Let $\sigma_i (\frac{3}{2})$ be the three generators of the irreducible spin 3/2 representation of $SU(2)$ (see http://easyspin.org/documentation/spinoperators.html for their explicit forms). Similarly, $\sigma_i (\frac{1}{2})$ are the usual Pauli matrices.

I have reasons to believe that it should be possible to perform a basis transformation such that the spin 3/2 matrices can be related to the spin 1/2 matrices as follows:

$U\sigma_i (\frac{3}{2})U^{-1}=\sigma_i(\frac{1}{2})\otimes B_i$

For some unitary $U$ and some matrices $B_i$. [EDIT: the single unitary $U$ must make this relation hold for all $i$] Evidently, $B_i$ must be 2 x 2 hermitian matrices with eigenvalues 1 and 3, modulo signs.

Is there any reason such a transformation should not be possible? If it is possible, is there an easy way to find a $U$ which satisfies this?

  • $\begingroup$ Isn't it obvious that $\sigma_z(3/2) = \sigma_z(1/2)\otimes\mathrm{diag}(1,3)$, and since every $\sigma_i$ can be brought into $\sigma_z$'s form by a unitary transformations, that's it? $\endgroup$
    – ACuriousMind
    Commented Feb 25, 2016 at 23:32
  • $\begingroup$ Yes, it is clear for $\sigma_z$. But the unitary $U$ must be independent of $i$ so that it brings all three of the 3/2 matrices into the desired forms simultaneously. So $U=I$ would work for $\sigma_z$, but not the other two. Unless I am misunderstanding your comment, I don't see that this solves the problem. $\endgroup$ Commented Feb 25, 2016 at 23:40
  • $\begingroup$ Why would you expect a single unitary operator to do it for all three? And why do you not just write out the Kronecker product and look if the resulting system of linear equations has solutions or not (it's not a shame to use a CAS to solve systems of equations)? $\endgroup$
    – ACuriousMind
    Commented Feb 25, 2016 at 23:44
  • $\begingroup$ The reason I expect this to be possible is quite complicated.. it has to do vaguely with the fact that, when a spin 3/2 irrep is restricted to a subgroup $\mathbb{Z}_2 \times \mathbb{Z}_2$ inside SO(3), it must be isomorphic to two copies of the pauli representation. You are right about using CAS; I was just hoping for a quicker, more insightful route, especially since my skill with CAS is somewhat lacking. $\endgroup$ Commented Feb 25, 2016 at 23:50

1 Answer 1


I can't claim this is a complete argument, but you might consider it... This is definitely not the standard coproduct, which, coincidentally, I have given to my students as a homework problem in the past.

Consider what Lie algebra both sides of your equation would satisfy: take the commutator of the left hand side, for a given i and j, so, e.g. 1 and 2. So, up to normalizations, the l.h.s. will be $iU \sigma_3(3/2)U^{-1}$.

Edit as per discussion. The r.h.s., I think, will be likewise $i\sigma_3(1/2)\otimes \{B_1,B_2\}$. Likewise for the other two components.

So the problem reduces to checking if $ \{B_i,B_j\}\propto \eta_{ijk}B_k$ can be solved for Bs with eigenvalues 1 and 3 or their opposites, in any combination. $\eta_{ijk}$ is symmetric in i and j, which must be different, otherwise it vanishes, and k must be the remaining index of the 3. All three ηs have the same value, so take them to be 1.

  • $\begingroup$ Thanks for your comment. Taking the commutators as you suggest, I find that $[B_i,B_j]=2 \epsilon_{ijk} B_k$ as you say. But I don't quite understand your argument re. what this implies for the eigenvalues. The eigenvalues need only be one of $\pm 3$ and one of $\pm 1$. Does the algebra of the B's imply this is not possible? Perhaps so; if I recall correctly the algebra alone enforces that the eigenvalues are $\pm \lambda$. $\endgroup$ Commented Feb 26, 2016 at 23:35
  • $\begingroup$ Apologies: I blew it! The 2 minus signs enforce that the 3 Bs close under anticommutation, not commutation... So they are not a doublet rep of SU(2), which, indeed, would dictate the eigenvalues are the opposite of each other, as you recall... So it has to be checked if there are or are not solutions to $\{ B_i,B_j\}\propto \epsilon_{ink} B_k$. ... $\endgroup$ Commented Feb 26, 2016 at 23:54
  • $\begingroup$ Oh, I made the same mistake! Regardless, this anticommutation relation will be helpful to solve/ disprove this. $\endgroup$ Commented Feb 26, 2016 at 23:59

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