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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.

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    $\begingroup$ 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? $\endgroup$
    – Rammus
    Apr 16 at 9:44
  • $\begingroup$ Sorry for the ambiguity, I've amended the question. $\endgroup$
    – narip
    Apr 16 at 10:25
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    $\begingroup$ So to confirm, for two qubits $J_x = \frac12 \sigma_x \otimes I + \frac12 I \otimes \sigma_x$? $\endgroup$
    – Rammus
    Apr 16 at 10:31
  • $\begingroup$ Yes, for two qubits, it is so. $\endgroup$
    – narip
    Apr 16 at 12:55
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We can transform the left hand side as follows

$$ \begin{align} \sum_{l=x,y,z}\langle J_l^2\rangle &= \sum_{l=x,y,z}\langle\psi|J_l^2|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}J_l^2\right)|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}\left(\sum_{i=1}^N\frac12\sigma_l^i\right)^2\right)|\psi\rangle \\ &= \langle\psi|\left(\sum_{l=x,y,z}\sum_{i,j=1}^N\frac14\sigma_l^i\sigma_l^j\right)|\psi\rangle \\ &= \frac14\langle\psi|\left(\sum_{l=x,y,z}\left(N + 2\sum_{i<j}\sigma_l^i\sigma_l^j\right)\right)|\psi\rangle \\ &= \frac14\langle\psi|\left(3N + 2\sum_{i<j}\sum_{l=x,y,z}\sigma_l^i\sigma_l^j\right)|\psi\rangle \\ &= \frac{3N}4+\frac12\langle\psi|\left(\sum_{i<j}\sigma_x^i\sigma_x^j+\sigma_y^i\sigma_y^j+\sigma_z^i\sigma_z^j\right)|\psi\rangle \\ &= \frac{3N}4+\frac12\sum_{i<j}\langle\psi|\sigma_x^i\sigma_x^j+\sigma_y^i\sigma_y^j+\sigma_z^i\sigma_z^j|\psi\rangle \end{align} $$

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,

$$ \begin{align} \sum_{l=x,y,z}\langle J_l^2\rangle &= \frac{3N}4+\frac12\sum_{i<j}\langle\psi|\sigma_x^i\sigma_x^j+\sigma_y^i\sigma_y^j+\sigma_z^i\sigma_z^j|\psi\rangle \\ &\le \frac{3N}4+\frac12\sum_{i<j}1 \\ &= \frac{3N}4+\frac12\frac{N(N-1)}2 \\ &= \frac{N(N+2)}4 \end{align} $$

which completes the proof.

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