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Appunti di Fisica Teorica - INFN

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Deriviamo lo stesso risultato in una maniera <strong>di</strong>versa, che passa per il<br />

calcolo esplicito dei caratteri χN(z) delle rappresentazioni a livello N. La<br />

(3.3) dà<br />

∞ ∞ ∞<br />

Zbosoni(q, z) = (zq) i (q/z) j q k ∞ ∞ ∞<br />

=<br />

q i+j+k z i−j<br />

(3.17)<br />

Pertanto<br />

i=0<br />

χ (bos)<br />

(z) =<br />

N<br />

j=0 k=0<br />

N N−i<br />

z i−j =<br />

i=0<br />

j=0<br />

z −N<br />

(1 − z) 2 (1 + z)<br />

N<br />

i=0<br />

i=0<br />

j=0 k=0<br />

i 1 − z−N−1+i<br />

z<br />

1 − z−1 = z<br />

N+1 1 − z 1 − z2(N+1)<br />

− z−N−1<br />

z − 1 1 − z 1 − z2 <br />

=<br />

z<br />

=<br />

−N<br />

(1 − z) 2 <br />

(1 + z)(z<br />

(1 + z)<br />

2N+2 − z N+1 ) + 1 − z 2N+2<br />

<br />

=<br />

<br />

=<br />

z 2N+3 − z N+1 − z N+2 <br />

+ 1 =<br />

= z−N (1 − z N+1 )(1 − z N+2 )<br />

(1 − z) 2 (1 + z)<br />

I numeri nN;j sono ottenuti pertanto calcolando gli integrali<br />

nN;j = 1<br />

<br />

dz<br />

2<br />

= 1<br />

<br />

dz (z − 1) (1 − z<br />

2 2πiz<br />

2j+1 )<br />

z1+j χN(z)<br />

= 1<br />

<br />

dz 1<br />

2 2πiz z1+j+N (z2j+1 − 1) (zN+1 − 1) (zN+2 − 1)<br />

(1 − z2 )<br />

= 1<br />

<br />

dz 1<br />

2 2πiz z1+j+N (1 − z2 <br />

)<br />

+z N+2 + z 2j+1 − z 2N+3 − z 2j+N+3 <br />

− 1 =<br />

= 1<br />

<br />

dz 1<br />

2 2πiz (1 − z2 <br />

z<br />

)<br />

j+N+3 − z j+1 + z −j +<br />

+z 1−j + z j−N − z N+2−j − z j+2 − z −N−j−1<br />

<br />

=<br />

= 1<br />

∞ dz<br />

<br />

z<br />

2 2πiz<br />

2k+j+N+3 − z 2k+j+1 + z 2k−j +<br />

2πiz (z1/2 − z −1/2 ) (z −(j+1/2) − z j+1/2 ) χN(z) =<br />

k=0<br />

13<br />

=<br />

z 2j+2N+4 − z 2j+N+2 + z N+1 +<br />

(3.18)

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