I am considering a system that has degenerate energies, degenerate $S^{2}$, and degenerate $S_{z}$, so how do I find the missing symmetry, probably rotational, that makes the states unique?

For example, $$ |1,1\rangle_1 = \frac{1}{2}\bigg[ |\uparrow \downarrow \uparrow \uparrow \rangle + |\downarrow \uparrow \uparrow \uparrow \rangle - |\uparrow \uparrow \uparrow \downarrow \rangle - |\uparrow \uparrow \downarrow \uparrow \rangle \bigg] $$ and $$ |1,1\rangle_2 = \frac{1}{\sqrt{2}}\bigg[ |\uparrow \uparrow \uparrow \downarrow \rangle - |\uparrow \uparrow \downarrow \uparrow \rangle \bigg] $$ and $$ |1,1\rangle_3 = \frac{1}{\sqrt{2}}\bigg[ |\uparrow \downarrow \uparrow \uparrow \rangle - |\downarrow \uparrow \uparrow \uparrow \rangle \bigg] $$

I want to do matrix representation, so I have the $16\times16$ matrix, but I'm not sure how to populate it. I'm assuming I find eigenvalues?

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    $\begingroup$ I strongly recommend absorbing the two - signs of the first line in the definitions of those states, which will flip the overall sign of the second line and will leave the third line alone. You then have the addition of 4 spin 1/2s to two singlets, one quintet, and three triplets, which you wrote. The first line is symmetric under all four site interchanges, the second antisymmetric for [3,4], and the third for [1,2]. Clebsch along. $\endgroup$ Jan 27 at 22:22
  • $\begingroup$ @CosmasZachos Unfortunately, these states popped out of me treating the 4 1/2-spin system as 2 singlet/triplet spin systems, so the coefficients must stay less I redo the preliminary work, which I don't think is necessary. This in fact gave me 2 singlets, 3 triplets, and 1 quintet. I'd like to just move forward from here and work with the block diagonal matrix to find the "missing" non-degenerate symmetry. But I don't know how. $\endgroup$ Jan 27 at 22:33
  • $\begingroup$ I'm not sure what you understand you want. It should be obvious to you that all three states you wrote will be annihilated by a raising operator, and all three are orthogonal to the one down state of the quintet. With the signs you already have, the first line is [1,3][2,4] and (12)(34). $\endgroup$ Jan 27 at 22:55
  • $\begingroup$ @CosmasZachos the energies are degenerate. The S_z 's are degenerate. The S^2 's are degenerate. I want to find a quantity that distinguishes these states. $\endgroup$ Jan 28 at 0:00
  • $\begingroup$ The projectors projecting out or in the permutation antisymmetries stated. The three states have dramatically different symmetries. Do you want details? $\endgroup$ Jan 28 at 0:37

2 Answers 2


Assuming you are looking at all four $|n,1\rangle$ states, and defining $$ |2,1\rangle_0 = \frac{1}{2}\bigg[ |\uparrow \downarrow \uparrow \uparrow \rangle + |\downarrow \uparrow \uparrow \uparrow \rangle + |\uparrow \uparrow \uparrow \downarrow \rangle + |\uparrow \uparrow \downarrow \uparrow \rangle \bigg] $$ as the only state that is not annihilated by a raising operator, so it is part of the quintet (spin 2), your four orthogonal states are differentiated by the projection operators where () means symmetrizing, and [] means antisymmetrizing, and the numbers denote original spin locations.

So, $$ P_{(1234)} |2,1\rangle_0 = |2,1\rangle_0 , ~~~0 ~~~\hbox {for the rest}, \\ P_{[1,3][2,4](12)(34)}|1,1\rangle_1= |1,1\rangle_1, ~~~0 ~~~\hbox {for the rest}, \\ P_{(12)[34]} |1,1\rangle_2= |1,1\rangle_2 , ~~~0 ~~~\hbox {for the rest}, \\ P_{[12](34)}|1,1\rangle_3= |1,1\rangle_3 , ~~~0 ~~~\hbox {for the rest}. $$ I 'm sure my notation is unconventional and redundant, but the proper notation is in discussions of Young tableaux for the symmetric croup. I'm just stating in Pidgin the symmetries and anti symmetries that uniquely distinguish your four states. Check they are automatically orthogonal to each other. (The 0th one, of course, is distinguished by its $S^2$ eigenvalue, so, ultimately, it is not part of your problem.)

You may construct the 4x4 orthogonal matrix that encodes that projection, acting on the relevant vectors $( |\downarrow \uparrow \uparrow \uparrow \rangle , |\uparrow \downarrow \uparrow \uparrow \rangle , |\uparrow \uparrow \downarrow \uparrow \rangle ,|\uparrow \uparrow \uparrow \downarrow \rangle )^T$, $$\begin {pmatrix} 1/2 & 1/2 &1/2 &1/2 \\ 1/2 &1/2 &-1/2 &-1/2 \\ 0&0&-1/\sqrt{2} & 1/\sqrt{2}\\ -1/\sqrt{2}& 1/\sqrt{2} &0&0 \end{pmatrix},$$ and manifestly displays the permutation symmetries in its rows.

  • $\begingroup$ I don't understand this. Sorry, I think this problem, though important to me, is proving beyond my Griffith's/Zettili's text skill. How is this a spin 2? Why are you using different notation that I originally provided? It's too confusing for me, and I appreciate the help thus far. $\endgroup$ Jan 28 at 1:39
  • $\begingroup$ Applying a raising operator to the zero state gets you the 4spin up state, visibly in the quintet, and o for the other three states. $\endgroup$ Jan 28 at 3:40

This is too long for a comment so…

What you want to do is check if the states transform as a linear combinations of themselves under the permutation group $S_4$ of $4$ objects. This group has $4!$ elements so it might be heavy going but since any element in $S_4$ is a product of transpositions (elements which permute only two particles and leave two alone), you are good enough to check for the transpositions $P_{12}, P_{13}, P_{14},P_{23},P_{24}, P_{34}$.

The hypothesis here is that your 3 basis states span an irreducible representation of $S_4$. Indeed, there is one irrep (denoted by the partition $\{3,1\}$ of $S_4$) that is of dimension 3, so the odds are in your favour.

If indeed it is the case that your states do for a basis for the irrep $\{3,1\}$, you might then try to organize them so they span an irrep of the subgroup $S_3$, which would permute the first three entries only, and then from there continue with the subgroup $S_2$ which permutes the first two entries. This would allow you to label your states by an irrep of $S_4$ (the irrep is probably $\{3,1\}$), simultaneously with an irrep of $S_3$ and simultaneously with an irrep of $S_2$.

If indeed your current set spans $\{3,1\}$ of $S_4$, you can obtain the $S_3$ irreps in $\{3,1\}$ using the standard branching rules, or by using the Young diagram method. The possible irreps of $S_3$ inside this irrep of $S_4$ are $\{3\}$ and $\{2,1\}$. $\{3\}$ is the fully symmetric irrep of $S_3$ and of dimension $1$ while $\{2,1\}$ is of mixed symmetry and of dimension $2$.

The procedure is tedious. Knowing that the irrep $\{3\}$ of $S_3$ will appear if what you have is $\{3,1\}$ of $S_4$, you might start by constructing a combo of the states that you have that will be fully symmetric under $S_3$, but it’s still tedious.

  • $\begingroup$ I am sturggling to follow. I apologize. As I stated in a prior comment, I'm at a Griffith's/Zettili text level, and this jargon is pushing those limits. I appreciate the help. Maybe you can explain how I would apply the permutation operator to the first state I wrote? The way I'm looking at this, and it could be way wrong, is that if these three are in an equilateral triangle and I rotate 120-degrees, then I permute them. Applying R once, gives me an asymmetric linear combination of the last two states I wrote. R again gives me a symmetric linear combination (with a -1 prefactor). $\endgroup$ Jan 28 at 1:46
  • $\begingroup$ There is not a single permutation operator but $4!$ of them. To apply $P_{12}$, which permutes particles $1$ and $2$, on your first state simply interchange the first and second entries of each ket in your linear combination. To apply $P_{24}$ on a state interchange the second and fourth entries of each ket in the combo for that state. Sorry to say that Griffiths is terrible when it comes to permutation symmetry. $\endgroup$ Jan 28 at 1:51
  • $\begingroup$ (It seems this is the problem of coupling $4$ spin-1/2 particles and decomposing the resulting states in simultaneous irreps of the permutation group and of $su(2)$. It’s not easy to do this without using the mathematical machinery of the permutation group… ) $\endgroup$ Jan 28 at 1:54
  • $\begingroup$ Yes, I guess I don't know what I want to do to get a "good quantum number" that isn't degenerate. I'm imagining (to learn this better) that I have three spins in a triangular array.(equilateral). So I rotate 1-2, 2-3, and 3-1 and do it by 120-degrees. What significance is this? $\endgroup$ Jan 28 at 2:05
  • $\begingroup$ You're approaching this in a way that might work for specific cases, but not in general. To your question "How do you find the missing symmetry", the technical answer is "By using the permutation group". See also physics.stackexchange.com/a/685428/36194 and physics.stackexchange.com/q/29443/36194 for additional examples of the complexity of the problem. $\endgroup$ Jan 28 at 2:26

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