Problem set 1
These are my solutions to the problem sets assigned to me during the graduate-level quantum mechanics course.
Problem 1 ¶ ∣ φ n ⟩ \left|\varphi_{n}\right\rangle ∣ φ n ⟩ H H H H H H ∣ φ n ⟩ \left|\varphi_{n}\right\rangle ∣ φ n ⟩ U ( m , n ) U\left(m,n\right) U ( m , n )
U ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣ U\left(m,n\right)=\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right| U ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣
a) Calculate the adjoint U † ( m , n ) U^{\dagger}\left(m,n\right) U † ( m , n ) U ( m , n ) U\left(m,n\right) U ( m , n )
b) Calculate the comutator [ H , U ( m , n ) ] \left[H,U\left(m,n\right)\right] [ H , U ( m , n ) ]
c) Prove the relation U ( m , n ) U † ( p , q ) = δ n q U ( m , p ) U\left(m,n\right)U^{\dagger}\left(p,q\right)=\delta_{nq}U\left(m,p\right) U ( m , n ) U † ( p , q ) = δ n q U ( m , p )
d) Calculate T r ( U ) Tr\left(U\right) T r ( U ) U ( m , n ) U\left(m,n\right) U ( m , n )
e) Let A A A A m n = ⟨ φ m ∣ A ∣ φ n ⟩ A_{mn}=\left\langle \varphi_{m}\left|A\right|\varphi_{n}\right\rangle A mn = ⟨ φ m ∣ A ∣ φ n ⟩ A = ∑ m , n A m n U ( m , n ) A=\sum_{m,n}A_{mn}U\left(m,n\right) A = ∑ m , n A mn U ( m , n )
f) Show that A p q = T r ( A U † ( p , q ) ) A_{pq}=Tr\left(AU^{\dagger}\left(p,q\right)\right) A pq = T r ( A U † ( p , q ) )
Escrevendo então o operador hermitiano como:
H = ∑ k N k ∣ φ k ⟩ ⟨ φ k ∣ H=\sum^{N}_{k}k\left|\varphi_{k}\right\rangle \left\langle \varphi_{k}\right| H = k ∑ N k ∣ φ k ⟩ ⟨ φ k ∣ a) Para calcular o adjunto partimos de:
U ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣ U\left(m,n\right)=\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right| U ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣ Multiplicando por um bra ⟨ α ∣ \left\langle \alpha\right| ⟨ α ∣
U ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣ ⟨ α ∣ U ( m , n ) = ⟨ α ∣ φ m ⟩ ⟨ φ n ∣ \begin{align*}
U\left(m,n\right) & =\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|\\
\left\langle \alpha\right|U\left(m,n\right) & =\left\langle \alpha|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|
\end{align*} U ( m , n ) ⟨ α ∣ U ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣ = ⟨ α ∣ φ m ⟩ ⟨ φ n ∣ Aplicando o adjunto:
U ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣ ( ⟨ α ∣ U ( m , n ) ) † = ( ⟨ α ∣ φ m ⟩ ⟨ φ n ∣ ) † U † ( m , n ) ∣ α ⟩ = ∣ φ n ⟩ ⟨ φ m ∣ α ⟩ U † ( m , n ) = ∣ φ n ⟩ ⟨ φ m ∣ \begin{align*}
U\left(m,n\right) & =\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|\\
\left(\left\langle \alpha\right|U\left(m,n\right)\right)^{\dagger} & =\left(\left\langle \alpha|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|\right)^{\dagger}\\
U^{\dagger}\left(m,n\right)\left|\alpha\right\rangle & =\left|\varphi_{n}\right\rangle \left\langle \varphi_{m}|\alpha\right\rangle \\
U^{\dagger}\left(m,n\right) & =\left|\varphi_{n}\right\rangle \left\langle \varphi_{m}\right|
\end{align*} U ( m , n ) ( ⟨ α ∣ U ( m , n ) ) † U † ( m , n ) ∣ α ⟩ U † ( m , n ) = ∣ φ m ⟩ ⟨ φ n ∣ = ( ⟨ α ∣ φ m ⟩ ⟨ φ n ∣ ) † = ∣ φ n ⟩ ⟨ φ m ∣ α ⟩ = ∣ φ n ⟩ ⟨ φ m ∣ Obtemos U † ( m , n ) = U ( n , m ) U^{\dagger}\left(m,n\right)=U\left(n,m\right) U † ( m , n ) = U ( n , m )
b)
O comutador pode ser escrito:
[ H , U ( m , n ) ] = H U ( m , n ) − U ( m , n ) H = ∑ k N k ∣ φ k ⟩ ⟨ φ k ∣ φ m ⟩ ⟨ φ n ∣ − ∑ k N k ∣ φ m ⟩ ⟨ φ n φ k ⟩ ⟨ φ k ∣ \begin{align*}
\left[H,U\left(m,n\right)\right] & =HU\left(m,n\right)-U\left(m,n\right)H\\
& =\sum^{N}_{k}k\left|\varphi_{k}\right\rangle \left\langle \varphi_{k}|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|-\sum^{N}_{k}k\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\varphi_{k}\right\rangle \left\langle \varphi_{k}\right|
\end{align*} [ H , U ( m , n ) ] = H U ( m , n ) − U ( m , n ) H = k ∑ N k ∣ φ k ⟩ ⟨ φ k ∣ φ m ⟩ ⟨ φ n ∣ − k ∑ N k ∣ φ m ⟩ ⟨ φ n φ k ⟩ ⟨ φ k ∣ Como ⟨ φ i ∣ φ j ⟩ = δ i j \left\langle \varphi_{i}|\varphi_{j}\right\rangle =\delta_{ij} ⟨ φ i ∣ φ j ⟩ = δ ij
[ H , U ( m , n ) ] = m ∣ φ m ⟩ ⟨ φ n ∣ − n ∣ φ m ⟩ ⟨ φ n ∣ = ( m − n ) ∣ φ m ⟩ ⟨ φ n ∣ \begin{align*}
\left[H,U\left(m,n\right)\right] & =m\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|-n\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|\\
& =\left(m-n\right)\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|
\end{align*} [ H , U ( m , n ) ] = m ∣ φ m ⟩ ⟨ φ n ∣ − n ∣ φ m ⟩ ⟨ φ n ∣ = ( m − n ) ∣ φ m ⟩ ⟨ φ n ∣ Então [ H , U ( m , n ) ] = ( m − n ) U ( m , n ) \left[H,U\left(m,n\right)\right]=\left(m-n\right)U\left(m,n\right) [ H , U ( m , n ) ] = ( m − n ) U ( m , n )
c)
Este é um cálculo bem direto, utilizando os resultados anteriores:
U ( m , n ) U † ( p , q ) = ∣ φ m ⟩ ⟨ φ n ∣ ( ∣ φ p ⟩ ⟨ φ q ∣ ) † = ∣ φ m ⟩ ⟨ φ n ∣ φ q ⟩ ⟨ φ p ∣ = δ n q ∣ φ m ⟩ ⟨ φ p ∣ \begin{align*}
U\left(m,n\right)U^{\dagger}\left(p,q\right) & =\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|\left(\left|\varphi_{p}\right\rangle \left\langle \varphi_{q}\right|\right)^{\dagger}\\
& =\left|\varphi_{m}\right\rangle \left\langle \varphi_{n}|\varphi_{q}\right\rangle \left\langle \varphi_{p}\right|\\
& =\delta_{nq}\left|\varphi_{m}\right\rangle \left\langle \varphi_{p}\right|
\end{align*} U ( m , n ) U † ( p , q ) = ∣ φ m ⟩ ⟨ φ n ∣ ( ∣ φ p ⟩ ⟨ φ q ∣ ) † = ∣ φ m ⟩ ⟨ φ n ∣ φ q ⟩ ⟨ φ p ∣ = δ n q ∣ φ m ⟩ ⟨ φ p ∣ Logo U ( m , n ) U † ( p , q ) = δ n q U ( m , p ) U\left(m,n\right)U^{\dagger}\left(p,q\right)=\delta_{nq}U\left(m,p\right) U ( m , n ) U † ( p , q ) = δ n q U ( m , p )
d)
Sendo que { ∣ φ n ⟩ } \left\{ \left|\varphi_{n}\right\rangle \right\} { ∣ φ n ⟩ }
T r ( U ( m , n ) ) = T r ( U ) = ∑ k N ⟨ φ k ∣ U ∣ φ k ⟩ = ∑ k N ⟨ φ k ∣ φ m ⟩ ⟨ φ n ∣ φ k ⟩ = ⟨ φ n ∣ φ m ⟩ \begin{align*}
Tr\left(U\left(m,n\right)\right) & =Tr\left(U\right)\\
& =\sum^{N}_{k}\left\langle \varphi_{k}\left|U\right|\varphi_{k}\right\rangle \\
& =\sum^{N}_{k}\left\langle \varphi_{k}|\varphi_{m}\right\rangle \left\langle \varphi_{n}|\varphi_{k}\right\rangle \\
& =\left\langle \varphi_{n}|\varphi_{m}\right\rangle
\end{align*} T r ( U ( m , n ) ) = T r ( U ) = k ∑ N ⟨ φ k ∣ U ∣ φ k ⟩ = k ∑ N ⟨ φ k ∣ φ m ⟩ ⟨ φ n ∣ φ k ⟩ = ⟨ φ n ∣ φ m ⟩ Obtemos T r ( U ( m , n ) ) = δ n m Tr\left(U\left(m,n\right)\right)=\delta_{nm} T r ( U ( m , n ) ) = δ nm
e)
Este resultado pode ser obtido de uma maneira bem direta sendo A m n = ⟨ φ m ∣ A ∣ φ n ⟩ A_{mn}=\left\langle \varphi_{m}\left|A\right|\varphi_{n}\right\rangle A mn = ⟨ φ m ∣ A ∣ φ n ⟩
A = ∑ m , n A m n U ( m , n ) = ∑ m , n ⟨ φ m ∣ A ∣ φ n ⟩ ∣ φ m ⟩ ⟨ φ n ∣ = ∑ m , n ∣ φ m ⟩ ⟨ φ m ∣ A ∣ φ n ⟩ ⟨ φ n ∣ = ( ∑ m ∣ φ m ⟩ ⟨ φ m ∣ ) A ( ∑ n ∣ φ n ⟩ ⟨ φ n ∣ ) = I A I = A \begin{align*}
A & =\sum_{m,n}A_{mn}U\left(m,n\right)\\
& =\sum_{m,n}\left\langle \varphi_{m}\left|A\right|\varphi_{n}\right\rangle \left|\varphi_{m}\right\rangle \left\langle \varphi_{n}\right|\\
& =\sum_{m,n}\left|\varphi_{m}\right\rangle \left\langle \varphi_{m}\left|A\right|\varphi_{n}\right\rangle \left\langle \varphi_{n}\right|\\
& =\left(\sum_{m}\left|\varphi_{m}\right\rangle \left\langle \varphi_{m}\right|\right)A\left(\sum_{n}\left|\varphi_{n}\right\rangle \left\langle \varphi_{n}\right|\right)\\
& =\mathbb{I}A\mathbb{I}\\
& =A
\end{align*} A = m , n ∑ A mn U ( m , n ) = m , n ∑ ⟨ φ m ∣ A ∣ φ n ⟩ ∣ φ m ⟩ ⟨ φ n ∣ = m , n ∑ ∣ φ m ⟩ ⟨ φ m ∣ A ∣ φ n ⟩ ⟨ φ n ∣ = ( m ∑ ∣ φ m ⟩ ⟨ φ m ∣ ) A ( n ∑ ∣ φ n ⟩ ⟨ φ n ∣ ) = I A I = A f)
A partir de pura manipulação dos termos obtemos que:
A p q = T r ( A U † ( p , q ) ) = T r ( A U ( q , p ) ) = T r ( ∑ m , n ⟨ φ m ∣ A ∣ φ n ⟩ ∣ φ m ⟩ ⟨ φ n ∣ φ q ⟩ ⟨ φ p ∣ ) = T r ( ∑ m ⟨ φ m ∣ A ∣ φ q ⟩ ∣ φ m ⟩ ⟨ φ p ∣ ) = ∑ k N ⟨ φ k ∣ ∑ m ⟨ φ m ∣ A ∣ φ q ⟩ ∣ φ m ⟩ ⟨ φ p ∣ ∣ φ k ⟩ = ∑ k N ∑ m ⟨ φ m ∣ A ∣ φ q ⟩ ⟨ φ k ∣ φ m ⟩ ⟨ φ p ∣ φ k ⟩ = ∑ m ⟨ φ m ∣ A ∣ φ q ⟩ ⟨ φ p ∣ φ m ⟩ = ⟨ φ p ∣ A ∣ φ q ⟩ \begin{align*}
A_{pq} & =Tr\left(AU^{\dagger}\left(p,q\right)\right)\\
& =Tr\left(AU\left(q,p\right)\right)\\
& =Tr\left(\sum_{m,n}\left\langle \varphi_{m}\left|A\right|\varphi_{n}\right\rangle \left|\varphi_{m}\right\rangle \left\langle \varphi_{n}|\varphi_{q}\right\rangle \left\langle \varphi_{p}\right|\right)\\
& =Tr\left(\sum_{m}\left\langle \varphi_{m}\left|A\right|\varphi_{q}\right\rangle \left|\varphi_{m}\right\rangle \left\langle \varphi_{p}\right|\right)\\
& =\sum^{N}_{k}\left\langle \varphi_{k}\left|\sum_{m}\left\langle \varphi_{m}\left|A\right|\varphi_{q}\right\rangle \left|\varphi_{m}\right\rangle \left\langle \varphi_{p}\right|\right|\varphi_{k}\right\rangle \\
& =\sum^{N}_{k}\sum_{m}\left\langle \varphi_{m}\left|A\right|\varphi_{q}\right\rangle \left\langle \varphi_{k}|\varphi_{m}\right\rangle \left\langle \varphi_{p}|\varphi_{k}\right\rangle \\
& =\sum_{m}\left\langle \varphi_{m}\left|A\right|\varphi_{q}\right\rangle \left\langle \varphi_{p}|\varphi_{m}\right\rangle \\
& =\left\langle \varphi_{p}\left|A\right|\varphi_{q}\right\rangle
\end{align*} A pq = T r ( A U † ( p , q ) ) = T r ( A U ( q , p ) ) = T r ( m , n ∑ ⟨ φ m ∣ A ∣ φ n ⟩ ∣ φ m ⟩ ⟨ φ n ∣ φ q ⟩ ⟨ φ p ∣ ) = T r ( m ∑ ⟨ φ m ∣ A ∣ φ q ⟩ ∣ φ m ⟩ ⟨ φ p ∣ ) = k ∑ N ⟨ φ k ∣ ∣ m ∑ ⟨ φ m ∣ A ∣ φ q ⟩ ∣ φ m ⟩ ⟨ φ p ∣ ∣ ∣ φ k ⟩ = k ∑ N m ∑ ⟨ φ m ∣ A ∣ φ q ⟩ ⟨ φ k ∣ φ m ⟩ ⟨ φ p ∣ φ k ⟩ = m ∑ ⟨ φ m ∣ A ∣ φ q ⟩ ⟨ φ p ∣ φ m ⟩ = ⟨ φ p ∣ A ∣ φ q ⟩ Logo A p q = ⟨ φ p ∣ A ∣ φ q ⟩ A_{pq}=\left\langle \varphi_{p}\left|A\right|\varphi_{q}\right\rangle A pq = ⟨ φ p ∣ A ∣ φ q ⟩
Problem 2 ¶ In a two-dimensional vector space, consider the operator whose matrix, in an orthonormal basis { ∣ 1 ⟩ , ∣ 2 ⟩ } \left\{ \left|1\right\rangle ,\left|2\right\rangle \right\} { ∣ 1 ⟩ , ∣ 2 ⟩ }
σ y = ( 0 − i i 0 ) \sigma_{y}=\left(\begin{array}{cc}
0 & -i\\
i & 0
\end{array}\right) σ y = ( 0 i − i 0 )
a) Is σ y \sigma_{y} σ y { ∣ 1 ⟩ , ∣ 2 ⟩ } \left\{ \left|1\right\rangle ,\left|2\right\rangle \right\} { ∣ 1 ⟩ , ∣ 2 ⟩ }
b) Calculate the matrices which represent the projectors onto these eigenvectors. Then verify that they satisfy the orthogonality and closure relations.
c) Same questions for the matrices:
M = ( 2 i 2 − i 2 3 ) M=\left(\begin{array}{cc}
2 & i\sqrt{2}\\
-i\sqrt{2} & 3
\end{array}\right) M = ( 2 − i 2 i 2 3 )
and, in a three-dimensional space
L y = ℏ 2 i ( 0 2 0 − 2 0 2 0 − 2 0 ) L_{y}=\frac{\hbar}{2i}\left(\begin{array}{ccc}
0 & \sqrt{2} & 0\\
-\sqrt{2} & 0 & \sqrt{2}\\
0 & -\sqrt{2} & 0
\end{array}\right) L y = 2 i ℏ ⎝ ⎛ 0 − 2 0 2 0 − 2 0 2 0 ⎠ ⎞
a)
Um operador é hermitiano se A = A T ‾ A=\overline{A^{T}} A = A T
σ y = ( 0 − i i 0 ) \sigma_{y}=\left(\begin{array}{cc}
0 & -i\\
i & 0
\end{array}\right) σ y = ( 0 i − i 0 ) E:
σ y † = ( 0 − i i 0 ) = σ y \sigma^{\dagger}_{y}=\left(\begin{array}{cc}
0 & -i\\
i & 0
\end{array}\right)=\sigma_{y} σ y † = ( 0 i − i 0 ) = σ y Logo é hermitiano. Para calcular os autovalores, fazemos:
det ∣ − λ − i i − λ ∣ = λ − 1 = 0 \det\left|\begin{array}{cc}
-\lambda & -i\\
i & -\lambda
\end{array}\right|=\lambda-1=0 det ∣ ∣ − λ i − i − λ ∣ ∣ = λ − 1 = 0 Então λ ± = ± 1 \lambda_{\pm}=\pm1 λ ± = ± 1
( − 1 − i i − 1 ) ( x 1 x 2 ) = ( 0 0 ) \left(\begin{array}{cc}
-1 & -i\\
i & -1
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right) ( − 1 i − i − 1 ) ( x 1 x 2 ) = ( 0 0 ) Logo − x 1 − i x 2 = 0 -x_{1}-ix_{2}=0 − x 1 − i x 2 = 0 x 1 = − i x 2 x_{1}=-ix_{2} x 1 = − i x 2 ( − i , 1 ) \left(-i,1\right) ( − i , 1 ) ∣ 1 ⟩ = ( 1 , 0 ) \left|1\right\rangle =\left(1,0\right) ∣ 1 ⟩ = ( 1 , 0 ) ∣ 2 ⟩ = ( 0 , 1 ) \left|2\right\rangle =\left(0,1\right) ∣ 2 ⟩ = ( 0 , 1 )
∣ λ + ⟩ = − i 2 ∣ 1 ⟩ + 1 2 ∣ 2 ⟩ \left|\lambda_{+}\right\rangle =-\frac{i}{\sqrt{2}}\left|1\right\rangle +\frac{1}{\sqrt{2}}\left|2\right\rangle ∣ λ + ⟩ = − 2 i ∣ 1 ⟩ + 2 1 ∣ 2 ⟩ De modo análogo para o outro autovalor:
( 1 − i i − 1 ) ( x 1 x 2 ) = ( 0 0 ) \left(\begin{array}{cc}
1 & -i\\
i & -1
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right) ( 1 i − i − 1 ) ( x 1 x 2 ) = ( 0 0 ) De onde obtemo que x 1 = i x 2 x_{1}=ix_{2} x 1 = i x 2 ( i , 1 ) \left(i,1\right) ( i , 1 )
∣ λ − ⟩ = i 2 ∣ 1 ⟩ + 1 2 ∣ 2 ⟩ \left|\lambda_{-}\right\rangle =\frac{i}{\sqrt{2}}\left|1\right\rangle +\frac{1}{\sqrt{2}}\left|2\right\rangle ∣ λ − ⟩ = 2 i ∣ 1 ⟩ + 2 1 ∣ 2 ⟩ b)
Calcuando então a representação matricial destes projetores:
P + = ∣ λ + ⟩ ⟨ λ + ∣ P − = ∣ λ − ⟩ ⟨ λ − ∣ = [ − i 2 1 2 ] [ i 2 1 2 ] = [ i 2 1 2 ] [ − i 2 1 2 ] P + = 1 2 [ 1 − i i 1 ] P − = 1 2 [ 1 i − i 1 ] \begin{align*}
P_{+} & =\left|\lambda_{+}\right\rangle \left\langle \lambda_{+}\right| & P_{-} & =\left|\lambda_{-}\right\rangle \left\langle \lambda_{-}\right|\\
& =\left[\begin{array}{c}
-\frac{i}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{array}\right]\left[\begin{array}{cc}
\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right] & & =\left[\begin{array}{c}
\frac{i}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{array}\right]\left[\begin{array}{cc}
-\frac{i}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right]\\
P_{+} & =\frac{1}{2}\left[\begin{array}{cc}
1 & -i\\
i & 1
\end{array}\right] & P_{-} & =\frac{1}{2}\left[\begin{array}{cc}
1 & i\\
-i & 1
\end{array}\right]
\end{align*} P + P + = ∣ λ + ⟩ ⟨ λ + ∣ = [ − 2 i 2 1 ] [ 2 i 2 1 ] = 2 1 [ 1 i − i 1 ] P − P − = ∣ λ − ⟩ ⟨ λ − ∣ = [ 2 i 2 1 ] [ − 2 i 2 1 ] = 2 1 [ 1 − i i 1 ] Temos então os projetores P + = 1 2 [ 1 − i i 1 ] P_{+}=\frac{1}{2}\left[\begin{array}{cc}
1 & -i\\
i & 1
\end{array}\right] P + = 2 1 [ 1 i − i 1 ] P − = 1 2 [ 1 i − i 1 ] P_{-}=\frac{1}{2}\left[\begin{array}{cc}
1 & i\\
-i & 1
\end{array}\right] P − = 2 1 [ 1 − i i 1 ]
P + + P − = 1 2 [ 1 − i i 1 ] + 1 2 [ 1 i − i 1 ] = 1 2 [ 2 0 0 2 ] = [ 1 0 0 1 ] E = I \begin{align*}
P_{+}+P_{-} & =\frac{1}{2}\left[\begin{array}{cc}
1 & -i\\
i & 1
\end{array}\right]+\frac{1}{2}\left[\begin{array}{cc}
1 & i\\
-i & 1
\end{array}\right]\\
& =\frac{1}{2}\left[\begin{array}{cc}
2 & 0\\
0 & 2
\end{array}\right]\\
& =\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]E\\
& =\mathbb{I}
\end{align*} P + + P − = 2 1 [ 1 i − i 1 ] + 2 1 [ 1 − i i 1 ] = 2 1 [ 2 0 0 2 ] = [ 1 0 0 1 ] E = I E a ortogononalidade:
P + ⋅ P − = 1 2 [ 1 − i i 1 ] ⋅ 1 2 [ 1 i − i 1 ] = 1 4 [ 0 0 0 0 ] \begin{align*}
P_{+}\cdot P_{-} & =\frac{1}{2}\left[\begin{array}{cc}
1 & -i\\
i & 1
\end{array}\right]\cdot\frac{1}{2}\left[\begin{array}{cc}
1 & i\\
-i & 1
\end{array}\right]\\
& =\frac{1}{4}\left[\begin{array}{cc}
0 & 0\\
0 & 0
\end{array}\right]
\end{align*} P + ⋅ P − = 2 1 [ 1 i − i 1 ] ⋅ 2 1 [ 1 − i i 1 ] = 4 1 [ 0 0 0 0 ] O mesmo se:
P − ⋅ P + = 1 2 [ 1 i − i 1 ] ⋅ 1 2 [ 1 − i i 1 ] = 1 4 [ 0 0 0 0 ] \begin{align*}
P_{-}\cdot P_{+} & =\frac{1}{2}\left[\begin{array}{cc}
1 & i\\
-i & 1
\end{array}\right]\cdot\frac{1}{2}\left[\begin{array}{cc}
1 & -i\\
i & 1
\end{array}\right]\\
& =\frac{1}{4}\left[\begin{array}{cc}
0 & 0\\
0 & 0
\end{array}\right]
\end{align*} P − ⋅ P + = 2 1 [ 1 − i i 1 ] ⋅ 2 1 [ 1 i − i 1 ] = 4 1 [ 0 0 0 0 ] c) Repetindo os mesmos passos para outras duas matrizes, primeiro para:
M = ( 2 i 2 − i 2 3 ) M=\left(\begin{array}{cc}
2 & i\sqrt{2}\\
-i\sqrt{2} & 3
\end{array}\right) M = ( 2 − i 2 i 2 3 ) Verificando se é hermitiano:
M † = ( 2 i 2 − i 2 3 ) = M M^{\dagger}=\left(\begin{array}{cc}
2 & i\sqrt{2}\\
-i\sqrt{2} & 3
\end{array}\right)=M M † = ( 2 − i 2 i 2 3 ) = M Cálculo de autovalores:
det ∣ 2 − λ i 2 − i 2 3 − λ ∣ = ( 2 − λ ) ( 3 − λ ) − 2 = 0 \det\left|\begin{array}{cc}
2-\lambda & i\sqrt{2}\\
-i\sqrt{2} & 3-\lambda
\end{array}\right|=\left(2-\lambda\right)\left(3-\lambda\right)-2=0 det ∣ ∣ 2 − λ − i 2 i 2 3 − λ ∣ ∣ = ( 2 − λ ) ( 3 − λ ) − 2 = 0 Então temos a seguinte equaçaõ de segundo grau λ 2 − 5 λ + 4 = 0 \lambda^{2}-5\lambda+4=0 λ 2 − 5 λ + 4 = 0 λ ∈ { 4 , 1 } \lambda\in\left\{ 4,1\right\} λ ∈ { 4 , 1 } λ 1 = 4 \lambda_{1}=4 λ 1 = 4
( − 2 i 2 − i 2 − 1 ) ( x 1 x 2 ) = ( 0 0 ) \left(\begin{array}{cc}
-2 & i\sqrt{2}\\
-i\sqrt{2} & -1
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right) ( − 2 − i 2 i 2 − 1 ) ( x 1 x 2 ) = ( 0 0 ) Da primeira linha temos que − 2 x 1 + i 2 x 2 = 0 -2x_{1}+i\sqrt{2}x_{2}=0 − 2 x 1 + i 2 x 2 = 0 x 1 = i x 2 2 x_{1}=\frac{ix_{2}}{\sqrt{2}} x 1 = 2 i x 2 − i 2 -i\sqrt{2} − i 2 x 2 = 1 x_{2}=1 x 2 = 1
∣ λ 1 ⟩ = 1 1 2 + 1 ( i 2 ∣ 1 ⟩ + ∣ 2 ⟩ ) = 2 3 ( i 2 ∣ 1 ⟩ + ∣ 2 ⟩ ) ∣ λ 1 ⟩ = i 3 ∣ 1 ⟩ + 2 3 ∣ 2 ⟩ \begin{align*}
\left|\lambda_{1}\right\rangle & =\frac{1}{\sqrt{\frac{1}{2}+1}}\left(\frac{i}{\sqrt{2}}\left|1\right\rangle +\left|2\right\rangle \right)=\sqrt{\frac{2}{3}}\left(\frac{i}{\sqrt{2}}\left|1\right\rangle +\left|2\right\rangle \right)\\
\left|\lambda_{1}\right\rangle & =\frac{i}{\sqrt{3}}\left|1\right\rangle +\sqrt{\frac{2}{3}}\left|2\right\rangle
\end{align*} ∣ λ 1 ⟩ ∣ λ 1 ⟩ = 2 1 + 1 1 ( 2 i ∣ 1 ⟩ + ∣ 2 ⟩ ) = 3 2 ( 2 i ∣ 1 ⟩ + ∣ 2 ⟩ ) = 3 i ∣ 1 ⟩ + 3 2 ∣ 2 ⟩ De modo análogo para o outro autovalor:
( 1 i 2 − i 2 2 ) ( x 1 x 2 ) = ( 0 0 ) \left(\begin{array}{cc}
1 & i\sqrt{2}\\
-i\sqrt{2} & 2
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right) ( 1 − i 2 i 2 2 ) ( x 1 x 2 ) = ( 0 0 ) Da primeira linha temos que x 1 + i 2 x 2 = 0 x_{1}+i\sqrt{2}x_{2}=0 x 1 + i 2 x 2 = 0 x 1 = − i 2 x 2 x_{1}=-i\sqrt{2}x_{2} x 1 = − i 2 x 2 x 2 = 1 x_{2}=1 x 2 = 1
∣ λ 2 ⟩ = 1 2 + 1 ( − i 2 ∣ 1 ⟩ + ∣ 2 ⟩ ) = 1 3 ( − i 2 ∣ 1 ⟩ + ∣ 2 ⟩ ) ∣ λ 2 ⟩ = − i 2 3 ∣ 1 ⟩ + 1 3 ∣ 2 ⟩ \begin{align*}
\left|\lambda_{2}\right\rangle & =\frac{1}{\sqrt{2+1}}\left(-i\sqrt{2}\left|1\right\rangle +\left|2\right\rangle \right)=\frac{1}{\sqrt{3}}\left(-i\sqrt{2}\left|1\right\rangle +\left|2\right\rangle \right)\\
\left|\lambda_{2}\right\rangle & =-i\sqrt{\frac{2}{3}}\left|1\right\rangle +\sqrt{\frac{1}{3}}\left|2\right\rangle
\end{align*} ∣ λ 2 ⟩ ∣ λ 2 ⟩ = 2 + 1 1 ( − i 2 ∣ 1 ⟩ + ∣ 2 ⟩ ) = 3 1 ( − i 2 ∣ 1 ⟩ + ∣ 2 ⟩ ) = − i 3 2 ∣ 1 ⟩ + 3 1 ∣ 2 ⟩ Calcuando então a representação matricial destes projetores:
P 1 = ∣ λ 1 ⟩ ⟨ λ 1 ∣ P 2 = ∣ λ 2 ⟩ ⟨ λ 2 ∣ = [ i 3 2 3 ] [ − i 3 2 3 ] = [ − i 2 3 1 3 ] [ i 2 3 1 3 ] P 1 = 1 3 [ 1 i 2 − i 2 2 ] P 2 = 1 3 [ 2 − i 2 i 2 1 ] \begin{align*}
P_{1} & =\left|\lambda_{1}\right\rangle \left\langle \lambda_{1}\right| & P_{2} & =\left|\lambda_{2}\right\rangle \left\langle \lambda_{2}\right|\\
& =\left[\begin{array}{c}
\frac{i}{\sqrt{3}}\\
\sqrt{\frac{2}{3}}
\end{array}\right]\left[\begin{array}{cc}
-\frac{i}{\sqrt{3}} & \sqrt{\frac{2}{3}}\end{array}\right] & & =\left[\begin{array}{c}
-i\sqrt{\frac{2}{3}}\\
\sqrt{\frac{1}{3}}
\end{array}\right]\left[\begin{array}{cc}
i\sqrt{\frac{2}{3}} & \sqrt{\frac{1}{3}}\end{array}\right]\\
P_{1} & =\frac{1}{3}\left[\begin{array}{cc}
1 & i\sqrt{2}\\
-i\sqrt{2} & 2
\end{array}\right] & P_{2} & =\frac{1}{3}\left[\begin{array}{cc}
2 & -i\sqrt{2}\\
i\sqrt{2} & 1
\end{array}\right]
\end{align*} P 1 P 1 = ∣ λ 1 ⟩ ⟨ λ 1 ∣ = [ 3 i 3 2 ] [ − 3 i 3 2 ] = 3 1 [ 1 − i 2 i 2 2 ] P 2 P 2 = ∣ λ 2 ⟩ ⟨ λ 2 ∣ = ⎣ ⎡ − i 3 2 3 1 ⎦ ⎤ [ i 3 2 3 1 ] = 3 1 [ 2 i 2 − i 2 1 ] Temos então os projetores P + = 1 3 [ 1 i 2 − i 2 2 ] P_{+}=\frac{1}{3}\left[\begin{array}{cc}
1 & i\sqrt{2}\\
-i\sqrt{2} & 2
\end{array}\right] P + = 3 1 [ 1 − i 2 i 2 2 ] P − = 1 3 [ 2 − i 2 i 2 1 ] P_{-}=\frac{1}{3}\left[\begin{array}{cc}
2 & -i\sqrt{2}\\
i\sqrt{2} & 1
\end{array}\right] P − = 3 1 [ 2 i 2 − i 2 1 ]
P 1 + P 2 = 1 3 [ 1 i 2 − i 2 2 ] + 1 3 [ 2 − i 2 i 2 1 ] = 1 3 [ 3 0 0 3 ] = [ 3 0 0 3 ] = I \begin{align*}
P_{1}+P_{2} & =\frac{1}{3}\left[\begin{array}{cc}
1 & i\sqrt{2}\\
-i\sqrt{2} & 2
\end{array}\right]+\frac{1}{3}\left[\begin{array}{cc}
2 & -i\sqrt{2}\\
i\sqrt{2} & 1
\end{array}\right]\\
& =\frac{1}{3}\left[\begin{array}{cc}
3 & 0\\
0 & 3
\end{array}\right]\\
& =\left[\begin{array}{cc}
3 & 0\\
0 & 3
\end{array}\right]\\
& =\mathbb{I}
\end{align*} P 1 + P 2 = 3 1 [ 1 − i 2 i 2 2 ] + 3 1 [ 2 i 2 − i 2 1 ] = 3 1 [ 3 0 0 3 ] = [ 3 0 0 3 ] = I E a ortogononalidade:
P 1 ⋅ P 2 = 1 3 [ 1 i 2 − i 2 2 ] ⋅ 1 3 [ 2 − i 2 i 2 1 ] = 1 9 [ 0 0 0 0 ] \begin{align*}
P_{1}\cdot P_{2} & =\frac{1}{3}\left[\begin{array}{cc}
1 & i\sqrt{2}\\
-i\sqrt{2} & 2
\end{array}\right]\cdot\frac{1}{3}\left[\begin{array}{cc}
2 & -i\sqrt{2}\\
i\sqrt{2} & 1
\end{array}\right]\\
& =\frac{1}{9}\left[\begin{array}{cc}
0 & 0\\
0 & 0
\end{array}\right]
\end{align*} P 1 ⋅ P 2 = 3 1 [ 1 − i 2 i 2 2 ] ⋅ 3 1 [ 2 i 2 − i 2 1 ] = 9 1 [ 0 0 0 0 ] O mesmo se:
P 2 ⋅ P 1 = 1 3 [ 2 − i 2 i 2 1 ] ⋅ 1 3 [ 1 i 2 − i 2 2 ] = 1 9 [ 0 0 0 0 ] \begin{align*}
P_{2}\cdot P_{1} & =\frac{1}{3}\left[\begin{array}{cc}
2 & -i\sqrt{2}\\
i\sqrt{2} & 1
\end{array}\right]\cdot\frac{1}{3}\left[\begin{array}{cc}
1 & i\sqrt{2}\\
-i\sqrt{2} & 2
\end{array}\right]\\
& =\frac{1}{9}\left[\begin{array}{cc}
0 & 0\\
0 & 0
\end{array}\right]
\end{align*} P 2 ⋅ P 1 = 3 1 [ 2 i 2 − i 2 1 ] ⋅ 3 1 [ 1 − i 2 i 2 2 ] = 9 1 [ 0 0 0 0 ] E por fim, para
L y = ℏ 2 i ( 0 2 0 − 2 0 2 0 − 2 0 ) L_{y}=\frac{\hbar}{2i}\left(\begin{array}{ccc}
0 & \sqrt{2} & 0\\
-\sqrt{2} & 0 & \sqrt{2}\\
0 & -\sqrt{2} & 0
\end{array}\right) L y = 2 i ℏ ⎝ ⎛ 0 − 2 0 2 0 − 2 0 2 0 ⎠ ⎞ Verificando se é hermitiano:
L y † = − ℏ 2 i ( 0 − 2 0 2 0 − 2 0 2 0 ) = ℏ 2 i ( 0 2 0 − 2 0 2 0 − 2 0 ) = L y L^{\dagger}_{y}=-\frac{\hbar}{2i}\left(\begin{array}{ccc}
0 & -\sqrt{2} & 0\\
\sqrt{2} & 0 & -\sqrt{2}\\
0 & \sqrt{2} & 0
\end{array}\right)=\frac{\hbar}{2i}\left(\begin{array}{ccc}
0 & \sqrt{2} & 0\\
-\sqrt{2} & 0 & \sqrt{2}\\
0 & -\sqrt{2} & 0
\end{array}\right)=L_{y} L y † = − 2 i ℏ ⎝ ⎛ 0 2 0 − 2 0 2 0 − 2 0 ⎠ ⎞ = 2 i ℏ ⎝ ⎛ 0 − 2 0 2 0 − 2 0 2 0 ⎠ ⎞ = L y Cálculo de autovalores:
∣ − λ ℏ 2 i 2 0 − ℏ 2 i 2 − λ ℏ 2 i 2 0 − ℏ 2 i 2 − λ ∣ = ( − λ 3 + ℏ 2 4 4 λ ) = ( 0 0 0 ) \left|\begin{array}{ccc}
-\lambda & \frac{\hbar}{2i}\sqrt{2} & 0\\
-\frac{\hbar}{2i}\sqrt{2} & -\lambda & \frac{\hbar}{2i}\sqrt{2}\\
0 & -\frac{\hbar}{2i}\sqrt{2} & -\lambda
\end{array}\right|=\left(-\lambda^{3}+\frac{\hbar^{2}}{4}4\lambda\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ∣ ∣ − λ − 2 i ℏ 2 0 2 i ℏ 2 − λ − 2 i ℏ 2 0 2 i ℏ 2 − λ ∣ ∣ = ( − λ 3 + 4 ℏ 2 4 λ ) = ⎝ ⎛ 0 0 0 ⎠ ⎞ Então temos a seguinte equaçaõ de segundo grau − λ ( λ 2 − ℏ 2 ) = 0 -\lambda\left(\lambda^{2}-\hbar^{2}\right)=0 − λ ( λ 2 − ℏ 2 ) = 0 λ 1 = 0 \lambda_{1}=0 λ 1 = 0 λ = ± ℏ \lambda=\pm\hbar λ = ± ℏ
( 0 ℏ 2 i 2 0 − ℏ 2 i 2 0 ℏ 2 i 2 0 − ℏ 2 i 2 0 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
0 & \frac{\hbar}{2i}\sqrt{2} & 0\\
-\frac{\hbar}{2i}\sqrt{2} & 0 & \frac{\hbar}{2i}\sqrt{2}\\
0 & -\frac{\hbar}{2i}\sqrt{2} & 0
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ⎝ ⎛ 0 − 2 i ℏ 2 0 2 i ℏ 2 0 − 2 i ℏ 2 0 2 i ℏ 2 0 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ Da primeira e últim linha temos que x 2 = 0 x_{2}=0 x 2 = 0 x 1 = x 3 x_{1}=x_{3} x 1 = x 3 x 1 = 1 2 x_{1}=\frac{1}{\sqrt{2}} x 1 = 2 1
∣ λ 1 ⟩ = 1 2 ∣ 1 ⟩ + 1 2 ∣ 3 ⟩ \begin{align*}
\left|\lambda_{1}\right\rangle & =\frac{1}{\sqrt{2}}\left|1\right\rangle +\frac{1}{\sqrt{2}}\left|3\right\rangle
\end{align*} ∣ λ 1 ⟩ = 2 1 ∣ 1 ⟩ + 2 1 ∣ 3 ⟩ De modo análogo para λ 2 = ℏ \lambda_{2}=\hbar λ 2 = ℏ
( − ℏ ℏ 2 i 2 0 − ℏ 2 i 2 − ℏ ℏ 2 i 2 0 − ℏ 2 i 2 − ℏ ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
-\hbar & \frac{\hbar}{2i}\sqrt{2} & 0\\
-\frac{\hbar}{2i}\sqrt{2} & -\hbar & \frac{\hbar}{2i}\sqrt{2}\\
0 & -\frac{\hbar}{2i}\sqrt{2} & -\hbar
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ⎝ ⎛ − ℏ − 2 i ℏ 2 0 2 i ℏ 2 − ℏ − 2 i ℏ 2 0 2 i ℏ 2 − ℏ ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ Então da primeira linha − ℏ x 1 + ℏ 2 i 2 x 2 = 0 -\hbar x_{1}+\frac{\hbar}{2i}\sqrt{2}x_{2}=0 − ℏ x 1 + 2 i ℏ 2 x 2 = 0 x 1 = x 2 2 i x_{1}=\frac{x_{2}}{\sqrt{2}i} x 1 = 2 i x 2 − ℏ 2 i 2 x 2 − ℏ x 3 = 0 -\frac{\hbar}{2i}\sqrt{2}x_{2}-\hbar x_{3}=0 − 2 i ℏ 2 x 2 − ℏ x 3 = 0 x 3 = − x 2 2 i x_{3}=-\frac{x_{2}}{\sqrt{2}i} x 3 = − 2 i x 2
− ℏ 2 i 2 ( x 2 2 i ) − ℏ x 2 + ℏ 2 i 2 ( − x 2 2 i ) = 0 + ℏ 2 x 2 − ℏ x 2 + ℏ 2 x 2 = 0 ( ℏ − ℏ ) x 2 = 0 \begin{align*}
-\frac{\hbar}{2i}\sqrt{2}\left(\frac{x_{2}}{\sqrt{2}i}\right)-\hbar x_{2}+\frac{\hbar}{2i}\sqrt{2}\left(-\frac{x_{2}}{\sqrt{2}i}\right) & =0\\
+\frac{\hbar}{2}x_{2}-\hbar x_{2}+\frac{\hbar}{2}x_{2} & =0\\
\left(\hbar-\hbar\right)x_{2} & =0
\end{align*} − 2 i ℏ 2 ( 2 i x 2 ) − ℏ x 2 + 2 i ℏ 2 ( − 2 i x 2 ) + 2 ℏ x 2 − ℏ x 2 + 2 ℏ x 2 ( ℏ − ℏ ) x 2 = 0 = 0 = 0 Então x 2 x_{2} x 2 L 2 L2 L 2 ( L 2 = 1 i 2 ( L 1 − L 3 ) ) \left(L_{2}=\frac{1}{i\sqrt{2}}\left(L_{1}-L_{3}\right)\right) ( L 2 = i 2 1 ( L 1 − L 3 ) ) x 2 = 2 i x_{2}=\sqrt{2}i x 2 = 2 i
∣ λ 2 ⟩ = 1 1 + 2 + 1 ( 1 ∣ 1 ⟩ + 2 i ∣ 2 ⟩ − 1 ∣ 3 ⟩ ) ∣ λ 2 ⟩ = 1 2 ∣ 1 ⟩ + i 2 ∣ 2 ⟩ − 1 2 ∣ 3 ⟩ \begin{align*}
\left|\lambda_{2}\right\rangle & =\frac{1}{\sqrt{1+2+1}}\left(1\left|1\right\rangle +\sqrt{2}i\left|2\right\rangle -1\left|3\right\rangle \right)\\
\left|\lambda_{2}\right\rangle & =\frac{1}{2}\left|1\right\rangle +\frac{i}{\sqrt{2}}\left|2\right\rangle -\frac{1}{2}\left|3\right\rangle
\end{align*} ∣ λ 2 ⟩ ∣ λ 2 ⟩ = 1 + 2 + 1 1 ( 1 ∣ 1 ⟩ + 2 i ∣ 2 ⟩ − 1 ∣ 3 ⟩ ) = 2 1 ∣ 1 ⟩ + 2 i ∣ 2 ⟩ − 2 1 ∣ 3 ⟩ E para o último auto valor λ 3 = − ℏ \lambda_{3}=-\hbar λ 3 = − ℏ
( ℏ ℏ 2 i 2 0 − ℏ 2 i 2 ℏ ℏ 2 i 2 0 − ℏ 2 i 2 ℏ ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
\hbar & \frac{\hbar}{2i}\sqrt{2} & 0\\
-\frac{\hbar}{2i}\sqrt{2} & \hbar & \frac{\hbar}{2i}\sqrt{2}\\
0 & -\frac{\hbar}{2i}\sqrt{2} & \hbar
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ⎝ ⎛ ℏ − 2 i ℏ 2 0 2 i ℏ 2 ℏ − 2 i ℏ 2 0 2 i ℏ 2 ℏ ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ Então da primeira linha ℏ x 1 + ℏ 2 i 2 x 2 = 0 \hbar x_{1}+\frac{\hbar}{2i}\sqrt{2}x_{2}=0 ℏ x 1 + 2 i ℏ 2 x 2 = 0 x 1 = − x 2 2 i x_{1}=-\frac{x_{2}}{\sqrt{2}i} x 1 = − 2 i x 2 − ℏ 2 i 2 x 2 + ℏ x 3 = 0 -\frac{\hbar}{2i}\sqrt{2}x_{2}+\hbar x_{3}=0 − 2 i ℏ 2 x 2 + ℏ x 3 = 0 x 3 = x 2 2 i x_{3}=\frac{x_{2}}{\sqrt{2}i} x 3 = 2 i x 2
− ℏ 2 i 2 ( − x 2 2 i ) + ℏ x 2 + ℏ 2 i 2 ( x 2 2 i ) = 0 − ℏ 2 x 2 + ℏ x 2 − ℏ 2 x 2 = 0 ( ℏ − ℏ ) x 2 = 0 \begin{align*}
-\frac{\hbar}{2i}\sqrt{2}\left(-\frac{x_{2}}{\sqrt{2}i}\right)+\hbar x_{2}+\frac{\hbar}{2i}\sqrt{2}\left(\frac{x_{2}}{\sqrt{2}i}\right) & =0\\
-\frac{\hbar}{2}x_{2}+\hbar x_{2}-\frac{\hbar}{2}x_{2} & =0\\
\left(\hbar-\hbar\right)x_{2} & =0
\end{align*} − 2 i ℏ 2 ( − 2 i x 2 ) + ℏ x 2 + 2 i ℏ 2 ( 2 i x 2 ) − 2 ℏ x 2 + ℏ x 2 − 2 ℏ x 2 ( ℏ − ℏ ) x 2 = 0 = 0 = 0 Análogo ao caso anterior, escolhendo x 2 = 2 i x_{2}=\sqrt{2}i x 2 = 2 i
∣ λ 3 ⟩ = 1 1 + 2 + 1 ( − ∣ 1 ⟩ + 2 i ∣ 2 ⟩ + ∣ 3 ⟩ ) ∣ λ 3 ⟩ = − 1 2 ∣ 1 ⟩ + i 2 ∣ 2 ⟩ + 1 2 ∣ 3 ⟩ \begin{align*}
\left|\lambda_{3}\right\rangle & =\frac{1}{\sqrt{1+2+1}}\left(-\left|1\right\rangle +\sqrt{2}i\left|2\right\rangle +\left|3\right\rangle \right)\\
\left|\lambda_{3}\right\rangle & =-\frac{1}{2}\left|1\right\rangle +\frac{i}{\sqrt{2}}\left|2\right\rangle +\frac{1}{2}\left|3\right\rangle
\end{align*} ∣ λ 3 ⟩ ∣ λ 3 ⟩ = 1 + 2 + 1 1 ( − ∣ 1 ⟩ + 2 i ∣ 2 ⟩ + ∣ 3 ⟩ ) = − 2 1 ∣ 1 ⟩ + 2 i ∣ 2 ⟩ + 2 1 ∣ 3 ⟩ Calcuando então a representação matricial destes projetores:
P 1 = ∣ λ 1 ⟩ ⟨ λ 1 ∣ P 2 = ∣ λ 2 ⟩ ⟨ λ 2 ∣ P 3 = ∣ λ 3 ⟩ ⟨ λ 3 ∣ = [ 1 2 0 1 2 ] [ 1 2 0 1 2 ] = [ 1 2 i 2 − 1 2 ] [ 1 2 − i 2 − 1 2 ] = [ − 1 2 i 2 1 2 ] [ − 1 2 − i 2 1 2 ] P 1 = 1 2 [ 1 0 1 0 0 0 1 0 1 ] P 2 = 1 4 [ 1 − 2 i 2 − 1 2 i 2 2 − 2 i 2 − 1 2 i 2 1 ] P 2 = 1 4 [ 1 2 i 2 − 1 − 2 i 2 2 2 i 2 − 1 − 2 i 2 1 ] \begin{align*}
P_{1} & =\left|\lambda_{1}\right\rangle \left\langle \lambda_{1}\right| & P_{2} & =\left|\lambda_{2}\right\rangle \left\langle \lambda_{2}\right| & P_{3} & =\left|\lambda_{3}\right\rangle \left\langle \lambda_{3}\right|\\
& =\left[\begin{array}{c}
\frac{1}{\sqrt{2}}\\
0\\
\frac{1}{\sqrt{2}}
\end{array}\right]\left[\begin{array}{ccc}
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\end{array}\right] & & =\left[\begin{array}{c}
\frac{1}{2}\\
\frac{i}{\sqrt{2}}\\
\frac{-1}{2}
\end{array}\right]\left[\begin{array}{ccc}
\frac{1}{2} & \frac{-i}{\sqrt{2}} & \frac{-1}{2}\end{array}\right] & & =\left[\begin{array}{c}
\frac{-1}{2}\\
\frac{i}{\sqrt{2}}\\
\frac{1}{2}
\end{array}\right]\left[\begin{array}{ccc}
\frac{-1}{2} & \frac{-i}{\sqrt{2}} & \frac{1}{2}\end{array}\right]\\
P_{1} & =\frac{1}{2}\left[\begin{array}{ccc}
1 & 0 & 1\\
0 & 0 & 0\\
1 & 0 & 1
\end{array}\right] & P_{2} & =\frac{1}{4}\left[\begin{array}{ccc}
1 & -\frac{2i}{\sqrt{2}} & -1\\
\frac{2i}{\sqrt{2}} & 2 & -\frac{2i}{\sqrt{2}}\\
-1 & \frac{2i}{\sqrt{2}} & 1
\end{array}\right] & P_{2} & =\frac{1}{4}\left[\begin{array}{ccc}
1 & \frac{2i}{\sqrt{2}} & -1\\
-\frac{2i}{\sqrt{2}} & 2 & \frac{2i}{\sqrt{2}}\\
-1 & \frac{-2i}{\sqrt{2}} & 1
\end{array}\right]
\end{align*} P 1 P 1 = ∣ λ 1 ⟩ ⟨ λ 1 ∣ = ⎣ ⎡ 2 1 0 2 1 ⎦ ⎤ [ 2 1 0 2 1 ] = 2 1 ⎣ ⎡ 1 0 1 0 0 0 1 0 1 ⎦ ⎤ P 2 P 2 = ∣ λ 2 ⟩ ⟨ λ 2 ∣ = ⎣ ⎡ 2 1 2 i 2 − 1 ⎦ ⎤ [ 2 1 2 − i 2 − 1 ] = 4 1 ⎣ ⎡ 1 2 2 i − 1 − 2 2 i 2 2 2 i − 1 − 2 2 i 1 ⎦ ⎤ P 3 P 2 = ∣ λ 3 ⟩ ⟨ λ 3 ∣ = ⎣ ⎡ 2 − 1 2 i 2 1 ⎦ ⎤ [ 2 − 1 2 − i 2 1 ] = 4 1 ⎣ ⎡ 1 − 2 2 i − 1 2 2 i 2 2 − 2 i − 1 2 2 i 1 ⎦ ⎤ Queremos demonstrar a relação de completude
P 1 + P 2 + P 3 = 1 2 [ 1 0 1 0 0 0 1 0 1 ] + 1 4 [ 1 − 2 i 2 − 1 2 i 2 2 − 2 i 2 − 1 2 i 2 1 ] + 1 4 [ 1 2 i 2 − 1 − 2 i 2 2 2 i 2 − 1 − 2 i 2 1 ] = 1 4 [ 4 0 0 0 4 0 0 0 4 ] = [ 1 0 0 0 1 0 0 0 2 ] = I \begin{align*}
P_{1}+P_{2}+P_{3} & =\frac{1}{2}\left[\begin{array}{ccc}
1 & 0 & 1\\
0 & 0 & 0\\
1 & 0 & 1
\end{array}\right]+\frac{1}{4}\left[\begin{array}{ccc}
1 & -\frac{2i}{\sqrt{2}} & -1\\
\frac{2i}{\sqrt{2}} & 2 & -\frac{2i}{\sqrt{2}}\\
-1 & \frac{2i}{\sqrt{2}} & 1
\end{array}\right]+\frac{1}{4}\left[\begin{array}{ccc}
1 & \frac{2i}{\sqrt{2}} & -1\\
-\frac{2i}{\sqrt{2}} & 2 & \frac{2i}{\sqrt{2}}\\
-1 & \frac{-2i}{\sqrt{2}} & 1
\end{array}\right]\\
& =\frac{1}{4}\left[\begin{array}{ccc}
4 & 0 & 0\\
0 & 4 & 0\\
0 & 0 & 4
\end{array}\right]\\
& =\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 2
\end{array}\right]\\
& =\mathbb{I}
\end{align*} P 1 + P 2 + P 3 = 2 1 ⎣ ⎡ 1 0 1 0 0 0 1 0 1 ⎦ ⎤ + 4 1 ⎣ ⎡ 1 2 2 i − 1 − 2 2 i 2 2 2 i − 1 − 2 2 i 1 ⎦ ⎤ + 4 1 ⎣ ⎡ 1 − 2 2 i − 1 2 2 i 2 2 − 2 i − 1 2 2 i 1 ⎦ ⎤ = 4 1 ⎣ ⎡ 4 0 0 0 4 0 0 0 4 ⎦ ⎤ = ⎣ ⎡ 1 0 0 0 1 0 0 0 2 ⎦ ⎤ = I E a ortogononalidade:
P 1 ⋅ P 2 = 1 4 [ 2 0 2 0 0 0 2 0 2 ] ⋅ 1 4 [ 1 − 2 i 2 − 1 2 i 2 2 − 2 i 2 − 1 2 i 2 1 ] P 2 ⋅ P 1 = 1 4 [ 1 − 2 i 2 − 1 2 i 2 2 − 2 i 2 − 1 2 i 2 1 ] ⋅ 1 4 [ 2 0 2 0 0 0 2 0 2 ] = 1 16 [ 0 0 0 0 0 0 0 0 0 ] = 1 16 [ 0 0 0 0 0 0 0 0 0 ] \begin{align*}
P_{1}\cdot P_{2} & =\frac{1}{4}\left[\begin{array}{ccc}
2 & 0 & 2\\
0 & 0 & 0\\
2 & 0 & 2
\end{array}\right]\cdot\frac{1}{4}\left[\begin{array}{ccc}
1 & -\frac{2i}{\sqrt{2}} & -1\\
\frac{2i}{\sqrt{2}} & 2 & -\frac{2i}{\sqrt{2}}\\
-1 & \frac{2i}{\sqrt{2}} & 1
\end{array}\right] & P_{2}\cdot P_{1} & =\frac{1}{4}\left[\begin{array}{ccc}
1 & -\frac{2i}{\sqrt{2}} & -1\\
\frac{2i}{\sqrt{2}} & 2 & -\frac{2i}{\sqrt{2}}\\
-1 & \frac{2i}{\sqrt{2}} & 1
\end{array}\right]\cdot\frac{1}{4}\left[\begin{array}{ccc}
2 & 0 & 2\\
0 & 0 & 0\\
2 & 0 & 2
\end{array}\right]\\
& =\frac{1}{16}\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right] & & =\frac{1}{16}\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]
\end{align*} P 1 ⋅ P 2 = 4 1 ⎣ ⎡ 2 0 2 0 0 0 2 0 2 ⎦ ⎤ ⋅ 4 1 ⎣ ⎡ 1 2 2 i − 1 − 2 2 i 2 2 2 i − 1 − 2 2 i 1 ⎦ ⎤ = 16 1 ⎣ ⎡ 0 0 0 0 0 0 0 0 0 ⎦ ⎤ P 2 ⋅ P 1 = 4 1 ⎣ ⎡ 1 2 2 i − 1 − 2 2 i 2 2 2 i − 1 − 2 2 i 1 ⎦ ⎤ ⋅ 4 1 ⎣ ⎡ 2 0 2 0 0 0 2 0 2 ⎦ ⎤ = 16 1 ⎣ ⎡ 0 0 0 0 0 0 0 0 0 ⎦ ⎤ P 1 ⋅ P 3 = 1 4 [ 2 0 2 0 0 0 2 0 2 ] ⋅ 1 4 [ 1 2 i 2 − 1 − 2 i 2 2 2 i 2 − 1 − 2 i 2 1 ] P 3 ⋅ P 1 = 1 4 [ 1 2 i 2 − 1 − 2 i 2 2 2 i 2 − 1 − 2 i 2 1 ] ⋅ 1 4 [ 2 0 2 0 0 0 2 0 2 ] = 1 16 [ 0 0 0 0 0 0 0 0 0 ] = 1 16 [ 0 0 0 0 0 0 0 0 0 ] \begin{align*}
P_{1}\cdot P_{3} & =\frac{1}{4}\left[\begin{array}{ccc}
2 & 0 & 2\\
0 & 0 & 0\\
2 & 0 & 2
\end{array}\right]\cdot\frac{1}{4}\left[\begin{array}{ccc}
1 & \frac{2i}{\sqrt{2}} & -1\\
-\frac{2i}{\sqrt{2}} & 2 & \frac{2i}{\sqrt{2}}\\
-1 & \frac{-2i}{\sqrt{2}} & 1
\end{array}\right] & P_{3}\cdot P_{1} & =\frac{1}{4}\left[\begin{array}{ccc}
1 & \frac{2i}{\sqrt{2}} & -1\\
-\frac{2i}{\sqrt{2}} & 2 & \frac{2i}{\sqrt{2}}\\
-1 & \frac{-2i}{\sqrt{2}} & 1
\end{array}\right]\cdot\frac{1}{4}\left[\begin{array}{ccc}
2 & 0 & 2\\
0 & 0 & 0\\
2 & 0 & 2
\end{array}\right]\\
& =\frac{1}{16}\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right] & & =\frac{1}{16}\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]
\end{align*} P 1 ⋅ P 3 = 4 1 ⎣ ⎡ 2 0 2 0 0 0 2 0 2 ⎦ ⎤ ⋅ 4 1 ⎣ ⎡ 1 − 2 2 i − 1 2 2 i 2 2 − 2 i − 1 2 2 i 1 ⎦ ⎤ = 16 1 ⎣ ⎡ 0 0 0 0 0 0 0 0 0 ⎦ ⎤ P 3 ⋅ P 1 = 4 1 ⎣ ⎡ 1 − 2 2 i − 1 2 2 i 2 2 − 2 i − 1 2 2 i 1 ⎦ ⎤ ⋅ 4 1 ⎣ ⎡ 2 0 2 0 0 0 2 0 2 ⎦ ⎤ = 16 1 ⎣ ⎡ 0 0 0 0 0 0 0 0 0 ⎦ ⎤ E por fim:
P 2 ⋅ P 3 = 1 4 [ 1 − 2 i 2 − 1 2 i 2 2 − 2 i 2 − 1 2 i 2 1 ] ⋅ 1 4 [ 1 2 i 2 − 1 − 2 i 2 2 2 i 2 − 1 − 2 i 2 1 ] P 3 ⋅ P 2 = 1 4 [ 1 2 i 2 − 1 − 2 i 2 2 2 i 2 − 1 − 2 i 2 1 ] ⋅ 1 4 [ 1 − 2 i 2 − 1 2 i 2 2 − 2 i 2 − 1 2 i 2 1 ] = 1 16 [ 1 − 2 i − 1 2 i 2 − 2 − 1 2 i 1 ] [ 1 2 i − 1 − 2 i 2 2 i − 1 2 i 1 ] = 1 16 [ 1 2 i − 1 − 2 i 2 2 i − 1 − 2 i 1 ] [ 1 − 2 i − 1 2 i 2 − 2 i − 1 2 i 1 ] = 1 16 [ 0 0 0 0 0 0 0 0 0 ] = 1 16 [ 0 0 0 0 0 0 0 0 0 ] \begin{align*}
P_{2}\cdot P_{3} & =\frac{1}{4}\left[\begin{array}{ccc}
1 & -\frac{2i}{\sqrt{2}} & -1\\
\frac{2i}{\sqrt{2}} & 2 & -\frac{2i}{\sqrt{2}}\\
-1 & \frac{2i}{\sqrt{2}} & 1
\end{array}\right]\cdot\frac{1}{4}\left[\begin{array}{ccc}
1 & \frac{2i}{\sqrt{2}} & -1\\
-\frac{2i}{\sqrt{2}} & 2 & \frac{2i}{\sqrt{2}}\\
-1 & \frac{-2i}{\sqrt{2}} & 1
\end{array}\right] & P_{3}\cdot P_{2} & =\frac{1}{4}\left[\begin{array}{ccc}
1 & \frac{2i}{\sqrt{2}} & -1\\
-\frac{2i}{\sqrt{2}} & 2 & \frac{2i}{\sqrt{2}}\\
-1 & \frac{-2i}{\sqrt{2}} & 1
\end{array}\right]\cdot\frac{1}{4}\left[\begin{array}{ccc}
1 & -\frac{2i}{\sqrt{2}} & -1\\
\frac{2i}{\sqrt{2}} & 2 & -\frac{2i}{\sqrt{2}}\\
-1 & \frac{2i}{\sqrt{2}} & 1
\end{array}\right]\\
& =\frac{1}{16}\left[\begin{array}{ccc}
1 & -\sqrt{2}i & -1\\
\sqrt{2}i & 2 & -\sqrt{2}\\
-1 & \sqrt{2}i & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & \sqrt{2}i & -1\\
-\sqrt{2}i & 2 & \sqrt{2}i\\
-1 & \sqrt{2}i & 1
\end{array}\right] & & =\frac{1}{16}\left[\begin{array}{ccc}
1 & \sqrt{2}i & -1\\
-\sqrt{2}i & 2 & \sqrt{2}i\\
-1 & -\sqrt{2}i & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & -\sqrt{2}i & -1\\
\sqrt{2}i & 2 & -\sqrt{2}i\\
-1 & \sqrt{2}i & 1
\end{array}\right]\\
& =\frac{1}{16}\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right] & & =\frac{1}{16}\left[\begin{array}{ccc}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right]
\end{align*} P 2 ⋅ P 3 = 4 1 ⎣ ⎡ 1 2 2 i − 1 − 2 2 i 2 2 2 i − 1 − 2 2 i 1 ⎦ ⎤ ⋅ 4 1 ⎣ ⎡ 1 − 2 2 i − 1 2 2 i 2 2 − 2 i − 1 2 2 i 1 ⎦ ⎤ = 16 1 ⎣ ⎡ 1 2 i − 1 − 2 i 2 2 i − 1 − 2 1 ⎦ ⎤ ⎣ ⎡ 1 − 2 i − 1 2 i 2 2 i − 1 2 i 1 ⎦ ⎤ = 16 1 ⎣ ⎡ 0 0 0 0 0 0 0 0 0 ⎦ ⎤ P 3 ⋅ P 2 = 4 1 ⎣ ⎡ 1 − 2 2 i − 1 2 2 i 2 2 − 2 i − 1 2 2 i 1 ⎦ ⎤ ⋅ 4 1 ⎣ ⎡ 1 2 2 i − 1 − 2 2 i 2 2 2 i − 1 − 2 2 i 1 ⎦ ⎤ = 16 1 ⎣ ⎡ 1 − 2 i − 1 2 i 2 − 2 i − 1 2 i 1 ⎦ ⎤ ⎣ ⎡ 1 2 i − 1 − 2 i 2 2 i − 1 − 2 i 1 ⎦ ⎤ = 16 1 ⎣ ⎡ 0 0 0 0 0 0 0 0 0 ⎦ ⎤ Problem 3 ¶ Consider a three-dimensional ket space. If a certain set of orthonormal kets say, { ∣ 1 ⟩ , ∣ 2 ⟩ , ∣ 3 ⟩ } \left\{ \left|1\right\rangle ,\left|2\right\rangle ,\left|3\right\rangle \right\} { ∣ 1 ⟩ , ∣ 2 ⟩ , ∣ 3 ⟩ } A A A B B B
A = ( a 0 0 0 − a 0 0 0 − a ) , B = ( b 0 0 0 0 − i b 0 i b 0 ) A=\left(\begin{array}{ccc}
a & 0 & 0\\
0 & -a & 0\\
0 & 0 & -a
\end{array}\right),\quad B=\left(\begin{array}{ccc}
b & 0 & 0\\
0 & 0 & -ib\\
0 & ib & 0
\end{array}\right) A = ⎝ ⎛ a 0 0 0 − a 0 0 0 − a ⎠ ⎞ , B = ⎝ ⎛ b 0 0 0 0 ib 0 − ib 0 ⎠ ⎞ with a a a b b b
a) Obviouly A A A B B B
b) Show that A A A B B B
c) Find a new set of orthonormal kets which are simultaneous eigenkets of both A A A B B B A A A B B B
3)
a)
Começamos calculando os autovalores de B:
det ∣ b − λ 0 0 0 − λ − i b 0 i b − λ ∣ = ( b − λ ) λ 2 − b 2 ( b − λ ) = 0 \det\left|\begin{array}{ccc}
b-\lambda & 0 & 0\\
0 & -\lambda & -ib\\
0 & ib & -\lambda
\end{array}\right|=\left(b-\lambda\right)\lambda^{2}-b^{2}\left(b-\lambda\right)=0 det ∣ ∣ b − λ 0 0 0 − λ ib 0 − ib − λ ∣ ∣ = ( b − λ ) λ 2 − b 2 ( b − λ ) = 0 Então:
( b − λ ) ( λ 2 − b 2 ) = 0 \left(b-\lambda\right)\left(\lambda^{2}-b^{2}\right)=0 ( b − λ ) ( λ 2 − b 2 ) = 0 Logo os autovalores são ( b , b , − b ) \left(b,b,-b\right) ( b , b , − b ) λ 2 \lambda^{2} λ 2 ± b \pm b ± b λ = b \lambda=b λ = b
b)
Se comutam, então [ A , B ] = 0 \left[A,B\right]=0 [ A , B ] = 0
[ A , B ] = A B − B A = ( a 0 0 0 − a 0 0 0 − a ) ( b 0 0 0 0 − i b 0 i b 0 ) − ( b 0 0 0 0 − i b 0 i b 0 ) ( a 0 0 0 − a 0 0 0 − a ) = ( a b 0 0 0 0 a i b 0 − a i b 0 ) − ( b a 0 0 0 0 i b a 0 − i b a 0 ) \begin{align*}
\left[A,B\right] & =AB-BA\\
& =\left(\begin{array}{ccc}
a & 0 & 0\\
0 & -a & 0\\
0 & 0 & -a
\end{array}\right)\left(\begin{array}{ccc}
b & 0 & 0\\
0 & 0 & -ib\\
0 & ib & 0
\end{array}\right)-\left(\begin{array}{ccc}
b & 0 & 0\\
0 & 0 & -ib\\
0 & ib & 0
\end{array}\right)\left(\begin{array}{ccc}
a & 0 & 0\\
0 & -a & 0\\
0 & 0 & -a
\end{array}\right)\\
& =\left(\begin{array}{ccc}
ab & 0 & 0\\
0 & 0 & aib\\
0 & -aib & 0
\end{array}\right)-\left(\begin{array}{ccc}
ba & 0 & 0\\
0 & 0 & iba\\
0 & -iba & 0
\end{array}\right)
\end{align*} [ A , B ] = A B − B A = ⎝ ⎛ a 0 0 0 − a 0 0 0 − a ⎠ ⎞ ⎝ ⎛ b 0 0 0 0 ib 0 − ib 0 ⎠ ⎞ − ⎝ ⎛ b 0 0 0 0 ib 0 − ib 0 ⎠ ⎞ ⎝ ⎛ a 0 0 0 − a 0 0 0 − a ⎠ ⎞ = ⎝ ⎛ ab 0 0 0 0 − aib 0 aib 0 ⎠ ⎞ − ⎝ ⎛ ba 0 0 0 0 − iba 0 iba 0 ⎠ ⎞ Logo [ A , B ] = 0 \left[A,B\right]=0 [ A , B ] = 0
c)
Por fim, queremos um conjunto de kets ortonormai que sejam autokets simultaneamente de ambos. Os autovetores de A A A a + = a a_{+}=a a + = a
( 0 0 0 0 − 2 a 0 0 0 − 2 a ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
0 & 0 & 0\\
0 & -2a & 0\\
0 & 0 & -2a
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ⎝ ⎛ 0 0 0 0 − 2 a 0 0 0 − 2 a ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ Então ∣ a + ⟩ = ( x , 0 , 0 ) \left|a_{+}\right\rangle =\left(x,0,0\right) ∣ a + ⟩ = ( x , 0 , 0 ) 0 ⋅ x 1 = 0 0\cdot x_{1}=0 0 ⋅ x 1 = 0 x 1 x_{1} x 1 − 2 a ⋅ x 2 = 0 -2a\cdot x_{2}=0 − 2 a ⋅ x 2 = 0 a ≠ 0 a\neq0 a = 0 x 2 = 0 x_{2}=0 x 2 = 0 x 3 x_{3} x 3 a − = a a_{-}=a a − = a
( 2 a 0 0 0 0 0 0 0 0 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
2a & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ⎝ ⎛ 2 a 0 0 0 0 0 0 0 0 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ O que nos retorna um autovetor do tipo ∣ a − ⟩ = ( 0 , x 1 , x 2 ) \left|a_{-}\right\rangle =\left(0,x_{1},x_{2}\right) ∣ a − ⟩ = ( 0 , x 1 , x 2 ) b + = b b_{+}=b b + = b
( 0 0 0 0 − b − i b 0 i b − b ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
0 & 0 & 0\\
0 & -b & -ib\\
0 & ib & -b
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ⎝ ⎛ 0 0 0 0 − b ib 0 − ib − b ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ De onde obtemos que − b x 2 − i b x 3 = 0 -bx_{2}-ibx_{3}=0 − b x 2 − ib x 3 = 0 x x = − i x 3 x_{x}=-ix_{3} x x = − i x 3 ∣ b + ⟩ = ( x 1 , − i x 3 , x 3 ) \left|b_{+}\right\rangle =\left(x_{1},-ix_{3},x_{3}\right) ∣ b + ⟩ = ( x 1 , − i x 3 , x 3 ) b − = − b b_{-}=-b b − = − b
( 2 b 0 0 0 b − i b 0 i b b ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
2b & 0 & 0\\
0 & b & -ib\\
0 & ib & b
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}
\end{array}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ⎝ ⎛ 2 b 0 0 0 b ib 0 − ib b ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ De onde obtemos que b x 2 − i b x 3 = 0 bx_{2}-ibx_{3}=0 b x 2 − ib x 3 = 0 x x = i x 3 x_{x}=ix_{3} x x = i x 3 ∣ b − ⟩ = ( 0 , i x 3 , x 3 ) \left|b_{-}\right\rangle =\left(0,ix_{3},x_{3}\right) ∣ b − ⟩ = ( 0 , i x 3 , x 3 )
∣ a + ⟩ = ( a 1 , 0 , 0 ) \left|a_{+}\right\rangle =\left(a_{1},0,0\right) ∣ a + ⟩ = ( a 1 , 0 , 0 ) a + = a a_{+}=a a + = a A A A
∣ a − ⟩ = ( 0 , a 2 , a 3 ) \left|a_{-}\right\rangle =\left(0,a_{2},a_{3}\right) ∣ a − ⟩ = ( 0 , a 2 , a 3 ) a − = − a a_{-}=-a a − = − a A A A
∣ b + ⟩ = ( b 1 , − i b 2 , b 2 ) \left|b_{+}\right\rangle =\left(b_{1},-ib_{2},b_{2}\right) ∣ b + ⟩ = ( b 1 , − i b 2 , b 2 ) b + = b b_{+}=b b + = b b b b
∣ b − ⟩ = ( 0 , i b 3 , b 3 ) \left|b_{-}\right\rangle =\left(0,ib_{3},b_{3}\right) ∣ b − ⟩ = ( 0 , i b 3 , b 3 ) b − = − b b_{-}=-b b − = − b B B B
Alteramos as notações de x i x_{i} x i a i a_{i} a i b i b_{i} b i ∣ a − ⟩ \left|a_{-}\right\rangle ∣ a − ⟩ ∣ b − ⟩ \left|b_{-}\right\rangle ∣ b − ⟩ a 2 = i b 3 a_{2}=ib_{3} a 2 = i b 3 a 3 = b 3 a_{3}=b_{3} a 3 = b 3
∣ a − , b − ⟩ = ( 0 , i b 3 , b 3 ) \left|a_{-},b_{-}\right\rangle =\left(0,ib_{3},b_{3}\right) ∣ a − , b − ⟩ = ( 0 , i b 3 , b 3 ) Para garantir a normalização então fazendo b 3 = 1 2 b_{3}=\frac{1}{\sqrt{2}} b 3 = 2 1
∣ a − , b − ⟩ = ( 0 , i 2 , 1 2 ) \left|a_{-},b_{-}\right\rangle =\left(0,\frac{i}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) ∣ a − , b − ⟩ = ( 0 , 2 i , 2 1 ) Uma vez que ⟨ a − , b − ∣ a − , b − ⟩ = 1 \sqrt{\left\langle a_{-},b_{-}|a_{-},b_{-}\right\rangle }=1 ⟨ a − , b − ∣ a − , b − ⟩ = 1 ∣ a + ⟩ \left|a_{+}\right\rangle ∣ a + ⟩ ∣ b + ⟩ \left|b_{+}\right\rangle ∣ b + ⟩ a 1 = b 1 a_{1}=b_{1} a 1 = b 1 b 2 = 0 b_{2}=0 b 2 = 0
∣ a + , b + ⟩ = ( b 1 , 0 , 0 ) \left|a_{+},b_{+}\right\rangle =\left(b_{1},0,0\right) ∣ a + , b + ⟩ = ( b 1 , 0 , 0 ) A normalização neste caso é trivial b 1 = 1 b_{1}=1 b 1 = 1
∣ a + , b + ⟩ = ( 1 , 0 , 0 ) \left|a_{+},b_{+}\right\rangle =\left(1,0,0\right) ∣ a + , b + ⟩ = ( 1 , 0 , 0 ) Por fim nos resta apenas os autovalores degenerados de cada matriz, ou seja, os autovetores ∣ a − ⟩ \left|a_{-}\right\rangle ∣ a − ⟩ ∣ b + ⟩ \left|b_{+}\right\rangle ∣ b + ⟩ b 1 = 0 b_{1}=0 b 1 = 0 a 2 = − i b 2 a_{2}=-ib_{2} a 2 = − i b 2 a 3 = b 2 a_{3}=b_{2} a 3 = b 2
∣ a − , b + ⟩ = ( 0 , − i b 2 , b 2 ) \left|a_{-},b_{+}\right\rangle =\left(0,-ib_{2},b_{2}\right) ∣ a − , b + ⟩ = ( 0 , − i b 2 , b 2 ) Fazendo novamente b 2 = 1 2 b_{2}=\frac{1}{\sqrt{2}} b 2 = 2 1
∣ a − , b + ⟩ = ( 0 , − i 2 , 1 2 ) \left|a_{-},b_{+}\right\rangle =\left(0,\frac{-i}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) ∣ a − , b + ⟩ = ( 0 , 2 − i , 2 1 ) Assim temos o seguinte conjunto de autovetores associados aos respectivos autovalores:
∣ a − , b − ⟩ = 1 2 ( 0 , i , 1 ) ∣ a + , b + ⟩ = ( 1 , 0 , 0 ) ∣ a − , b + ⟩ = 1 2 ( 0 , − i , 1 ) \left|a_{-},b_{-}\right\rangle =\frac{1}{\sqrt{2}}\left(0,i,1\right)\qquad\left|a_{+},b_{+}\right\rangle =\left(1,0,0\right)\qquad\left|a_{-},b_{+}\right\rangle =\frac{1}{\sqrt{2}}\left(0,-i,1\right) ∣ a − , b − ⟩ = 2 1 ( 0 , i , 1 ) ∣ a + , b + ⟩ = ( 1 , 0 , 0 ) ∣ a − , b + ⟩ = 2 1 ( 0 , − i , 1 ) Além de serem normalizados são ortogonais umavez que ⟨ a − , b − ∣ a + , b + ⟩ = ⟨ a − , b − ∣ a − , b + ⟩ = ⟨ a − , b + ∣ a + , b + ⟩ = 0 \left\langle a_{-},b_{-}|a_{+},b_{+}\right\rangle =\left\langle a_{-},b_{-}|a_{-},b_{+}\right\rangle =\left\langle a_{-},b_{+}|a_{+},b_{+}\right\rangle =0 ⟨ a − , b − ∣ a + , b + ⟩ = ⟨ a − , b − ∣ a − , b + ⟩ = ⟨ a − , b + ∣ a + , b + ⟩ = 0
Problem 4 ¶ The σ x \sigma_{x} σ x
σ x = ( 0 1 1 0 ) \sigma_{x}=\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right) σ x = ( 0 1 1 0 ) Prove the relation:
e i α σ x = I cos α + i σ x sin α e^{i\alpha\sigma_{x}}=\mathbb{I}\cos\alpha+i\sigma_{x}\sin\alpha e i α σ x = I cos α + i σ x sin α where I \mathbb{I} I 2 × 2 2\times2 2 × 2
Primeir verificaremos se σ x \sigma_{x} σ x
σ x † = ( 0 1 1 0 ) = σ x \sigma^{\dagger}_{x}=\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right)=\sigma_{x} σ x † = ( 0 1 1 0 ) = σ x Logo então f ( σ x ) = ∑ i f ( λ i ) ∣ λ i ⟩ ⟨ λ i ∣ f\left(\sigma_{x}\right)=\sum_{i}f\left(\lambda_{i}\right)\left|\lambda_{i}\right\rangle \left\langle \lambda_{i}\right| f ( σ x ) = ∑ i f ( λ i ) ∣ λ i ⟩ ⟨ λ i ∣ λ ± = ± 1 \lambda_{\pm}=\pm1 λ ± = ± 1 λ + = + 1 \lambda_{+}=+1 λ + = + 1
( − 1 1 1 − 1 ) ( x 1 x 2 ) = ( 0 0 ) \left(\begin{array}{cc}
-1 & 1\\
1 & -1
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right) ( − 1 1 1 − 1 ) ( x 1 x 2 ) = ( 0 0 ) Temos então x 1 = x 2 x_{1}=x_{2} x 1 = x 2 ∣ λ + ⟩ = ( 1 2 , 1 2 ) \left|\lambda_{+}\right\rangle =\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right) ∣ λ + ⟩ = ( 2 1 , 2 1 ) λ − = − 1 \lambda_{-}=-1 λ − = − 1
( 1 1 1 1 ) ( x 1 x 2 ) = ( 0 0 ) \left(\begin{array}{cc}
1 & 1\\
1 & 1
\end{array}\right)\left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right) ( 1 1 1 1 ) ( x 1 x 2 ) = ( 0 0 ) Um resultado similar onde x 1 = − x 2 x_{1}=-x_{2} x 1 = − x 2 ∣ λ − ⟩ = ( − 1 2 , 1 2 ) . \left|\lambda_{-}\right\rangle =\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right). ∣ λ − ⟩ = ( − 2 1 , 2 1 ) .
σ x = λ + ∣ λ + ⟩ ⟨ λ i + ∣ + λ − ∣ λ − ⟩ ⟨ λ i − ∣ = ( + 1 ) ( 1 2 1 2 ) ( 1 2 1 2 ) + ( − 1 ) ( − 1 2 1 2 ) ( − 1 2 1 2 ) = ( + 1 ) ( 1 2 1 2 1 2 1 2 ) + ( − 1 ) ( 1 2 − 1 2 − 1 2 1 2 ) \begin{align*}
\sigma_{x} & =\lambda_{+}\left|\lambda_{+}\right\rangle \left\langle \lambda_{i+}\right|+\lambda_{-}\left|\lambda_{-}\right\rangle \left\langle \lambda_{i-}\right|\\
& =\left(+1\right)\left(\begin{array}{c}
\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{array}\right)\left(\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)+\left(-1\right)\left(\begin{array}{c}
-\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{array}\right)\left(\begin{array}{cc}
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)\\
& =\left(+1\right)\left(\begin{array}{cc}
\frac{1}{2} & \frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right)+\left(-1\right)\left(\begin{array}{cc}
\frac{1}{2} & -\frac{1}{2}\\
-\frac{1}{2} & \frac{1}{2}
\end{array}\right)
\end{align*} σ x = λ + ∣ λ + ⟩ ⟨ λ i + ∣ + λ − ∣ λ − ⟩ ⟨ λ i − ∣ = ( + 1 ) ( 2 1 2 1 ) ( 2 1 2 1 ) + ( − 1 ) ( − 2 1 2 1 ) ( − 2 1 2 1 ) = ( + 1 ) ( 2 1 2 1 2 1 2 1 ) + ( − 1 ) ( 2 1 − 2 1 − 2 1 2 1 ) Sendo então f ( x ) = e i α x f\left(x\right)=e^{i\alpha x} f ( x ) = e i αx
e i α σ x = e i α ( 1 2 1 2 1 2 1 2 ) + e − i α ( 1 2 − 1 2 − 1 2 1 2 ) = ( e i α + e − i α 2 e i α − e − i α 2 e i α − e − i α 2 e i α + e − i α 2 ) = ( e i α + e − i α 2 0 0 e i α + e − i α 2 ) + ( 0 e i α − e − i α 2 e i α − e − i α 2 0 ) = e i α + e − i α 2 ( 1 0 0 1 ) + e i α − e − i α 2 ( 0 1 1 0 ) = e i α + e − i α 2 I + e i α − e − i α 2 σ x \begin{align*}
e^{i\alpha\sigma_{x}} & =e^{i\alpha}\left(\begin{array}{cc}
\frac{1}{2} & \frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right)+e^{-i\alpha}\left(\begin{array}{cc}
\frac{1}{2} & -\frac{1}{2}\\
-\frac{1}{2} & \frac{1}{2}
\end{array}\right)\\
& =\left(\begin{array}{cc}
\frac{e^{i\alpha}+e^{-i\alpha}}{2} & \frac{e^{i\alpha}-e^{-i\alpha}}{2}\\
\frac{e^{i\alpha}-e^{-i\alpha}}{2} & \frac{e^{i\alpha}+e^{-i\alpha}}{2}
\end{array}\right)\\
& =\left(\begin{array}{cc}
\frac{e^{i\alpha}+e^{-i\alpha}}{2} & 0\\
0 & \frac{e^{i\alpha}+e^{-i\alpha}}{2}
\end{array}\right)+\left(\begin{array}{cc}
0 & \frac{e^{i\alpha}-e^{-i\alpha}}{2}\\
\frac{e^{i\alpha}-e^{-i\alpha}}{2} & 0
\end{array}\right)\\
& =\frac{e^{i\alpha}+e^{-i\alpha}}{2}\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right)+\frac{e^{i\alpha}-e^{-i\alpha}}{2}\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right)\\
& =\frac{e^{i\alpha}+e^{-i\alpha}}{2}\mathbb{I}+\frac{e^{i\alpha}-e^{-i\alpha}}{2}\sigma_{x}
\end{align*} e i α σ x = e i α ( 2 1 2 1 2 1 2 1 ) + e − i α ( 2 1 − 2 1 − 2 1 2 1 ) = ( 2 e i α + e − i α 2 e i α − e − i α 2 e i α − e − i α 2 e i α + e − i α ) = ( 2 e i α + e − i α 0 0 2 e i α + e − i α ) + ( 0 2 e i α − e − i α 2 e i α − e − i α 0 ) = 2 e i α + e − i α ( 1 0 0 1 ) + 2 e i α − e − i α ( 0 1 1 0 ) = 2 e i α + e − i α I + 2 e i α − e − i α σ x Usando então as relações trigonométricas cos ( α ) = e i α + e − i α 2 \cos\left(\alpha\right)=\frac{e^{i\alpha}+e^{-i\alpha}}{2} cos ( α ) = 2 e i α + e − i α sin ( α ) = e i α − e − i α 2 i \sin\left(\alpha\right)=\frac{e^{i\alpha}-e^{-i\alpha}}{2i} sin ( α ) = 2 i e i α − e − i α
e i α σ x = cos α I + i σ x sin ( α ) e^{i\alpha\sigma_{x}}=\mathbb{\cos\alpha I}+i\sigma_{x}\sin\left(\alpha\right) e i α σ x = cos α I + i σ x sin ( α )