# What is the physical meaning of the zeroness of the antidiagonal of the matrix representation of $\textbf{n}\cdot\textbf{S}$ operator?

I encountered a problem where I had to use $$\textbf{n}\cdot{\textbf{S}}$$. It was found to be:

What does it mean physically, that the antidiagonal of this matrix is 0, for any $$\textbf{n}$$?

All spin matrices for integer spin j have this property in the spherical basis (not in the Cartesian one, of course).

It is easy to see that. They are $$(2j+1)\times (2j+1)$$-dimensional matrices.

$$S_z$$ is diagonal, but its central eigenvalue is 0 (for integer spins).

$$S_+$$ is a raising operator by one, so only the first parallel above the diagonal is nonzero; however, it has length 2 j, so it does not "shadow" the central diagonal 0, and so the second, third, ... parallels vanish, yielding 0s. The analog argument holds for $$S_-$$, and hence the linear combinations of raising and lowering operators.

You are done.

As indicated, this is a fluke property of the spherical basis. In the Cartesian basis, you'd have nonzero antidiagonals for $$S_y$$. I don't see much physics in a basis-dependent statement, then.

• I don't see much physics in a basis-dependent statement, then. Well, the spherical basis isn't a basis whatever. It's a basis of eigenvectors of a component of $\vec S$. – Elio Fabri Mar 27 at 16:04
• @Elio Fabri. Correct. A basis then defines the orientation of operators (here, matrices) that act on it, and hence defines a basis for these matrices. Changing the basis of the vectors similarity-transforms the matrices to the corresponding new basis. So, then, contrast $S_z$ in the spherical basis to $L_z$ in the Cartesian basis, as in the first week of angular momentum treatments. – Cosmas Zachos Mar 27 at 16:33