Problem set 2
These are my solutions to the problem sets assigned to me during the graduate-level quantum mechanics course.
Problem 1 ¶ 1) Suppose a 2 × 2 2\times2 2 × 2 X X X X = a 0 + σ ⋅ a X=a_{0}+\sigma\cdot\boldsymbol{a} X = a 0 + σ ⋅ a a 0 a_{0} a 0 a 1 , 2 , 3 a_{1,2,3} a 1 , 2 , 3
a) How are a 0 a_{0} a 0 a k a_{k} a k k = 1 , 2 , 3 k=1,2,3 k = 1 , 2 , 3 T r ( X ) Tr\left(X\right) T r ( X ) T r ( σ k X ) Tr\left(\sigma_{k}X\right) T r ( σ k X )
b) Obtain a0 and ak in terms of the matrix elements Xij.
Tendo então uma matriz X X X 2 × 2 2\times2 2 × 2
X = ( a 0 + σ → ⋅ a → ) σ 0 X=\left(a_{0}+\overrightarrow{\sigma}\cdot\overrightarrow{a}\right)\sigma_{0} X = ( a 0 + σ ⋅ a ) σ 0 Sendo σ → = ( σ 1 , σ 2 , σ 3 ) \overrightarrow{\sigma}=\left(\sigma_{1},\sigma_{2},\sigma_{3}\right) σ = ( σ 1 , σ 2 , σ 3 ) σ k \sigma_{k} σ k σ 0 \sigma_{0} σ 0 a k a_{k} a k ( k = 0 , 1 , 2 , 3 ) \left(k=0,1,2,3\right) ( k = 0 , 1 , 2 , 3 ) a → = ( a 1 , a 2 , a 3 ) \overrightarrow{a}=\left(a_{1},a_{2},a_{3}\right) a = ( a 1 , a 2 , a 3 )
X = ( a 0 + σ 1 a 1 + σ 2 a 2 + σ 3 a 3 ) σ 0 = a 0 σ 0 + σ 1 a 1 σ 0 + σ 2 a 2 σ 0 + a 3 σ 3 σ 0 = a 0 σ 0 + σ 1 a 1 + σ 2 a 2 + a 3 σ 3 = ∑ k = 0 3 a k σ k \begin{align*}
X= & \left(a_{0}+\sigma_{1}a_{1}+\sigma_{2}a_{2}+\sigma_{3}a_{3}\right)\sigma_{0}\\
= & a_{0}\sigma_{0}+\sigma_{1}a_{1}\sigma_{0}+\sigma_{2}a_{2}\sigma_{0}+a_{3}\sigma_{3}\sigma_{0}\\
= & a_{0}\sigma_{0}+\sigma_{1}a_{1}+\sigma_{2}a_{2}+a_{3}\sigma_{3}\\
= & \sum^{3}_{k=0}a_{k}\sigma_{k}
\end{align*} X = = = = ( a 0 + σ 1 a 1 + σ 2 a 2 + σ 3 a 3 ) σ 0 a 0 σ 0 + σ 1 a 1 σ 0 + σ 2 a 2 σ 0 + a 3 σ 3 σ 0 a 0 σ 0 + σ 1 a 1 + σ 2 a 2 + a 3 σ 3 k = 0 ∑ 3 a k σ k Pela seguinte propriedade do traçoç T r ( A + B ) = T r ( A ) + T r ( B ) Tr\left(A+B\right)=Tr\left(A\right)+Tr\left(B\right) T r ( A + B ) = T r ( A ) + T r ( B ) T r ( c A ) = c T r ( A ) Tr\left(cA\right)=cTr\left(A\right) T r ( c A ) = c T r ( A ) A , B A,B A , B c c c
T r ( X ) = T r ( ∑ k = 0 3 a k σ k ) = ∑ k = 0 3 a k T r ( σ k ) \begin{align*}
Tr\left(X\right)= & Tr\left(\sum^{3}_{k=0}a_{k}\sigma_{k}\right)\\
= & \sum^{3}_{k=0}a_{k}Tr\left(\sigma_{k}\right)
\end{align*} T r ( X ) = = T r ( k = 0 ∑ 3 a k σ k ) k = 0 ∑ 3 a k T r ( σ k ) Lembrando das matrizes de Pauli:
σ 0 = ( 1 0 0 1 ) , σ 1 = ( 0 1 1 0 ) , σ 2 = ( 0 − i i 0 ) , σ 3 = ( 1 0 0 − 1 ) \sigma_{0}=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right),\sigma_{1}=\left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right),\sigma_{2}=\left(\begin{array}{cc}
0 & -i\\
i & 0
\end{array}\right),\sigma_{3}=\left(\begin{array}{cc}
1 & 0\\
0 & -1
\end{array}\right) σ 0 = ( 1 0 0 1 ) , σ 1 = ( 0 1 1 0 ) , σ 2 = ( 0 i − i 0 ) , σ 3 = ( 1 0 0 − 1 ) Então temos que:
T r ( σ 0 ) = 2 T r k ≠ 0 ( σ k ) = 0 Tr\left(\sigma_{0}\right)=2\qquad Tr_{k\neq0}\left(\sigma_{k}\right)=0 T r ( σ 0 ) = 2 T r k = 0 ( σ k ) = 0 Logo:
T r ( X ) = 2 a 0 Tr\left(X\right)=2a_{0} T r ( X ) = 2 a 0 Agora precisamos calcular primeiramente ( σ k X ) \left(\sigma_{k}X\right) ( σ k X ) σ i σ j = δ i j σ 0 + i ϵ i j k σ k \sigma_{i}\sigma_{j}=\delta_{ij}\sigma_{0}+i\epsilon_{ijk}\sigma_{k} σ i σ j = δ ij σ 0 + i ϵ ijk σ k
σ i X = σ i ( ∑ j = 0 3 a j σ j ) = ∑ j = 0 3 a j σ i σ j = ∑ j = 0 3 a j ( δ i j σ 0 + i ϵ i j k σ k ) = ∑ j = 0 3 ( δ i j a j σ 0 + i ϵ i j k a j σ k ) \begin{align*}
\sigma_{i}X= & \sigma_{i}\left(\sum^{3}_{j=0}a_{j}\sigma_{j}\right)\\
= & \sum^{3}_{j=0}a_{j}\sigma_{i}\sigma_{j}\\
= & \sum^{3}_{j=0}a_{j}\left(\delta_{ij}\sigma_{0}+i\epsilon_{ijk}\sigma_{k}\right)\\
= & \sum^{3}_{j=0}\left(\delta_{ij}a_{j}\sigma_{0}+i\epsilon_{ijk}a_{j}\sigma_{k}\right)
\end{align*} σ i X = = = = σ i ( j = 0 ∑ 3 a j σ j ) j = 0 ∑ 3 a j σ i σ j j = 0 ∑ 3 a j ( δ ij σ 0 + i ϵ ijk σ k ) j = 0 ∑ 3 ( δ ij a j σ 0 + i ϵ ijk a j σ k ) De acordo com a condições propostas no enunciado temos que i = 1 , 2 , 3 i=1,2,3 i = 1 , 2 , 3
T r ( σ i X ) = ∑ j = 0 3 ( δ i j a j T r ( σ 0 ) + i ϵ i j k a j T r ( σ k ) ) = ∑ j = 0 3 δ i j 2 a j = 2 a i \begin{align*}
Tr\left(\sigma_{i}X\right)= & \sum^{3}_{j=0}\left(\delta_{ij}a_{j}Tr\left(\sigma_{0}\right)+i\epsilon_{ijk}a_{j}Tr\left(\sigma_{k}\right)\right)\\
= & \sum^{3}_{j=0}\delta_{ij}2a_{j}\\
= & 2a_{i}
\end{align*} T r ( σ i X ) = = = j = 0 ∑ 3 ( δ ij a j T r ( σ 0 ) + i ϵ ijk a j T r ( σ k ) ) j = 0 ∑ 3 δ ij 2 a j 2 a i Temos então as seguintes relações:
T r ( X ) = 2 a 0 T r ( σ k X ) = 2 a k Tr\left(X\right)=2a_{0}\qquad Tr\left(\sigma_{k}X\right)=2a_{k} T r ( X ) = 2 a 0 T r ( σ k X ) = 2 a k b)
Dos resultados anteriores sabemos que:
a 0 = T r ( X ) 2 a k = T r ( σ k X ) 2 a_{0}=\frac{Tr\left(X\right)}{2}\qquad a_{k}=\frac{Tr\left(\sigma_{k}X\right)}{2} a 0 = 2 T r ( X ) a k = 2 T r ( σ k X ) Escrevendo então X X X
X = ( X 11 X 12 X 21 X 22 ) X=\left(\begin{array}{cc}
X_{11} & X_{12}\\
X_{21} & X_{22}
\end{array}\right) X = ( X 11 X 21 X 12 X 22 ) Pode-se perceber que:
a 0 = T r ( X ) 2 = X 11 + X 22 2 a_{0}=\frac{Tr\left(X\right)}{2}=\frac{X_{11}+X_{22}}{2} a 0 = 2 T r ( X ) = 2 X 11 + X 22 E calculando o produto das matrizes:
σ 1 X = ( 0 1 1 0 ) ( X 11 X 12 X 21 X 22 ) = ( X 21 X 22 X 11 X 12 ) σ 2 X = ( 0 − i i 0 ) ( X 11 X 12 X 21 X 22 ) = ( − i X 21 − i X 22 i X 11 i X 12 ) σ 1 X = ( 1 0 0 − 1 ) ( X 11 X 12 X 21 X 22 ) = ( X 11 X 12 − X 21 − X 22 ) \begin{alignat}{2}
\sigma_{1}X= & \left(\begin{array}{cc}
0 & 1\\
1 & 0
\end{array}\right)\left(\begin{array}{cc}
X_{11} & X_{12}\\
X_{21} & X_{22}
\end{array}\right)= & \left(\begin{array}{cc}
X_{21} & X_{22}\\
X_{11} & X_{12}
\end{array}\right)\\
\sigma_{2}X= & \left(\begin{array}{cc}
0 & -i\\
i & 0
\end{array}\right)\left(\begin{array}{cc}
X_{11} & X_{12}\\
X_{21} & X_{22}
\end{array}\right)= & \left(\begin{array}{cc}
-iX_{21} & -iX_{22}\\
iX_{11} & iX_{12}
\end{array}\right)\\
\sigma_{1}X= & \left(\begin{array}{cc}
1 & 0\\
0 & -1
\end{array}\right)\left(\begin{array}{cc}
X_{11} & X_{12}\\
X_{21} & X_{22}
\end{array}\right)= & \left(\begin{array}{cc}
X_{11} & X_{12}\\
-X_{21} & -X_{22}
\end{array}\right)
\end{alignat} σ 1 X = σ 2 X = σ 1 X = ( 0 1 1 0 ) ( X 11 X 21 X 12 X 22 ) = ( 0 i − i 0 ) ( X 11 X 21 X 12 X 22 ) = ( 1 0 0 − 1 ) ( X 11 X 21 X 12 X 22 ) = ( X 21 X 11 X 22 X 12 ) ( − i X 21 i X 11 − i X 22 i X 12 ) ( X 11 − X 21 X 12 − X 22 ) Logo, podemos obter quase que diretamente:
a 1 = 1 2 T r ( σ 1 X ) = X 21 + X 12 2 a 2 = 1 2 T r ( σ 2 X ) = i X 12 − X 21 2 a 3 = 1 2 T r ( σ 3 X ) = X 11 − X 22 2 \begin{alignat}{2}
a_{1}= & \frac{1}{2}Tr\left(\sigma_{1}X\right)= & \frac{X_{21}+X_{12}}{2}\\
a_{2}= & \frac{1}{2}Tr\left(\sigma_{2}X\right)= & i\frac{X_{12}-X_{21}}{2}\\
a_{3}= & \frac{1}{2}Tr\left(\sigma_{3}X\right)= & \frac{X_{11}-X_{22}}{2}
\end{alignat} a 1 = a 2 = a 3 = 2 1 T r ( σ 1 X ) = 2 1 T r ( σ 2 X ) = 2 1 T r ( σ 3 X ) = 2 X 21 + X 12 i 2 X 12 − X 21 2 X 11 − X 22 Problem 2 ¶ The Hamiltonian operator for a two-state system is given by
H = a ( ∣ 1 ⟩ ⟨ 1 ∣ − ∣ 2 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 2 ∣ + ∣ 2 ⟩ ⟨ 1 ∣ ) H=a\left(\left|1\right\rangle \left\langle 1\right|-\left|2\right\rangle \left\langle 2\right|+\left|1\right\rangle \left\langle 2\right|+\left|2\right\rangle \left\langle 1\right|\right) H = a ( ∣ 1 ⟩ ⟨ 1 ∣ − ∣ 2 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 2 ∣ + ∣ 2 ⟩ ⟨ 1 ∣ ) where a a a ∣ 1 ⟩ \left|1\right\rangle ∣ 1 ⟩ ∣ 2 ⟩ \left|2\right\rangle ∣ 2 ⟩
Temos o seguinte hamiltoniano para um sistema de dois estados:
H = a ( ∣ 1 ⟩ ⟨ 1 ∣ − ∣ 2 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 2 ∣ + ∣ 2 ⟩ ⟨ 1 ∣ ) H=a\left(\left|1\right\rangle \left\langle 1\right|-\left|2\right\rangle \left\langle 2\right|+\left|1\right\rangle \left\langle 2\right|+\left|2\right\rangle \left\langle 1\right|\right) H = a ( ∣ 1 ⟩ ⟨ 1 ∣ − ∣ 2 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 2 ∣ + ∣ 2 ⟩ ⟨ 1 ∣ ) Sendo ∣ 1 ⟩ = ( 1 0 ) T \left|1\right\rangle =\left(\begin{array}{cc}
1 & 0\end{array}\right)^{T} ∣ 1 ⟩ = ( 1 0 ) T ∣ 2 ⟩ = ( 0 1 ) T \left|2\right\rangle =\left(\begin{array}{cc}
0 & 1\end{array}\right)^{T} ∣ 2 ⟩ = ( 0 1 ) T
H = a [ ( 1 0 ) ( 1 0 ) − ( 0 1 ) ( 0 1 ) + ( 1 0 ) ( 0 1 ) + ( 0 1 ) ( 1 0 ) ] = a [ ( 1 0 0 0 ) − ( 0 0 0 1 ) + ( 0 1 0 0 ) + ( 0 0 1 0 ) ] = a ( 1 1 1 − 1 ) \begin{align*}
H= & a\left[\left(\begin{array}{c}
1\\
0
\end{array}\right)\left(\begin{array}{cc}
1 & 0\end{array}\right)-\left(\begin{array}{c}
0\\
1
\end{array}\right)\left(\begin{array}{cc}
0 & 1\end{array}\right)+\left(\begin{array}{c}
1\\
0
\end{array}\right)\left(\begin{array}{cc}
0 & 1\end{array}\right)+\left(\begin{array}{c}
0\\
1
\end{array}\right)\left(\begin{array}{cc}
1 & 0\end{array}\right)\right]\\
= & a\left[\left(\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right)-\left(\begin{array}{cc}
0 & 0\\
0 & 1
\end{array}\right)+\left(\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right)+\left(\begin{array}{cc}
0 & 0\\
1 & 0
\end{array}\right)\right]\\
= & a\left(\begin{array}{cc}
1 & 1\\
1 & -1
\end{array}\right)
\end{align*} H = = = a [ ( 1 0 ) ( 1 0 ) − ( 0 1 ) ( 0 1 ) + ( 1 0 ) ( 0 1 ) + ( 0 1 ) ( 1 0 ) ] a [ ( 1 0 0 0 ) − ( 0 0 0 1 ) + ( 0 0 1 0 ) + ( 0 1 0 0 ) ] a ( 1 1 1 − 1 ) Os autovalores são então dados pela raíz do determinante:
∣ H − λ I ∣ = ∣ a − λ a a − ( a + λ ) ∣ = 0 0 = − ( a − λ ) ( a + λ ) − a 2 0 = − ( a 2 − λ 2 ) − a 2 0 = − a 2 + λ 2 − a 2 0 = − 2 a 2 + λ 2 λ 2 = 2 a 2 λ = ± a 2 \begin{align*}
\left|H-\lambda\mathbb{I}\right|= & \left|\begin{array}{cc}
a-\lambda & a\\
a & -\left(a+\lambda\right)
\end{array}\right|=0\\
0= & -\left(a-\lambda\right)\left(a+\lambda\right)-a^{2}\\
0= & -\left(a^{2}-\lambda^{2}\right)-a^{2}\\
0= & -a^{2}+\lambda^{2}-a^{2}\\
0= & -2a^{2}+\lambda^{2}\\
\lambda^{2}= & 2a^{2}\\
\lambda & =\pm a\sqrt{2}
\end{align*} ∣ H − λ I ∣ = 0 = 0 = 0 = 0 = λ 2 = λ ∣ ∣ a − λ a a − ( a + λ ) ∣ ∣ = 0 − ( a − λ ) ( a + λ ) − a 2 − ( a 2 − λ 2 ) − a 2 − a 2 + λ 2 − a 2 − 2 a 2 + λ 2 2 a 2 = ± a 2 Calculando então o autovetor para λ + = a 2 \lambda_{+}=a\sqrt{2} λ + = a 2
( H − λ I ) ∣ λ + ⟩ = 0 ( a ( 1 − 2 ) a a − a ( 1 + 2 ) ) ( x 1 x 2 ) = ( 0 0 ) \begin{align*}
\left(H-\lambda\mathbb{I}\right)\left|\lambda_{+}\right\rangle & =\boldsymbol{0}\\
\left(\begin{array}{cc}
a\left(1-\sqrt{2}\right) & a\\
a & -a\left(1+\sqrt{2}\right)
\end{array}\right) & \left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right)
\end{align*} ( H − λ I ) ∣ λ + ⟩ ( a ( 1 − 2 ) a a − a ( 1 + 2 ) ) = 0 ( x 1 x 2 ) = ( 0 0 ) Como:
− a ( 1 + 2 ) ⋅ ( 1 − 2 ) = − a ( 1 − 2 ) = a -a\left(1+\sqrt{2}\right)\cdot\left(1-\sqrt{2}\right)=-a\left(1-2\right)=a − a ( 1 + 2 ) ⋅ ( 1 − 2 ) = − a ( 1 − 2 ) = a As linhas são linearmente dependentes. Então da primeira linha podemos obter que:
0 = a ( 1 − 2 ) x 1 + a x 2 0 = ( 1 − 2 ) x 1 + x 2 x 2 = ( 2 − 1 ) x 1 \begin{align*}
0 & =a\left(1-\sqrt{2}\right)x_{1}+ax_{2}\\
0 & =\left(1-\sqrt{2}\right)x_{1}+x_{2}\\
x_{2} & =\left(\sqrt{2}-1\right)x_{1}
\end{align*} 0 0 x 2 = a ( 1 − 2 ) x 1 + a x 2 = ( 1 − 2 ) x 1 + x 2 = ( 2 − 1 ) x 1 Logo o autovetor normalizado pode ser escrito:
∣ λ + ⟩ = ( x 1 ∣ 1 ⟩ + ( 2 − 1 ) x 1 ∣ 2 ⟩ ) x 1 2 + ( 2 − 1 ) 2 x 1 2 \left|\lambda_{+}\right\rangle =\frac{\left(x_{1}\left|1\right\rangle +\left(\sqrt{2}-1\right)x_{1}\left|2\right\rangle \right)}{\sqrt{x^{2}_{1}+\left(\sqrt{2}-1\right)^{2}x^{2}_{1}}} ∣ λ + ⟩ = x 1 2 + ( 2 − 1 ) 2 x 1 2 ( x 1 ∣ 1 ⟩ + ( 2 − 1 ) x 1 ∣ 2 ⟩ ) Escolhendo x 1 = 1 x_{1}=1 x 1 = 1
∣ λ + ⟩ = ( ∣ 1 ⟩ + ( 2 − 1 ) ∣ 2 ⟩ ) 1 + 2 − 2 2 + 1 = 1 4 − 2 2 ( ∣ 1 ⟩ + ( 2 − 1 ) ∣ 2 ⟩ ) \left|\lambda_{+}\right\rangle =\frac{\left(\left|1\right\rangle +\left(\sqrt{2}-1\right)\left|2\right\rangle \right)}{\sqrt{1+2-2\sqrt{2}+1}}=\frac{1}{\sqrt{4-2\sqrt{2}}}\left(\left|1\right\rangle +\left(\sqrt{2}-1\right)\left|2\right\rangle \right) ∣ λ + ⟩ = 1 + 2 − 2 2 + 1 ( ∣ 1 ⟩ + ( 2 − 1 ) ∣ 2 ⟩ ) = 4 − 2 2 1 ( ∣ 1 ⟩ + ( 2 − 1 ) ∣ 2 ⟩ ) E para λ − = − a 2 \lambda_{-}=-a\sqrt{2} λ − = − a 2
( H − λ I ) ∣ λ + ⟩ = 0 ( a ( 1 + 2 ) a a a ( 2 − 1 ) ) ( x 1 x 2 ) = ( 0 0 ) \begin{align*}
\left(H-\lambda\mathbb{I}\right)\left|\lambda_{+}\right\rangle & =\boldsymbol{0}\\
\left(\begin{array}{cc}
a\left(1+\sqrt{2}\right) & a\\
a & a\left(\sqrt{2}-1\right)
\end{array}\right) & \left(\begin{array}{c}
x_{1}\\
x_{2}
\end{array}\right)=\left(\begin{array}{c}
0\\
0
\end{array}\right)
\end{align*} ( H − λ I ) ∣ λ + ⟩ ( a ( 1 + 2 ) a a a ( 2 − 1 ) ) = 0 ( x 1 x 2 ) = ( 0 0 ) Como:
− a ( 1 + 2 ) ⋅ ( 1 − 2 ) = − a ( 1 − 2 ) = a -a\left(1+\sqrt{2}\right)\cdot\left(1-\sqrt{2}\right)=-a\left(1-2\right)=a − a ( 1 + 2 ) ⋅ ( 1 − 2 ) = − a ( 1 − 2 ) = a As linhas novamente são linearmente dependentes. De modo análogo:
0 = a ( 1 + 2 ) x 1 + a x 2 0 = ( 1 + 2 ) x 1 + x 2 x 2 = − ( 1 + 2 ) x 1 \begin{align*}
0 & =a\left(1+\sqrt{2}\right)x_{1}+ax_{2}\\
0 & =\left(1+\sqrt{2}\right)x_{1}+x_{2}\\
x_{2} & =-\left(1+\sqrt{2}\right)x_{1}
\end{align*} 0 0 x 2 = a ( 1 + 2 ) x 1 + a x 2 = ( 1 + 2 ) x 1 + x 2 = − ( 1 + 2 ) x 1 Logo o autovetor normalizado pode ser escrito:
∣ λ − ⟩ = ( x 1 ∣ 1 ⟩ − ( 1 + 2 ) x 1 ∣ 2 ⟩ ) x 1 2 + ( 2 + 1 ) 2 x 1 2 \left|\lambda_{-}\right\rangle =\frac{\left(x_{1}\left|1\right\rangle -\left(1+\sqrt{2}\right)x_{1}\left|2\right\rangle \right)}{\sqrt{x^{2}_{1}+\left(\sqrt{2}+1\right)^{2}x^{2}_{1}}} ∣ λ − ⟩ = x 1 2 + ( 2 + 1 ) 2 x 1 2 ( x 1 ∣ 1 ⟩ − ( 1 + 2 ) x 1 ∣ 2 ⟩ ) Escolhendo x 1 = 1 x_{1}=1 x 1 = 1
∣ λ − ⟩ = ( ∣ 1 ⟩ − ( 2 + 1 ) ∣ 2 ⟩ ) 1 + 2 + 2 2 + 1 = 1 4 + 2 2 ( ∣ 1 ⟩ − ( 2 + 1 ) ∣ 2 ⟩ ) \left|\lambda_{-}\right\rangle =\frac{\left(\left|1\right\rangle -\left(\sqrt{2}+1\right)\left|2\right\rangle \right)}{\sqrt{1+2+2\sqrt{2}+1}}=\frac{1}{\sqrt{4+2\sqrt{2}}}\left(\left|1\right\rangle -\left(\sqrt{2}+1\right)\left|2\right\rangle \right) ∣ λ − ⟩ = 1 + 2 + 2 2 + 1 ( ∣ 1 ⟩ − ( 2 + 1 ) ∣ 2 ⟩ ) = 4 + 2 2 1 ( ∣ 1 ⟩ − ( 2 + 1 ) ∣ 2 ⟩ ) Desta forma, reescrevendo os autovetores utilizando um resultado numérico, obtemos que:
∣ λ + ⟩ = 0.92 ∣ 1 ⟩ + 0.38 ∣ 2 ⟩ ∣ λ + ⟩ = 0.38 ∣ 1 ⟩ − 0.92 ∣ 2 ⟩ \left|\lambda_{+}\right\rangle =0.92\left|1\right\rangle +0.38\left|2\right\rangle \qquad\left|\lambda_{+}\right\rangle =0.38\left|1\right\rangle -0.92\left|2\right\rangle ∣ λ + ⟩ = 0.92 ∣ 1 ⟩ + 0.38 ∣ 2 ⟩ ∣ λ + ⟩ = 0.38 ∣ 1 ⟩ − 0.92 ∣ 2 ⟩ Problem 3 ¶ A certain observable in quantum mechanics has a 3 × 3 3\times3 3 × 3
1 2 ( 0 1 0 1 0 1 0 1 0 ) \frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 1\\
0 & 1 & 0
\end{array}\right) 2 1 ⎝ ⎛ 0 1 0 1 0 1 0 1 0 ⎠ ⎞ a) Find the normalized eigenvectors of this observable and the corresponding eigenvalues. Is there any degeneracy?
b) Give a physical example where all this is relevant.
Temos a seguinte matriz:
1 2 ( 0 1 0 1 0 1 0 1 0 ) \frac{1}{\sqrt{2}}\left(\begin{array}{ccc}
0 & 1 & 0\\
1 & 0 & 1\\
0 & 1 & 0
\end{array}\right) 2 1 ⎝ ⎛ 0 1 0 1 0 1 0 1 0 ⎠ ⎞ Calculando entao os autovalores:
∣ − λ 1 2 0 1 2 − λ 1 2 0 1 2 − λ ∣ = ( − λ 3 + λ 4 ) = ( 0 0 0 ) \left|\begin{array}{ccc}
-\lambda & \frac{1}{\sqrt{2}} & 0\\
\frac{1}{\sqrt{2}} & -\lambda & \frac{1}{\sqrt{2}}\\
0 & \frac{1}{\sqrt{2}} & -\lambda
\end{array}\right|=\left(-\lambda^{3}+\frac{\lambda}{4}\right)=\left(\begin{array}{c}
0\\
0\\
0
\end{array}\right) ∣ ∣ − λ 2 1 0 2 1 − λ 2 1 0 2 1 − λ ∣ ∣ = ( − λ 3 + 4 λ ) = ⎝ ⎛ 0 0 0 ⎠ ⎞ A partir da seguinte equação de segundo grau − λ ( λ 2 − 1 ) = 0 -\lambda\left(\lambda^{2}-1\right)=0 − λ ( λ 2 − 1 ) = 0 λ 1 = 0 \lambda_{1}=0 λ 1 = 0 λ 2 , 3 = ± 1 \lambda_{2,3}=\pm1 λ 2 , 3 = ± 1
( 0 1 2 0 1 2 0 1 2 0 1 2 0 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
0 & \frac{1}{\sqrt{2}} & 0\\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\\
0 & \frac{1}{\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 1 0 2 1 0 2 1 0 2 1 0 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ Da primeira e última 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 0 1 2 ) T \begin{align*}
\left|\lambda_{1}\right\rangle & =\left(\begin{array}{ccc}
-\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}\end{array}\right)^{T}
\end{align*} ∣ λ 1 ⟩ = ( − 2 1 0 2 1 ) T De modo análogo para λ 2 = 1 \lambda_{2}=1 λ 2 = 1
( − 1 1 2 0 1 2 − 1 1 2 0 1 2 − 1 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
-1 & \frac{1}{\sqrt{2}} & 0\\
\frac{1}{\sqrt{2}} & -1 & \frac{1}{\sqrt{2}}\\
0 & \frac{1}{\sqrt{2}} & -1
\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) ⎝ ⎛ − 1 2 1 0 2 1 − 1 2 1 0 2 1 − 1 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ Da primeira linha − x 1 + x 2 2 = 0 -x_{1}+\frac{x_{2}}{\sqrt{2}}=0 − x 1 + 2 x 2 = 0 x 1 = x 2 2 x_{1}=\frac{x_{2}}{\sqrt{2}} x 1 = 2 x 2 x 2 2 − x 3 = 0 \frac{x_{2}}{\sqrt{2}}-x_{3}=0 2 x 2 − x 3 = 0 x 3 = x 2 2 x_{3}=\frac{x_{2}}{\sqrt{2}} x 3 = 2 x 2
1 2 ( x 2 2 ) − x 2 + 1 2 ( x 2 2 ) = 0 x 2 2 − x 2 + x 2 2 = 0 ( 1 − 1 ) x 2 = 0 \begin{align*}
\frac{1}{\sqrt{2}}\left(\frac{x_{2}}{\sqrt{2}}\right)-x_{2}+\frac{1}{\sqrt{2}}\left(\frac{x_{2}}{\sqrt{2}}\right) & =0\\
\frac{x_{2}}{2}-x_{2}+\frac{x_{2}}{2} & =0\\
\left(1-1\right)x_{2} & =0
\end{align*} 2 1 ( 2 x 2 ) − x 2 + 2 1 ( 2 x 2 ) 2 x 2 − x 2 + 2 x 2 ( 1 − 1 ) x 2 = 0 = 0 = 0 Então x 2 x_{2} x 2 L 2 L2 L 2 ( L 2 = − 1 2 ( L 1 + L 3 ) ) \left(L_{2}=-\frac{1}{\sqrt{2}}\left(L_{1}+L_{3}\right)\right) ( L 2 = − 2 1 ( L 1 + L 3 ) ) x 2 = 2 x_{2}=\sqrt{2} x 2 = 2
∣ λ 2 ⟩ 1 1 + 2 + 1 ( 1 2 1 ) T ∣ λ 2 ⟩ = 1 2 ( 1 2 1 ) T \begin{align*}
\left|\lambda_{2}\right\rangle & \frac{1}{\sqrt{1+2+1}}\left(\begin{array}{ccc}
1 & \sqrt{2} & 1\end{array}\right)^{T}\\
\left|\lambda_{2}\right\rangle & =\frac{1}{2}\left(\begin{array}{ccc}
1 & \sqrt{2} & 1\end{array}\right)^{T}
\end{align*} ∣ λ 2 ⟩ ∣ λ 2 ⟩ 1 + 2 + 1 1 ( 1 2 1 ) T = 2 1 ( 1 2 1 ) T E por fim λ 3 = − 1 \lambda_{3}=-1 λ 3 = − 1
( 1 1 2 0 1 2 1 1 2 0 1 2 1 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) \left(\begin{array}{ccc}
1 & \frac{1}{\sqrt{2}} & 0\\
\frac{1}{\sqrt{2}} & 1 & \frac{1}{\sqrt{2}}\\
0 & \frac{1}{\sqrt{2}} & 1
\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) ⎝ ⎛ 1 2 1 0 2 1 1 2 1 0 2 1 1 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ Então da primeira linha x 1 + x 2 2 = 0 x_{1}+\frac{x_{2}}{\sqrt{2}}=0 x 1 + 2 x 2 = 0 x 1 = − x 2 2 x_{1}=-\frac{x_{2}}{\sqrt{2}} x 1 = − 2 x 2 x 2 2 + x 3 = 0 \frac{x_{2}}{\sqrt{2}}+x_{3}=0 2 x 2 + x 3 = 0 x 3 = − x 2 2 x_{3}=-\frac{x_{2}}{\sqrt{2}} x 3 = − 2 x 2
1 2 ( − x 2 2 ) + x 2 + 1 2 ( − x 2 2 ) = 0 − x 2 2 + x 2 − x 2 2 = 0 ( 1 − 1 ) x 2 = 0 \begin{align*}
\frac{1}{\sqrt{2}}\left(-\frac{x_{2}}{\sqrt{2}}\right)+x_{2}+\frac{1}{\sqrt{2}}\left(-\frac{x_{2}}{\sqrt{2}}\right) & =0\\
-\frac{x_{2}}{2}+x_{2}-\frac{x_{2}}{2} & =0\\
\left(1-1\right)x_{2} & =0
\end{align*} 2 1 ( − 2 x 2 ) + x 2 + 2 1 ( − 2 x 2 ) − 2 x 2 + x 2 − 2 x 2 ( 1 − 1 ) x 2 = 0 = 0 = 0 Em situação análoga ao caso anterior, escolhemos: x 2 = − 2 x_{2}=-\sqrt{2} x 2 = − 2
∣ λ 3 ⟩ 1 1 + 2 + 1 ( 1 − 2 1 ) T ∣ λ 3 ⟩ = 1 2 ( 1 − 2 1 ) T \begin{align*}
\left|\lambda_{3}\right\rangle & \frac{1}{\sqrt{1+2+1}}\left(\begin{array}{ccc}
1 & -\sqrt{2} & 1\end{array}\right)^{T}\\
\left|\lambda_{3}\right\rangle & =\frac{1}{2}\left(\begin{array}{ccc}
1 & -\sqrt{2} & 1\end{array}\right)^{T}
\end{align*} ∣ λ 3 ⟩ ∣ λ 3 ⟩ 1 + 2 + 1 1 ( 1 − 2 1 ) T = 2 1 ( 1 − 2 1 ) T Logo os autovetores são:
∣ λ 1 ⟩ = ( − 1 2 0 + 1 2 ) ∣ λ 2 ⟩ = 1 2 ( 1 2 1 ) ∣ λ 3 ⟩ = 1 2 ( 1 − 2 1 ) \left|\lambda_{1}\right\rangle =\left(\begin{array}{c}
-\frac{1}{\sqrt{2}}\\
0\\
+\frac{1}{\sqrt{2}}
\end{array}\right)\qquad\left|\lambda_{2}\right\rangle =\frac{1}{2}\left(\begin{array}{c}
1\\
\sqrt{2}\\
1
\end{array}\right)\qquad\left|\lambda_{3}\right\rangle =\frac{1}{2}\left(\begin{array}{c}
1\\
-\sqrt{2}\\
1
\end{array}\right) ∣ λ 1 ⟩ = ⎝ ⎛ − 2 1 0 + 2 1 ⎠ ⎞ ∣ λ 2 ⟩ = 2 1 ⎝ ⎛ 1 2 1 ⎠ ⎞ ∣ λ 3 ⟩ = 2 1 ⎝ ⎛ 1 − 2 1 ⎠ ⎞ b)
Com a diferença de uma constante, esta matriz pode ser usada para representar o momento angular com spin unitário na direção x x x L x L_{x} L x S x S_{x} S x
Problem 4 ¶ Find the linear combination of ∣ + ⟩ \left|+\right\rangle ∣ + ⟩ ∣ − ⟩ \left|-\right\rangle ∣ − ⟩
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ \left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ Verify explicity that for the linear combination you found, the uncertainty relation for S x S_{x} S x S y S_{y} S y
Um estado qualquer formado pela combinação dos vetores ∣ + ⟩ \left|+\right\rangle ∣ + ⟩ ∣ − ⟩ \left|-\right\rangle ∣ − ⟩
∣ α ⟩ = a ∣ + ⟩ + b e i β ∣ − ⟩ \left|\alpha\right\rangle =a\left|+\right\rangle +be^{i\beta}\left|-\right\rangle ∣ α ⟩ = a ∣ + ⟩ + b e i β ∣ − ⟩ Onde a a a b b b a 2 + b 2 = 1 a^{2}+b^{2}=1 a 2 + b 2 = 1
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ \left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ Os operadores S x S_{x} S x S y S_{y} S y
S x = ℏ 2 ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) S y = i ℏ 2 ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) S_{x}=\frac{\hbar}{2}\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\qquad S_{y}=\frac{i\hbar}{2}\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right) S x = 2 ℏ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) S y = 2 i ℏ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) Lembrando que:
⟨ ( Δ S i ) 2 ⟩ ≡ ⟨ S i 2 ⟩ − ⟨ S i ⟩ 2 ⟨ S i ⟩ = ⟨ α ∣ S i ∣ α ⟩ \left\langle \left(\Delta S_{i}\right)^{2}\right\rangle \equiv\left\langle S^{2}_{i}\right\rangle -\left\langle S_{i}\right\rangle ^{2}\qquad\left\langle S_{i}\right\rangle =\left\langle \alpha\left|S_{i}\right|\alpha\right\rangle ⟨ ( Δ S i ) 2 ⟩ ≡ ⟨ S i 2 ⟩ − ⟨ S i ⟩ 2 ⟨ S i ⟩ = ⟨ α ∣ S i ∣ α ⟩ Começando então calculando S i 2 S^{2}_{i} S i 2 I = ∣ + ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ − ∣ \mathbb{I}=\left|+\right\rangle \left\langle +\right|+\left|-\right\rangle \left\langle -\right| I = ∣ + ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ − ∣
S x 2 = ℏ 2 ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ℏ 2 ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) = ℏ 2 4 ( ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ + ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ) = ℏ 2 4 ( ∣ + ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ − ∣ ) = ℏ 2 4 \begin{align*}
S^{2}_{x} & =\frac{\hbar}{2}\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\frac{\hbar}{2}\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\\
& =\frac{\hbar^{2}}{4}\left(\left|+\right\rangle \left\langle -|+\right\rangle \left\langle -\right|+\left|+\right\rangle \left\langle -|-\right\rangle \left\langle +\right|+\left|-\right\rangle \left\langle +|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +|-\right\rangle \left\langle +\right|\right)\\
& =\frac{\hbar^{2}}{4}\left(\left|+\right\rangle \left\langle +\right|+\left|-\right\rangle \left\langle -\right|\right)\\
& =\frac{\hbar^{2}}{4}
\end{align*} S x 2 = 2 ℏ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) 2 ℏ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) = 4 ℏ 2 ( ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ + ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ) = 4 ℏ 2 ( ∣ + ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ − ∣ ) = 4 ℏ 2 E:
S y 2 = i ℏ 2 ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) i ℏ 2 ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) = − ℏ 2 4 [ + ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ − ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ] = − ℏ 2 4 [ − ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ] = ℏ 2 4 [ ∣ + ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ − ∣ ] = ℏ 2 4 \begin{align*}
S^{2}_{y} & =\frac{i\hbar}{2}\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\frac{i\hbar}{2}\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\\
& =-\frac{\hbar^{2}}{4}\left[+\left|+\right\rangle \left\langle -|+\right\rangle \left\langle -\right|-\left|+\right\rangle \left\langle -|-\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle +|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +|-\right\rangle \left\langle +\right|\right]\\
& =-\frac{\hbar^{2}}{4}\left[-\left|+\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle -\right|\right]\\
& =\frac{\hbar^{2}}{4}\left[\left|+\right\rangle \left\langle +\right|+\left|-\right\rangle \left\langle -\right|\right]\\
& =\frac{\hbar^{2}}{4}
\end{align*} S y 2 = 2 i ℏ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) 2 i ℏ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) = − 4 ℏ 2 [ + ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ − ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ] = − 4 ℏ 2 [ − ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ] = 4 ℏ 2 [ ∣ + ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ − ∣ ] = 4 ℏ 2 Logo, temos que S x 2 = S y 2 = ℏ 2 4 = S 2 S^{2}_{x}=S^{2}_{y}=\frac{\hbar^{2}}{4}=S^{2} S x 2 = S y 2 = 4 ℏ 2 = S 2 S = ℏ 2 4 S=\frac{\hbar^{2}}{4} S = 4 ℏ 2 S x S_{x} S x S y S_{y} S y
⟨ S 2 ⟩ = ⟨ α ∣ S 2 ∣ α ⟩ = ⟨ α ∣ ℏ 2 4 ∣ α ⟩ = ℏ 2 4 ⟨ α ∣ α ⟩ = ℏ 2 4 \begin{align*}
\left\langle S^{2}\right\rangle & =\left\langle \alpha\left|S^{2}\right|\alpha\right\rangle =\left\langle \alpha\left|\frac{\hbar^{2}}{4}\right|\alpha\right\rangle =\frac{\hbar^{2}}{4}\left\langle \alpha|\alpha\right\rangle =\frac{\hbar^{2}}{4}
\end{align*} ⟨ S 2 ⟩ = ⟨ α ∣ ∣ S 2 ∣ ∣ α ⟩ = ⟨ α ∣ ∣ 4 ℏ 2 ∣ ∣ α ⟩ = 4 ℏ 2 ⟨ α ∣ α ⟩ = 4 ℏ 2 E para o segundo termo:
⟨ α ∣ S i ∣ α ⟩ = ( ⟨ + ∣ a + ⟨ − ∣ b e − i β ) S i ( a ∣ + ⟩ + b e i β ∣ − ⟩ ) = a 2 ⟨ + ∣ S i ∣ + ⟩ + a b e i β ⟨ + ∣ S i ∣ − ⟩ + a b e − i β ⟨ − ∣ S i ∣ + ⟩ + b 2 ⟨ − ∣ S i ∣ − ⟩ \begin{align*}
\left\langle \alpha\left|S_{i}\right|\alpha\right\rangle & =\left(\left\langle +\right|a+\left\langle -\right|be^{-i\beta}\right)S_{i}\left(a\left|+\right\rangle +be^{i\beta}\left|-\right\rangle \right)\\
& =a^{2}\left\langle +\right|S_{i}\left|+\right\rangle +abe^{i\beta}\left\langle +\right|S_{i}\left|-\right\rangle +abe^{-i\beta}\left\langle -\right|S_{i}\left|+\right\rangle +b^{2}\left\langle -\right|S_{i}\left|-\right\rangle
\end{align*} ⟨ α ∣ S i ∣ α ⟩ = ( ⟨ + ∣ a + ⟨ − ∣ b e − i β ) S i ( a ∣ + ⟩ + b e i β ∣ − ⟩ ) = a 2 ⟨ + ∣ S i ∣ + ⟩ + ab e i β ⟨ + ∣ S i ∣ − ⟩ + ab e − i β ⟨ − ∣ S i ∣ + ⟩ + b 2 ⟨ − ∣ S i ∣ − ⟩ Para S x S_{x} S x
⟨ α ∣ S x ∣ α ⟩ = ℏ 2 ( a 2 ⟨ + ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ + a b e i β ⟨ + ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ + a b ⟨ − ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ e − i β + b 2 ⟨ − ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ ) = ℏ 2 ( a 2 ( ⟨ + ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ + ⟩ ) + a b e i β ( ⟨ + ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ − ⟩ ) + a b ( ⟨ − ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ + ⟩ ) e − i β + b 2 ( ⟨ − ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ − ⟩ ) ) = ℏ 2 ( a b e i β + a b e − i β ) = ℏ a b ( e i β + e − i β ) 2 = ℏ a b cos β \begin{align*}
\left\langle \alpha\left|S_{x}\right|\alpha\right\rangle = & \frac{\hbar}{2}\left(a^{2}\left\langle +\right|\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|+\right\rangle +abe^{i\beta}\left\langle +\right|\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|-\right\rangle +\right.\\
& \left.ab\left\langle -\right|\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|+\right\rangle e^{-i\beta}+b^{2}\left\langle -\right|\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|-\right\rangle \right)\\
= & \frac{\hbar}{2}\left(a^{2}\left(\left\langle +|+\right\rangle \left\langle -|+\right\rangle +\left\langle +|-\right\rangle \left\langle +|+\right\rangle \right)+abe^{i\beta}\left(\left\langle +|+\right\rangle \left\langle -|-\right\rangle +\left\langle +|-\right\rangle \left\langle +|-\right\rangle \right)+\right.\\
& \left.ab\left(\left\langle -|+\right\rangle \left\langle -|+\right\rangle +\left\langle -|-\right\rangle \left\langle +|+\right\rangle \right)e^{-i\beta}+b^{2}\left(\left\langle -|+\right\rangle \left\langle -|-\right\rangle +\left\langle -|-\right\rangle \left\langle +|-\right\rangle \right)\right)\\
= & \frac{\hbar}{2}\left(abe^{i\beta}+abe^{-i\beta}\right)\\
= & \hbar ab\frac{\left(e^{i\beta}+e^{-i\beta}\right)}{2}\\
= & \hbar ab\cos\beta
\end{align*} ⟨ α ∣ S x ∣ α ⟩ = = = = = 2 ℏ ( a 2 ⟨ + ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ + ab e i β ⟨ + ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ + ab ⟨ − ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ e − i β + b 2 ⟨ − ∣ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ ) 2 ℏ ( a 2 ( ⟨ + ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ + ⟩ ) + ab e i β ( ⟨ + ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ − ⟩ ) + ab ( ⟨ − ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ + ⟩ ) e − i β + b 2 ( ⟨ − ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ − ⟩ ) ) 2 ℏ ( ab e i β + ab e − i β ) ℏ ab 2 ( e i β + e − i β ) ℏ ab cos β Para S y S_{y} S y
⟨ α ∣ S y ∣ α ⟩ = i ℏ 2 ( a 2 ⟨ + ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ + a b e i β ⟨ + ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ + a b ⟨ − ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ e − i β + b 2 ⟨ − ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ ) = i ℏ 2 ( a 2 ( − ⟨ + ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ + ⟩ ) + a b e i β ( − ⟨ + ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ − ⟩ ) + a b ( − ⟨ − ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ + ⟩ ) e − i β + b 2 ( − ⟨ − ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ − ⟩ ) ) = i ℏ 2 ( − a b e i β + a b e − i β ) = ℏ a b i ( − e i β + e − i β ) 2 = ℏ a b ( e i β − e − i β ) 2 i = ℏ a b sin β \begin{align*}
\left\langle \alpha\left|S_{y}\right|\alpha\right\rangle = & \frac{i\hbar}{2}\left(a^{2}\left\langle +\right|\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|+\right\rangle +abe^{i\beta}\left\langle +\right|\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|-\right\rangle +\right.\\
& \left.ab\left\langle -\right|\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|+\right\rangle e^{-i\beta}+b^{2}\left\langle -\right|\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left|-\right\rangle \right)\\
= & \frac{i\hbar}{2}\left(a^{2}\left(-\left\langle +|+\right\rangle \left\langle -|+\right\rangle +\left\langle +|-\right\rangle \left\langle +|+\right\rangle \right)+abe^{i\beta}\left(-\left\langle +|+\right\rangle \left\langle -|-\right\rangle +\left\langle +|-\right\rangle \left\langle +|-\right\rangle \right)+\right.\\
& \left.ab\left(-\left\langle -|+\right\rangle \left\langle -|+\right\rangle +\left\langle -|-\right\rangle \left\langle +|+\right\rangle \right)e^{-i\beta}+b^{2}\left(-\left\langle -|+\right\rangle \left\langle -|-\right\rangle +\left\langle -|-\right\rangle \left\langle +|-\right\rangle \right)\right)\\
= & \frac{i\hbar}{2}\left(-abe^{i\beta}+abe^{-i\beta}\right)\\
= & \hbar ab\frac{i\left(-e^{i\beta}+e^{-i\beta}\right)}{2}\\
= & \hbar ab\frac{\left(e^{i\beta}-e^{-i\beta}\right)}{2i}\\
& =\hbar ab\sin\beta
\end{align*} ⟨ α ∣ S y ∣ α ⟩ = = = = = 2 i ℏ ( a 2 ⟨ + ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ + ab e i β ⟨ + ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ + ab ⟨ − ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ + ⟩ e − i β + b 2 ⟨ − ∣ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ∣ − ⟩ ) 2 i ℏ ( a 2 ( − ⟨ + ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ + ⟩ ) + ab e i β ( − ⟨ + ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ + ∣ − ⟩ ⟨ + ∣ − ⟩ ) + ab ( − ⟨ − ∣ + ⟩ ⟨ − ∣ + ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ + ⟩ ) e − i β + b 2 ( − ⟨ − ∣ + ⟩ ⟨ − ∣ − ⟩ + ⟨ − ∣ − ⟩ ⟨ + ∣ − ⟩ ) ) 2 i ℏ ( − ab e i β + ab e − i β ) ℏ ab 2 i ( − e i β + e − i β ) ℏ ab 2 i ( e i β − e − i β ) = ℏ ab sin β Logo temos:
⟨ S x ⟩ = ℏ a b cos β ⟨ S y ⟩ = ℏ a b sin β → ⟨ S x ⟩ 2 = ( ℏ a b cos β ) 2 ⟨ S y ⟩ 2 = ( ℏ a b sin β ) 2 \left\langle S_{x}\right\rangle =\hbar ab\cos\beta\qquad\left\langle S_{y}\right\rangle =\hbar ab\sin\beta\rightarrow\left\langle S_{x}\right\rangle ^{2}=\left(\hbar ab\cos\beta\right)^{2}\qquad\left\langle S_{y}\right\rangle ^{2}=\left(\hbar ab\sin\beta\right)^{2} ⟨ S x ⟩ = ℏ ab cos β ⟨ S y ⟩ = ℏ ab sin β → ⟨ S x ⟩ 2 = ( ℏ ab cos β ) 2 ⟨ S y ⟩ 2 = ( ℏ ab sin β ) 2 Substituindo então as relações de dispersão:
⟨ ( Δ S x ) 2 ⟩ = ⟨ S x 2 ⟩ − ⟨ S x ⟩ 2 ⟨ ( Δ S y ) 2 ⟩ = ⟨ S y 2 ⟩ − ⟨ S y ⟩ 2 = ℏ 2 4 − ( ℏ a b cos β ) 2 = ℏ 2 4 − ( ℏ a b sin β ) 2 \begin{align*}
\left\langle \left(\Delta S_{x}\right)^{2}\right\rangle = & \left\langle S^{2}_{x}\right\rangle -\left\langle S_{x}\right\rangle ^{2} & \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle = & \left\langle S^{2}_{y}\right\rangle -\left\langle S_{y}\right\rangle ^{2}\\
= & \frac{\hbar^{2}}{4}-\left(\hbar ab\cos\beta\right)^{2} & = & \frac{\hbar^{2}}{4}-\left(\hbar ab\sin\beta\right)^{2}
\end{align*} ⟨ ( Δ S x ) 2 ⟩ = = ⟨ S x 2 ⟩ − ⟨ S x ⟩ 2 4 ℏ 2 − ( ℏ ab cos β ) 2 ⟨ ( Δ S y ) 2 ⟩ = = ⟨ S y 2 ⟩ − ⟨ S y ⟩ 2 4 ℏ 2 − ( ℏ ab sin β ) 2 Então temos o seguinte produto da incerteza:
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = ( ℏ 2 4 − ( ℏ a b cos β ) 2 ) ( ℏ 2 4 − ( ℏ a b sin β ) 2 ) = ℏ 4 [ 1 16 − ( a b sin β ) 2 4 − ( a b cos β ) 2 4 + ( a b cos β ) 2 ( a b sin β ) 2 ] = ℏ 4 [ 1 16 − ( a b 2 ) 2 + ( a b ) 4 ( 2 cos β sin β 2 ) 2 ] \begin{align*}
\left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle & =\left(\frac{\hbar^{2}}{4}-\left(\hbar ab\cos\beta\right)^{2}\right)\left(\frac{\hbar^{2}}{4}-\left(\hbar ab\sin\beta\right)^{2}\right)\\
& =\hbar^{4}\left[\frac{1}{16}-\frac{\left(ab\sin\beta\right)^{2}}{4}-\frac{\left(ab\cos\beta\right)^{2}}{4}+\left(ab\cos\beta\right)^{2}\left(ab\sin\beta\right)^{2}\right]\\
& =\hbar^{4}\left[\frac{1}{16}-\left(\frac{ab}{2}\right)^{2}+\left(ab\right)^{4}\left(\frac{2\cos\beta\sin\beta}{2}\right)^{2}\right]
\end{align*} ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = ( 4 ℏ 2 − ( ℏ ab cos β ) 2 ) ( 4 ℏ 2 − ( ℏ ab sin β ) 2 ) = ℏ 4 [ 16 1 − 4 ( ab sin β ) 2 − 4 ( ab cos β ) 2 + ( ab cos β ) 2 ( ab sin β ) 2 ] = ℏ 4 [ 16 1 − ( 2 ab ) 2 + ( ab ) 4 ( 2 2 cos β sin β ) 2 ] E usando a identidade trigonométrica sin ( 2 x ) = 2 sin x cos x \sin\left(2x\right)=2\sin x\cos x sin ( 2 x ) = 2 sin x cos x
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = ℏ 4 [ 1 16 − ( a b 2 ) 2 + ( a b ) 4 ( sin ( 2 β ) 2 ) 2 ] \begin{align*}
\left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle & =\hbar^{4}\left[\frac{1}{16}-\left(\frac{ab}{2}\right)^{2}+\left(ab\right)^{4}\left(\frac{\sin\left(2\beta\right)}{2}\right)^{2}\right]
\end{align*} ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = ℏ 4 [ 16 1 − ( 2 ab ) 2 + ( ab ) 4 ( 2 sin ( 2 β ) ) 2 ] Como os termos ( a b ) 2 \left(ab\right)^{2} ( ab ) 2 sin ( 2 β ) = 1 \sin\left(2\beta\right)=1 sin ( 2 β ) = 1 − π ≤ β ≤ + π -\pi\leq\beta\leq+\pi − π ≤ β ≤ + π
2 β = ± π 2 → β = π 4 2\beta=\pm\frac{\pi}{2}\rightarrow\beta=\frac{\pi}{4} 2 β = ± 2 π → β = 4 π Nesta situação temos que:
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = ℏ 4 16 + ℏ 4 ( a b 2 ) 2 [ ( a b ) 2 − 1 ] \left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle =\frac{\hbar^{4}}{16}+\hbar^{4}\left(\frac{ab}{2}\right)^{2}\left[\left(ab\right)^{2}-1\right] ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = 16 ℏ 4 + ℏ 4 ( 2 ab ) 2 [ ( ab ) 2 − 1 ] Pela condição de normalização temos que b 2 = 1 − a 2 b^{2}=1-a^{2} b 2 = 1 − a 2
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = ℏ 4 16 [ 1 + [ 2 a 2 ( 1 − a 2 ) ] 2 − 4 a 2 ( 1 − a 2 ) ] = ℏ 4 16 [ 1 − 2 a 2 ( 1 − a 2 ) ] 2 \begin{align*}
\left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle & =\frac{\hbar^{4}}{16}\left[1+\left[2a^{2}\left(1-a^{2}\right)\right]^{2}-4a^{2}\left(1-a^{2}\right)\right]\\
& =\frac{\hbar^{4}}{16}\left[1-2a^{2}\left(1-a^{2}\right)\right]^{2}
\end{align*} ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = 16 ℏ 4 [ 1 + [ 2 a 2 ( 1 − a 2 ) ] 2 − 4 a 2 ( 1 − a 2 ) ] = 16 ℏ 4 [ 1 − 2 a 2 ( 1 − a 2 ) ] 2 Novamente pelas condições de normalização temos que 0 ≤ a 2 ≤ 1 0\leq a^{2}\leq1 0 ≤ a 2 ≤ 1 ( 1 − a 2 ) ≥ 0 \left(1-a^{2}\right)\geq0 ( 1 − a 2 ) ≥ 0 2 a 2 ( 1 − a 2 ) ≥ 0 2a^{2}\left(1-a^{2}\right)\geq0 2 a 2 ( 1 − a 2 ) ≥ 0 a 2 ( 1 − a 2 ) = 0 a^{2}\left(1-a^{2}\right)=0 a 2 ( 1 − a 2 ) = 0 a = ± 1 a=\pm1 a = ± 1 a = 0 a=0 a = 0
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = ℏ 4 16 \left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle =\frac{\hbar^{4}}{16} ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ = 16 ℏ 4 relembrando nosso estado geral:
∣ α ⟩ = a ∣ + ⟩ + b e i β ∣ − ⟩ \left|\alpha\right\rangle =a\left|+\right\rangle +be^{i\beta}\left|-\right\rangle ∣ α ⟩ = a ∣ + ⟩ + b e i β ∣ − ⟩ Ou ainda:
∣ α ⟩ = a ∣ + ⟩ + 1 − a 2 e i β ∣ − ⟩ \left|\alpha\right\rangle =a\left|+\right\rangle +\sqrt{1-a^{2}}e^{i\beta}\left|-\right\rangle ∣ α ⟩ = a ∣ + ⟩ + 1 − a 2 e i β ∣ − ⟩ Então o máximo de incerteza ocorre quando a = 0 a=0 a = 0 β = ± π 4 \beta=\pm\frac{\pi}{4} β = ± 4 π
∣ α 1 ⟩ = ± e ± π 4 i ∣ − ⟩ \left|\alpha_{1}\right\rangle =\pm e^{\pm\frac{\pi}{4}i}\left|-\right\rangle ∣ α 1 ⟩ = ± e ± 4 π i ∣ − ⟩ Ou se a = ± 1 a=\pm1 a = ± 1
∣ α 2 ⟩ = ± ∣ + ⟩ \left|\alpha_{2}\right\rangle =\pm\left|+\right\rangle ∣ α 2 ⟩ = ± ∣ + ⟩ Por fim, só nos resta verificar explicitamente que estas combinações lineares que encontramos não violam a relação de incerteza. Isto é, uma vez que a relação de incerteza para S x S_{x} S x S y S_{y} S y
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ ≥ 1 4 ∣ ⟨ [ S x , S y ] ⟩ ∣ 2 \left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle \geq\frac{1}{4}\left|\left\langle \left[S_{x},S_{y}\right]\right\rangle \right|^{2} ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ ≥ 4 1 ∣ ⟨ [ S x , S y ] ⟩ ∣ 2 Queremos demonstrar que :
ℏ 4 16 ≥ 1 4 ∣ ⟨ [ S x , S y ] ⟩ ∣ 2 \frac{\hbar^{4}}{16}\geq\frac{1}{4}\left|\left\langle \left[S_{x},S_{y}\right]\right\rangle \right|^{2} 16 ℏ 4 ≥ 4 1 ∣ ⟨ [ S x , S y ] ⟩ ∣ 2 Calculando então o comutador entre S x S_{x} S x S y S_{y} S y
[ S x , S y ] = S x S y − S y S x = ℏ 2 ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) i ℏ 2 ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) − ℏ 2 i ℏ 2 ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) = i ℏ 2 4 ( − ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ + ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ) − i ℏ 2 4 ( − ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ − ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ) = i ℏ 2 4 ( 2 ∣ + ⟩ ⟨ + ∣ − 2 ∣ − ⟩ ⟨ − ∣ ) = i ℏ [ ℏ 2 ( ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ) ] = i ℏ S z \begin{align*}
\left[S_{x},S_{y}\right]= & S_{x}S_{y}-S_{y}S_{x}\\
= & \frac{\hbar}{2}\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\frac{i\hbar}{2}\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)-\frac{\hbar}{2}\frac{i\hbar}{2}\left(-\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\left(\left|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +\right|\right)\\
= & \frac{i\hbar^{2}}{4}\left(-\left|+\right\rangle \left\langle -|+\right\rangle \left\langle -\right|+\left|+\right\rangle \left\langle -|-\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle +|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +|-\right\rangle \left\langle +\right|\right)-\\
& \frac{i\hbar^{2}}{4}\left(-\left|+\right\rangle \left\langle -|+\right\rangle \left\langle -\right|-\left|+\right\rangle \left\langle -|-\right\rangle \left\langle +\right|+\left|-\right\rangle \left\langle +|+\right\rangle \left\langle -\right|+\left|-\right\rangle \left\langle +|-\right\rangle \left\langle +\right|\right)\\
= & \frac{i\hbar^{2}}{4}\left(2\left|+\right\rangle \left\langle +\right|-2\left|-\right\rangle \left\langle -\right|\right)\\
= & i\hbar\left[\frac{\hbar}{2}\left(\left|+\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle -\right|\right)\right]\\
= & i\hbar S_{z}
\end{align*} [ S x , S y ] = = = = = = S x S y − S y S x 2 ℏ ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) 2 i ℏ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) − 2 ℏ 2 i ℏ ( − ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) ( ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ ) 4 i ℏ 2 ( − ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ + ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ) − 4 i ℏ 2 ( − ∣ + ⟩ ⟨ − ∣ + ⟩ ⟨ − ∣ − ∣ + ⟩ ⟨ − ∣ − ⟩ ⟨ + ∣ + ∣ − ⟩ ⟨ + ∣ + ⟩ ⟨ − ∣ + ∣ − ⟩ ⟨ + ∣ − ⟩ ⟨ + ∣ ) 4 i ℏ 2 ( 2 ∣ + ⟩ ⟨ + ∣ − 2 ∣ − ⟩ ⟨ − ∣ ) i ℏ [ 2 ℏ ( ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ) ] i ℏ S z Usando então ∣ α 1 ⟩ \left|\alpha_{1}\right\rangle ∣ α 1 ⟩
⟨ [ S x , S y ] ⟩ 1 = ⟨ α 1 ∣ i ℏ S z ∣ α 1 ⟩ = i ℏ 2 2 [ ⟨ − ∣ e ∓ π 4 i ( ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ) e ± π 4 i ∣ − ⟩ ] = i ℏ 2 2 [ ( ⟨ − ∣ + ⟩ ⟨ + ∣ − ⟩ − ⟨ − ∣ − ⟩ ⟨ − ∣ − ⟩ ) ] = − i ℏ 2 2 \begin{align*}
\left\langle \left[S_{x},S_{y}\right]\right\rangle _{1}= & \left\langle \alpha_{1}\left|i\hbar S_{z}\right|\alpha_{1}\right\rangle \\
= & \frac{i\hbar^{2}}{2}\left[\left\langle -\right|e^{\mp\frac{\pi}{4}i}\left(\left|+\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle -\right|\right)e^{\pm\frac{\pi}{4}i}\left|-\right\rangle \right]\\
= & \frac{i\hbar^{2}}{2}\left[\left(\left\langle -|+\right\rangle \left\langle +|-\right\rangle -\left\langle -|-\right\rangle \left\langle -|-\right\rangle \right)\right]\\
= & -\frac{i\hbar^{2}}{2}
\end{align*} ⟨ [ S x , S y ] ⟩ 1 = = = = ⟨ α 1 ∣ i ℏ S z ∣ α 1 ⟩ 2 i ℏ 2 [ ⟨ − ∣ e ∓ 4 π i ( ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ) e ± 4 π i ∣ − ⟩ ] 2 i ℏ 2 [ ( ⟨ − ∣ + ⟩ ⟨ + ∣ − ⟩ − ⟨ − ∣ − ⟩ ⟨ − ∣ − ⟩ ) ] − 2 i ℏ 2 Usando então ∣ α 2 ⟩ \left|\alpha_{2}\right\rangle ∣ α 2 ⟩
⟨ [ S x , S y ] ⟩ 2 = ⟨ α 2 ∣ i ℏ S z ∣ α 2 ⟩ = i ℏ 2 2 [ ⟨ + ∣ ± ( ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ) ± ∣ + ⟩ ] = i ℏ 2 2 [ ( ⟨ + ∣ + ⟩ ⟨ + ∣ + ⟩ − ⟨ + ∣ − ⟩ ⟨ − ∣ + ⟩ ) ] = i ℏ 2 2 \begin{align*}
\left\langle \left[S_{x},S_{y}\right]\right\rangle _{2}= & \left\langle \alpha_{2}\left|i\hbar S_{z}\right|\alpha_{2}\right\rangle \\
= & \frac{i\hbar^{2}}{2}\left[\left\langle +\right|\pm\left(\left|+\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle -\right|\right)\pm\left|+\right\rangle \right]\\
= & \frac{i\hbar^{2}}{2}\left[\left(\left\langle +|+\right\rangle \left\langle +|+\right\rangle -\left\langle +|-\right\rangle \left\langle -|+\right\rangle \right)\right]\\
= & \frac{i\hbar^{2}}{2}
\end{align*} ⟨ [ S x , S y ] ⟩ 2 = = = = ⟨ α 2 ∣ i ℏ S z ∣ α 2 ⟩ 2 i ℏ 2 [ ⟨ + ∣ ± ( ∣ + ⟩ ⟨ + ∣ − ∣ − ⟩ ⟨ − ∣ ) ± ∣ + ⟩ ] 2 i ℏ 2 [ ( ⟨ + ∣ + ⟩ ⟨ + ∣ + ⟩ − ⟨ + ∣ − ⟩ ⟨ − ∣ + ⟩ ) ] 2 i ℏ 2 Então independente de qual combinação linear escolhermos, ficamos com:
∣ ⟨ [ S x , S y ] ⟩ ∣ = ℏ 2 2 \left|\left\langle \left[S_{x},S_{y}\right]\right\rangle \right|=\frac{\hbar^{2}}{2} ∣ ⟨ [ S x , S y ] ⟩ ∣ = 2 ℏ 2 Consequentemente:
∣ ⟨ [ S x , S y ] ⟩ ∣ 2 = ℏ 4 4 \left|\left\langle \left[S_{x},S_{y}\right]\right\rangle \right|^{2}=\frac{\hbar^{4}}{4} ∣ ⟨ [ S x , S y ] ⟩ ∣ 2 = 4 ℏ 4 Logo a relação de incerteza é:
⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ ≥ ℏ 4 16 \left\langle \left(\Delta S_{x}\right)^{2}\right\rangle \left\langle \left(\Delta S_{y}\right)^{2}\right\rangle \geq\frac{\hbar^{4}}{16} ⟨ ( Δ S x ) 2 ⟩ ⟨ ( Δ S y ) 2 ⟩ ≥ 16 ℏ 4 Portanto ela não é violada, uma vez que:
ℏ 4 16 ≥ ℏ 4 16 \frac{\hbar^{4}}{16}\geq\frac{\hbar^{4}}{16} 16 ℏ 4 ≥ 16 ℏ 4 É uma igualdade válida.