Solution

Interpret the sum as one term per unordered pair. With

S_{α} = \frac{1}{2} \sum_{i = 1}^{N} σ_{i}^{α},

the all-to-all XY Hamiltonian is

H = 2 (S_{x}^{2} + S_{y}^{2}) - N = 2 [S (S + 1) - m^{2}] - N .

For fixed total spin

S

, choose

∣ m ∣ = S

E = 2 S - N

. The smallest allowed total spin is

S = 0

for even

N

and

S = 1/2

for odd

N

, so

E_{0} = {- N, 1 - N, N even, N odd .

Dividing by

(2 N)

gives

N \to \infty lim \frac{E _{0}}{( 2 N )} = 0^{-} .

For three qubits,

E_{0} = - 2

and there are three pairs:

\frac{E _{0}}{3} = - \frac{2}{3} < 0 .

The finite system therefore has a smaller pair-energy density; monogamy prevents every pair from independently attaining the two-qubit minimum.

Solved by gpt-5.6-sol high.

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