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entropia di entanglement in teorie invarianti conformi bidimensionali

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Appen<strong>di</strong>ce C<br />

OPE tra operatori <strong>di</strong> vertice<br />

Vogliamo provare la formula:<br />

con 〈ϕ(z, ¯z)ϕ(0)〉 = − log |z| 2 .<br />

Abbiamo:<br />

∞ (iγ1) k (iγ2) j<br />

k=0<br />

j=0<br />

=<br />

=<br />

=<br />

k!<br />

: e iγ1ϕ(z,¯z) : : e iγ2ϕ(0) : = |z| 2γ1γ2 : e iγ1ϕ(z,¯z)+iγ2ϕ(0) : , (C.1)<br />

j!<br />

∞ (iγ1) k (iγ2) j<br />

k=0<br />

j=0<br />

m<strong>in</strong>(k,j) <br />

h=0<br />

∞<br />

h=0<br />

k!<br />

j!<br />

: ϕ k (z, ¯z) : : ϕ j (0) : (C.2)<br />

m<strong>in</strong>(k,j) <br />

h=0<br />

<br />

2γ1γ2<br />

h<br />

log |z|<br />

h!<br />

<br />

2γ1γ2<br />

h<br />

log |z|<br />

h!<br />

∞<br />

k=0<br />

j=0<br />

∞<br />

k<br />

h<br />

m≡k−h=0<br />

n≡j−h=0<br />

<br />

j<br />

h! : ϕ<br />

h<br />

k−h (z, ¯z)ϕ j−h (0) : log |z| −2<br />

(iγ1) k−h (iγ2) j−h<br />

(k − h)!(j − h)!<br />

(iγ1) m (iγ2) n<br />

m!n!<br />

(C.3)<br />

: ϕ k−h (z, ¯z)ϕ(0) j−h : (C.4)<br />

: ϕ m (z, ¯z)ϕ(0) n : (C.5)<br />

= |z| 2γ1γ2 : e iγ1ϕ(z,¯z)+iγ2ϕ(0) : . (C.6)<br />

107

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