Extracting $\mathbf b$ from $M = a I + \mathbf b \cdot \mathbf S$, when $S_i$ are higher spin matrices?

This is a cross-post of a question that I posted on the Math SE, that did not get any answers there. It is fundamentally a mathematics question, but it pertains to spin matrices, which many Physics SE users have a lot of knowledge about, so I thought I might try here. My end use case is also in physics, rather than pure mathematics.

The original question

Any $$2 \times 2$$ hermitian matrix $$M$$ can be written $$M = a I + \mathbf b \cdot \boldsymbol \sigma,\tag{1}$$ where $$a \in \mathbb R$$, $$\mathbf b \in \mathbb R^3$$, $$\boldsymbol \sigma$$ is the Pauli vector, and $$\mathbf b \cdot \boldsymbol \sigma := \sum b_i \sigma_i$$. This is because the identity matrix and the Pauli matrices make up a basis for the real vector space of $$2 \times 2$$ hermitian matrices. Moreover, given $$M$$, one can extract $$a$$ and $$\mathbf b$$ via $$a = \frac{1}{2} \mathrm{tr}(M), \quad b_i = \frac{1}{2} \mathrm{tr}(\sigma_i M)\tag{2}$$ because the Pauli matrices are traceless and obey the identity $$\sigma_j \sigma_k = \delta_{jk} I + i \epsilon_{jkl} \sigma_l,\tag{3}$$ where we use implicit summation over the repeated index $$l$$.

A generalization of the Pauli matrices that is of particular interest to physicists is the higher spin matrices. Hence, say we replace the Pauli vector in (1) with a "vector" $$\mathbf S$$ of $$d \times d$$ spin matrices, yielding $$M = a I + \mathbf b \cdot \mathbf S,\tag{4}$$ where now, of course, $$M$$ and $$I$$ are $$d \times d$$. (This is clearly not a general $$d \times d$$ hermitian matrix for $$d > 2$$, but more restricted.) Is there then some simple way, similar to (2), to extract $$\mathbf b$$ given $$M$$?

I tried the cases $$d = 3, 4$$ in Mathematica and found that, indeed, $$b_i \propto \; \mathrm{tr}(S_i M).\tag{5}$$ So I suspect this works in general, but my initial attempts at a simple proof have failed. The identity (3) does not hold for the higher spin matrices, because, while they do obey $$[S_j, S_k] = i \epsilon_{jkl} S_l\tag{6}$$ much like the Pauli matrices (up to normalization), they do not obey the same anticommutation relations.

1 Answer

Because the higher spin matrices generate a representation of $$SO(3)$$, we have : $$\operatorname{Tr}(S_iS_j) \propto \delta_{ij} \tag{1}$$

(because $$\delta_{ij}$$ is the only invariant rank 2 tensor)

To find the normalization constant in $$(1)$$, contract $$i$$ and $$j$$ and express the RHS in terms of the total spin.

On top of that, we know that $$\operatorname{Tr}(S_i) = 0$$. Therefore : $$\operatorname{Tr}(S_i(aI + B_jS_j)) = B_j\operatorname{Tr}(S_iS_j)\propto B_i$$

Edit

Equation $$(1)$$ is a classical result from the theory of simple Lie algebras (that the Killing form is positive definite (or negative definite if you choose anti-hermitian generators as mathematicians do)). Therefore you can choose generators so that $$(1)$$ is satisfied.

Now, in a given representation $$R$$, let us write $$\newcommand{\Tr}{\operatorname{Tr}} \Tr_R(S_iS_j) = T(R)\delta_{ij}$$ where $$T$$ is called the intex of the representation. $$\mathbf S^2 = S_iS_i = C(R)$$ is the quadratic Casimir (which I called total spin above).

Contracting $$i$$ and $$j$$ above we get : $$\dim(R)C(R) = T(R)\dim(G)$$

In the case of $$SU(2)$$ or $$SO(3)$$, the spin $$J$$ representation has $$\dim(R) = 2J+1$$, $$C(G) = J(J+1)$$ so we get : $$T(R) = \frac{J(J+1)(2J+1)}{3}$$

• Could you elaborate on how to obtain equation $(1)$, please? Apr 23 at 19:18
• Thank you for the answer @SolubleFish! Just like @JasonFunderberker above I wonder if you could elaborate on how to obtain equation (1). I know about the homomorphism between SU(2) and SO(3), but I don't see how that implies (1). I'm also not exactly shure what you mean by "total spin" in this context; do you mean $\mathbf S^2$?
– ummg
Apr 23 at 20:46
• Great, thanks for the edit. Apr 24 at 7:32