A vector function of an angular momentum vector $\mathbf{S}$ must be given by a multiple of $\mathbf{S}$?

Question

I've been reading Ballentine's Quantum Mechanics, A Modern Development and a statement made in Chapter 3 has been puzzling for me.

In Chapter 3 of his book, Ballentine derives the kinematics and dynamics of quantum mechanics through Gallilean spacetime symmetries. In deriving the dynamics for a particle with spin, he has this to say (in the following $\mathbf{S}$ is the 3-vector of spin operators):

The only function of $\mathbf{S}$ which is a 3-vector is a multiple of $\mathbf{S}$. [We have that $\mathbf{S}\times\mathbf{S} = i\mathbf{S}$ so no new vector operator can be formed by taking higher powers of $\mathbf{S}$.]

Likewise, he makes the same statement about scalar functions of $\mathbf{S}$, where he claims that the only scalar functions of $\mathbf{S}$ are multiples of $\mathbf{S}\cdot \mathbf{S}$.

In what sense are the only vector functions of $\mathbf{S}$ given by multiples of $\mathbf{S}$? More generally, why does Ballentine claim that the only functions of $\mathbf{S}$ which are 3-vectors themselves have to be given by cross-products of $\mathbf{S}$? Isn't it conceivable that we have some more general function of $\mathbf{S}$ which can also be a 3-vector? Likewise, why does he claim that the only scalar function of $\mathbf{S}$ has to be given by the dot product? Even without resorting to arbitrary scalar functions of $\mathbf{S}$, wouldn't any inner product serve as an appropriate scalar function of $\mathbf{S}$?

Answer 1

Ballentine tacitly assumes that the components of the spin vector $\mathbf{S}$ are the generators of the Lie algebra of $\rm SU(2)$ in an irreducible representation. This implies that the scalar product $\mathbf{S} \cdot \mathbf{S}$ is a multiple of the identity operator, $\mathbf{S}\cdot \mathbf{S}= s(s+1) \mathbf{1}$, with the possible weights $s = 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots$ characterizing the irreducible representations. As a consequence, the most general form of a scalar operator being a function of $\mathbf{S}$ is given by $f(\mathbf{S} \cdot \mathbf{S})= f\left(s(s+1)\right) \mathbf{1}$, where $f$ is an arbitrary function. Thus $f(\mathbf{S} \cdot \mathbf{S})$ is just a multiple of the identity. Likewise, the most general form of a $3$-vector operator being a function of $\mathbf{S}$ is given by $f(\mathbf{S} \cdot \mathbf{S}) \, \mathbf{S}= f\!\left(s(s+1)\right) \, \mathbf{S}$, being indeed just a multiple of $\mathbf{S}$ .

Answer 2

Any (analytical) function $f(x)$ can be written as a Taylor expansion,

$$ f(x) = \sum_{k=0}^\infty a_k x^k $$

As you see, in this expansion, only powers of $x$ and some constants appear.

If we want to build a function of $\mathbf{S}$ which gives us a vector, the only way we can do it using only $\mathbf{S}$ is taking its vector product with himself, things like $\mathbf{S}\times\mathbf{S}$ or $(\mathbf{S}\times\mathbf{S})\times\mathbf{S}$, etc. But at the end of the day, after computing all this products, you will get a constant times $\mathbf{S}$, because $\mathbf{S}\times\mathbf{S}=i\mathbf{S}$.

On the other hand, the only scalar that you can build using only $\mathbf{S}$ is its scalar product with himself. Another combination, for example $3S_x+2S_y$ is not a scalar, because when you transform the coordinates, the quantity changes. So, in the Taylor expansion you will only have powers of $\mathbf{S}\cdot\mathbf{S}$.

Well, of course you could have something like $\mathbf{S}\cdot[\mathbf{S}\times(\mathbf{S}\times\mathbf{S})]$ and more complicated things, but is easy to see that this just gives $\mathbf{S}\cdot\mathbf{S}$.

What I don't see right now is why the most general function is a multiple of the scalar product. Maybe some vectorial relations allow us to use the vectorial product formula to reduce the powers of the scalar product.