# How to prove identity $\langle \mathbf{S} \rangle^2 + \langle \mathbf{Q} \rangle^2 = 4/3$ for any spin-1 wavefunction?

Is there an easy way to prove that for an arbitrary wavefunction of spin-one

$$\langle \mathbf{S} \rangle^2 + \langle \mathbf{Q} \rangle^2 = 4/3$$

where $$\mathbf{S} = (S_x, S_y, S_z)$$ for spin-1, quadrupolar operator $$\mathbf{Q} = (S_x^2 - S_y^2, \frac{1}{\sqrt{3}}(2S_z^2 - S_x^2 - S_y^2), S_x S_y + S_y S_x, S_y S_z + S_z S_y, S_z S_x + S_x S_z)$$.

I use Mathematica code to check that equality is true for arbitrary wavefunction I give. How to prove it explicitly (or intuitively) without the brute force method?

• It seems to me that these operator squares are incomparable dimension-wise. Have you set $\hbar$ to 1? It is also unclear to me how the square of that 6-vector (2x3 matrix?) Is supposed to look like. Anyway, you could try to rewrite everything in terms of annihilation and creation operators and see how this acts on an arbitrary spin one state. There should be some simplifications since an action of three consecutive ladder operators annihilates any state in the triplet. – Herr_Mitesch May 11 at 9:18
• @Herr_Mitesch Yes choose $\hbar = 1$ – maplemaple May 11 at 10:44
• Work in the spherical basis, so, then, reexpress $S_x, S_y$ in terms of $S_\pm$. Then there are only 3 eigenvectors of $S_z$ to check, with straightforward results. The way you have written your expression up is a nightmare. – Cosmas Zachos May 11 at 18:17