two angular momenta). Here we see that a real orthogonal matrix works for 3 spins-1/2, and we even have more than one possible real solutions. I'd like to mention that my choice looks more balanced compared with those in the lecture notes. For my choice, I believe some linear combination of the operators $S_1\cdot S_2$, $S_2\cdot S_3$, and $S_1\cdot S_3$ will give the corresponding constant of motion that resolve the degeneracy. $\endgroup$ – Enigma Dec 19 '18 at 8:43

Reading over the previous answers, perhaps the answer is simple. We know, in atoms, for example, that it is impossible to have three electrons with the same orbital state. You're question is essentially asking if we have one orbital and we put three electrons into it what state will it be in?

The answer is that, due to Pauli exclusion, all 3 electrons need to be in different spin states if they all want to live in the same orbital state. But there are only 2 states available to the single electron so this is impossible. Hence why the OP and other answers do not find an anti-symmetric subspace within the space of three spin 1/2 particles.

If your single particle Hilbert space has dimension $D$ then it is only possible to have up to $N=D$ indentical Fermions in the system.


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