Problem set 5
These are my solutions to the problem sets assigned to me during the graduate-level quantum mechanics course.
Problem 1 ¶ The operator D ( z ) = exp [ z a † − z ∗ a ] D\left(z\right)=\exp\left[za^{\dagger}-z^{*}a\right] D ( z ) = exp [ z a † − z ∗ a ]
D ( q 0 , p 0 ) Q D − 1 ( q 0 , p 0 ) = Q − q 0 I D ( q 0 , p 0 ) P D − 1 ( q 0 , p 0 ) = P − p 0 I \begin{align*}
D\left(q_{0},p_{0}\right)QD^{-1}\left(q_{0},p_{0}\right) & =Q-q_{0}\mathbb{I}\\
D\left(q_{0},p_{0}\right)PD^{-1}\left(q_{0},p_{0}\right) & =P-p_{0}\mathbb{I}
\end{align*} D ( q 0 , p 0 ) Q D − 1 ( q 0 , p 0 ) D ( q 0 , p 0 ) P D − 1 ( q 0 , p 0 ) = Q − q 0 I = P − p 0 I
where D ( z ) = exp [ i ( p 0 Q − q 0 P ) / ℏ ] D\left(z\right)=\exp\left[i\left(p_{0}Q-q_{0}P\right)/\hbar\right] D ( z ) = exp [ i ( p 0 Q − q 0 P ) /ℏ ] D ( z 1 ) D ( z 1 ) D\left(z_{1}\right)D\left(z_{1}\right) D ( z 1 ) D ( z 1 ) D ( z 1 + z 2 ) D\left(z_{1}+z_{2}\right) D ( z 1 + z 2 )
Temos o operador D ( z ) = exp ( z a † − z ∗ a ) D\left(z\right)=\exp\left(za^{\dagger}-z^{*}a\right) D ( z ) = exp ( z a † − z ∗ a ) D ( z 1 ) D ( z 2 ) D\left(z_{1}\right)D\left(z_{2}\right) D ( z 1 ) D ( z 2 ) [ a , a † ] = I \left[a,a^{\dagger}\right]=\mathbb{I} [ a , a † ] = I Lima (2013)
e t ( A + B ) = e t A e t B e − t 2 2 ! [ A , B ] e t 3 3 ! ( 2 [ B , [ A , B ] ] + [ A , [ A , B ] ] ) … e^{t\left(A+B\right)}=e^{tA}e^{tB}e^{-\frac{t^{2}}{2!}\left[A,B\right]}e^{\frac{t^{3}}{3!}\left(2\left[B,\left[A,B\right]\right]+\left[A,\left[A,B\right]\right]\right)}\dots e t ( A + B ) = e t A e tB e − 2 ! t 2 [ A , B ] e 3 ! t 3 ( 2 [ B , [ A , B ] ] + [ A , [ A , B ] ] ) … Logo:
D ( z 1 + z 2 ) = exp ( ( z 1 + z 2 ) a † − ( z 1 + z 2 ) ∗ a ) = exp ( ( z 1 + z 2 ) a † − ( z 1 ∗ + z 2 ∗ ) a ) = exp ( [ z 1 a † − z 1 ∗ a ] + [ z 2 a † − z 2 ∗ a ] ) \begin{align*}
D\left(z_{1}+z_{2}\right) & =\exp\left(\left(z_{1}+z_{2}\right)a^{\dagger}-\left(z_{1}+z_{2}\right)^{*}a\right)\\
& =\exp\left(\left(z_{1}+z_{2}\right)a^{\dagger}-\left(z^{*}_{1}+z^{*}_{2}\right)a\right)\\
& =\exp\left(\left[z_{1}a^{\dagger}-z^{*}_{1}a\right]+\left[z_{2}a^{\dagger}-z^{*}_{2}a\right]\right)
\end{align*} D ( z 1 + z 2 ) = exp ( ( z 1 + z 2 ) a † − ( z 1 + z 2 ) ∗ a ) = exp ( ( z 1 + z 2 ) a † − ( z 1 ∗ + z 2 ∗ ) a ) = exp ( [ z 1 a † − z 1 ∗ a ] + [ z 2 a † − z 2 ∗ a ] ) Sendo A = z 1 a † − z 1 ∗ a A=z_{1}a^{\dagger}-z^{*}_{1}a A = z 1 a † − z 1 ∗ a B = z 2 a † − z 2 ∗ a B=z_{2}a^{\dagger}-z^{*}_{2}a B = z 2 a † − z 2 ∗ a
[ z 1 a † − z 1 ∗ a , z 2 a † − z 2 ∗ a ] = ( z 1 a † − z 1 ∗ a ) ( z 2 a † − z 2 ∗ a ) − ( z 2 a † − z 2 ∗ a ) ( z 1 a † − z 1 ∗ a ) [ A , B ] = ( z 2 z 1 a † a † − z 2 ∗ z 1 a † a − z 2 z 1 ∗ a a † + z 1 ∗ a z 2 ∗ a ) − ( z 1 z 2 a † a † − z 1 ∗ z 2 a † a − z 2 ∗ a z 1 a † + z 2 ∗ a z 1 ∗ a ) = − z 2 ∗ z 1 a † a − z 2 z 1 ∗ a a † + z 1 ∗ z 2 a † a + z 2 ∗ a z 1 a † = ( z 1 ∗ z 2 − z 2 ∗ z 1 ) a † a + ( z 2 ∗ z 1 − z 2 z 1 ∗ ) a a † = ( z 2 ∗ z 1 − z 2 z 1 ∗ ) a a † − ( z 2 ∗ z 1 − z 1 ∗ z 2 ) a † a = ( z 2 ∗ z 1 − z 2 z 1 ∗ ) ( a a † − a † a ) = ( z 2 ∗ z 1 − z 2 z 1 ∗ ) [ a , a † ] = z 2 ∗ z 1 − z 2 z 1 ∗ \begin{align*}
\left[z_{1}a^{\dagger}-z^{*}_{1}a,z_{2}a^{\dagger}-z^{*}_{2}a\right] & =\left(z_{1}a^{\dagger}-z^{*}_{1}a\right)\left(z_{2}a^{\dagger}-z^{*}_{2}a\right)-\left(z_{2}a^{\dagger}-z^{*}_{2}a\right)\left(z_{1}a^{\dagger}-z^{*}_{1}a\right)\\
\left[A,B\right] & =\left(z_{2}z_{1}a^{\dagger}a^{\dagger}-z^{*}_{2}z_{1}a^{\dagger}a-z_{2}z^{*}_{1}aa^{\dagger}+z^{*}_{1}az^{*}_{2}a\right)-\left(z_{1}z_{2}a^{\dagger}a^{\dagger}-z^{*}_{1}z_{2}a^{\dagger}a-z^{*}_{2}az_{1}a^{\dagger}+z^{*}_{2}az^{*}_{1}a\right)\\
& =-z^{*}_{2}z_{1}a^{\dagger}a-z_{2}z^{*}_{1}aa^{\dagger}+z^{*}_{1}z_{2}a^{\dagger}a+z^{*}_{2}az_{1}a^{\dagger}\\
& =\left(z^{*}_{1}z_{2}-z^{*}_{2}z_{1}\right)a^{\dagger}a+\left(z^{*}_{2}z_{1}-z_{2}z^{*}_{1}\right)aa^{\dagger}\\
& =\left(z^{*}_{2}z_{1}-z_{2}z^{*}_{1}\right)aa^{\dagger}-\left(z^{*}_{2}z_{1}-z^{*}_{1}z_{2}\right)a^{\dagger}a\\
& =\left(z^{*}_{2}z_{1}-z_{2}z^{*}_{1}\right)\left(aa^{\dagger}-a^{\dagger}a\right)\\
& =\left(z^{*}_{2}z_{1}-z_{2}z^{*}_{1}\right)\left[a,a^{\dagger}\right]\\
& =z^{*}_{2}z_{1}-z_{2}z^{*}_{1}
\end{align*} [ z 1 a † − z 1 ∗ a , z 2 a † − z 2 ∗ a ] [ A , B ] = ( z 1 a † − z 1 ∗ a ) ( z 2 a † − z 2 ∗ a ) − ( z 2 a † − z 2 ∗ a ) ( z 1 a † − z 1 ∗ a ) = ( z 2 z 1 a † a † − z 2 ∗ z 1 a † a − z 2 z 1 ∗ a a † + z 1 ∗ a z 2 ∗ a ) − ( z 1 z 2 a † a † − z 1 ∗ z 2 a † a − z 2 ∗ a z 1 a † + z 2 ∗ a z 1 ∗ a ) = − z 2 ∗ z 1 a † a − z 2 z 1 ∗ a a † + z 1 ∗ z 2 a † a + z 2 ∗ a z 1 a † = ( z 1 ∗ z 2 − z 2 ∗ z 1 ) a † a + ( z 2 ∗ z 1 − z 2 z 1 ∗ ) a a † = ( z 2 ∗ z 1 − z 2 z 1 ∗ ) a a † − ( z 2 ∗ z 1 − z 1 ∗ z 2 ) a † a = ( z 2 ∗ z 1 − z 2 z 1 ∗ ) ( a a † − a † a ) = ( z 2 ∗ z 1 − z 2 z 1 ∗ ) [ a , a † ] = z 2 ∗ z 1 − z 2 z 1 ∗ Então [ B , [ A , B ] ] = [ A , [ A , B ] ] = 0 \left[B,\left[A,B\right]\right]=\left[A,\left[A,B\right]\right]=0 [ B , [ A , B ] ] = [ A , [ A , B ] ] = 0
D ( z 1 + z 2 ) = e ( z 1 a † − z 1 ∗ a ) a † e ( z 2 a † − z 2 ∗ a ) a e − 1 2 ( z 2 ∗ z 1 − z 2 z 1 ∗ ) \begin{align*}
D\left(z_{1}+z_{2}\right) & =e^{\left(z_{1}a^{\dagger}-z^{*}_{1}a\right)a^{\dagger}}e^{\left(z_{2}a^{\dagger}-z^{*}_{2}a\right)a}e^{-\frac{1}{2}\left(z^{*}_{2}z_{1}-z_{2}z^{*}_{1}\right)}
\end{align*} D ( z 1 + z 2 ) = e ( z 1 a † − z 1 ∗ a ) a † e ( z 2 a † − z 2 ∗ a ) a e − 2 1 ( z 2 ∗ z 1 − z 2 z 1 ∗ ) E pela definiçãoD ( z ) = exp ( z a † − z ∗ a ) D\left(z\right)=\exp\left(za^{\dagger}-z^{*}a\right) D ( z ) = exp ( z a † − z ∗ a )
D ( z 1 + z 2 ) = e 1 2 ( z 2 z 1 ∗ − z 2 ∗ z 1 ) D ( z 1 ) D ( z 2 ) D\left(z_{1}+z_{2}\right)=e^{\frac{1}{2}\left(z_{2}z^{*}_{1}-z^{*}_{2}z_{1}\right)}D\left(z_{1}\right)D\left(z_{2}\right) D ( z 1 + z 2 ) = e 2 1 ( z 2 z 1 ∗ − z 2 ∗ z 1 ) D ( z 1 ) D ( z 2 ) Abrindo os números imaginários z i = u i + i v i z_{i}=u_{i}+iv_{i} z i = u i + i v i
z 2 z 1 ∗ − z 2 ∗ z 1 = ( u 2 + i v 2 ) ( u 1 − i v 1 ) − ( u 2 − i v 2 ) ( u 1 + i v 1 ) = ( u 2 u 1 − i u 2 v 1 + i v 2 u 1 + v 2 v 1 ) − ( u 2 u 1 + i u 2 v 1 − i v 2 u 1 + v 1 v 2 ) = 2 i ( v 2 u 1 − u 2 v 1 ) \begin{align*}
z_{2}z^{*}_{1}-z^{*}_{2}z_{1} & =\left(u_{2}+iv_{2}\right)\left(u_{1}-iv_{1}\right)-\left(u_{2}-iv_{2}\right)\left(u_{1}+iv_{1}\right)\\
& =\left(u_{2}u_{1}-iu_{2}v_{1}+iv_{2}u_{1}+v_{2}v_{1}\right)-\left(u_{2}u_{1}+iu_{2}v_{1}-iv_{2}u_{1}+v_{1}v_{2}\right)\\
& =2i\left(v_{2}u_{1}-u_{2}v_{1}\right)
\end{align*} z 2 z 1 ∗ − z 2 ∗ z 1 = ( u 2 + i v 2 ) ( u 1 − i v 1 ) − ( u 2 − i v 2 ) ( u 1 + i v 1 ) = ( u 2 u 1 − i u 2 v 1 + i v 2 u 1 + v 2 v 1 ) − ( u 2 u 1 + i u 2 v 1 − i v 2 u 1 + v 1 v 2 ) = 2 i ( v 2 u 1 − u 2 v 1 ) De orma que:
D ( z 1 + z 2 ) = e i ( v 2 u 1 − u 2 v 1 ) D ( z 1 ) D ( z 2 ) D\left(z_{1}+z_{2}\right)=e^{i\left(v_{2}u_{1}-u_{2}v_{1}\right)}D\left(z_{1}\right)D\left(z_{2}\right) D ( z 1 + z 2 ) = e i ( v 2 u 1 − u 2 v 1 ) D ( z 1 ) D ( z 2 ) Logo dois doslocamentos na sequência é quase a mesma coisa que fazer um único deslocamento, a única diferença é uma fase. Outra forma de resolver esta questão é usando a fórmula:
e t ( A + B ) = e t A e t B e − t 2 2 ! [ A , B ] e t 3 3 ! ( 2 [ B , [ A , B ] ] + [ A , [ A , B ] ] ) … e^{t\left(A+B\right)}=e^{tA}e^{tB}e^{-\frac{t^{2}}{2!}\left[A,B\right]}e^{\frac{t^{3}}{3!}\left(2\left[B,\left[A,B\right]\right]+\left[A,\left[A,B\right]\right]\right)}\dots e t ( A + B ) = e t A e tB e − 2 ! t 2 [ A , B ] e 3 ! t 3 ( 2 [ B , [ A , B ] ] + [ A , [ A , B ] ] ) … Então:
D ( z ) = exp ( [ z a † ] + [ − z ∗ a ] ) = e z a † e − z ∗ a e − 1 2 [ z a † , − z ∗ a ] e 1 3 ! ( 2 [ − z ∗ a , [ z a † , − z ∗ a ] ] + [ z a † , [ z a † , − z ∗ a ] ] ) … \begin{align*}
D\left(z\right) & =\exp\left(\left[za^{\dagger}\right]+\left[-z^{*}a\right]\right)\\
& =e^{za^{\dagger}}e^{-z^{*}a}e^{-\frac{1}{2}\left[za^{\dagger},-z^{*}a\right]}e^{\frac{1}{3!}\left(2\left[-z^{*}a,\left[za^{\dagger},-z^{*}a\right]\right]+\left[za^{\dagger},\left[za^{\dagger},-z^{*}a\right]\right]\right)}\dots
\end{align*} D ( z ) = exp ( [ z a † ] + [ − z ∗ a ] ) = e z a † e − z ∗ a e − 2 1 [ z a † , − z ∗ a ] e 3 ! 1 ( 2 [ − z ∗ a , [ z a † , − z ∗ a ] ] + [ z a † , [ z a † , − z ∗ a ] ] ) … E como:
[ z a † , − z ∗ a ] = z a † ( − z ∗ a ) − ( − z ∗ a ) z a † = ∣ z ∣ 2 [ a , a † ] = ∣ z ∣ 2 \begin{align*}
\left[za^{\dagger},-z^{*}a\right] & =za^{\dagger}\left(-z^{*}a\right)-\left(-z^{*}a\right)za^{\dagger}=\left|z\right|^{2}\left[a,a^{\dagger}\right]=\left|z\right|^{2}
\end{align*} [ z a † , − z ∗ a ] = z a † ( − z ∗ a ) − ( − z ∗ a ) z a † = ∣ z ∣ 2 [ a , a † ] = ∣ z ∣ 2 Ficamos com:
D ( z ) = e z a † e − z ∗ a e − 1 2 ∣ z ∣ 2 e 1 3 ! ( 2 [ − z ∗ a , ∣ z ∣ 2 ] + [ z a † , ∣ z ∣ 2 ] ) … = e z a † e − z ∗ a e − ∣ z ∣ 2 / 2 \begin{align*}
D\left(z\right) & =e^{za^{\dagger}}e^{-z^{*}a}e^{-\frac{1}{2}\left|z\right|^{2}}e^{\frac{1}{3!}\left(2\left[-z^{*}a,\left|z\right|^{2}\right]+\left[za^{\dagger},\left|z\right|^{2}\right]\right)}\dots\\
& =e^{za^{\dagger}}e^{-z^{*}a}e^{-\left|z\right|^{2}/2}
\end{align*} D ( z ) = e z a † e − z ∗ a e − 2 1 ∣ z ∣ 2 e 3 ! 1 ( 2 [ − z ∗ a , ∣ z ∣ 2 ] + [ z a † , ∣ z ∣ 2 ] ) … = e z a † e − z ∗ a e − ∣ z ∣ 2 /2 Ainda podemos fazer:
D ( z ) = exp ( [ − z ∗ a ] + [ z a † ] ) = e − z ∗ a e z a † e − 1 2 [ − z ∗ a , z a † ] e 1 3 ( 2 [ z a † , [ − z ∗ a , z a † ] ] + [ − z ∗ a , [ − z ∗ a , z a † ] ] ) … = e − z ∗ a e z a † e 1 2 ∣ z ∣ 2 e t 3 3 ! ( 2 [ z a † , − ∣ z ∣ 2 ] + [ − z ∗ a , − ∣ z ∣ 2 ] ) = e − z ∗ a e z a † e ∣ z ∣ 2 / 2 \begin{align*}
D\left(z\right) & =\exp\left(\left[-z^{*}a\right]+\left[za^{\dagger}\right]\right)\\
& =e^{-z^{*}a}e^{za^{\dagger}}e^{-\frac{1}{2}\left[-z^{*}a,za^{\dagger}\right]}e^{\frac{1}{3}\left(2\left[za^{\dagger},\left[-z^{*}a,za^{\dagger}\right]\right]+\left[-z^{*}a,\left[-z^{*}a,za^{\dagger}\right]\right]\right)}\dots\\
& =e^{-z^{*}a}e^{za^{\dagger}}e^{\frac{1}{2}\left|z\right|^{2}}e^{\frac{t^{3}}{3!}\left(2\left[za^{\dagger},-\left|z\right|^{2}\right]+\left[-z^{*}a,-\left|z\right|^{2}\right]\right)}\\
& =e^{-z^{*}a}e^{za^{\dagger}}e^{\left|z\right|^{2}/2}
\end{align*} D ( z ) = exp ( [ − z ∗ a ] + [ z a † ] ) = e − z ∗ a e z a † e − 2 1 [ − z ∗ a , z a † ] e 3 1 ( 2 [ z a † , [ − z ∗ a , z a † ] ] + [ − z ∗ a , [ − z ∗ a , z a † ] ] ) … = e − z ∗ a e z a † e 2 1 ∣ z ∣ 2 e 3 ! t 3 ( 2 [ z a † , − ∣ z ∣ 2 ] + [ − z ∗ a , − ∣ z ∣ 2 ] ) = e − z ∗ a e z a † e ∣ z ∣ 2 /2 Onde usamos o fato de que [ A , B ] = A B − B A = − ( B A − A B ) = − [ B , A ] \left[A,B\right]=AB-BA=-\left(BA-AB\right)=-\left[B,A\right] [ A , B ] = A B − B A = − ( B A − A B ) = − [ B , A ]
D ( z ) = e z a † e − z ∗ a e − ∣ z ∣ 2 / 2 = e − z ∗ a e z a † e ∣ z ∣ 2 / 2 D\left(z\right)=e^{za^{\dagger}}e^{-z^{*}a}e^{-\left|z\right|^{2}/2}=e^{-z^{*}a}e^{za^{\dagger}}e^{\left|z\right|^{2}/2} D ( z ) = e z a † e − z ∗ a e − ∣ z ∣ 2 /2 = e − z ∗ a e z a † e ∣ z ∣ 2 /2 E então:
D ( z 1 + z 2 ) = exp ( [ ( z 1 + z 2 ) a † ] + [ − ( z 1 + z 2 ) ∗ a ] ) = exp ( [ ( z 1 + z 2 ) a † ] + [ − ( z 1 ∗ + z 2 ∗ ) a ] ) = e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 1 2 [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] e t 3 3 ! ( 2 [ − ( z 1 ∗ + z 2 ∗ ) a , [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] ] + [ ( z 1 + z 2 ) a † , [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] ] ) … \begin{align*}
D\left(z_{1}+z_{2}\right) & =\exp\left(\left[\left(z_{1}+z_{2}\right)a^{\dagger}\right]+\left[-\left(z_{1}+z_{2}\right)^{*}a\right]\right)\\
& =\exp\left(\left[\left(z_{1}+z_{2}\right)a^{\dagger}\right]+\left[-\left(z^{*}_{1}+z^{*}_{2}\right)a\right]\right)\\
& =e^{\left(z_{1}+z_{2}\right)a^{\dagger}}e^{-\left(z^{*}_{1}+z^{*}_{2}\right)a}e^{-\frac{1}{2}\left[\left(z_{1}+z_{2}\right)a^{\dagger},-\left(z^{*}_{1}+z^{*}_{2}\right)a\right]}e^{\frac{t^{3}}{3!}\left(2\left[-\left(z^{*}_{1}+z^{*}_{2}\right)a,\left[\left(z_{1}+z_{2}\right)a^{\dagger},-\left(z^{*}_{1}+z^{*}_{2}\right)a\right]\right]+\left[\left(z_{1}+z_{2}\right)a^{\dagger},\left[\left(z_{1}+z_{2}\right)a^{\dagger},-\left(z^{*}_{1}+z^{*}_{2}\right)a\right]\right]\right)\dots}
\end{align*} D ( z 1 + z 2 ) = exp ( [ ( z 1 + z 2 ) a † ] + [ − ( z 1 + z 2 ) ∗ a ] ) = exp ( [ ( z 1 + z 2 ) a † ] + [ − ( z 1 ∗ + z 2 ∗ ) a ] ) = e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 2 1 [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] e 3 ! t 3 ( 2 [ − ( z 1 ∗ + z 2 ∗ ) a , [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] ] + [ ( z 1 + z 2 ) a † , [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] ] ) … E então:
[ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] = − ( z 1 + z 2 ) a † ( z 1 ∗ + z 2 ∗ ) a − ( − ( z 1 ∗ + z 2 ∗ ) a ( z 1 + z 2 ) a † ) = − ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) a † a + ( z 1 ∗ + z 2 ∗ ) ( z 1 + z 2 ) a a † = ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) ( a a † − a † a ) = ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) [ a , a † ] = ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) \begin{align*}
\left[\left(z_{1}+z_{2}\right)a^{\dagger},-\left(z^{*}_{1}+z^{*}_{2}\right)a\right] & =-\left(z_{1}+z_{2}\right)a^{\dagger}\left(z^{*}_{1}+z^{*}_{2}\right)a-\left(-\left(z^{*}_{1}+z^{*}_{2}\right)a\left(z_{1}+z_{2}\right)a^{\dagger}\right)\\
& =-\left(z_{1}+z_{2}\right)\left(z^{*}_{1}+z^{*}_{2}\right)a^{\dagger}a+\left(z^{*}_{1}+z^{*}_{2}\right)\left(z_{1}+z_{2}\right)aa^{\dagger}\\
& =\left(z_{1}+z_{2}\right)\left(z^{*}_{1}+z^{*}_{2}\right)\left(aa^{\dagger}-a^{\dagger}a\right)\\
& =\left(z_{1}+z_{2}\right)\left(z^{*}_{1}+z^{*}_{2}\right)\left[a,a^{\dagger}\right]\\
& =\left(z_{1}+z_{2}\right)\left(z^{*}_{1}+z^{*}_{2}\right)
\end{align*} [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] = − ( z 1 + z 2 ) a † ( z 1 ∗ + z 2 ∗ ) a − ( − ( z 1 ∗ + z 2 ∗ ) a ( z 1 + z 2 ) a † ) = − ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) a † a + ( z 1 ∗ + z 2 ∗ ) ( z 1 + z 2 ) a a † = ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) ( a a † − a † a ) = ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) [ a , a † ] = ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) Dessa forma análogo ao caso anterior temos:
[ − ( z 1 ∗ + z 2 ∗ ) a , [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] ] = [ − ( z 1 ∗ + z 2 ∗ ) a , ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) ] = 0 \left[-\left(z^{*}_{1}+z^{*}_{2}\right)a,\left[\left(z_{1}+z_{2}\right)a^{\dagger},-\left(z^{*}_{1}+z^{*}_{2}\right)a\right]\right]=\left[-\left(z^{*}_{1}+z^{*}_{2}\right)a,\left(z_{1}+z_{2}\right)\left(z^{*}_{1}+z^{*}_{2}\right)\right]=0 [ − ( z 1 ∗ + z 2 ∗ ) a , [ ( z 1 + z 2 ) a † , − ( z 1 ∗ + z 2 ∗ ) a ] ] = [ − ( z 1 ∗ + z 2 ∗ ) a , ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) ] = 0 E então:
D ( z 1 + z 2 ) = e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 1 2 ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) = e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 1 2 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + z 1 z 2 ∗ + z 2 z 1 ∗ ) = e − 1 2 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 1 2 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) \begin{align*}
D\left(z_{1}+z_{2}\right) & =e^{\left(z_{1}+z_{2}\right)a^{\dagger}}e^{-\left(z^{*}_{1}+z^{*}_{2}\right)a}e^{-\frac{1}{2}\left(z_{1}+z_{2}\right)\left(z^{*}_{1}+z^{*}_{2}\right)}\\
& =e^{\left(z_{1}+z_{2}\right)a^{\dagger}}e^{-\left(z^{*}_{1}+z^{*}_{2}\right)a}e^{-\frac{1}{2}\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1}z^{*}_{2}+z_{2}z^{*}_{1}\right)}\\
& =e^{-\frac{1}{2}\left(z_{1}z^{*}_{2}+z_{2}z^{*}_{1}\right)}e^{\left(z_{1}+z_{2}\right)a^{\dagger}}e^{-\left(z^{*}_{1}+z^{*}_{2}\right)a}e^{-\frac{1}{2}\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)}
\end{align*} D ( z 1 + z 2 ) = e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 2 1 ( z 1 + z 2 ) ( z 1 ∗ + z 2 ∗ ) = e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 2 1 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + z 1 z 2 ∗ + z 2 z 1 ∗ ) = e − 2 1 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 2 1 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) E abrindo novamente:
e ( z 1 + z 2 ) a = e z 1 a e z 2 a e − 1 2 [ z 1 a , z 2 a ] ⋯ = e z 1 a e z 2 a \begin{align*}
e^{\left(z_{1}+z_{2}\right)a} & =e^{z_{1}a}e^{z_{2}a}e^{-\frac{1}{2}\left[z_{1}a,z_{2}a\right]}\dots=e^{z_{1}a}e^{z_{2}a}
\end{align*} e ( z 1 + z 2 ) a = e z 1 a e z 2 a e − 2 1 [ z 1 a , z 2 a ] ⋯ = e z 1 a e z 2 a Então:
D ( z 1 + z 2 ) = e − 1 2 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 1 2 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) = e − 1 2 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e z 1 a † [ e z 2 a † e − z 1 ∗ a ] e − z 2 ∗ a e − 1 2 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) \begin{align*}
D\left(z_{1}+z_{2}\right) & =e^{-\frac{1}{2}\left(z_{1}z^{*}_{2}+z_{2}z^{*}_{1}\right)}e^{\left(z_{1}+z_{2}\right)a^{\dagger}}e^{-\left(z^{*}_{1}+z^{*}_{2}\right)a}e^{-\frac{1}{2}\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)}\\
& =e^{-\frac{1}{2}\left(z_{1}z^{*}_{2}+z_{2}z^{*}_{1}\right)}e^{z_{1}a^{\dagger}}\left[e^{z_{2}a^{\dagger}}e^{-z^{*}_{1}a}\right]e^{-z^{*}_{2}a}e^{-\frac{1}{2}\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)}
\end{align*} D ( z 1 + z 2 ) = e − 2 1 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e ( z 1 + z 2 ) a † e − ( z 1 ∗ + z 2 ∗ ) a e − 2 1 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) = e − 2 1 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e z 1 a † [ e z 2 a † e − z 1 ∗ a ] e − z 2 ∗ a e − 2 1 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) Queremos agora inverter e z 2 a † e − z 1 ∗ a → e − z 1 ∗ a e z 2 a † e^{z_{2}a^{\dagger}}e^{-z^{*}_{1}a}\rightarrow e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}} e z 2 a † e − z 1 ∗ a → e − z 1 ∗ a e z 2 a † e ( A + B ) = e A e B e − 1 2 [ A , B ] e^{\left(A+B\right)}=e^{A}e^{B}e^{-\frac{1}{2}\left[A,B\right]} e ( A + B ) = e A e B e − 2 1 [ A , B ]
e ( z 2 a † − z 1 ∗ a ) = e z 2 a † e − z 1 ∗ a e − 1 2 [ z 2 a † , − z 1 ∗ a ] = e z 2 a † e − z 1 ∗ a e − z 1 ∗ z 2 2 e ( − z 1 ∗ a + z 2 a † ) = e − z 1 ∗ a e z 2 a † e − 1 2 [ − z 1 ∗ a , z 2 a † ] = e − z 1 ∗ a e z 2 a † e z 1 ∗ z 2 2 \begin{align*}
e^{\left(z_{2}a^{\dagger}-z^{*}_{1}a\right)} & =e^{z_{2}a^{\dagger}}e^{-z^{*}_{1}a}e^{-\frac{1}{2}\left[z_{2}a^{\dagger},-z^{*}_{1}a\right]}=e^{z_{2}a^{\dagger}}e^{-z^{*}_{1}a}e^{-\frac{z^{*}_{1}z_{2}}{2}}\\
e^{\left(-z^{*}_{1}a+z_{2}a^{\dagger}\right)} & =e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}}e^{-\frac{1}{2}\left[-z^{*}_{1}a,z_{2}a^{\dagger}\right]}=e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}}e^{\frac{z^{*}_{1}z_{2}}{2}}
\end{align*} e ( z 2 a † − z 1 ∗ a ) e ( − z 1 ∗ a + z 2 a † ) = e z 2 a † e − z 1 ∗ a e − 2 1 [ z 2 a † , − z 1 ∗ a ] = e z 2 a † e − z 1 ∗ a e − 2 z 1 ∗ z 2 = e − z 1 ∗ a e z 2 a † e − 2 1 [ − z 1 ∗ a , z 2 a † ] = e − z 1 ∗ a e z 2 a † e 2 z 1 ∗ z 2 Logo:
e − z 1 ∗ a e z 2 a † e z 1 ∗ z 2 2 = e z 2 a † e − z 1 ∗ a e − z 1 ∗ z 2 2 e − z 1 ∗ a e z 2 a † e z 1 ∗ z 2 = e z 2 a † e − z 1 ∗ a \begin{align*}
e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}}e^{\frac{z^{*}_{1}z_{2}}{2}} & =e^{z_{2}a^{\dagger}}e^{-z^{*}_{1}a}e^{-\frac{z^{*}_{1}z_{2}}{2}}\\
e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}}e^{z^{*}_{1}z_{2}} & =e^{z_{2}a^{\dagger}}e^{-z^{*}_{1}a}
\end{align*} e − z 1 ∗ a e z 2 a † e 2 z 1 ∗ z 2 e − z 1 ∗ a e z 2 a † e z 1 ∗ z 2 = e z 2 a † e − z 1 ∗ a e − 2 z 1 ∗ z 2 = e z 2 a † e − z 1 ∗ a Substituindo então:
D ( z 1 + z 2 ) = e − 1 2 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e z 1 a † e − z 1 ∗ a e z 2 a † e z 1 ∗ z 2 e − z 2 ∗ a e − 1 2 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) = e − 1 2 ( z 2 z 1 ∗ − z 1 z 2 ∗ ) e z 1 a † e − z 1 ∗ a e z 2 a † e − z 2 ∗ a e − 1 2 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) \begin{align*}
D\left(z_{1}+z_{2}\right) & =e^{-\frac{1}{2}\left(z_{1}z^{*}_{2}+z_{2}z^{*}_{1}\right)}e^{z_{1}a^{\dagger}}e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}}e^{z^{*}_{1}z_{2}}e^{-z^{*}_{2}a}e^{-\frac{1}{2}\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)}\\
& =e^{-\frac{1}{2}\left(z_{2}z^{*}_{1}-z_{1}z^{*}_{2}\right)}e^{z_{1}a^{\dagger}}e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}}e^{-z^{*}_{2}a}e^{-\frac{1}{2}\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)}
\end{align*} D ( z 1 + z 2 ) = e − 2 1 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) e z 1 a † e − z 1 ∗ a e z 2 a † e z 1 ∗ z 2 e − z 2 ∗ a e − 2 1 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) = e − 2 1 ( z 2 z 1 ∗ − z 1 z 2 ∗ ) e z 1 a † e − z 1 ∗ a e z 2 a † e − z 2 ∗ a e − 2 1 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) Então o produto vai ser dado por:
D ( z 1 ) D ( z 2 ) = e z 1 a † e − z 1 ∗ a e − 1 2 ∣ z 1 ∣ 2 e z 2 a † e − z 2 ∗ a e − 1 2 ∣ z 2 ∣ 2 = e z 1 a † e − z 1 ∗ a e z 2 a † e − z 2 ∗ a e − 1 2 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) = e 1 2 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) D ( z 1 + z 2 ) \begin{align*}
D\left(z_{1}\right)D\left(z_{2}\right) & =e^{z_{1}a^{\dagger}}e^{-z^{*}_{1}a}e^{-\frac{1}{2}\left|z_{1}\right|^{2}}e^{z_{2}a^{\dagger}}e^{-z^{*}_{2}a}e^{-\frac{1}{2}\left|z_{2}\right|^{2}}\\
& =e^{z_{1}a^{\dagger}}e^{-z^{*}_{1}a}e^{z_{2}a^{\dagger}}e^{-z^{*}_{2}a}e^{-\frac{1}{2}\left(\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}\right)}\\
& =e^{\frac{1}{2}\left(z_{1}z^{*}_{2}+z_{2}z^{*}_{1}\right)}D\left(z_{1}+z_{2}\right)
\end{align*} D ( z 1 ) D ( z 2 ) = e z 1 a † e − z 1 ∗ a e − 2 1 ∣ z 1 ∣ 2 e z 2 a † e − z 2 ∗ a e − 2 1 ∣ z 2 ∣ 2 = e z 1 a † e − z 1 ∗ a e z 2 a † e − z 2 ∗ a e − 2 1 ( ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 ) = e 2 1 ( z 1 z 2 ∗ + z 2 z 1 ∗ ) D ( z 1 + z 2 ) Problem 2 ¶ If the initial state vector of a harmonic oscillator is the coherent state, ∣ Ψ ( 0 ) ⟩ = ∣ z ⟩ \left|\Psi\left(0\right)\right\rangle =\left|z\right\rangle ∣ Ψ ( 0 ) ⟩ = ∣ z ⟩ ∣ Ψ ( t ) ⟩ \left|\Psi\left(t\right)\right\rangle ∣ Ψ ( t ) ⟩ ( q 0 , p 0 ) \left(q_{0},p_{0}\right) ( q 0 , p 0 )
A demonstração que um estado coerente permanece coerente, e que sua evolução no tempo corresponde a uma órbita clássica no espaço ( q 0 , p 0 ) \left(q_{0},p_{0}\right) ( q 0 , p 0 ) ∣ α ⟩ \left|\alpha\right\rangle ∣ α ⟩ a ^ \widehat{a} a a ^ ∣ α ⟩ = α ∣ α ⟩ \widehat{a}\left|\alpha\right\rangle =\alpha\left|\alpha\right\rangle a ∣ α ⟩ = α ∣ α ⟩ a ^ \widehat{a} a
a ^ = m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ \widehat{a}=\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p} a = 2ℏ mω x + 2ℏ mω i p Então, se α \alpha α ⟨ α ∣ a ^ ∣ α ⟩ = α \left\langle \alpha\left|\widehat{a}\right|\alpha\right\rangle =\alpha ⟨ α ∣ a ∣ α ⟩ = α x 0 = ⟨ α ∣ x ^ ∣ α ⟩ x_{0}=\left\langle \alpha\left|\widehat{x}\right|\alpha\right\rangle x 0 = ⟨ α ∣ x ∣ α ⟩ p 0 = ⟨ α ∣ p ^ ∣ α ⟩ p_{0}=\left\langle \alpha\left|\widehat{p}\right|\alpha\right\rangle p 0 = ⟨ α ∣ p ∣ α ⟩
α = ⟨ α ∣ a ^ ∣ α ⟩ = ⟨ α ∣ m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ ∣ α ⟩ = m ω 2 ℏ ⟨ α ∣ x ^ ∣ α ⟩ + i 2 ℏ m ω ⟨ α ∣ p ^ ∣ α ⟩ = m ω 2 ℏ x 0 + i 2 ℏ m ω p 0 \begin{align}
\alpha & =\left\langle \alpha\left|\widehat{a}\right|\alpha\right\rangle \\
& =\left\langle \alpha\left|\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right|\alpha\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}\left\langle \alpha\left|\widehat{x}\right|\alpha\right\rangle +\frac{i}{\sqrt{2\hbar m\omega}}\left\langle \alpha\left|\widehat{p}\right|\alpha\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}x_{0}+\frac{i}{\sqrt{2\hbar m\omega}}p_{0}
\end{align} α = ⟨ α ∣ a ∣ α ⟩ = ⟨ α ∣ ∣ 2ℏ mω x + 2ℏ mω i p ∣ ∣ α ⟩ = 2ℏ mω ⟨ α ∣ x ∣ α ⟩ + 2ℏ mω i ⟨ α ∣ p ∣ α ⟩ = 2ℏ mω x 0 + 2ℏ mω i p 0 Queremos reescrever U ^ ∣ α ⟩ \widehat{U}\left|\alpha\right\rangle U ∣ α ⟩ U ^ = U ^ ( t − t 0 ) = exp [ − i H ( t − t 0 ) ℏ ] \widehat{U}=\widehat{U}\left(t-t_{0}\right)=\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right] U = U ( t − t 0 ) = exp [ ℏ − i H ( t − t 0 ) ] U ^ † a ^ U ^ \widehat{U}^{\dagger}\widehat{a}\widehat{U} U † a U
U ^ † a ^ U ^ = exp [ i ( t − t 0 ) ℏ H ^ ] a ^ exp [ − i ( t − t 0 ) ℏ H ^ ] = e λ A ^ B ^ e − λ A ^ \begin{align*}
\widehat{U}^{\dagger}\widehat{a}\widehat{U} & =\exp\left[\frac{i\left(t-t_{0}\right)}{\hbar}\widehat{H}\right]\widehat{a}\exp\left[-\frac{i\left(t-t_{0}\right)}{\hbar}\widehat{H}\right]\\
& =e^{\lambda\widehat{A}}\widehat{B}e^{-\lambda\widehat{A}}
\end{align*} U † a U = exp [ ℏ i ( t − t 0 ) H ] a exp [ − ℏ i ( t − t 0 ) H ] = e λ A B e − λ A Onde A ^ = H ^ \widehat{A}=\widehat{H} A = H B ^ = a ^ \widehat{B}=\widehat{a} B = a λ = i ( t − t 0 ) ℏ \lambda=\frac{i\left(t-t_{0}\right)}{\hbar} λ = ℏ i ( t − t 0 )
e λ A ^ B ^ e − λ A ^ = B ^ + λ [ A ^ , B ^ ] + λ 2 2 ! [ A ^ , [ A ^ , B ^ ] ] + λ 3 3 ! [ A ^ , [ A ^ , [ A ^ , B ^ ] ] ] + … e^{\lambda\widehat{A}}\widehat{B}e^{-\lambda\widehat{A}}=\widehat{B}+\lambda\left[\widehat{A},\widehat{B}\right]+\frac{\lambda^{2}}{2!}\left[\widehat{A},\left[\widehat{A},\widehat{B}\right]\right]+\frac{\lambda^{3}}{3!}\left[\widehat{A},\left[\widehat{A},\left[\widehat{A},\widehat{B}\right]\right]\right]+\dots e λ A B e − λ A = B + λ [ A , B ] + 2 ! λ 2 [ A , [ A , B ] ] + 3 ! λ 3 [ A , [ A , [ A , B ] ] ] + … E como:
a ^ † a ^ = ( m ω 2 ℏ x ^ − i 2 ℏ m ω p ^ ) ( m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ ) = ( m ω 2 ℏ x ^ 2 + i 2 ℏ x ^ p ^ − i 2 ℏ p ^ x ^ + 1 2 ℏ m ω p ^ 2 ) = m ω 2 ℏ x ^ 2 + i 2 ℏ ( x ^ p ^ − p ^ x ^ ) + 1 2 ℏ m ω p ^ 2 = m ω 2 ℏ x ^ 2 + i 2 ℏ [ x ^ , p ^ ] + 1 2 ℏ m ω p ^ 2 \begin{align*}
\widehat{a}^{\dagger}\widehat{a} & =\left(\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}-\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right)\left(\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right)\\
& =\left(\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{i}{2\hbar}\widehat{x}\widehat{p}-\frac{i}{2\hbar}\widehat{p}\widehat{x}+\frac{1}{2\hbar m\omega}\widehat{p}^{2}\right)\\
& =\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{i}{2\hbar}\left(\widehat{x}\widehat{p}-\widehat{p}\widehat{x}\right)+\frac{1}{2\hbar m\omega}\widehat{p}^{2}\\
& =\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{i}{2\hbar}\left[\widehat{x},\widehat{p}\right]+\frac{1}{2\hbar m\omega}\widehat{p}^{2}
\end{align*} a † a = ( 2ℏ mω x − 2ℏ mω i p ) ( 2ℏ mω x + 2ℏ mω i p ) = ( 2ℏ mω x 2 + 2ℏ i x p − 2ℏ i p x + 2ℏ mω 1 p 2 ) = 2ℏ mω x 2 + 2ℏ i ( x p − p x ) + 2ℏ mω 1 p 2 = 2ℏ mω x 2 + 2ℏ i [ x , p ] + 2ℏ mω 1 p 2 Lembrando que [ x ^ , p ^ ] = i ℏ \left[\widehat{x},\widehat{p}\right]=i\hbar [ x , p ] = i ℏ
a ^ † a ^ = m ω 2 ℏ x ^ 2 + 1 2 ℏ m ω p ^ 2 − 1 2 = 1 ℏ ω ( 1 2 m p ^ 2 + m ω 2 2 x ^ 2 − ℏ ω 2 ) \begin{align*}
\widehat{a}^{\dagger}\widehat{a} & =\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{1}{2\hbar m\omega}\widehat{p}^{2}-\frac{1}{2}\\
& =\frac{1}{\hbar\omega}\left(\frac{1}{2m}\widehat{p}^{2}+\frac{m\omega^{2}}{2}\widehat{x}^{2}-\frac{\hbar\omega}{2}\right)
\end{align*} a † a = 2ℏ mω x 2 + 2ℏ mω 1 p 2 − 2 1 = ℏ ω 1 ( 2 m 1 p 2 + 2 m ω 2 x 2 − 2 ℏ ω ) O Hamiltoniano para um oscilador harmônico pode ser escrito como:
H ^ = 1 2 m p ^ 2 + 1 2 m ω 2 x ^ 2 = ℏ ω ( a ^ † a ^ + 1 2 ) \begin{align*}
\widehat{H} & =\frac{1}{2m}\widehat{p}^{2}+\frac{1}{2}m\omega^{2}\widehat{x}^{2}\\
& =\hbar\omega\left(\widehat{a}^{\dagger}\widehat{a}+\frac{1}{2}\right)
\end{align*} H = 2 m 1 p 2 + 2 1 m ω 2 x 2 = ℏ ω ( a † a + 2 1 ) Definindo o operador número N ^ = a ^ † a ^ \widehat{N}=\widehat{a}^{\dagger}\widehat{a} N = a † a
H ^ = ℏ ω ( N ^ + 1 2 ) \widehat{H}=\hbar\omega\left(\widehat{N}+\frac{1}{2}\right) H = ℏ ω ( N + 2 1 ) E como e [ N ^ , a ^ ] = − a ^ \left[\widehat{N},\widehat{a}\right]=-\widehat{a} [ N , a ] = − a ω ℏ 2 \frac{\omega\hbar}{2} 2 ω ℏ
e λ A ^ B ^ e − λ A ^ = a ^ + λ ℏ ω [ N ^ , a ^ ] + ( λ ℏ ω ) 2 2 ! [ N ^ , [ N ^ , a ^ ] ] + ( λ ℏ ω ) 3 3 ! [ N ^ , [ N ^ , [ N ^ , a ^ ] ] ] + … = a ^ + ( − λ ℏ ω ) a ^ − ( λ ℏ ω ) 2 2 ! [ N ^ , a ^ ] + ( − λ ℏ ω ) 3 3 ! [ N ^ , [ N ^ , a ^ ] ] + … = a ^ + ( − λ ℏ ω ) a ^ + ( λ ℏ ω ) 2 2 ! a ^ − ( − λ ℏ ω ) 3 3 ! [ N ^ , a ^ ] + … = a ^ + ( − λ ℏ ω ) a ^ + ( λ ℏ ω ) 2 2 ! a ^ + ( − λ ℏ ω ) 3 3 ! a ^ + … = a ^ [ 1 + ( − λ ℏ ω ) + ( λ ℏ ω ) 2 2 ! + ( − λ ℏ ω ) 3 3 ! + … ] = a ^ e − λ ℏ ω \begin{align*}
e^{\lambda\widehat{A}}\widehat{B}e^{-\lambda\widehat{A}} & =\widehat{a}+\lambda\hbar\omega\left[\widehat{N},\widehat{a}\right]+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\left[\widehat{N},\left[\widehat{N},\widehat{a}\right]\right]+\frac{\left(\lambda\hbar\omega\right)^{3}}{3!}\left[\widehat{N},\left[\widehat{N},\left[\widehat{N},\widehat{a}\right]\right]\right]+\dots\\
& =\widehat{a}+\left(-\lambda\hbar\omega\right)\widehat{a}-\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\left[\widehat{N},\widehat{a}\right]+\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}\left[\widehat{N},\left[\widehat{N},\widehat{a}\right]\right]+\dots\\
& =\widehat{a}+\left(-\lambda\hbar\omega\right)\widehat{a}+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\widehat{a}-\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}\left[\widehat{N},\widehat{a}\right]+\dots\\
& =\widehat{a}+\left(-\lambda\hbar\omega\right)\widehat{a}+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\widehat{a}+\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}\widehat{a}+\dots\\
& =\widehat{a}\left[1+\left(-\lambda\hbar\omega\right)+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}+\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}+\dots\right]\\
& =\widehat{a}e^{-\lambda\hbar\omega}
\end{align*} e λ A B e − λ A = a + λ ℏ ω [ N , a ] + 2 ! ( λ ℏ ω ) 2 [ N , [ N , a ] ] + 3 ! ( λ ℏ ω ) 3 [ N , [ N , [ N , a ] ] ] + … = a + ( − λ ℏ ω ) a − 2 ! ( λ ℏ ω ) 2 [ N , a ] + 3 ! ( − λ ℏ ω ) 3 [ N , [ N , a ] ] + … = a + ( − λ ℏ ω ) a + 2 ! ( λ ℏ ω ) 2 a − 3 ! ( − λ ℏ ω ) 3 [ N , a ] + … = a + ( − λ ℏ ω ) a + 2 ! ( λ ℏ ω ) 2 a + 3 ! ( − λ ℏ ω ) 3 a + … = a [ 1 + ( − λ ℏ ω ) + 2 ! ( λ ℏ ω ) 2 + 3 ! ( − λ ℏ ω ) 3 + … ] = a e − λ ℏ ω E como λ = i ( t − t 0 ) ℏ \lambda=\frac{i\left(t-t_{0}\right)}{\hbar} λ = ℏ i ( t − t 0 )
U ^ † a ^ U ^ = a ^ e − i ω ( t − t 0 ) \widehat{U}^{\dagger}\widehat{a}\widehat{U}=\widehat{a}e^{-i\omega\left(t-t_{0}\right)} U † a U = a e − iω ( t − t 0 ) Aplicando o operador U ^ \widehat{U} U
a ^ = U ^ † a ^ U ^ e i ω ( t − t 0 ) U ^ a ^ = U ^ U ^ † a ^ U ^ e i ω ( t − t 0 ) = a ^ U ^ e i ω ( t − t 0 ) \begin{align*}
\widehat{a} & =\widehat{U}^{\dagger}\widehat{a}\widehat{U}e^{i\omega\left(t-t_{0}\right)}\\
\widehat{U}\widehat{a} & =\widehat{U}\widehat{U}^{\dagger}\widehat{a}\widehat{U}e^{i\omega\left(t-t_{0}\right)}\\
& =\widehat{a}\widehat{U}e^{i\omega\left(t-t_{0}\right)}
\end{align*} a U a = U † a U e iω ( t − t 0 ) = U U † a U e iω ( t − t 0 ) = a U e iω ( t − t 0 ) Mmanipulando a ^ ∣ α ⟩ = α ∣ α ⟩ \widehat{a}\left|\alpha\right\rangle =\alpha\left|\alpha\right\rangle a ∣ α ⟩ = α ∣ α ⟩
a ^ ∣ α ⟩ = α ∣ α ⟩ U ^ a ^ ∣ α ⟩ = α U ^ ∣ α ⟩ a ^ e i ω ( t − t 0 ) U ^ ∣ α ⟩ = α U ^ ∣ α ⟩ a ^ ( U ^ ∣ α ⟩ ) = ( α e i ω ( t − t 0 ) ) ( U ^ ∣ α ⟩ ) a ^ ∣ α ( t ) ⟩ = α ( t ) ∣ α ( t ) ⟩ \begin{align*}
\widehat{a}\left|\alpha\right\rangle & =\alpha\left|\alpha\right\rangle \\
\widehat{U}\widehat{a}\left|\alpha\right\rangle & =\alpha\widehat{U}\left|\alpha\right\rangle \\
\widehat{a}e^{i\omega\left(t-t_{0}\right)}\widehat{U}\left|\alpha\right\rangle & =\alpha\widehat{U}\left|\alpha\right\rangle \\
\widehat{a}\left(\widehat{U}\left|\alpha\right\rangle \right) & =\left(\alpha e^{i\omega\left(t-t_{0}\right)}\right)\left(\widehat{U}\left|\alpha\right\rangle \right)\\
\widehat{a}\left|\alpha\left(t\right)\right\rangle & =\alpha\left(t\right)\left|\alpha\left(t\right)\right\rangle
\end{align*} a ∣ α ⟩ U a ∣ α ⟩ a e iω ( t − t 0 ) U ∣ α ⟩ a ( U ∣ α ⟩ ) a ∣ α ( t ) ⟩ = α ∣ α ⟩ = α U ∣ α ⟩ = α U ∣ α ⟩ = ( α e iω ( t − t 0 ) ) ( U ∣ α ⟩ ) = α ( t ) ∣ α ( t ) ⟩ Isto significa que o estado coerente permanece um estado coerente durante a evolução do tempo com um autovalor dependente do tempo, que é a primeira coisa demonstração que estávamos buscando. Agora fazendo:
⟨ α ( t ) ∣ a ^ ∣ α ( t ) ⟩ = ⟨ α ∣ U ^ † a ^ U ^ ∣ α ⟩ = ⟨ α ∣ a ^ e − i ω ( t − t 0 ) ∣ α ⟩ = e − i ω ( t − t 0 ) ⟨ α ∣ a ^ ∣ α ⟩ = e − i ω ( t − t 0 ) α \begin{align}
\left\langle \alpha\left(t\right)\left|\widehat{a}\right|\alpha\left(t\right)\right\rangle & =\left\langle \alpha\left|\widehat{U}^{\dagger}\widehat{a}\widehat{U}\right|\alpha\right\rangle \\
& =\left\langle \alpha\left|\widehat{a}e^{-i\omega\left(t-t_{0}\right)}\right|\alpha\right\rangle \\
& =e^{-i\omega\left(t-t_{0}\right)}\left\langle \alpha\left|\widehat{a}\right|\alpha\right\rangle \\
& =e^{-i\omega\left(t-t_{0}\right)}\alpha
\end{align} ⟨ α ( t ) ∣ a ∣ α ( t ) ⟩ = ⟨ α ∣ ∣ U † a U ∣ ∣ α ⟩ = ⟨ α ∣ ∣ a e − iω ( t − t 0 ) ∣ ∣ α ⟩ = e − iω ( t − t 0 ) ⟨ α ∣ a ∣ α ⟩ = e − iω ( t − t 0 ) α E usando a definição de a ^ \widehat{a} a
⟨ α ( t ) ∣ a ^ ∣ α ( t ) ⟩ = ⟨ α ( t ) ∣ m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ ∣ α ( t ) ⟩ = m ω 2 ℏ ⟨ α ( t ) ∣ x ^ ∣ α ( t ) ⟩ + i 2 ℏ m ω ⟨ α ( t ) ∣ p ^ ∣ α ( t ) ⟩ = m ω 2 ℏ x ( t ) + i 2 ℏ m ω p ( t ) \begin{align}
\left\langle \alpha\left(t\right)\left|\widehat{a}\right|\alpha\left(t\right)\right\rangle & =\left\langle \alpha\left(t\right)\left|\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right|\alpha\left(t\right)\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}\left\langle \alpha\left(t\right)\left|\widehat{x}\right|\alpha\left(t\right)\right\rangle +\frac{i}{\sqrt{2\hbar m\omega}}\left\langle \alpha\left(t\right)\left|\widehat{p}\right|\alpha\left(t\right)\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}x\left(t\right)+\frac{i}{\sqrt{2\hbar m\omega}}p\left(t\right)
\end{align} ⟨ α ( t ) ∣ a ∣ α ( t ) ⟩ = ⟨ α ( t ) ∣ ∣ 2ℏ mω x + 2ℏ mω i p ∣ ∣ α ( t ) ⟩ = 2ℏ mω ⟨ α ( t ) ∣ x ∣ α ( t ) ⟩ + 2ℏ mω i ⟨ α ( t ) ∣ p ∣ α ( t ) ⟩ = 2ℏ mω x ( t ) + 2ℏ mω i p ( t ) Igualando então 37 38
α = e i ω ( t − t 0 ) ( m ω 2 ℏ x ( t ) + i 2 ℏ m ω p ( t ) ) \alpha=e^{i\omega\left(t-t_{0}\right)}\left(\sqrt{\frac{m\omega}{2\hbar}}x\left(t\right)+\frac{i}{\sqrt{2\hbar m\omega}}p\left(t\right)\right) α = e iω ( t − t 0 ) ( 2ℏ mω x ( t ) + 2ℏ mω i p ( t ) ) Para facilitar, vamos denotar τ = t − t 0 \tau=t-t_{0} τ = t − t 0 ( x 0 , p 0 ) \left(x_{0},p_{0}\right) ( x 0 , p 0 )
x ( τ ) ≡ x 0 cos ( ω τ ) + p 0 m ω sin ( ω τ ) p ( τ ) ≡ − m ω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) \begin{align*}
x\left(\tau\right) & \equiv x_{0}\cos\left(\omega\tau\right)+\frac{p_{0}}{m\omega}\sin\left(\omega\tau\right)\\
p\left(\tau\right) & \equiv-m\omega x_{0}\sin\left(\omega\tau\right)+p_{0}\cos\left(\omega\tau\right)
\end{align*} x ( τ ) p ( τ ) ≡ x 0 cos ( ω τ ) + mω p 0 sin ( ω τ ) ≡ − mω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) Pois:
α = e i ω τ [ m ω 2 ℏ ( x 0 cos ( ω τ ) + p 0 m ω sin ( ω τ ) ) + i 2 ℏ m ω ( − m ω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) ) ] = e i ω τ [ m ω 2 ℏ x 0 cos ( ω τ ) + p 0 2 m ω sin ( ω τ ) − i m ω 2 ℏ x 0 sin ( ω τ ) + p 0 2 ℏ m ω cos ( ω τ ) ] = e i ω τ [ m ω 2 ℏ x 0 ( e i ω τ + e − i ω τ 2 ) + p 0 2 m ω ( e i ω τ − e − i ω τ 2 i ) − i m ω 2 ℏ x 0 ( e i ω τ − e − i ω τ 2 i ) + p 0 2 ℏ m ω ( e i ω τ + e − i ω τ 2 ) ] = [ m ω 2 ℏ x 0 ( e 2 i ω τ + 1 2 ) − i p 0 2 m ω ( e 2 i ω τ − 1 2 ) − m ω 2 ℏ x 0 ( e 2 i ω τ − 1 2 ) + p 0 2 ℏ m ω ( e 2 i ω τ + 1 2 ) ] = m ω 2 ℏ x 0 + i p 0 2 m ω \begin{align*}
\alpha & =e^{i\omega\tau}\left[\sqrt{\frac{m\omega}{2\hbar}}\left(x_{0}\cos\left(\omega\tau\right)+\frac{p_{0}}{m\omega}\sin\left(\omega\tau\right)\right)+\frac{i}{\sqrt{2\hbar m\omega}}\left(-m\omega x_{0}\sin\left(\omega\tau\right)+p_{0}\cos\left(\omega\tau\right)\right)\right]\\
& =e^{i\omega\tau}\left[\sqrt{\frac{m\omega}{2\hbar}}x_{0}\cos\left(\omega\tau\right)+\frac{p_{0}}{\sqrt{2 m\omega}}\sin\left(\omega\tau\right)-i\sqrt{\frac{m\omega}{2\hbar}}x_{0}\sin\left(\omega\tau\right)+\frac{p_{0}}{\sqrt{2\hbar m\omega}}\cos\left(\omega\tau\right)\right]\\
& =e^{i\omega\tau}\left[\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{i\omega\tau}+e^{-i\omega\tau}}{2}\right)+\frac{p_{0}}{\sqrt{2 m\omega}}\left(\frac{e^{i\omega\tau}-e^{-i\omega\tau}}{2i}\right)-i\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{i\omega\tau}-e^{-i\omega\tau}}{2i}\right)+\frac{p_{0}}{\sqrt{2\hbar m\omega}}\left(\frac{e^{i\omega\tau}+e^{-i\omega\tau}}{2}\right)\right]\\
& =\left[\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{2i\omega\tau}+1}{2}\right)-\frac{ip_{0}}{\sqrt{2 m\omega}}\left(\frac{e^{2i\omega\tau}-1}{2}\right)-\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{2i\omega\tau}-1}{2}\right)+\frac{p_{0}}{\sqrt{2\hbar m\omega}}\left(\frac{e^{2i\omega\tau}+1}{2}\right)\right]\\
& =\sqrt{\frac{m\omega}{2\hbar}}x_{0}+\frac{ip_{0}}{\sqrt{2 m\omega}}
\end{align*} α = e iω τ [ 2ℏ mω ( x 0 cos ( ω τ ) + mω p 0 sin ( ω τ ) ) + 2ℏ mω i ( − mω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) ) ] = e iω τ [ 2ℏ mω x 0 cos ( ω τ ) + 2 mω p 0 sin ( ω τ ) − i 2ℏ mω x 0 sin ( ω τ ) + 2ℏ mω p 0 cos ( ω τ ) ] = e iω τ [ 2ℏ mω x 0 ( 2 e iω τ + e − iω τ ) + 2 mω p 0 ( 2 i e iω τ − e − iω τ ) − i 2ℏ mω x 0 ( 2 i e iω τ − e − iω τ ) + 2ℏ mω p 0 ( 2 e iω τ + e − iω τ ) ] = [ 2ℏ mω x 0 ( 2 e 2 iω τ + 1 ) − 2 mω i p 0 ( 2 e 2 iω τ − 1 ) − 2ℏ mω x 0 ( 2 e 2 iω τ − 1 ) + 2ℏ mω p 0 ( 2 e 2 iω τ + 1 ) ] = 2ℏ mω x 0 + 2 mω i p 0 Que é o resultado obtido inicialmente na equação 26
Problem 3 ¶ Consider a harmonic oscillator with frequency ω \omega ω m m m
H = ( N + 1 2 ) ℏ ω H=\left(N+\frac{1}{2}\right)\hbar\omega H = ( N + 2 1 ) ℏ ω where N ≡ a † a N\equiv a^{\dagger}a N ≡ a † a with N ∣ n ⟩ = n ∣ n ⟩ N\left|n\right\rangle =n\left|n\right\rangle N ∣ n ⟩ = n ∣ n ⟩ and we recall that the lowering and raising operators satisfy
a ∣ n ⟩ = n ∣ n − 1 ⟩ a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ \begin{align*}
a\left|n\right\rangle & =\sqrt{n}\left|n-1\right\rangle \\
a^{\dagger}\left|n\right\rangle & =\sqrt{n+1}\left|n+1\right\rangle
\end{align*} a ∣ n ⟩ a † ∣ n ⟩ = n ∣ n − 1 ⟩ = n + 1 ∣ n + 1 ⟩ Assuming that at the timet = 0 t=0 t = 0 ∣ n ⟩ \left|n\right\rangle ∣ n ⟩
a ∣ z ⟩ = z ∣ z ⟩ a\left|z\right\rangle =z\left|z\right\rangle a ∣ z ⟩ = z ∣ z ⟩ answer:
a) What is the value of ⟨ z ∣ N ∣ z ⟩ \left\langle z\left|N\right|z\right\rangle ⟨ z ∣ N ∣ z ⟩ z = 1 2 exp ( i π / 4 ) z=\frac{1}{2}\exp\left(i\pi/4\right) z = 2 1 exp ( iπ /4 ) ∣ z ⟩ \left|z\right\rangle ∣ z ⟩
b) Assuming that at t = 0 t=0 t = 0 ∣ 0 ⟩ \left|0\right\rangle ∣ 0 ⟩
at time t = 1 / 10 s t=1/10s t = 1/10 s ω = 5 π s − 1 \omega=5\pi s^{-1} ω = 5 π s − 1
c) What is the value of c n c_{n} c n n n n z z z ∣ z ⟩ = ∑ n = 0 + ∞ c n ∣ n ⟩ \left|z\right\rangle =\sum^{+\infty}_{n=0}c_{n}\left|n\right\rangle ∣ z ⟩ = ∑ n = 0 + ∞ c n ∣ n ⟩ (expanded in the eigenstates ∣ n ⟩ \left|n\right\rangle ∣ n ⟩ N N N
(Remember that e x = ∑ n = 0 + ∞ x n / n ! e^{x}=\sum^{+\infty}_{n=0}x^{n}/n! e x = ∑ n = 0 + ∞ x n / n !
d) Use the result of the previous item and calculate the numerical value of ∣ ⟨ z ′ ∣ z ⟩ ∣ 2 \left|\left\langle z'|z\right\rangle \right|^{2} ∣ ⟨ z ′ ∣ z ⟩ ∣ 2 z = 1 2 exp ( i π / 4 ) z=\frac{1}{2}\exp\left(i\pi/4\right) z = 2 1 exp ( iπ /4 )
and z ′ = 1 5 exp ( i π / 4 ) z'=\frac{1}{5}\exp\left(i\pi/4\right) z ′ = 5 1 exp ( iπ /4 )
Temos então o seguinte hamiltoniano de um oscilador harmônico com frequência ω \omega ω
H = ( N + 1 2 ) ℏ ω H=\left(N+\frac{1}{2}\right)\hbar\omega H = ( N + 2 1 ) ℏ ω Onde N ≡ a † a N\equiv a^{\dagger}a N ≡ a † a N ∣ n ⟩ = n ∣ n ⟩ N\left|n\right\rangle =n\left|n\right\rangle N ∣ n ⟩ = n ∣ n ⟩ a a a a † a^{\dagger} a †
a ∣ n ⟩ = n ∣ n − 1 ⟩ a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ \begin{align*}
a\left|n\right\rangle & =\sqrt{n}\left|n-1\right\rangle \\
a^{\dagger}\left|n\right\rangle & =\sqrt{n+1}\left|n+1\right\rangle
\end{align*} a ∣ n ⟩ a † ∣ n ⟩ = n ∣ n − 1 ⟩ = n + 1 ∣ n + 1 ⟩ Assumimos também que no tempo t = 0 t=0 t = 0 ∣ z ⟩ \left|z\right\rangle ∣ z ⟩
a ∣ z ⟩ = z ∣ z ⟩ a\left|z\right\rangle =z\left|z\right\rangle a ∣ z ⟩ = z ∣ z ⟩ Queremos então:
a) O valor ⟨ z ∣ N ∣ z ⟩ \left\langle z\left|N\right|z\right\rangle ⟨ z ∣ N ∣ z ⟩ z = 1 2 exp ( i π / 4 ) z=\frac{1}{2}\exp\left(i\pi/4\right) z = 2 1 exp ( iπ /4 ) ∣ z ⟩ \left|z\right\rangle ∣ z ⟩ N N N
⟨ z ∣ N ∣ z ⟩ = ⟨ z ∣ a † a ∣ z ⟩ = ( a ∣ z ⟩ ) † a ∣ z ⟩ = ( z ∣ z ⟩ ) † z ∣ z ⟩ = ⟨ z ∣ z ∗ z ∣ z ⟩ = ∣ z ∣ 2 ⟨ z ∣ z ⟩ = ∣ z ∣ 2 \begin{align*}
\left\langle z\left|N\right|z\right\rangle & =\left\langle z\left|a^{\dagger}a\right|z\right\rangle \\
& =\left(a\left|z\right\rangle \right)^{\dagger}a\left|z\right\rangle \\
& =\left(z\left|z\right\rangle \right)^{\dagger}z\left|z\right\rangle \\
& =\left\langle z\left|z^{*}z\right|z\right\rangle \\
& =\left|z\right|^{2}\left\langle z|z\right\rangle \\
& =\left|z\right|^{2}
\end{align*} ⟨ z ∣ N ∣ z ⟩ = ⟨ z ∣ ∣ a † a ∣ ∣ z ⟩ = ( a ∣ z ⟩ ) † a ∣ z ⟩ = ( z ∣ z ⟩ ) † z ∣ z ⟩ = ⟨ z ∣ z ∗ z ∣ z ⟩ = ∣ z ∣ 2 ⟨ z ∣ z ⟩ = ∣ z ∣ 2 Então sendo z = 1 2 exp ( i π / 4 ) z=\frac{1}{2}\exp\left(i\pi/4\right) z = 2 1 exp ( iπ /4 )
⟨ z ∣ N ∣ z ⟩ = 1 2 exp ( i π / 4 ) 1 2 exp ( − i π / 4 ) \begin{align*}
\left\langle z\left|N\right|z\right\rangle & =\frac{1}{2}\exp\left(i\pi/4\right)\frac{1}{2}\exp\left(-i\pi/4\right)
\end{align*} ⟨ z ∣ N ∣ z ⟩ = 2 1 exp ( iπ /4 ) 2 1 exp ( − iπ /4 ) Logo temos:
⟨ z ∣ N ∣ z ⟩ = 1 4 \left\langle z\left|N\right|z\right\rangle =\frac{1}{4} ⟨ z ∣ N ∣ z ⟩ = 4 1 b) Assumindo que em t = 0 t=0 t = 0 ∣ 0 ⟩ \left|0\right\rangle ∣ 0 ⟩ t = 1 / 10 t=1/10 t = 1/10 ω = 5 π s − 1 \omega=5\pi s^{-1} ω = 5 π s − 1 U ^ = exp [ − i H t ℏ ] \widehat{U}=\exp\left[\frac{-iHt}{\hbar}\right] U = exp [ ℏ − i H t ]
∣ ψ ( t ) ⟩ = U ^ ∣ 0 ⟩ = exp [ − i H t ℏ ] ∣ 0 ⟩ = exp [ − i ( N + 1 2 ) ℏ ω t ℏ ] ∣ 0 ⟩ = exp [ − i 1 2 ω t ] exp [ − i N ω t ℏ ] ∣ 0 ⟩ = exp [ − i 1 2 ω t ] exp [ − i ω t ℏ 0 ] ∣ 0 ⟩ = exp [ − i 1 2 ω t ] ∣ 0 ⟩ = [ cos ( − ω t 2 ) + sin ( − ω t 2 ) ] ∣ 0 ⟩ = [ cos ( ω t 2 ) − sin ( ω t 2 ) ] ∣ 0 ⟩ \begin{align*}
\left|\psi\left(t\right)\right\rangle & =\widehat{U}\left|0\right\rangle \\
& =\exp\left[\frac{-iHt}{\hbar}\right]\left|0\right\rangle \\
& =\exp\left[\frac{-i\left(N+\frac{1}{2}\right)\hbar\omega t}{\hbar}\right]\left|0\right\rangle \\
& =\exp\left[-i\frac{1}{2}\omega t\right]\exp\left[\frac{-iN\omega t}{\hbar}\right]\left|0\right\rangle \\
& =\exp\left[-i\frac{1}{2}\omega t\right]\exp\left[\frac{-i\omega t}{\hbar}0\right]\left|0\right\rangle \\
& =\exp\left[-i\frac{1}{2}\omega t\right]\left|0\right\rangle \\
& =\left[\cos\left(-\frac{\omega t}{2}\right)+\sin\left(-\frac{\omega t}{2}\right)\right]\left|0\right\rangle \\
& =\left[\cos\left(\frac{\omega t}{2}\right)-\sin\left(\frac{\omega t}{2}\right)\right]\left|0\right\rangle
\end{align*} ∣ ψ ( t ) ⟩ = U ∣ 0 ⟩ = exp [ ℏ − i H t ] ∣ 0 ⟩ = exp [ ℏ − i ( N + 2 1 ) ℏ ω t ] ∣ 0 ⟩ = exp [ − i 2 1 ω t ] exp [ ℏ − i N ω t ] ∣ 0 ⟩ = exp [ − i 2 1 ω t ] exp [ ℏ − iω t 0 ] ∣ 0 ⟩ = exp [ − i 2 1 ω t ] ∣ 0 ⟩ = [ cos ( − 2 ω t ) + sin ( − 2 ω t ) ] ∣ 0 ⟩ = [ cos ( 2 ω t ) − sin ( 2 ω t ) ] ∣ 0 ⟩ Onde usamos que F ( A ) ∣ a ⟩ = F ( a ) ∣ a ⟩ F\left(A\right)\left|a\right\rangle =F\left(a\right)\left|a\right\rangle F ( A ) ∣ a ⟩ = F ( a ) ∣ a ⟩
∣ ψ ( t ) ⟩ = [ cos ( 5 π 10 ⋅ 2 ) − i sin ( 5 π 10 ⋅ 2 ) ] ∣ 0 ⟩ \begin{align*}
\left|\psi\left(t\right)\right\rangle & =\left[\cos\left(\frac{5\pi}{10\cdot2}\right)-i\sin\left(\frac{5\pi}{10\cdot2}\right)\right]\left|0\right\rangle
\end{align*} ∣ ψ ( t ) ⟩ = [ cos ( 10 ⋅ 2 5 π ) − i sin ( 10 ⋅ 2 5 π ) ] ∣ 0 ⟩ O estado que buscamos é então:
∣ ψ ( t ) ⟩ = ( 1 − i 2 ) ∣ 0 ⟩ \begin{align*}
\left|\psi\left(t\right)\right\rangle & =\left(\frac{1-i}{\sqrt{2}}\right)\left|0\right\rangle
\end{align*} ∣ ψ ( t ) ⟩ = ( 2 1 − i ) ∣ 0 ⟩ c) Queremos agora o valor de c n c_{n} c n n n n z z z N N N ∣ z ⟩ = ∑ n = 0 + ∞ c n ∣ n ⟩ \left|z\right\rangle =\sum^{+\infty}_{n=0}c_{n}\left|n\right\rangle ∣ z ⟩ = ∑ n = 0 + ∞ c n ∣ n ⟩ e x = ∑ n = 0 + ∞ x n / n ! e^{x}=\sum^{+\infty}_{n=0}x^{n}/n! e x = ∑ n = 0 + ∞ x n / n !
∣ z ⟩ = ∑ n = 0 + ∞ ⟨ n ∣ z ⟩ ∣ n ⟩ \begin{align*}
\left|z\right\rangle & =\sum^{+\infty}_{n=0}\left\langle n|z\right\rangle \left|n\right\rangle
\end{align*} ∣ z ⟩ = n = 0 ∑ + ∞ ⟨ n ∣ z ⟩ ∣ n ⟩ E partindo de a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ a^{\dagger}\left|n\right\rangle =\sqrt{n+1}\left|n+1\right\rangle a † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩
a † ∣ 0 ⟩ = ∣ 1 ⟩ → ∣ 1 ⟩ = a † 1 ∣ 0 ⟩ a † ∣ 1 ⟩ = 2 ∣ 2 ⟩ → ∣ 2 ⟩ = a † 2 ∣ 1 ⟩ = ( a † ) 2 2 ⋅ 1 ∣ 0 ⟩ a † ∣ 2 ⟩ = 3 ∣ 3 ⟩ → ∣ 3 ⟩ = a † 3 ∣ 2 ⟩ = ( a † ) 3 3 ⋅ 2 ⋅ 1 ∣ 0 ⟩ ⋮ ∣ n ⟩ = ( a † ) n n ! ∣ 0 ⟩ \begin{align*}
a^{\dagger}\left|0\right\rangle & =\left|1\right\rangle & \rightarrow\left|1\right\rangle & =\frac{a^{\dagger}}{\sqrt{1}}\left|0\right\rangle \\
a^{\dagger}\left|1\right\rangle & =\sqrt{2}\left|2\right\rangle & \rightarrow\left|2\right\rangle & =\frac{a^{\dagger}}{\sqrt{2}}\left|1\right\rangle =\frac{\left(a^{\dagger}\right)^{2}}{\sqrt{2\cdot1}}\left|0\right\rangle \\
a^{\dagger}\left|2\right\rangle & =\sqrt{3}\left|3\right\rangle & \rightarrow\left|3\right\rangle & =\frac{a^{\dagger}}{\sqrt{3}}\left|2\right\rangle =\frac{\left(a^{\dagger}\right)^{3}}{\sqrt{3\cdot2\cdot1}}\left|0\right\rangle \\
& & \vdots\\
& & \left|n\right\rangle & =\frac{\left(a^{\dagger}\right)^{n}}{\sqrt{n!}}\left|0\right\rangle
\end{align*} a † ∣ 0 ⟩ a † ∣ 1 ⟩ a † ∣ 2 ⟩ = ∣ 1 ⟩ = 2 ∣ 2 ⟩ = 3 ∣ 3 ⟩ → ∣ 1 ⟩ → ∣ 2 ⟩ → ∣ 3 ⟩ ⋮ ∣ n ⟩ = 1 a † ∣ 0 ⟩ = 2 a † ∣ 1 ⟩ = 2 ⋅ 1 ( a † ) 2 ∣ 0 ⟩ = 3 a † ∣ 2 ⟩ = 3 ⋅ 2 ⋅ 1 ( a † ) 3 ∣ 0 ⟩ = n ! ( a † ) n ∣ 0 ⟩ Temos então:
∣ z ⟩ = ∑ n = 0 + ∞ ( ( a † ) n n ! ∣ 0 ⟩ ) † ∣ z ⟩ ∣ n ⟩ = ∑ n = 0 + ∞ ⟨ 0 ∣ ( a ) n n ! ∣ z ⟩ ∣ n ⟩ = ∑ n = 0 + ∞ 1 n ! ⟨ 0 ∣ a n ∣ z ⟩ ∣ n ⟩ = ∑ n = 0 + ∞ z n n ! ⟨ 0 ∣ z ⟩ ∣ n ⟩ \begin{align*}
\left|z\right\rangle & =\sum^{+\infty}_{n=0}\left(\frac{\left(a^{\dagger}\right)^{n}}{\sqrt{n!}}\left|0\right\rangle \right)^{\dagger}\left|z\right\rangle \left|n\right\rangle \\
& =\sum^{+\infty}_{n=0}\left\langle 0\right|\frac{\left(a\right)^{n}}{\sqrt{n!}}\left|z\right\rangle \left|n\right\rangle \\
& =\sum^{+\infty}_{n=0}\frac{1}{\sqrt{n!}}\left\langle 0\right|a^{n}\left|z\right\rangle \left|n\right\rangle \\
& =\sum^{+\infty}_{n=0}\frac{z^{n}}{\sqrt{n!}}\left\langle 0|z\right\rangle \left|n\right\rangle
\end{align*} ∣ z ⟩ = n = 0 ∑ + ∞ ( n ! ( a † ) n ∣ 0 ⟩ ) † ∣ z ⟩ ∣ n ⟩ = n = 0 ∑ + ∞ ⟨ 0 ∣ n ! ( a ) n ∣ z ⟩ ∣ n ⟩ = n = 0 ∑ + ∞ n ! 1 ⟨ 0 ∣ a n ∣ z ⟩ ∣ n ⟩ = n = 0 ∑ + ∞ n ! z n ⟨ 0∣ z ⟩ ∣ n ⟩ Onde usamos que a ∣ z ⟩ = z ∣ z ⟩ a\left|z\right\rangle =z\left|z\right\rangle a ∣ z ⟩ = z ∣ z ⟩ F ( A ) ∣ a ⟩ = F ( a ) ∣ a ⟩ F\left(A\right)\left|a\right\rangle =F\left(a\right)\left|a\right\rangle F ( A ) ∣ a ⟩ = F ( a ) ∣ a ⟩ c n = ⟨ n ∣ z ⟩ → c 0 = ⟨ 0 ∣ z ⟩ c_{n}=\left\langle n|z\right\rangle \rightarrow c_{0}=\left\langle 0|z\right\rangle c n = ⟨ n ∣ z ⟩ → c 0 = ⟨ 0∣ z ⟩
c n = z n n ! c 0 c_{n}=\frac{z^{n}}{\sqrt{n!}}c_{0} c n = n ! z n c 0 Precisamos então idetnficar c 0 c_{0} c 0
⟨ z ∣ z ⟩ = ( ∑ m = 0 + ∞ ⟨ m ∣ c m ∗ ) ( ∑ n = 0 + ∞ c n ∣ n ⟩ ) 1 = ∑ m , n = 0 + ∞ c m ∗ c n ⟨ m ∣ n ⟩ = ∑ m , n = 0 + ∞ c m ∗ c n δ m n = ∑ n = 0 + ∞ ∣ c n ∣ 2 \begin{align*}
\left\langle z|z\right\rangle & =\left(\sum^{+\infty}_{m=0}\left\langle m\right|c^{*}_{m}\right)\left(\sum^{+\infty}_{n=0}c_{n}\left|n\right\rangle \right)\\
1 & =\sum^{+\infty}_{m,n=0}c^{*}_{m}c_{n}\left\langle m|n\right\rangle \\
& =\sum^{+\infty}_{m,n=0}c^{*}_{m}c_{n}\delta_{mn}\\
& =\sum^{+\infty}_{n=0}\left|c_{n}\right|^{2}
\end{align*} ⟨ z ∣ z ⟩ 1 = ( m = 0 ∑ + ∞ ⟨ m ∣ c m ∗ ) ( n = 0 ∑ + ∞ c n ∣ n ⟩ ) = m , n = 0 ∑ + ∞ c m ∗ c n ⟨ m ∣ n ⟩ = m , n = 0 ∑ + ∞ c m ∗ c n δ mn = n = 0 ∑ + ∞ ∣ c n ∣ 2 Substituindo então:
1 = ∑ n = 0 + ∞ ∣ z n n ! c 0 ∣ 2 = ∣ c 0 ∣ 2 ∑ n = 0 + ∞ ( ∣ z ∣ ) n n ! = ∣ c 0 ∣ 2 ∑ n = 0 + ∞ ( ∣ z ∣ 2 ) n n ! = ∣ c 0 ∣ 2 e ∣ z ∣ 2 \begin{align*}
1 & =\sum^{+\infty}_{n=0}\left|\frac{z^{n}}{\sqrt{n!}}c_{0}\right|^{2}\\
& =\left|c_{0}\right|^{2}\sum^{+\infty}_{n=0}\frac{\left(\left|z\right|\right)^{n}}{n!}\\
& =\left|c_{0}\right|^{2}\sum^{+\infty}_{n=0}\frac{\left(\left|z\right|^{2}\right)^{n}}{n!}\\
& =\left|c_{0}\right|^{2}e^{\left|z\right|^{2}}
\end{align*} 1 = n = 0 ∑ + ∞ ∣ ∣ n ! z n c 0 ∣ ∣ 2 = ∣ c 0 ∣ 2 n = 0 ∑ + ∞ n ! ( ∣ z ∣ ) n = ∣ c 0 ∣ 2 n = 0 ∑ + ∞ n ! ( ∣ z ∣ 2 ) n = ∣ c 0 ∣ 2 e ∣ z ∣ 2 Onde usamos que e x = ∑ n = 0 + ∞ x n / n ! e^{x}=\sum^{+\infty}_{n=0}x^{n}/n! e x = ∑ n = 0 + ∞ x n / n !
∣ c 0 ∣ = e − ∣ z ∣ 2 / 2 \left|c_{0}\right|=e^{-\left|z\right|^{2}/2} ∣ c 0 ∣ = e − ∣ z ∣ 2 /2 Assim o valor que buscamos para c n c_{n} c n
c n = z n n ! e − ∣ z ∣ 2 / 2 c_{n}=\frac{z^{n}}{\sqrt{n!}}e^{-\left|z\right|^{2}/2} c n = n ! z n e − ∣ z ∣ 2 /2 Uma vez que e − ∣ z ∣ 2 / 2 e^{-\left|z\right|^{2}/2} e − ∣ z ∣ 2 /2
d)
Usando o resultado do item anterior, vamos calcular ∣ ⟨ z ′ ∣ z ⟩ ∣ 2 \left|\left\langle z'|z\right\rangle \right|^{2} ∣ ⟨ z ′ ∣ z ⟩ ∣ 2 z = 1 2 e i π / 4 z=\frac{1}{2}e^{i\pi/4} z = 2 1 e iπ /4 z ′ = 1 4 e i π / 4 z'=\frac{1}{4}e^{i\pi/4} z ′ = 4 1 e iπ /4
⟨ z ′ ∣ z ⟩ = ( ∑ n ′ = 0 + ∞ ⟨ n ′ ∣ c n ′ ) ( ∑ n = 0 + ∞ c n ∣ n ⟩ ) = ∑ n , n ′ = 0 + ∞ c n ′ ∗ c n ⟨ n ′ n ⟩ = ∑ n , n ′ = 0 + ∞ ( ( z ′ n ′ ) ∗ n ′ ! e − ∣ z ′ ∣ 2 / 2 ) ( z n n ! e − ∣ z ∣ 2 / 2 ) δ n ′ n = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) / 2 ∑ n = 0 + ∞ ( ( z ′ n ′ ) ∗ n ! ) ( z n n ! ) = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) / 2 ∑ n = 0 + ∞ ( ( z ′ ∗ z ) n n ! ) = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) / 2 ∑ n = 0 + ∞ ( ( 1 4 e − i π / 4 1 2 e i π / 4 ) n n ! ) = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) / 2 ∑ n = 0 + ∞ ( ( 1 / 8 ) n n ! ) \begin{align*}
\left\langle z'|z\right\rangle & =\left(\sum^{+\infty}_{n'=0}\left\langle n'\right|c_{n'}\right)\left(\sum^{+\infty}_{n=0}c_{n}\left|n\right\rangle \right)\\
& =\sum^{+\infty}_{n,n'=0}c^{*}_{n'}c_{n}\left\langle n'n\right\rangle \\
& =\sum^{+\infty}_{n,n'=0}\left(\frac{\left(z'^{n'}\right)^{*}}{\sqrt{n'!}}e^{-\left|z'\right|^{2}/2}\right)\left(\frac{z^{n}}{\sqrt{n!}}e^{-\left|z\right|^{2}/2}\right)\delta_{n'n}\\
& =e^{-\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)/2}\sum^{+\infty}_{n=0}\left(\frac{\left(z'^{n'}\right)^{*}}{\sqrt{n!}}\right)\left(\frac{z^{n}}{\sqrt{n!}}\right)\\
& =e^{-\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)/2}\sum^{+\infty}_{n=0}\left(\frac{\left(z'^{*}z\right){}^{n}}{n!}\right)\\
& =e^{-\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)/2}\sum^{+\infty}_{n=0}\left(\frac{\left(\frac{1}{4}e^{-i\pi/4}\frac{1}{2}e^{i\pi/4}\right){}^{n}}{n!}\right)\\
& =e^{-\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)/2}\sum^{+\infty}_{n=0}\left(\frac{\left(1/8\right){}^{n}}{n!}\right)
\end{align*} ⟨ z ′ ∣ z ⟩ = ( n ′ = 0 ∑ + ∞ ⟨ n ′ ∣ c n ′ ) ( n = 0 ∑ + ∞ c n ∣ n ⟩ ) = n , n ′ = 0 ∑ + ∞ c n ′ ∗ c n ⟨ n ′ n ⟩ = n , n ′ = 0 ∑ + ∞ ⎝ ⎛ n ′ ! ( z ′ n ′ ) ∗ e − ∣ z ′ ∣ 2 /2 ⎠ ⎞ ( n ! z n e − ∣ z ∣ 2 /2 ) δ n ′ n = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) /2 n = 0 ∑ + ∞ ⎝ ⎛ n ! ( z ′ n ′ ) ∗ ⎠ ⎞ ( n ! z n ) = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) /2 n = 0 ∑ + ∞ ( n ! ( z ′ ∗ z ) n ) = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) /2 n = 0 ∑ + ∞ ( n ! ( 4 1 e − iπ /4 2 1 e iπ /4 ) n ) = e − ( ∣ z ∣ 2 + ∣ z ′ ∣ 2 ) /2 n = 0 ∑ + ∞ ( n ! ( 1/8 ) n ) Novamente usando e x = ∑ n = 0 + ∞ x n / n ! e^{x}=\sum^{+\infty}_{n=0}x^{n}/n! e x = ∑ n = 0 + ∞ x n / n ! ∣ z ∣ 2 = 1 4 , ∣ z ′ ∣ 2 = 1 16 \left|z\right|^{2}=\frac{1}{4},\left|z'\right|^{2}=\frac{1}{16} ∣ z ∣ 2 = 4 1 , ∣ z ′ ∣ 2 = 16 1
⟨ z ′ ∣ z ⟩ = exp [ − ( 1 4 + 1 16 ) 1 2 ] exp [ 1 8 ] = exp [ − ( 4 32 + 1 32 ) + 4 32 ] = exp [ − 1 16 ] \begin{align*}
\left\langle z'|z\right\rangle & =\exp\left[-\left(\frac{1}{4}+\frac{1}{16}\right)\frac{1}{2}\right]\exp\left[\frac{1}{8}\right]\\
& =\exp\left[-\left(\frac{4}{32}+\frac{1}{32}\right)+\frac{4}{32}\right]\\
& =\exp\left[-\frac{1}{16}\right]
\end{align*} ⟨ z ′ ∣ z ⟩ = exp [ − ( 4 1 + 16 1 ) 2 1 ] exp [ 8 1 ] = exp [ − ( 32 4 + 32 1 ) + 32 4 ] = exp [ − 16 1 ] Logo:
⟨ z ′ ∣ z ⟩ = exp [ − ( 1 4 + 1 16 ) 1 2 ] exp [ 1 8 ] = exp [ − ( 4 32 + 1 32 ) + 4 32 ] = exp [ − 1 32 ] \begin{align*}
\left\langle z'|z\right\rangle & =\exp\left[-\left(\frac{1}{4}+\frac{1}{16}\right)\frac{1}{2}\right]\exp\left[\frac{1}{8}\right]\\
& =\exp\left[-\left(\frac{4}{32}+\frac{1}{32}\right)+\frac{4}{32}\right]\\
& =\exp\left[-\frac{1}{32}\right]
\end{align*} ⟨ z ′ ∣ z ⟩ = exp [ − ( 4 1 + 16 1 ) 2 1 ] exp [ 8 1 ] = exp [ − ( 32 4 + 32 1 ) + 32 4 ] = exp [ − 32 1 ] Fazendo o quadrado temos
∣ ⟨ z ′ ∣ z ⟩ ∣ 2 = exp [ − 1 16 ] \left|\left\langle z'|z\right\rangle \right|^{2}=\exp\left[-\frac{1}{16}\right] ∣ ⟨ z ′ ∣ z ⟩ ∣ 2 = exp [ − 16 1 ] Ou em valores numéricos ∣ ⟨ z ′ ∣ z ⟩ ∣ 2 = 0.94 \left|\left\langle z'|z\right\rangle \right|^{2}=0.94 ∣ ⟨ z ′ ∣ z ⟩ ∣ 2 = 0.94
Lima, P. C. de. (2013). Equações Diferenciais Parciais I . http://150.164.25.15/ lima/apostilas/EDP1-aulas-22-8-13.pdf