Solución

(a) $B=A^\dagger$ si y sólo si para dos estados cualesquiera $\vert\psi\rangle$ y $\vert\phi\rangle$

$$\langle\psi\vert B\vert\phi\rangle=\langle\phi\vert A\vert\psi\rangle^*$$ (4.1)

Basta verficarlo para elementos de una base:

$$\langle(\cdots,n_\alpha,\cdots)\vert a_\alpha\vert (\cdots,n^\prime_\alpha,\cdots)\rangle^* = \zeta^{\nu_\alpha^\prime} \sqrt{n_\alpha^\prime} \langle(\cdots,n_\alpha,\cdots)\vert(\cdots,n^\prime_\alpha-1,\cdots)\rangle^*$$

$$= \zeta^{\nu_\alpha^\prime} \sqrt{n_\alpha^\prime} \left(\prod_{\ga... ...delta_{n_\gamma n_\gamma^\prime}\right) \delta_{n_\alpha n^\prime_\alpha-1} \,.$$ (4.2)

Por otro lado

$$\langle(\cdots,n^\prime_\alpha,\cdots)\vert a_\alpha^\dagger\vert (\cdots,n_\alpha,\cdots)\rangle = \zeta^{\nu_\alpha} \sqrt{n_\alpha+1} \langle(\cdots,n_\alpha^\prime,\cdots)\vert(\cdots,n_\alpha+1,\cdots)\rangle$$

$$= \zeta^{\nu_\alpha} \sqrt{n_\alpha+1} \left(\prod_{\gamma\not=\a... ...delta_{n_\gamma^\prime n_\gamma}\right) \delta_{n_\alpha^\prime n_\alpha+1} \,.$$ (4.3)

En estas expresiones

$$\nu_\alpha^\prime=\sum_{\gamma<\alpha}n_\alpha^\prime \,,\qquad \nu_\alpha=\sum_{\gamma<\alpha}n_\alpha \,.$$ (4.4)

Los dos elementos de matriz coinciden teniendo en cuenta las deltas de Kronecker. En particular $\nu_\alpha^\prime= \nu_\alpha$ ya que $n_{\gamma<\alpha}=n_{\gamma<\alpha}^\prime$ .

(b) Queremos verificar la relación

$$a_\alpha a_\beta^\dagger= \zeta a_\beta^\dagger a_\alpha +\delta_{\alpha\beta}$$ (4.5)

(b.1) Caso $\alpha\not=\beta$

$$a_\alpha a_\beta^\dagger \vert(\cdots,n_\gamma,\cdots)\rangle = \zeta^{\nu_\beta} \sqrt{n_\beta+1} a_\alpha\vert(\cdots,n_\beta+1,\cdots)\rangle$$ (4.6)

$$= \zeta^{\nu_\beta+\nu_\alpha^\prime} \sqrt{n_\beta+1}\sqrt{n_\alpha} \vert(\cdots,n_\alpha-1,\cdots,n_\beta+1,\cdots)\rangle$$

con

$$\nu_\alpha^\prime= \left\{ \begin{matrix}\nu_\alpha+1, & \beta<\a... ...begin{matrix}\zeta, & \beta<\alpha \cr 1, & \alpha < \beta \end{matrix} \right.$$ (4.7)

por otro lado

$$a_\beta^\dagger a_\alpha \vert(\cdots,n_\gamma,\cdots)\rangle = \zeta^{\nu_\alpha} \sqrt{n_\alpha} a_\beta^\dagger\vert(\cdots,n_\alpha-1,\cdots)\rangle$$ (4.8)

$$= \zeta^{\nu_\alpha+\nu_\beta^\prime} \sqrt{n_\alpha}\sqrt{n_\beta+1} \vert(\cdots,n_\alpha-1,\cdots,n_\beta+1,\cdots)\rangle$$

con

$$\nu_\beta^\prime= \left\{ \begin{matrix}\nu_\beta, & \beta<\alpha... ...begin{matrix}1, & \beta<\alpha \cr \zeta, & \alpha < \beta \end{matrix} \right.$$ (4.9)

( $\zeta=\pm 1$ ). Las expresiones coinciden salvo un factor $\zeta$ sobre cualquier elemento de la base, es decir,

$$a_\alpha a_\beta^\dagger= \zeta a_\beta^\dagger a_\alpha \qquad (\alpha\not=\beta) \,.$$ (4.10)

(b.2) Caso $\alpha=\beta$ .

Para bosones $(\zeta=+1)$ ,

$$a_\alpha a_\alpha^\dagger\vert(\cdots )\rangle = \sqrt{n_\alpha+1}a_\alpha\vert(\cdots,n_\alpha+1,\cdots)\rangle = (n_\alpha+1)\vert(\cdots)\rangle$$

$$a_\alpha^\dagger a_\alpha\vert(\cdots )\rangle = \sqrt{n_\alpha}a_\alpha^\dagger\vert(\cdots,n_\alpha-1,\cdots)\rangle = n_\alpha\vert(\cdots)\rangle$$ (4.11)

es decir,

$$a_\alpha a_\alpha^\dagger= a_\alpha^\dagger a_\alpha+1 \qquad ( ) \,.$$ (4.12)

Para fermiones $(\zeta=-1)$ ,

$$a_\alpha a_\alpha^\dagger\vert(\cdots,n_\alpha=0,\cdots )\rangle = (-1)^{\nu_\alpha}a_\alpha\vert(\cdots,n_\alpha=1,\cdots)\rangle = \vert(\cdots,n_\alpha=0,\cdots)\rangle$$

$$a_\alpha^\dagger a_\alpha\vert(\cdots,n_\alpha=0,\cdots )\rangle =$$ 0 (4.13)

es decir

$$(a_\alpha a_\alpha^\dagger + a_\alpha^\dagger a_\alpha-1) \vert(\cdots,n_\alpha=0,\cdots )\rangle =0$$ (4.14)

análogamente

$$a_\alpha a_\alpha^\dagger\vert(\cdots,n_\alpha=1,\cdots )\rangle =$$ 0 (4.15)

$$a_\alpha^\dagger a_\alpha\vert(\cdots,n_\alpha=1,\cdots )\rangle = (-1)^{\nu_\alpha}a_\alpha^\dagger\vert(\cdots,n_\alpha=0,\cdots)\rangle = \vert(\cdots,n_\alpha=1,\cdots)\rangle$$

y por tanto

$$(a_\alpha a_\alpha^\dagger + a_\alpha^\dagger a_\alpha-1) \vert(\cdots,n_\alpha=1,\cdots )\rangle =0$$ (4.16)

En todo caso

$$a_\alpha a_\alpha^\dagger = -a_\alpha^\dagger a_\alpha+1 \qquad ( ) \,.$$ (4.17)