Solución

Sea

$$A:= \int \vert\phi\rangle\langle\phi\vert \prod_\alpha e^{-\phi_\alpha^*\phi_\alpha} \frac{d\phi_\alpha^* d\phi_\alpha}{2\pi i} \,.$$ (4.34)

Reexpresamos los estados coherentes en la base de números de ocupación:

$$\vert\phi\rangle= \sum_{\{n_\alpha\}}\prod_\alpha \frac{\phi_\alpha^{n_\alpha}}{\sqrt{n_\alpha !}} \vert(n_1,\ldots,n_\alpha,\ldots)\rangle \,,$$ (4.35)

es decir

$$A = \sum_{\{n_\alpha\}} \sum_{\{n^\prime_\alpha\}} \vert(n_1,\ldots)\rangle \langle(n^\prime_1,\ldots)\vert \prod_\alpha I_\alpha$$ (4.36)

donde

$$I_\alpha:= \int \frac{\phi_\alpha^{n_\alpha}}{\sqrt{n_\alpha !}} ... ...}} e^{-\phi_\alpha^*\phi_\alpha} \frac{d\phi_\alpha^* d\phi_\alpha}{2\pi i} \,.$$ (4.37)

Esta integral se puede hacer en coordenadas polares

$$\phi_\alpha=r e^{i\varphi},\quad \phi^*_\alpha=r e^{-i\varphi},\quad \frac{d\phi_\alpha^* d\phi_\alpha}{2i} =dx\,dy=rdr\,d\varphi \,,$$ (4.38)

$$I_\alpha = \int \frac{rdrd\varphi}{\pi}e^{-r^2} \frac{ r^{n_\alph... ...{i(n_\alpha-n^\prime_\alpha)\varphi}} {\sqrt{n_\alpha !n_\alpha^\prime ! }} \,.$$ (4.39)

Usando las relaciones

$$\int_0^{2\pi}d\varphi \, e^{in\varphi}= 2\pi\delta_{n,0} \,,\qquad \int_0^\infty dx x^n e^{-x} =n!$$ (4.40)

se obtiene

$$I_\alpha= \frac{1}{n_\alpha!}\frac{1}{\pi} 2\pi\delta_{n_\alpha,n... ...lpha} \delta_{n_\alpha,n^\prime_\alpha} = \delta_{n_\alpha,n^\prime_\alpha} \,,$$ (4.41)

que implica

$$A= \sum_{\{n_\alpha\}} \vert(n_1,\ldots)\rangle \langle(n_1,\ldots)\vert =1 \,.$$ (4.42)