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 Incoming links: On-site spin-square normal-ordering identity

Let

D_{n} = \overset{n}{^}_{n ↑} \overset{n}{^}_{n ↓}

. On a site, empty and doubly occupied states have spin zero, while either singly occupied state has spin

1/2

. Thus

\hat{S}_{n}^{2} = \frac{3}{4} (\overset{n}{^}_{n} - 2 D_{n}), \hat{S}_{n}^{2} - \frac{3}{4} \overset{n}{^}_{n} = - \frac{3}{2} D_{n} =: \hat{S}_{n}^{2} : .

3 U \sum_{n} D_{n}

3 U

in this parameterization, rather than the

U

convention often used for the Hubbard model.

ξ_{k} = ϵ_{k} - μ

Z = \int D (\overset{ˉ}{ψ}, ψ) e^{- S}, S = \int_{0}^{β} d τ [k σ \sum \overset{ˉ}{ψ}_{k σ} (\partial_{τ} + ξ_{k}) ψ_{k σ} + 3 U n \sum \overset{ˉ}{ψ}_{n ↑} \overset{ˉ}{ψ}_{n ↓} ψ_{n ↓} ψ_{n ↑}] .

The independent fields have antiperiodic thermal boundary conditions. Normal ordering is crucial: the explicit

3 \overset{n}{^} /4

subtraction removes the spin-square contraction, so there is no extra chemical-potential shift from it.

Use

ℏ = k_{B} = 1

S_{n}^{2} = \frac{3}{4} (n_{n} - 2 n_{n ↑} n_{n ↓}), : S_{n}^{2} := S_{n}^{2} - \frac{3}{4} n_{n} = - \frac{3}{2} n_{n ↑} n_{n ↓} .

Indeed the empty and doubly occupied states have spin zero, while either singly occupied state has

S^{2} = 3/4

. Thus the interaction in normal-ordered form is $3 U \sum_{n} c_{n ↑}^{†} c_{n ↓}^{†} c_{n ↓} c_{n ↑}$ . There is no residual

3 U n /2

chemical-potential shift: that contraction has cancelled.

ξ_{k} = ϵ_{k} - μ

and

D_{n} = \overset{ˉ}{ψ}_{n ↑} \overset{ˉ}{ψ}_{n ↓} ψ_{n ↓} ψ_{n ↑}

Z = \int_{AP} D (\overset{ˉ}{ψ}, ψ) e^{- S_{0} - 3 U \int_{0}^{β} d τ \sum_{n} D_{n}}, S_{0} = \int_{0}^{β} d τ kσ \sum \overset{ˉ}{ψ}_{kσ} (\partial_{τ} + ξ_{k}) ψ_{kσ} .

0, β

. Adjacent time-slice labels in the normal-ordered symbol fix the continuum prescription and thermal normalization. The interaction coefficient is

3 U

, not the

U

of a differently normalized Hubbard model.