# How to prove the fundamental equation in the theory of angular momentum $\sum_{l=x,y,z}\langle J_l^2\rangle\le\frac{N(N+2)}{4}$?

How to prove the inequality$$\sum_{l=x,y,z}\langle J_l^2\rangle\le\frac{N(N+2)}{4}$$ where $$J_l = \mathop{\Sigma}_{i=1}^N \frac{1}{2}\sigma_l^{i}$$, and $$\sigma_l^i$$ is pauli matrix acting on the $$i$$th qubit, $$N$$ stands for the number of qubits, and $$\langle\,.\rangle$$ denotes the average over $$N$$ qubits? The $$N$$ qubits may be entangled.

• At the moment your question is ambiguous. Exactly how do you define $J$? You say that it is a "Pauli matrix". But if you have $2$ qubits then I'm assuming you mean $J_x = \frac12 \sigma_x \otimes \sigma_x$. But this can't be correct as it would seem the bound wouldn't grow with $N$ if this is how you define $J$. Can you add some more detail to your question? Apr 16 at 9:44
• Sorry for the ambiguity, I've amended the question. Apr 16 at 10:25
• So to confirm, for two qubits $J_x = \frac12 \sigma_x \otimes I + \frac12 I \otimes \sigma_x$? Apr 16 at 10:31
• Yes, for two qubits, it is so. Apr 16 at 12:55

We can transform the left hand side as follows

\begin{align} \sum_{l=x,y,z}\langle J_l^2\rangle &= \sum_{l=x,y,z}\langle\psi|J_l^2|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}J_l^2\right)|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}\left(\sum_{i=1}^N\frac12\sigma_l^i\right)^2\right)|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}\sum_{i,j=1}^N\frac14\sigma_l^i\sigma_l^j\right)|\psi\rangle \\ &= \frac14\langle\psi|\left(\sum_{l=x,y,z}\left(N + 2\sum_{i

where $$|\psi\rangle$$ is a possibly entangled state of $$N$$ qubits. We can bound the sum in the last equation from above by exploiting the fact that for any Hermitian operator $$A$$ the real number $$\langle\psi|A|\psi\rangle$$ is less than or equal to the largest eigenvalue of $$A$$, see Rayleigh quotient. Also, the eigenvalues of the operator $$\sigma_x^i\sigma_x^j+\sigma_y^i\sigma_y^j+\sigma_z^i\sigma_z^j$$ are independent of $$i$$ and $$j$$. In the computational basis, the matrix of the operator restricted to qubits $$i$$ and $$j$$ takes the form

$$\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 2 & 0 \\ 0 & 2 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}.$$

The characteristic polynomial of the center $$2\times 2$$ block is $$(x+1)^2-4=x^2+2x-3$$ so $$\sigma_x^i\sigma_x^j+\sigma_y^i\sigma_y^j+\sigma_z^i\sigma_z^j$$ has eigenvalues $$+1$$ and $$-3$$. Therefore,

\begin{align} \sum_{l=x,y,z}\langle J_l^2\rangle &= \frac{3N}4+\frac12\sum_{i

which completes the proof.