Solución ejercicio clase #18. 2021-2

 a) $$ \langle{\Psi} | = a^{*} \langle{1} | - b^{*} \langle{2} | + a^{*} \langle{3} |  $$

$$ \langle{\Phi} | = b^{*} \langle{1} |  + a^{*} \langle{2} |  $$

b) $$ \langle{\Psi} | \Phi \rangle = a^{*}b \langle{1} | 1 \rangle - b^{*}a \langle{2} | 2 \rangle = a^{*}b -  b^{*}a$$

$$ \langle{\Phi} | \Psi \rangle = b^{*}a \langle{1} | 1 \rangle - a^{*}b \langle{2} | 2 \rangle = b^{*}a -  a^{*}b=\langle{\Psi} | \Phi \rangle^{*}$$

c) $$ A=|\Phi \rangle \langle \Psi | = ba^{*}|1\rangle \langle 1| - bb^{*}|1\rangle \langle 2| + ba^{*}|1\rangle \langle 3| + aa^{*}|2\rangle \langle 1| - ab^{*}|2\rangle \langle 2| + aa^{*}|2\rangle \langle 3| $$

$$[A] = \begin{pmatrix}ba^{*}&-|b|^2& ba^{*}\\|a|^{2}&-ab^{*}&|a|^{2}\\0&0&0 \end{pmatrix}$$

d)  Sea  $$B=|\Phi \rangle \langle \Phi | = bb^{*}|1\rangle \langle 1| + ba^{*}|1\rangle \langle 2| + ab^{*}|2\rangle \langle 1| + aa^{*} |2\rangle \langle 2| = B^{\dagger}$$

$$[B] = \begin{pmatrix}|b|^{2}&ba^{*}& 0\\ab^{*}&|a|^{2}&0\\0&0&0 \end{pmatrix}= [B]^{\dagger}$$

Observamos que [B] es hermitiana

Sea $$ C=|\Psi \rangle \langle \Psi | = |a|^2 |1\rangle \langle 1| + ab^{*}|1\rangle \langle 2| + |a|^2 |1\rangle \langle 3| + ba^{*}|2\rangle \langle 1|  + |b|^{2}|2\rangle \langle 2| + ba^{*} |2\rangle \langle 3| $$

$$+ |a|^{2}|3\rangle \langle 1|+ ab^{*}|3\rangle \langle 2| + |a|^{2}|3\rangle \langle 3|=C^{\dagger}$$

$$[C] = \begin{pmatrix}|a|^{2}&ab^{*}& |a|^{2}\\ba^{*}&|b|^{2}&ba^{*}\\|a|^{2}&ab^{*}&|a|^{2} \end{pmatrix}= [C]^{\dagger}$$

Observamos que [C] es hermitiana    

  

Como conclusión tenemos que  en forma matricial $[Q] = [B] + [C] = [B]^{\dagger} + [C]^{\dagger}=[Q]^{\dagger} $, en su forma de operadores notamos $Q = B + C = B^{\dagger} + C^{\dagger}=Q^{\dagger} $