My question is very similar to the question here, however I want to ask about some specifics. I am having trouble keeping track of exactly how we are "tensor-producting" or "direct-producting" the different vector spaces.

Suppose we have a single particle $A$. "Good quantum numbers" for this particle are the particle's total spin and it's z-spin. It turns out that a single particle has a fixed total spin, so the vector space of this quantum number is not very interesting and is just one dimensional, but we will call it $V^2_A$. The vector space of the particle's $z$-spin is a 2-dimensional vector space $V^Z_A$.

Let $S^2_A$ denote the spin operator in this vector space, and let $S^Z_A$ denote the $z$-spin operator.

Suppose we have another particle $B$. Let $S^2_B$ denote the spin operator in this vector space, and let $S^Z_B$ denote the $z$-spin operator. The operators $S^2_{A},$ $S^2_{B}$ and $S^Z_{A},$ $S^Z_{B}$ are isomorphic but they act on different vector spaces.

The state of particle $A$ lives in the vector space $V^2_A \otimes V^Z_A.$ The state of particle $B$ lives in the vector space $V^2_B \otimes V^Z_B.$ The "multiparticle" state lives in the vector space $(V^2_A \otimes V^Z_A) \otimes (V^2_B \otimes V^Z_B).$

The detail which seems to get brushed under the rug is that now we define new operators, the "total" spin, $S^2_{A+B},$ and the "total" spin $S^Z_{A+B}$ but I am not exactly sure how these are defined in terms of the proceeding operators.

Naively, I would expect them to be something like: $$S^2_{A+B} \equiv (S^2_A \otimes I_B) + (I_A \otimes S^2_B)$$ but this is not correct.

The notation $\frac{1}{2} \otimes \frac{1}{2} = 1 \oplus 0$ then means something like the following:

"The tensor product $(V^2_A \otimes V^Z_A) \otimes (V^2_B \otimes V^Z_B)$ is formed from two vector spaces which each have eigenvalue $1/2$ under the total spin operator. The result is a new vector space which has eigenvalues $1$ or $0$ under the newly defined total spin operator."

Is this correct? If so, how are the new spin operators defined?

  • $\begingroup$ Have you ever Kronecker -multiplied two spin doublet representations together? (= added two spin 1/2 s together?) $\endgroup$ Commented Apr 29, 2021 at 22:03
  • $\begingroup$ @CosmasZachos The "outer product' ? $\endgroup$
    – Jbag1212
    Commented Apr 29, 2021 at 22:16
  • 3
    $\begingroup$ Well, they are all tensor products, but "outer product" is hostile, confusing, and counterproductive. What we do in physics is Kronecker multiplication: tensor multiplying two matrices to get a larger matrix, which, e.g. here. is in block form: a 3x3 block and a 1x1 block. The question you link details and illustrates it all. Using the identities for the Casimir as you do can only lead to grief and confusion. $\endgroup$ Commented Apr 29, 2021 at 22:46
  • 1
    $\begingroup$ @CosmasZachos Yes, I am familiar with the Kronecker product. Are my $S^2_{A+B}$ and $S^Z_{A+B}$ just $S^2_A \otimes S^2_B$ and $S^Z_A \otimes S^Z_B$?. This doesn't make sense, because $S^2_A \otimes S^2_B$ is one-dimensional. I don't know what you mean by "Casimir" $\endgroup$
    – Jbag1212
    Commented Apr 29, 2021 at 23:09
  • 2
    $\begingroup$ Related : Total spin of two spin- 1/2 particles. $\endgroup$
    – Frobenius
    Commented Apr 30, 2021 at 4:05

3 Answers 3


You've taken such a dramatically wrong left turn in framing the problem, that it is hard to join you and pull you back. Instead, for the specific problem you are examining, the addition of two spin 1/2s to a spin 1 and a spin 0, I'll define all terms in the conventional manner for general representations that you see illustrated in your linked question.

Particle A "lives" on a 2d vector space $V_A$, period. Both $S_A^z$ and $S_A^2$ act on just it. The operators acting on it are 2×2 matrices, $S^z_A=\sigma_3/2$, $S^x_A$ and $S^y_A$, obeying the all-important su(2) Lie algebra $[S_A^i,S_A^j]= i\epsilon^{ijk}S_A^k$.

The quadratic Casimir invariant of this Lie algebra is the operator $$S_A^2=(S^{x}_A)^2+ (S^{y}_A)^2+(S^z_A)^2= s_A(s_A+1)1\!\!1 ~.$$ In spin physics one calls the $s_A$ part of the eigenvalue "total spin", and the (two) eigenvalues of $S^z_A~~$ z-spin. These are the "good quantum numbers" of particle A represented on the 2-vector (spinor) $V_A$. ($s_A$ is the eigenvalue of the rarely discussed operator $\sqrt{S^2_A+1/4} -1/2$ acting on the entire 2d spinor.)

Repeat for particle B, which just happens to be in the doublet representation as well, in your example, so it is described by the above paragraph identically with B supplanting A.

Now, the fundamental feature of Lie algebra reps is that the coproduct operators, $$ \Delta(S^i) \equiv S^i_A \otimes 1\!\!1_B + 1\!\!1_A \otimes S^i_B , $$ the Kronecker product of these two representations, 4×4 matrices acting on the 4-vectors $V_A\otimes V_B$, obeys the very same su(2) Lie algebra that A and B matrices do. There are at least half a dozen questions on this site explaining this, e.g. here. We call this "addition of spin A to spin B" in elementary physics, as a reflection of the + sign above.

For example, here, explicitly work out your composite representation $$ \Delta(S^z)=\operatorname{diag}(1,0,0,-1),\\ \Delta^2=\Delta(S^x)^2+\Delta(S^y)^2+\Delta(S^z)^2 =\begin{pmatrix}2&0&0&0\\ 0&1&1&0\\ 0&1&1&0\\0&0&0&2 \end{pmatrix}~. $$

Since this (composite) coproduct representation Δ obeys that very same Lie algebra, the above quadratic Casimir $\Delta^2$ must necessarily commute with all three generators above, although it is not proportional to the 4×4 identity matrix. Your own expression for it is only one piece of this operator, but incorrect/incomplete, as you surmised: You skipped the cross terms $2S_A^i\otimes S_B^i$, $$ \bbox[yellow]{\Delta(S^i)\Delta(S^i) = (S^i_A)^2 \otimes 1\!\!1_B + 2S_A^i\otimes S_B^i+1\!\!1_A \otimes (S^i_B)^2 } ~. $$

As evident in the Casimir, this 4×4 representation is reducible. That is, you see there is a 3d invariant subspace, and a 1d invariant subspace. When you redefine your basis (mixing the 2nd and 3rd entries to diagonalize the Casimir) to neatly split them apart (Clebsching into disjoint block matrices$^\natural$), the now diagonal Casimir is proportional to the identity in each subspace, with eigenvalue 2=1(1+1) and 0, respectively. That is, the respective blocks have total spin 1 (triplet) and spin 0. The singlet representation, spin 0, is, of course, null matrices acting on the 1d space and representing the group most trivially by 0, 0, 0.

This details what your title summarizes. I explicitly exemplify the triplet and the singlet in my answer to your linked question. You might study the attachment at the bottom of my answer to this question for a further example.

$^\natural$From your Clebsch-Gordan table, you readily see the orthogonal matrix
$\small O=\begin{pmatrix}1&0&0&0\\ 0&1&0 &0 \\ 0&0&0&1\\0& 0&1&0 \end{pmatrix}~ \begin{pmatrix}1&0&0&0\\ 0&1/\sqrt 2&1/\sqrt 2&0\\ 0&1/\sqrt 2&-1/\sqrt 2&0 \\0&0&0&1 \end{pmatrix}~ $ similarity-transforms $O \Delta^2 O^T=\operatorname{diag}(2,2,2,0)= 2~ 1\!\!1 _3\oplus 0~ 1\!\!1_1$.

  • $\begingroup$ One thing just to clarify: you say the quadratic Casimir must necessarily commute with all three generators. I understand that this follows from the fact that it obeys the same Lie algebra. But is it really appropriate to call this a quadratic Casimir? By Schur's Lemma, mustn't the quadratic Casimir necessarily be proportional to the identity? $\endgroup$
    – Jbag1212
    Commented Apr 30, 2021 at 16:19
  • 2
    $\begingroup$ Indeed, for irreducible representations, much unlike $\Delta^2$, the quadratic Casimir invariant is always proportional to the identity, which you should be able to identify in each irreducible block after Clebsching. The Casimir invariant is in the universal Lie algebra (so, not the Lie algebra, and not the group). $\endgroup$ Commented Apr 30, 2021 at 16:50
  • $\begingroup$ I just came across this topic. Very helpful but one question. The CG tables show your 3rd and 4th matrix rows in reverse order. Using your matrix does indeed yield the block diagonal form but using the CG tables as is does not. What is the rationale for switching the last 2 rows? $\endgroup$ Commented Apr 10, 2023 at 13:18
  • $\begingroup$ Thanks for responding. So it is just a matter of trial and error to find the correct order of rows/columns and find an orthogonal matrix that produces the block diagonal form? $\endgroup$ Commented Apr 11, 2023 at 16:51
  • $\begingroup$ I factored the similarity transformation to a Clebsch matrix and a 3-4 permutation matrix in the footnote. Now you can first Clebsch and then permute 3-4 labels... $\endgroup$ Commented Apr 12, 2023 at 20:16

For a 2-particle system, it is natural to define \begin{align} S_x=S_x^{(1)} + S_x^{(2)} := S_x \otimes \mathbb{1}_{2\times 2} +\mathbb{1}_{2\times 2}\otimes S_x \end{align} and so forth for $S_y$ and $S_z$. These would act on states $\vert +\rangle\vert +\rangle,\vert +\rangle\vert -\rangle, \vert - \rangle\vert +\rangle, \vert -\rangle\vert -\rangle$. The matrix representation of these operators are \begin{align} S_x&=\frac{1}{2}\left( \begin{array}{cccc} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ \end{array} \right)\, ,\\ S_y&= \frac{1}{2}\left( \begin{array}{cccc} 0 & -i & -i & 0 \\ i & 0 & 0 & -i \\ i & 0 & 0 & -i \\ 0 & i & i & 0 \\ \end{array} \right)\, ,\\ S_z&= \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ \end{array} \right)\, . \end{align} There is a (common) unitary transformation $T$, given by \begin{align} T=\frac{1}{\sqrt{2}}\left( \begin{array}{cccc} \sqrt{2} & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & \sqrt{2} \\ 0 & 1 & -1 & 0 \\ \end{array} \right) \end{align} that will bring these matrices to block diagonal form: there will be one 3-dimensional block and one 1-dimensional block. For instance: \begin{align} T\cdot S_x\cdot T^{-1}= \frac{1}{\sqrt{2}} \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\, . \end{align} The $3\times 3$ block is identical to the matrix representation of $S_x$ for spin $S=1$ states, and the $1\times 1$ block is identical to the matrix representation of $S_x$ for an $S=0$ state. Thus the total space splits into a direct sum of $S=1$ and $S=0$ subspaces.

Hence, $\frac{1}{2}\otimes \frac{1}{2}=1\oplus 0$.


The reason why $S^2$ does not satisfy

\begin{align} S^2_{A+B} \equiv (S^2_A \otimes I_B) + (I_A \otimes S^2_B) \end{align}

is because Total Angular Momentum Squared (TAMS) does not have the addition property: the TAMS of a system is not equal to the TAMS of its parts.

As an example of this, take two classical particles orbiting the same point with opposite angular momenta. The TAMS of the system is zero, but the TAMS of each particle is a positive finite quantity, so their sum cannot be zero.

The spin projection $S^z$, on the other hand, does have this property. For this operator your guess would be correct:

\begin{align} S^z_{A+B} \equiv (S^z_A \otimes I_B) + (I_A \otimes S^z_B). \end{align}

This expression is also true for the other axes ($x$ and $y$) as well. This allows one to calculate the correct expression for $S^2_{A+B}$ through its definition:

\begin{align} S^2_{A+B} = (S^x_{A+B})^2 + (S^y_{A+B})^2 + (S^z_{A+B})^2. \end{align}


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