Problem set 6
These are my solutions to the problem sets assigned to me during the graduate-level quantum mechanics course.
Problem 1 ¶ Note that if one associates with the coordinates x μ x^{\mu} x μ
X = ( x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 − x 3 ) = x μ σ μ X=\left(\begin{array}{cc}
x^{0}+x^{3} & x^{1}-ix^{2}\\
x^{1}+ix^{2} & x^{0}-x^{3}
\end{array}\right)=x^{\mu}\sigma_{\mu} X = ( x 0 + x 3 x 1 + i x 2 x 1 − i x 2 x 0 − x 3 ) = x μ σ μ then det X = D ( X ) = g μ ν x μ x ν \det X=D\left(X\right)=g_{\mu\nu}x^{\mu}x^{\nu} det X = D ( X ) = g μν x μ x ν has the value D ( X ) = g μ ν x μ x ν D\left(X\right)=g_{\mu\nu}x^{\mu}x^{\nu} D ( X ) = g μν x μ x ν
a) If the matrix
W = ( α β γ δ ) W=\left(\begin{array}{cc}
\alpha & \beta\\
\gamma & \delta
\end{array}\right) W = ( α γ β δ ) with complex elements α \alpha α β \beta β γ \gamma γ δ \delta δ
W ∗ = ( α ‾ γ ‾ β ‾ δ ‾ ) W^{*}=\left(\begin{array}{cc}
\overline{\alpha} & \overline{\gamma}\\
\overline{\beta} & \overline{\delta}
\end{array}\right) W ∗ = ( α β γ δ ) is its Hermitian adjoint, show that
defines a proper Lorentz transformation x ′ μ = Λ ν μ x ν x^{'\mu}=\Lambda^{\mu}_{\nu}x^{\nu} x ′ μ = Λ ν μ x ν , (i.e., show that det Λ = + 1 , Λ 0 0 ≥ 1 \det\Lambda=+1,\Lambda^{0}_{0}\ge1 det Λ = + 1 , Λ 0 0 ≥ 1 , and that D ( X ) = D ( X ′ ) D\left(X\right)=D\left(X'\right) D ( X ) = D ( X ′ )
Temos:
X = ( x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 − x 3 ) = x μ σ μ X=\left(\begin{array}{cc}
x^{0}+x^{3} & x^{1}-ix^{2}\\
x^{1}+ix^{2} & x^{0}-x^{3}
\end{array}\right)=x^{\mu}\sigma_{\mu} X = ( x 0 + x 3 x 1 + i x 2 x 1 − i x 2 x 0 − x 3 ) = x μ σ μ e det X = D ( X ) = g μ ν x μ x ν \det X=D\left(X\right)=g_{\mu\nu}x^{\mu}x^{\nu} det X = D ( X ) = g μν x μ x ν
W = ( α β γ δ ) W=\left(\begin{array}{cc}
\alpha & \beta\\
\gamma & \delta
\end{array}\right) W = ( α γ β δ ) E sua adjunta hermitiana é:
W ∗ = ( α ‾ γ ‾ β ‾ δ ‾ ) W^{*}=\left(\begin{array}{cc}
\overline{\alpha} & \overline{\gamma}\\
\overline{\beta} & \overline{\delta}
\end{array}\right) W ∗ = ( α β γ δ ) Então queremos mostrar que:
determine uma transormação de Lorentz x ′ μ = Λ ν μ x ν x^{'\mu}=\Lambda^{\mu}_{\nu}x^{\nu} x ′ μ = Λ ν μ x ν det Λ = + 1 \det\Lambda=+1 det Λ = + 1 Λ 0 0 ≥ 1 \Lambda^{0}_{0}\ge1 Λ 0 0 ≥ 1 D ( X ) = D ( X ′ ) D\left(X\right)=D\left(X'\right) D ( X ) = D ( X ′ )
Um vetor de espaço-tempo contravariante de 4 dimensões pode ser dado por x μ = ( c t , x ) = ( x 0 , x 1 , x 2 , x 3 ) x^{\mu}=\left(ct,\boldsymbol{x}\right)=\left(x^{0},x^{1},x^{2},x^{3}\right) x μ = ( c t , x ) = ( x 0 , x 1 , x 2 , x 3 )
x μ = g μ ν x ν x_{\mu}=g_{\mu\nu}x^{\nu} x μ = g μν x ν Onde o tensor métrico é responsável por subir ou baixar os índices:
g μ ν = ( 1 0 0 0 0 − 1 0 0 0 0 − 1 0 0 0 0 − 1 ) g_{\mu\nu}=\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{array}\right) g μν = ⎝ ⎛ 1 0 0 0 0 − 1 0 0 0 0 − 1 0 0 0 0 − 1 ⎠ ⎞ O produto interno então é dado entre o vetor covariante e contravariante:
x μ x μ = x μ x μ x_{\mu}x^{\mu}=x^{\mu}x_{\mu} x μ x μ = x μ x μ Dessa forma temos que D ( X ) = g μ ν x μ x ν = x ν x ν = x ν x ν D\left(X\right)=g_{\mu\nu}x^{\mu}x^{\nu}=x_{\nu}x^{\nu}=x^{\nu}x_{\nu} D ( X ) = g μν x μ x ν = x ν x ν = x ν x ν
x ′ μ = Λ ν μ x ν x^{'\mu}=\Lambda^{\mu}_{\nu}x^{\nu} x ′ μ = Λ ν μ x ν Temo então o produto escalar
x ′ μ g μ ν x ′ ν = x β Λ β μ g α β Λ α μ x α = x μ g α β x ν \begin{align*}
x^{'\mu}g_{\mu\nu}x^{'\nu} & =x^{\beta}\Lambda^{\mu}_{\beta}g_{\alpha\beta}\Lambda^{\mu}_{\alpha}x^{\alpha}\\
& =x^{\mu}g_{\alpha\beta}x^{\nu}
\end{align*} x ′ μ g μν x ′ ν = x β Λ β μ g α β Λ α μ x α = x μ g α β x ν Se demandamos que seja preservado o produto interno esalar de Minkowski:
Λ α μ g μ ν Λ β μ = g α β \Lambda^{\mu}_{\alpha}g_{\mu\nu}\Lambda^{\mu}_{\beta}=g_{\alpha\beta} Λ α μ g μν Λ β μ = g α β Então agora voltando às nossas matrizes, podemos ver que, utilizando o Wolfram Mathematica:
Det{[}\{\{x0~+~x3,~x1~-~I~x2\},~\{x1~+~I~x2,~x0~-~x3\}\}{]}D ( X ) = x ν x ν = x ν g μ ν x μ = ( x 0 ) 2 − ( x 1 ) 2 − ( x 2 ) 2 − ( x 3 ) 2 \begin{align*}
D\left(X\right) & =x_{\nu}x^{\nu}\\
& =x_{\nu}g_{\mu\nu}x_{\mu}\\
& =\left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}
\end{align*} D ( X ) = x ν x ν = x ν g μν x μ = ( x 0 ) 2 − ( x 1 ) 2 − ( x 2 ) 2 − ( x 3 ) 2 E:
X~:=~\{\{x0~+~x3,~x1~-~I~x2\},~\{x1~+~I~x2,~x0~-~x3\}\};
W~:=~\{\{a,~b\},~\{g,~d\}\};
WC~:=~\{\{Conjugate{[}a{]},~Conjugate{[}g{]}\},\{Conjugate{[}b{]},~Conjugate{[}d{]}\}\};
Simplify{[}Det{[}W~.~X~.~WC{]}{]}D ( X ′ ) = x ν ′ x ′ ν D ( X ′ ) = x ν ′ g μ ν x μ ′ D ( W X W ∗ ) = x β Λ β μ g μ ν Λ α μ x α = ( α δ − β γ ) [ ( x 0 ) 2 − ( x 1 ) 2 − ( x 2 ) 2 − ( x 3 ) 2 ] ( α ‾ δ ‾ − β ‾ γ ‾ ) \begin{align*}
D\left(X'\right) & =x^{'}_{\nu}x^{'\nu}\\
D\left(X'\right) & =x'_{\nu}g_{\mu\nu}x'_{\mu}\\
D\left(WXW^{*}\right) & =x^{\beta}\Lambda^{\mu}_{\beta}g_{\mu\nu}\Lambda^{\mu}_{\alpha}x^{\alpha}\\
& =\left(\alpha\delta-\beta\gamma\right)\left[\left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}\right]\left(\overline{\alpha}\overline{\delta}-\overline{\beta}\overline{\gamma}\right)
\end{align*} D ( X ′ ) D ( X ′ ) D ( W X W ∗ ) = x ν ′ x ′ ν = x ν ′ g μν x μ ′ = x β Λ β μ g μν Λ α μ x α = ( α δ − β γ ) [ ( x 0 ) 2 − ( x 1 ) 2 − ( x 2 ) 2 − ( x 3 ) 2 ] ( α δ − β γ ) Temos então, como queríamos demonstrar, que:
D ( X ′ ) = ( x 0 ) 2 − ( x 1 ) 2 − ( x 2 ) 2 − ( x 3 ) 2 = D ( X ) D\left(X'\right)=\left(x^{0}\right)^{2}-\left(x^{1}\right)^{2}-\left(x^{2}\right)^{2}-\left(x^{3}\right)^{2}=D\left(X\right) D ( X ′ ) = ( x 0 ) 2 − ( x 1 ) 2 − ( x 2 ) 2 − ( x 3 ) 2 = D ( X ) Tendo então X \mathbb{X} X 2 × 2 2\times2 2 × 2
X = { W , ∈ M a t 2 × 2 ( C ) : X = X † } \mathbb{X}=\left\{ W,\in Mat_{2\times2}\left(\mathbb{C}\right):X=X^{\dagger}\right\} X = { W , ∈ M a t 2 × 2 ( C ) : X = X † } Isso pode ser considerado uma forma alternativa de R 1 , 3 \mathbb{R}^{1,3} R 1 , 3
R 1 , 3 → X , ( x 0 , x 1 , x 2 , x 3 ) ↦ X = ( x 0 + x 3 x 1 − i x 2 x 1 + i x 2 x 0 − x 3 ) \mathbb{R}^{1,3}\rightarrow\mathbb{X},\quad\left(x^{0},x^{1},x^{2},x^{3}\right)\mapsto X=\left(\begin{array}{cc}
x^{0}+x^{3} & x^{1}-ix^{2}\\
x^{1}+ix^{2} & x^{0}-x^{3}
\end{array}\right) R 1 , 3 → X , ( x 0 , x 1 , x 2 , x 3 ) ↦ X = ( x 0 + x 3 x 1 + i x 2 x 1 − i x 2 x 0 − x 3 ) Lembrando que det X = x ν x ν \det X=x_{\nu}x^{\nu} det X = x ν x ν R 1 , 3 \mathbb{R}^{1,3} R 1 , 3 X \mathbb{X} X ( x 0 , x 1 , x 2 , x 3 ) \left(x^{0},x^{1},x^{2},x^{3}\right) ( x 0 , x 1 , x 2 , x 3 ) 2 × 2 2\times2 2 × 2 X = x μ σ μ X=x^{\mu}\sigma_{\mu} X = x μ σ μ
⟨ A , B ⟩ = T r ( A † B ) \left\langle A,B\right\rangle =Tr\left(A^{\dagger}B\right) ⟨ A , B ⟩ = T r ( A † B ) E a norma de Frobenius ∥ A ∥ = ⟨ A , A ⟩ \left\Vert A\right\Vert =\sqrt{\left\langle A,A\right\rangle } ∥ A ∥ = ⟨ A , A ⟩
⟨ σ μ , σ ν ⟩ = δ μ ν \left\langle \sigma_{\mu},\sigma_{\nu}\right\rangle =\delta_{\mu\nu} ⟨ σ μ , σ ν ⟩ = δ μν Desde que definimos o produto interno como: ⟨ A , B ⟩ = 1 2 T r ( A † B ) \left\langle A,B\right\rangle =\frac{1}{2}Tr\left(A^{\dagger}B\right) ⟨ A , B ⟩ = 2 1 T r ( A † B ) Lima (2013) X \mathbb{X} X R 1 , 3 \mathbb{R}^{1,3} R 1 , 3
X ↦ ( ⟨ X , σ 0 ⟩ , ⟨ X , σ 1 ⟩ , ⟨ X , σ 2 ⟩ , ⟨ X , σ 3 ⟩ ) X\mapsto\left(\left\langle X,\sigma_{0}\right\rangle ,\left\langle X,\sigma_{1}\right\rangle ,\left\langle X,\sigma_{2}\right\rangle ,\left\langle X,\sigma_{3}\right\rangle \right) X ↦ ( ⟨ X , σ 0 ⟩ , ⟨ X , σ 1 ⟩ , ⟨ X , σ 2 ⟩ , ⟨ X , σ 3 ⟩ ) Calculando para X = x μ σ μ X=x^{\mu}\sigma_{\mu} X = x μ σ μ
X ↦ ( x 0 , x 1 , x 2 , x 3 ) X\mapsto\left(x^{0},x^{1},x^{2},x^{3}\right) X ↦ ( x 0 , x 1 , x 2 , x 3 ) Então temos que:
X ′ ↦ ( x ′ 0 , x ′ 1 , x ′ 2 , x ′ 3 ) ↦ ( Λ ν 0 x ν , Λ ν 1 x ν , Λ ν 2 x ν , Λ ν 3 x ν ) ↦ ( ⟨ X ′ , σ 0 ⟩ , ⟨ X ′ , σ 1 ⟩ , ⟨ X ′ , σ 2 ⟩ , ⟨ X ′ , σ 3 ⟩ ) ↦ ( ⟨ W X W ∗ , σ 0 ⟩ , ⟨ W X W ∗ , σ 1 ⟩ , ⟨ W X W ∗ , σ 2 ⟩ , ⟨ W X W ∗ , σ 3 ⟩ ) \begin{align*}
X' & \mapsto\left(x^{'0},x^{'1},x^{'2},x^{'3}\right)\\
& \mapsto\left(\Lambda^{0}_{\nu}x^{\nu},\Lambda^{1}_{\nu}x^{\nu},\Lambda^{2}_{\nu}x^{\nu},\Lambda^{3}_{\nu}x^{\nu}\right)\\
& \mapsto\left(\left\langle X',\sigma_{0}\right\rangle ,\left\langle X',\sigma_{1}\right\rangle ,\left\langle X',\sigma_{2}\right\rangle ,\left\langle X',\sigma_{3}\right\rangle \right)\\
& \mapsto\left(\left\langle WXW^{*},\sigma_{0}\right\rangle ,\left\langle WXW^{*},\sigma_{1}\right\rangle ,\left\langle WXW^{*},\sigma_{2}\right\rangle ,\left\langle WXW^{*},\sigma_{3}\right\rangle \right)
\end{align*} X ′ ↦ ( x ′ 0 , x ′ 1 , x ′ 2 , x ′ 3 ) ↦ ( Λ ν 0 x ν , Λ ν 1 x ν , Λ ν 2 x ν , Λ ν 3 x ν ) ↦ ( ⟨ X ′ , σ 0 ⟩ , ⟨ X ′ , σ 1 ⟩ , ⟨ X ′ , σ 2 ⟩ , ⟨ X ′ , σ 3 ⟩ ) ↦ ( ⟨ W X W ∗ , σ 0 ⟩ , ⟨ W X W ∗ , σ 1 ⟩ , ⟨ W X W ∗ , σ 2 ⟩ , ⟨ W X W ∗ , σ 3 ⟩ ) Abrindo então o primero termo:
Λ ν 0 x ν = ⟨ W X W ∗ , σ 0 ⟩ = ⟨ W σ 0 W ∗ x 0 + W σ 1 W ∗ x 1 + W σ 2 W ∗ x 2 + W σ 3 W ∗ x 3 , σ 0 ⟩ = ⟨ W σ 0 W ∗ x 0 + W σ 1 W ∗ x 1 + W σ 2 W ∗ x 2 + W σ 3 W ∗ x 3 , σ 0 ⟩ = ⟨ W σ 0 W ∗ , σ 0 ⟩ x 0 + ⟨ W σ 1 W ∗ , σ 0 ⟩ x 1 + ⟨ W σ 2 W ∗ , σ 0 ⟩ x 2 + ⟨ W σ 3 W ∗ , σ 0 ⟩ x 3 = ⟨ W σ ν W ∗ , σ 0 ⟩ x ν \begin{align*}
\Lambda^{0}_{\nu}x^{\nu} & =\left\langle WXW^{*},\sigma_{0}\right\rangle \\
& =\left\langle W\sigma_{0}W^{*}x^{0}+W\sigma_{1}W^{*}x^{1}+W\sigma_{2}W^{*}x^{2}+W\sigma_{3}W^{*}x^{3},\sigma_{0}\right\rangle \\
& =\left\langle W\sigma_{0}W^{*}x^{0}+W\sigma_{1}W^{*}x^{1}+W\sigma_{2}W^{*}x^{2}+W\sigma_{3}W^{*}x^{3},\sigma_{0}\right\rangle \\
& =\left\langle W\sigma_{0}W^{*},\sigma_{0}\right\rangle x^{0}+\left\langle W\sigma_{1}W^{*},\sigma_{0}\right\rangle x^{1}+\left\langle W\sigma_{2}W^{*},\sigma_{0}\right\rangle x^{2}+\left\langle W\sigma_{3}W^{*},\sigma_{0}\right\rangle x^{3}\\
& =\left\langle W\sigma_{\nu}W^{*},\sigma_{0}\right\rangle x^{\nu}
\end{align*} Λ ν 0 x ν = ⟨ W X W ∗ , σ 0 ⟩ = ⟨ W σ 0 W ∗ x 0 + W σ 1 W ∗ x 1 + W σ 2 W ∗ x 2 + W σ 3 W ∗ x 3 , σ 0 ⟩ = ⟨ W σ 0 W ∗ x 0 + W σ 1 W ∗ x 1 + W σ 2 W ∗ x 2 + W σ 3 W ∗ x 3 , σ 0 ⟩ = ⟨ W σ 0 W ∗ , σ 0 ⟩ x 0 + ⟨ W σ 1 W ∗ , σ 0 ⟩ x 1 + ⟨ W σ 2 W ∗ , σ 0 ⟩ x 2 + ⟨ W σ 3 W ∗ , σ 0 ⟩ x 3 = ⟨ W σ ν W ∗ , σ 0 ⟩ x ν Temos então que:
Λ ν 0 = ⟨ W σ ν W ∗ , σ 0 ⟩ \Lambda^{0}_{\nu}=\left\langle W\sigma_{\nu}W^{*},\sigma_{0}\right\rangle Λ ν 0 = ⟨ W σ ν W ∗ , σ 0 ⟩ Ou ainda mais geral:
Λ ν μ = ⟨ W σ ν W ∗ , σ μ ⟩ \Lambda^{\mu}_{\nu}=\left\langle W\sigma_{\nu}W^{*},\sigma_{\mu}\right\rangle Λ ν μ = ⟨ W σ ν W ∗ , σ μ ⟩ Logo:
Λ = ( Λ 0 0 Λ 1 0 Λ 2 0 Λ 3 0 Λ 0 1 Λ 1 1 Λ 2 1 Λ 3 1 Λ 0 2 Λ 1 2 Λ 2 2 Λ 3 2 Λ 0 3 Λ 1 3 Λ 2 3 Λ 3 3 ) = ( ⟨ W σ 0 W ∗ , σ 0 ⟩ ⟨ W σ 1 W ∗ , σ 0 ⟩ ⟨ W σ 2 W ∗ , σ 0 ⟩ ⟨ W σ 3 W ∗ , σ 0 ⟩ ⟨ W σ 0 W ∗ , σ 1 ⟩ ⟨ W σ 1 W ∗ , σ 1 ⟩ ⟨ W σ 2 W ∗ , σ 1 ⟩ ⟨ W σ 3 W ∗ , σ 1 ⟩ ⟨ W σ 0 W ∗ , σ 2 ⟩ ⟨ W σ 1 W ∗ , σ 2 ⟩ ⟨ W σ 2 W ∗ , σ 2 ⟩ ⟨ W σ 3 W ∗ , σ 2 ⟩ ⟨ W σ 0 W ∗ , σ 3 ⟩ ⟨ W σ 1 W ∗ , σ 3 ⟩ ⟨ W σ 2 W ∗ , σ 3 ⟩ ⟨ W σ 3 W ∗ , σ 3 ⟩ ) \Lambda=\left(\begin{array}{cccc}
\Lambda^{0}_{0} & \Lambda^{0}_{1} & \Lambda^{0}_{2} & \Lambda^{0}_{3}\\
\Lambda^{1}_{0} & \Lambda^{1}_{1} & \Lambda^{1}_{2} & \Lambda^{1}_{3}\\
\Lambda^{2}_{0} & \Lambda^{2}_{1} & \Lambda^{2}_{2} & \Lambda^{2}_{3}\\
\Lambda^{3}_{0} & \Lambda^{3}_{1} & \Lambda^{3}_{2} & \Lambda^{3}_{3}
\end{array}\right)=\left(\begin{array}{cccc}
\left\langle W\sigma_{0}W^{*},\sigma_{0}\right\rangle & \left\langle W\sigma_{1}W^{*},\sigma_{0}\right\rangle & \left\langle W\sigma_{2}W^{*},\sigma_{0}\right\rangle & \left\langle W\sigma_{3}W^{*},\sigma_{0}\right\rangle \\
\left\langle W\sigma_{0}W^{*},\sigma_{1}\right\rangle & \left\langle W\sigma_{1}W^{*},\sigma_{1}\right\rangle & \left\langle W\sigma_{2}W^{*},\sigma_{1}\right\rangle & \left\langle W\sigma_{3}W^{*},\sigma_{1}\right\rangle \\
\left\langle W\sigma_{0}W^{*},\sigma_{2}\right\rangle & \left\langle W\sigma_{1}W^{*},\sigma_{2}\right\rangle & \left\langle W\sigma_{2}W^{*},\sigma_{2}\right\rangle & \left\langle W\sigma_{3}W^{*},\sigma_{2}\right\rangle \\
\left\langle W\sigma_{0}W^{*},\sigma_{3}\right\rangle & \left\langle W\sigma_{1}W^{*},\sigma_{3}\right\rangle & \left\langle W\sigma_{2}W^{*},\sigma_{3}\right\rangle & \left\langle W\sigma_{3}W^{*},\sigma_{3}\right\rangle
\end{array}\right) Λ = ⎝ ⎛ Λ 0 0 Λ 0 1 Λ 0 2 Λ 0 3 Λ 1 0 Λ 1 1 Λ 1 2 Λ 1 3 Λ 2 0 Λ 2 1 Λ 2 2 Λ 2 3 Λ 3 0 Λ 3 1 Λ 3 2 Λ 3 3 ⎠ ⎞ = ⎝ ⎛ ⟨ W σ 0 W ∗ , σ 0 ⟩ ⟨ W σ 0 W ∗ , σ 1 ⟩ ⟨ W σ 0 W ∗ , σ 2 ⟩ ⟨ W σ 0 W ∗ , σ 3 ⟩ ⟨ W σ 1 W ∗ , σ 0 ⟩ ⟨ W σ 1 W ∗ , σ 1 ⟩ ⟨ W σ 1 W ∗ , σ 2 ⟩ ⟨ W σ 1 W ∗ , σ 3 ⟩ ⟨ W σ 2 W ∗ , σ 0 ⟩ ⟨ W σ 2 W ∗ , σ 1 ⟩ ⟨ W σ 2 W ∗ , σ 2 ⟩ ⟨ W σ 2 W ∗ , σ 3 ⟩ ⟨ W σ 3 W ∗ , σ 0 ⟩ ⟨ W σ 3 W ∗ , σ 1 ⟩ ⟨ W σ 3 W ∗ , σ 2 ⟩ ⟨ W σ 3 W ∗ , σ 3 ⟩ ⎠ ⎞ Ou como T r ( A B ) = T r ( B A ) Tr\left(AB\right)=Tr\left(BA\right) T r ( A B ) = T r ( B A )
Λ = ( ⟨ σ 0 , W σ 0 W ∗ ⟩ ⟨ σ 0 , W σ 1 W ∗ ⟩ ⟨ σ 0 , W σ 2 W ∗ ⟩ ⟨ σ 0 , W σ 3 W ∗ ⟩ ⟨ σ 1 , W σ 0 W ∗ ⟩ ⟨ σ 1 , W σ 1 W ∗ ⟩ ⟨ σ 1 , W σ 2 W ∗ ⟩ ⟨ σ 1 , W σ 3 W ∗ ⟩ ⟨ σ 2 , W σ 0 W ∗ ⟩ ⟨ σ 2 , W σ 1 W ∗ ⟩ ⟨ σ 2 , W σ 2 W ∗ ⟩ ⟨ σ 2 , W σ 3 W ∗ ⟩ ⟨ σ 3 , W σ 0 W ∗ ⟩ ⟨ σ 3 , W σ 1 W ∗ ⟩ ⟨ σ 3 , W σ 2 W ∗ ⟩ ⟨ σ 3 , W σ 3 W ∗ ⟩ ) \Lambda=\left(\begin{array}{cccc}
\left\langle \sigma_{0},W\sigma_{0}W^{*}\right\rangle & \left\langle \sigma_{0},W\sigma_{1}W^{*}\right\rangle & \left\langle \sigma_{0},W\sigma_{2}W^{*}\right\rangle & \left\langle \sigma_{0},W\sigma_{3}W^{*}\right\rangle \\
\left\langle \sigma_{1},W\sigma_{0}W^{*}\right\rangle & \left\langle \sigma_{1},W\sigma_{1}W^{*}\right\rangle & \left\langle \sigma_{1},W\sigma_{2}W^{*}\right\rangle & \left\langle \sigma_{1},W\sigma_{3}W^{*}\right\rangle \\
\left\langle \sigma_{2},W\sigma_{0}W^{*}\right\rangle & \left\langle \sigma_{2},W\sigma_{1}W^{*}\right\rangle & \left\langle \sigma_{2},W\sigma_{2}W^{*}\right\rangle & \left\langle \sigma_{2},W\sigma_{3}W^{*}\right\rangle \\
\left\langle \sigma_{3},W\sigma_{0}W^{*}\right\rangle & \left\langle \sigma_{3},W\sigma_{1}W^{*}\right\rangle & \left\langle \sigma_{3},W\sigma_{2}W^{*}\right\rangle & \left\langle \sigma_{3},W\sigma_{3}W^{*}\right\rangle
\end{array}\right) Λ = ⎝ ⎛ ⟨ σ 0 , W σ 0 W ∗ ⟩ ⟨ σ 1 , W σ 0 W ∗ ⟩ ⟨ σ 2 , W σ 0 W ∗ ⟩ ⟨ σ 3 , W σ 0 W ∗ ⟩ ⟨ σ 0 , W σ 1 W ∗ ⟩ ⟨ σ 1 , W σ 1 W ∗ ⟩ ⟨ σ 2 , W σ 1 W ∗ ⟩ ⟨ σ 3 , W σ 1 W ∗ ⟩ ⟨ σ 0 , W σ 2 W ∗ ⟩ ⟨ σ 1 , W σ 2 W ∗ ⟩ ⟨ σ 2 , W σ 2 W ∗ ⟩ ⟨ σ 3 , W σ 2 W ∗ ⟩ ⟨ σ 0 , W σ 3 W ∗ ⟩ ⟨ σ 1 , W σ 3 W ∗ ⟩ ⟨ σ 2 , W σ 3 W ∗ ⟩ ⟨ σ 3 , W σ 3 W ∗ ⟩ ⎠ ⎞ Usando o Wolfram Mathematica, podemos calcular o determinante:
O0~:=~\{\{1,~0\},~~\{0,~1\}\};~
O1~:=~\{\{0,~1\},~~\{1,~0\}\};~
O2~:=~\{\{0,~-I\},~\{I,~0\}\};~
O3~:=~\{\{1,~0\},~\{0,~-1\}\};
Simplify{[}Det{[}\{
\{Tr{[}O0.W.O0.WC{]}/2,Tr{[}O0.W.O1.WC{]}/2,~Tr{[}O0.W.O2.WC{]}/2,Tr{[}O0.W.O3.WC{]}/2\},
\{Tr{[}O1.W.O0.WC{]}/2,Tr{[}O1.W.O1.WC{]}/2,~Tr{[}O1.W.O2.WC{]}/2,Tr{[}O1.W.O3.WC{]}/2\},~
\{Tr{[}O2.W.O0.WC{]}/2,Tr{[}O2.W.O1.WC{]}/2,~Tr{[}O2.W.O2.WC{]}/2,Tr{[}O2.W.O3.WC{]}/2\},~
\{Tr{[}O3.W.O0.WC{]}/2,Tr{[}O3.W.O1.WC{]}/2,~Tr{[}O3.W.O2.WC{]}/2,Tr{[}O3.W.O3.WC{]}/2\}\}{]}{]}det ( Λ ) = ( α δ − β γ ) 2 ( α ‾ δ ‾ − β ‾ γ ‾ ) 2 = + 1 \det\left(\Lambda\right)=\left(\alpha\delta-\beta\gamma\right)^{2}\left(\overline{\alpha}\overline{\delta}-\overline{\beta}\overline{\gamma}\right)^{2}=+1 det ( Λ ) = ( α δ − β γ ) 2 ( α δ − β γ ) 2 = + 1 Como queríamoss demonstrar. Agora calculando o primeiro termo:
Λ 0 0 = , W σ 0 W ∗ = 1 2 T r ( σ 0 W σ 0 W ∗ ) = 1 2 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) = 1 2 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) \begin{align*}
\Lambda^{0}_{0} & =,W\sigma_{0}W^{*}\\
& =\frac{1}{2}Tr\left(\sigma_{0}W\sigma_{0}W^{*}\right)\\
& =\frac{1}{2}\left(\left|\alpha\right|^{2}+\left|\delta\right|^{2}+\left|\beta\right|^{2}+\left|\gamma\right|^{2}\right)\\
& =\frac{1}{2}\left(\left|\alpha\right|^{2}+\left|\delta\right|^{2}+\left|\beta\right|^{2}+\left|\gamma\right|^{2}\right)
\end{align*} Λ 0 0 = , W σ 0 W ∗ = 2 1 T r ( σ 0 W σ 0 W ∗ ) = 2 1 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) = 2 1 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) Como vimos em aula 32, as constantes Λ β α \Lambda^{\alpha}_{\beta} Λ β α
Λ Γ α Λ γ β η α β = η Γ γ Λ 0 α Λ 0 β η α β = η 00 ( Λ 0 0 ) 2 + ( Λ 0 1 ) 2 + ( Λ 0 2 ) 2 + ( Λ 0 3 ) = 1 ( Λ 0 0 ) 2 = 1 + ∑ i = 1 3 ( Λ 0 i ) 2 \begin{align*}
\Lambda^{\alpha}_{\varGamma} & \Lambda^{\beta}_{\gamma}\eta_{\alpha\beta}=\eta_{\varGamma\gamma}\\
\Lambda^{\alpha}_{0} & \Lambda^{\beta}_{0}\eta_{\alpha\beta}=\eta_{00}\\
\left(\Lambda^{0}_{0}\right)^{2}+\left(\Lambda^{1}_{0}\right)^{2}+\left(\Lambda^{2}_{0}\right)^{2}+\left(\Lambda^{3}_{0}\right) & =1\\
\left(\Lambda^{0}_{0}\right)^{2}= & 1+\sum^{3}_{i=1}\left(\Lambda^{i}_{0}\right)^{2}
\end{align*} Λ Γ α Λ 0 α ( Λ 0 0 ) 2 + ( Λ 0 1 ) 2 + ( Λ 0 2 ) 2 + ( Λ 0 3 ) ( Λ 0 0 ) 2 = Λ γ β η α β = η Γ γ Λ 0 β η α β = η 00 = 1 1 + i = 1 ∑ 3 ( Λ 0 i ) 2 Onde usamos γ = Γ = 0 \gamma=\Gamma=0 γ = Γ = 0 η α β \eta_{\alpha\beta} η α β
FullSimplify{[}Expand{[}(Tr{[}O1.W.O0.WC{]}/2)\textasciicircum 2+(Tr{[}O2.W.O0.WC{]}/2)\textasciicircum 2+(Tr{[}O3.W.O0.WC{]}/2)\textasciicircum 2{]}{]}∑ i = 1 3 ( Λ 0 i ) 2 = 1 4 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( α ‾ ( − δ ‾ ( α δ − β γ − β γ ) + α ∣ β ∣ 2 + α ∣ γ ∣ 2 ) + β ‾ ( γ ‾ ( α δ + α δ − β γ ) + β ∣ δ ∣ 2 ) + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = 1 4 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( α ‾ δ ‾ β γ − ( δ ‾ α ‾ − γ ‾ β ‾ ) + α δ γ ‾ β ‾ + ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ α ∣ 2 ∣ β ∣ 2 + ∣ β ∣ 2 ∣ δ ∣ 2 + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = 1 4 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( α ‾ δ ‾ β γ + α δ γ ‾ β ‾ − 1 + ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ α ∣ 2 ∣ β ∣ 2 + ∣ β ∣ 2 ∣ δ ∣ 2 + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = 1 4 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( ∣ α ∣ 2 ∣ δ ∣ 2 + ∣ β ∣ 2 ∣ γ ∣ 2 − 2 + ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ α ∣ 2 ∣ β ∣ 2 + ∣ β ∣ 2 ∣ δ ∣ 2 + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = − 1 + 1 4 ( ∣ α ∣ 4 + 2 ∣ α ∣ 2 ∣ δ ∣ 2 + 2 ∣ α ∣ 2 ∣ β ∣ 2 + 2 ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ δ ∣ 4 + 2 ∣ β ∣ 2 ∣ δ ∣ 2 + 2 ∣ δ ∣ 2 ∣ γ ∣ 2 + ∣ β ∣ 4 + 2 ∣ β ∣ 2 ∣ γ ∣ 2 + ∣ γ ∣ 4 ) = − 1 + [ 1 2 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) ] 2 \begin{align*}
\sum^{3}_{i=1}\left(\Lambda^{i}_{0}\right)^{2} & =\frac{1}{4}\left(\left(\left|\alpha\right|^{4}+\left|\beta\right|^{4}+\left|\delta\right|^{4}+\left|\gamma\right|^{4}+2\left(\overline{\alpha}\left(-\overline{\delta}(\alpha\delta-\beta\gamma-\beta\gamma)+\alpha\left|\beta\right|^{2}+\alpha\left|\gamma\right|^{2}\right)+\overline{\beta}\left(\overline{\gamma}(\alpha\delta+\alpha\delta-\beta\gamma)+\beta\left|\delta\right|^{2}\right)+\left|\delta\right|^{2}\left|\gamma\right|^{2}\right)\right)\right)\\
& =\frac{1}{4}\left(\left(\left|\alpha\right|^{4}+\left|\beta\right|^{4}+\left|\delta\right|^{4}+\left|\gamma\right|^{4}+2\left(\overline{\alpha}\overline{\delta}\beta\gamma-\left(\overline{\delta}\overline{\alpha}-\overline{\gamma}\overline{\beta}\right)+\alpha\delta\overline{\gamma}\overline{\beta}+\left|\alpha\right|^{2}\left|\gamma\right|^{2}+\left|\alpha\right|^{2}\left|\beta\right|^{2}+\left|\beta\right|^{2}\left|\delta\right|^{2}+\left|\delta\right|^{2}\left|\gamma\right|^{2}\right)\right)\right)\\
& =\frac{1}{4}\left(\left(\left|\alpha\right|^{4}+\left|\beta\right|^{4}+\left|\delta\right|^{4}+\left|\gamma\right|^{4}+2\left(\overline{\alpha}\overline{\delta}\beta\gamma+\alpha\delta\overline{\gamma}\overline{\beta}-1+\left|\alpha\right|^{2}\left|\gamma\right|^{2}+\left|\alpha\right|^{2}\left|\beta\right|^{2}+\left|\beta\right|^{2}\left|\delta\right|^{2}+\left|\delta\right|^{2}\left|\gamma\right|^{2}\right)\right)\right)\\
& =\frac{1}{4}\left(\left(\left|\alpha\right|^{4}+\left|\beta\right|^{4}+\left|\delta\right|^{4}+\left|\gamma\right|^{4}+2\left(\left|\alpha\right|^{2}\left|\delta\right|^{2}+\left|\beta\right|^{2}\left|\gamma\right|^{2}-2+\left|\alpha\right|^{2}\left|\gamma\right|^{2}+\left|\alpha\right|^{2}\left|\beta\right|^{2}+\left|\beta\right|^{2}\left|\delta\right|^{2}+\left|\delta\right|^{2}\left|\gamma\right|^{2}\right)\right)\right)\\
& =-1+\frac{1}{4}\left(\left|\alpha\right|^{4}+2\left|\alpha\right|^{2}\left|\delta\right|^{2}+2\left|\alpha\right|^{2}\left|\beta\right|^{2}+2\left|\alpha\right|^{2}\left|\gamma\right|^{2}+\left|\delta\right|^{4}+2\left|\beta\right|^{2}\left|\delta\right|^{2}+2\left|\delta\right|^{2}\left|\gamma\right|^{2}+\left|\beta\right|^{4}+2\left|\beta\right|^{2}\left|\gamma\right|^{2}+\left|\gamma\right|^{4}\right)\\
& =-1+\left[\frac{1}{2}\left(\left|\alpha\right|^{2}+\left|\delta\right|^{2}+\left|\beta\right|^{2}+\left|\gamma\right|^{2}\right)\right]^{2}
\end{align*} i = 1 ∑ 3 ( Λ 0 i ) 2 = 4 1 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( α ( − δ ( α δ − β γ − β γ ) + α ∣ β ∣ 2 + α ∣ γ ∣ 2 ) + β ( γ ( α δ + α δ − β γ ) + β ∣ δ ∣ 2 ) + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = 4 1 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( α δ β γ − ( δ α − γ β ) + α δ γ β + ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ α ∣ 2 ∣ β ∣ 2 + ∣ β ∣ 2 ∣ δ ∣ 2 + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = 4 1 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( α δ β γ + α δ γ β − 1 + ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ α ∣ 2 ∣ β ∣ 2 + ∣ β ∣ 2 ∣ δ ∣ 2 + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = 4 1 ( ( ∣ α ∣ 4 + ∣ β ∣ 4 + ∣ δ ∣ 4 + ∣ γ ∣ 4 + 2 ( ∣ α ∣ 2 ∣ δ ∣ 2 + ∣ β ∣ 2 ∣ γ ∣ 2 − 2 + ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ α ∣ 2 ∣ β ∣ 2 + ∣ β ∣ 2 ∣ δ ∣ 2 + ∣ δ ∣ 2 ∣ γ ∣ 2 ) ) ) = − 1 + 4 1 ( ∣ α ∣ 4 + 2 ∣ α ∣ 2 ∣ δ ∣ 2 + 2 ∣ α ∣ 2 ∣ β ∣ 2 + 2 ∣ α ∣ 2 ∣ γ ∣ 2 + ∣ δ ∣ 4 + 2 ∣ β ∣ 2 ∣ δ ∣ 2 + 2 ∣ δ ∣ 2 ∣ γ ∣ 2 + ∣ β ∣ 4 + 2 ∣ β ∣ 2 ∣ γ ∣ 2 + ∣ γ ∣ 4 ) = − 1 + [ 2 1 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) ] 2 Onde usamos o fato de que − 1 = ( α δ − β γ ) ( γ ‾ β ‾ − α ‾ δ ‾ ) = α ‾ δ ‾ β γ − ∣ α ∣ 2 ∣ δ ∣ 2 − ∣ β ∣ 2 ∣ γ ∣ 2 + α δ γ ‾ β ‾ -1=\left(\alpha\delta-\beta\gamma\right)\left(\overline{\gamma}\overline{\beta}-\overline{\alpha}\overline{\delta}\right)=\overline{\alpha}\overline{\delta}\beta\gamma-\left|\alpha\right|^{2}\left|\delta\right|^{2}-\left|\beta\right|^{2}\left|\gamma\right|^{2}+\alpha\delta\overline{\gamma}\overline{\beta} − 1 = ( α δ − β γ ) ( γ β − α δ ) = α δ β γ − ∣ α ∣ 2 ∣ δ ∣ 2 − ∣ β ∣ 2 ∣ γ ∣ 2 + α δ γ β
∑ i = 1 3 ( Λ 0 i ) 2 ≥ 0 ⟶ [ 1 2 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) ] 2 ≥ 1 \sum^{3}_{i=1}\left(\Lambda^{i}_{0}\right)^{2}\geq0\longrightarrow\left[\frac{1}{2}\left(\left|\alpha\right|^{2}+\left|\delta\right|^{2}+\left|\beta\right|^{2}+\left|\gamma\right|^{2}\right)\right]^{2}\geq1 i = 1 ∑ 3 ( Λ 0 i ) 2 ≥ 0 ⟶ [ 2 1 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) ] 2 ≥ 1 Então:
∣ 1 2 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) ∣ ≥ 1 \left|\frac{1}{2}\left(\left|\alpha\right|^{2}+\left|\delta\right|^{2}+\left|\beta\right|^{2}+\left|\gamma\right|^{2}\right)\right|\geq1 ∣ ∣ 2 1 ( ∣ α ∣ 2 + ∣ δ ∣ 2 + ∣ β ∣ 2 + ∣ γ ∣ 2 ) ∣ ∣ ≥ 1 Uma vez que se ∣ x ∣ > 1 ( ∣ x ∣ < 1 ) , \left|x\right|>1\left(\left|x\right|<1\right), ∣ x ∣ > 1 ( ∣ x ∣ < 1 ) , x 2 > 1 ( x 2 < 1 ) x^{2}>1\left(x^{2}<1\right) x 2 > 1 ( x 2 < 1 )
\[
\Lambda^{0}_{0}=\frac{1}{2}\left(\left|\alpha\right|^{2}+\left|\delta\right|^{2}+\left|\beta\right|^{2}+\left|\gamma\right|^{2}\right)
\]Concluímos então que:
∣ Λ 0 0 ∣ ≥ 1 \left|\Lambda^{0}_{0}\right|\geq1 ∣ ∣ Λ 0 0 ∣ ∣ ≥ 1 Problem 2 ¶ Operate on ( i γ μ ∂ μ − m ) ψ = 0 \left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi=0 ( i γ μ ∂ μ − m ) ψ = 0 γ μ ∂ μ \gamma^{\mu}\partial_{\mu} γ μ ∂ μ ψ i \psi_{i} ψ i ( □ 2 + m 2 ) ψ i = 0 \left(\square^{2}+m^{2}\right)\psi_{i}=0 ( □ 2 + m 2 ) ψ i = 0
Primeiro temos que o oprador d’Alembertiano é dado por:
□ 2 = η α β ∂ ∂ x β ∂ ∂ x α = ∇ 2 − ∂ 2 ∂ t 2 \square^{2}=\eta^{\alpha\beta}\frac{\partial}{\partial x^{\beta}}\frac{\partial}{\partial x^{\alpha}}=\nabla^{2}-\frac{\partial^{2}}{\partial t^{2}} □ 2 = η α β ∂ x β ∂ ∂ x α ∂ = ∇ 2 − ∂ t 2 ∂ 2 Sendo a função onda ψ \psi ψ
( i γ μ ∂ μ − m ) ψ = 0 \left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi=0 ( i γ μ ∂ μ − m ) ψ = 0 Vamos operar γ ν ∂ ν \gamma^{\nu}\partial_{\nu} γ ν ∂ ν
∂ μ = ∂ ∂ x μ = ( ∂ ∂ t , ∇ ) \partial_{\mu}=\frac{\partial}{\partial x^{\mu}}=\left(\frac{\partial}{\partial t},\boldsymbol{\nabla}\right) ∂ μ = ∂ x μ ∂ = ( ∂ t ∂ , ∇ ) E como temos
p → − i ℏ ∇ , E = i ℏ ∂ ∂ t , p μ = ( E , − p ) \boldsymbol{p}\rightarrow-i\hbar\nabla,\quad E=i\hbar\frac{\partial}{\partial t},\qquad p_{\mu}=\left(E,-\boldsymbol{p}\right) p → − i ℏ∇ , E = i ℏ ∂ t ∂ , p μ = ( E , − p ) Então vamos escrever γ ν i ∂ ν = γ ν p ν / ℏ \gamma^{\nu}i\partial_{\nu}=\gamma^{\nu}p_{\nu}/\hbar γ ν i ∂ ν = γ ν p ν /ℏ ℏ = 1 \hbar=1 ℏ = 1
( i γ μ ∂ μ − m ) ψ = 0 ( γ μ p μ − m ) ψ = 0 γ μ p μ ψ = − m ψ \begin{align*}
\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi & =0\\
\left(\gamma^{\mu}p_{\mu}-m\right)\psi & =0\\
\gamma^{\mu}p_{\mu}\psi & =-m\psi
\end{align*} ( i γ μ ∂ μ − m ) ψ ( γ μ p μ − m ) ψ γ μ p μ ψ = 0 = 0 = − m ψ E ainda sendo γ μ p μ = p \gamma^{\mu}p_{\mu}=\cancel{p} γ μ p μ = p
p ψ = − m ψ \cancel{p}\psi=-m\psi p ψ = − m ψ Operando então: − i p -i\cancel{p} − i p
i p p ψ = − i p m ψ \begin{align*}
i\cancel{p}\cancel{p} & \psi=-i\cancel{p}m\psi
\end{align*} i p p ψ = − i p m ψ E uma vez que na aula 37, na equação 159 vimos que γ μ γ η + γ η γ μ = I 2 g μ η \gamma^{\mu}\gamma^{\eta}+\gamma^{\eta}\gamma^{\mu}=\mathbb{I}2g^{\mu\eta} γ μ γ η + γ η γ μ = I 2 g μ η
p p = γ μ p μ γ ν p ν = γ μ γ ν p μ p ν = 1 2 ( γ μ γ ν + γ ν γ μ ) p μ p ν = I 1 2 2 g μ η p μ p ν = I p μ p ν = I p 2 \begin{align*}
\cancel{p}\cancel{p} & =\gamma^{\mu}p_{\mu}\gamma^{\nu}p_{\nu}\\
& =\gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu}\\
& =\frac{1}{2}\left(\gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}\right)p_{\mu}p_{\nu}\\
& =\mathbb{I}\frac{1}{2}2g^{\mu\eta}p_{\mu}p_{\nu}\\
& =\mathbb{I}p^{\mu}p_{\nu}\\
& =\mathbb{I}p^{2}
\end{align*} p p = γ μ p μ γ ν p ν = γ μ γ ν p μ p ν = 2 1 ( γ μ γ ν + γ ν γ μ ) p μ p ν = I 2 1 2 g μ η p μ p ν = I p μ p ν = I p 2 Lembando que [ p i , p j ] = 0 \left[p_{i},p_{j}\right]=0 [ p i , p j ] = 0 γ μ γ ν p μ p ν = γ ν γ μ p ν p μ = γ ν γ μ p μ p ν \gamma^{\mu}\gamma^{\nu}p_{\mu}p_{\nu}=\gamma^{\nu}\gamma^{\mu}p_{\nu}p_{\mu}=\gamma^{\nu}\gamma^{\mu}p_{\mu}p_{\nu} γ μ γ ν p μ p ν = γ ν γ μ p ν p μ = γ ν γ μ p μ p ν γ ν \gamma^{\nu} γ ν 42
p 2 ψ = m 2 ψ p μ p μ ψ = m 2 ψ ( i ∂ μ ) ( i ∂ μ ) ψ = m 2 ψ − ( ∂ 2 ∂ t 2 − ∇ 2 ) ψ = m 2 ψ − □ 2 ψ = m 2 ψ \begin{align*}
p^{2}\psi & =m^{2}\psi\\
p^{\mu}p_{\mu}\psi & =m^{2}\psi\\
\left(i\partial^{\mu}\right)\left(i\partial_{\mu}\right)\psi & =m^{2}\psi\\
-\left(\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}\right)\psi & =m^{2}\psi\\
-\square^{2}\psi & =m^{2}\psi
\end{align*} p 2 ψ p μ p μ ψ ( i ∂ μ ) ( i ∂ μ ) ψ − ( ∂ t 2 ∂ 2 − ∇ 2 ) ψ − □ 2 ψ = m 2 ψ = m 2 ψ = m 2 ψ = m 2 ψ = m 2 ψ Ou então, como queríamos mostrar:
( □ 2 + m 2 ) ψ = 0 \left(\square^{2}+m^{2}\right)\psi=0 ( □ 2 + m 2 ) ψ = 0 Ou para cada uma das componentes de ψ \psi ψ
( □ 2 + m 2 ) ψ i = 0 \left(\square^{2}+m^{2}\right)\psi_{i}=0 ( □ 2 + m 2 ) ψ i = 0 Problem 3 ¶ Show that the helicity operator
h ^ = ∑ ^ ⋅ p ^ ≡ ( σ ⋅ p ^ 0 0 σ ⋅ p ^ ) \widehat{h}=\widehat{\sum}\cdot\widehat{\boldsymbol{p}}\equiv\left(\begin{array}{cc}
\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}} & 0\\
0 & \boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}
\end{array}\right) h = ∑ ⋅ p ≡ ( σ ⋅ p 0 0 σ ⋅ p ) commutes with the Dirac Hamiltonian H ^ = α ⋅ p + β m \widehat{H}=\alpha\cdot\boldsymbol{p}+\beta m H = α ⋅ p + β m p ^ \widehat{\boldsymbol{p}} p p / ∣ p ∣ \boldsymbol{p}/\left|\boldsymbol{p}\right| p / ∣ p ∣ 1 2 σ ⋅ p ^ \frac{1}{2}\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}} 2 1 σ ⋅ p
O operador de helicidade é dado por:
h ^ = ∑ ^ ⋅ p ^ ≡ ( σ ⋅ p ^ 0 0 σ ⋅ p ^ ) \widehat{h}=\widehat{\sum}\cdot\widehat{\boldsymbol{p}}\equiv\left(\begin{array}{cc}
\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}} & 0\\
0 & \boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}
\end{array}\right) h = ∑ ⋅ p ≡ ( σ ⋅ p 0 0 σ ⋅ p ) Onde p ^ = p / ∣ p ∣ \widehat{\boldsymbol{p}}=\boldsymbol{p}/\left|\boldsymbol{p}\right| p = p / ∣ p ∣ h ^ \widehat{h} h H ^ = α ⋅ p + β m \widehat{H}=\alpha\cdot\boldsymbol{p}+\beta m H = α ⋅ p + β m
Temos da aula 35 e equações 101 e 102 que:
α i = ( 0 σ i σ i 0 ) , β = ( 1 0 0 − 1 ) \alpha_{i}=\left(\begin{array}{cc}
0 & \sigma_{i}\\
\sigma_{i} & 0
\end{array}\right),\qquad\beta=\left(\begin{array}{cc}
\boldsymbol{1} & 0\\
0 & -\boldsymbol{1}
\end{array}\right) α i = ( 0 σ i σ i 0 ) , β = ( 1 0 0 − 1 ) Então:
H ^ = α ⋅ p + β m = ( 0 σ ⋅ p σ ⋅ p 0 ) + ( m 1 0 0 − m 1 ) = ( m σ ⋅ p σ ⋅ p m ) \begin{align*}
\widehat{H} & =\alpha\cdot\boldsymbol{p}+\beta m=\left(\begin{array}{cc}
0 & \boldsymbol{\sigma}\cdot\boldsymbol{p}\\
\boldsymbol{\sigma}\cdot\boldsymbol{p} & 0
\end{array}\right)+\left(\begin{array}{cc}
m\boldsymbol{1} & 0\\
0 & -m\boldsymbol{1}
\end{array}\right)=\left(\begin{array}{cc}
\boldsymbol{m} & \boldsymbol{\sigma}\cdot\boldsymbol{p}\\
\boldsymbol{\sigma}\cdot\boldsymbol{p} & \boldsymbol{m}
\end{array}\right)
\end{align*} H = α ⋅ p + β m = ( 0 σ ⋅ p σ ⋅ p 0 ) + ( m 1 0 0 − m 1 ) = ( m σ ⋅ p σ ⋅ p m ) Então podemos verificar explicitamente que:
[ H ^ , h ^ ] = ( m σ ⋅ p σ ⋅ p m ) ( σ ⋅ p ^ 0 0 σ ⋅ p ^ ) − ( σ ⋅ p ^ 0 0 σ ⋅ p ^ ) ( m σ ⋅ p σ ⋅ p m ) = ( m ( σ ⋅ p ^ ) ( σ ⋅ p ) ( σ ⋅ p ^ ) ( σ ⋅ p ) ( σ ⋅ p ^ ) m ( σ ⋅ p ^ ) ) − ( ( σ ⋅ p ^ ) m ( σ ⋅ p ^ ) ( σ ⋅ p ) ( σ ⋅ p ^ ) ( σ ⋅ p ) ( σ ⋅ p ^ ) m ) \begin{align*}
\left[\widehat{H},\widehat{h}\right] & =\left(\begin{array}{cc}
\boldsymbol{m} & \boldsymbol{\sigma}\cdot\boldsymbol{p}\\
\boldsymbol{\sigma}\cdot\boldsymbol{p} & \boldsymbol{m}
\end{array}\right)\left(\begin{array}{cc}
\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}} & 0\\
0 & \boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}
\end{array}\right)-\left(\begin{array}{cc}
\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}} & 0\\
0 & \boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}
\end{array}\right)\left(\begin{array}{cc}
\boldsymbol{m} & \boldsymbol{\sigma}\cdot\boldsymbol{p}\\
\boldsymbol{\sigma}\cdot\boldsymbol{p} & \boldsymbol{m}
\end{array}\right)\\
& =\left(\begin{array}{cc}
\boldsymbol{m}\left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right) & \left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)\left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right)\\
\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)\left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right) & \boldsymbol{m}\left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right)
\end{array}\right)-\left(\begin{array}{cc}
\left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right)\boldsymbol{m} & \left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right)\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)\\
\left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right)\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right) & \left(\boldsymbol{\sigma}\cdot\widehat{\boldsymbol{p}}\right)\boldsymbol{m}
\end{array}\right)
\end{align*} [ H , h ] = ( m σ ⋅ p σ ⋅ p m ) ( σ ⋅ p 0 0 σ ⋅ p ) − ( σ ⋅ p 0 0 σ ⋅ p ) ( m σ ⋅ p σ ⋅ p m ) = ( m ( σ ⋅ p ) ( σ ⋅ p ) ( σ ⋅ p ) ( σ ⋅ p ) ( σ ⋅ p ) m ( σ ⋅ p ) ) − ( ( σ ⋅ p ) m ( σ ⋅ p ) ( σ ⋅ p ) ( σ ⋅ p ) ( σ ⋅ p ) ( σ ⋅ p ) m ) E como queríamos demonstrar:
[ H ^ , h ^ ] = 0 \left[\widehat{H},\widehat{h}\right]=0 [ H , h ] = 0 Problem 4 ¶ We have already seen that the equation H u = ( α ⋅ p + β m ) u = E u Hu=\left(\boldsymbol{\alpha}\cdot\boldsymbol{p}+\beta m\right)u=Eu H u = ( α ⋅ p + β m ) u = E u p ≠ 0 \boldsymbol{p}\neq0 p = 0
σ ⋅ p u B = ( E c − m c ) u A σ ⋅ p u A = ( E c + m c ) u B \begin{align*}
\boldsymbol{\sigma}\cdot\boldsymbol{p}u_{B} & =\left(\frac{E}{c}-mc\right)u_{A}\\
\boldsymbol{\sigma}\cdot\boldsymbol{p}u_{A} & =\left(\frac{E}{c}+mc\right)u_{B}
\end{align*} σ ⋅ p u B σ ⋅ p u A = ( c E − m c ) u A = ( c E + m c ) u B where u u u u A u_{A} u A u B u_{B} u B
a) For a nonrelativistic electron of velocity v v v u A u_{A} u A u B u_{B} u B v / c v/c v / c ψ A \psi_{A} ψ A ψ B \psi_{B} ψ B ψ \psi ψ
b) In the nonrelativistic limit, show that the Dirac equation for a electron (charge − e -e − e A μ = ( A 0 , A ) A^{\mu}=\left(A^{0},\boldsymbol{A}\right) A μ = ( A 0 , A )
( 1 2 m ( P + e A ) 2 + e 2 m σ ⋅ B − e A 0 ) ψ A = E N R ψ A \left(\frac{1}{2m}\left(\boldsymbol{P}+e\boldsymbol{A}\right)^{2}+\frac{e}{2m}\boldsymbol{\sigma}\cdot\boldsymbol{B}-eA^{0}\right)\psi_{A}=E_{NR}\psi_{A} ( 2 m 1 ( P + e A ) 2 + 2 m e σ ⋅ B − e A 0 ) ψ A = E NR ψ A where the magnetic field B = ∇ × A \boldsymbol{B}=\nabla\times\boldsymbol{A} B = ∇ × A E N R = E − m E_{NR}=E-m E NR = E − m ∣ e A 0 ∣ ≪ m \left|eA^{0}\right|\ll m ∣ ∣ e A 0 ∣ ∣ ≪ m
Hint: Make the substitution P μ → P μ + e A μ P^{\mu}\rightarrow P^{\mu}+eA^{\mu} P μ → P μ + e A μ ψ A , B \psi_{A,B} ψ A , B ψ B \psi_{B} ψ B P × A + A × P = i ∇ × A \boldsymbol{P}\times\boldsymbol{A}+\boldsymbol{A}\times\boldsymbol{P}=i\nabla\times\boldsymbol{A} P × A + A × P = i ∇ × A P = − i ∇ \boldsymbol{P}=-i\nabla P = − i ∇
Não usando unidades naturais pois queremos a razão v / c v/c v / c
σ ⋅ p u B = ( E c − m c ) u A σ ⋅ p u A = ( E c + m c ) u B \begin{align}
\boldsymbol{\sigma}\cdot\boldsymbol{p}u_{B} & =\left(\frac{E}{c}-mc\right)u_{A}\\
\boldsymbol{\sigma}\cdot\boldsymbol{p}u_{A} & =\left(\frac{E}{c}+mc\right)u_{B}
\end{align} σ ⋅ p u B σ ⋅ p u A = ( c E − m c ) u A = ( c E + m c ) u B Ou manipulando:
u A = σ ⋅ p ( E c − m c ) u B u B = σ ⋅ p ( E c + m c ) u A \begin{align}
u_{A} & =\frac{\boldsymbol{\sigma}\cdot\boldsymbol{p}}{\left(\frac{E}{c}-mc\right)}u_{B}\\
u_{B} & =\frac{\boldsymbol{\sigma}\cdot\boldsymbol{p}}{\left(\frac{E}{c}+mc\right)}u_{A}
\end{align} u A u B = ( c E − m c ) σ ⋅ p u B = ( c E + m c ) σ ⋅ p u A Pegando então a segunda equação, e calculando o módulo de u B u_{B} u B
∣ u B ∣ = 1 ( E c + m c ) ∣ ( σ ⋅ p ) u A ∣ \left|u_{B}\right|=\frac{1}{\left(\frac{E}{c}+mc\right)}\left|\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)u_{A}\right| ∣ u B ∣ = ( c E + m c ) 1 ∣ ( σ ⋅ p ) u A ∣ E abrindo:
σ ⋅ p = p k σ k → ( σ ⋅ p ) ⋅ ( σ ⋅ p ) = p k σ k p j σ j = p k p j ( σ k σ j ) = p j p j = ∣ p ∣ 2 \boldsymbol{\sigma}\cdot\boldsymbol{p}=p_{k}\sigma_{k}\rightarrow\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)\cdot\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)=p_{k}\sigma_{k}p_{j}\sigma_{j}=p_{k}p_{j}\left(\sigma_{k}\sigma_{j}\right)=p_{j}p_{j}=\left|p\right|^{2} σ ⋅ p = p k σ k → ( σ ⋅ p ) ⋅ ( σ ⋅ p ) = p k σ k p j σ j = p k p j ( σ k σ j ) = p j p j = ∣ p ∣ 2 E como ∣ v ∣ = v ⋅ v \left|\boldsymbol{v}\right|=\sqrt{\boldsymbol{v}\cdot\boldsymbol{v}} ∣ v ∣ = v ⋅ v ∣ σ ⋅ p ∣ = p j p j = ∣ p ∣ \left|\boldsymbol{\sigma}\cdot\boldsymbol{p}\right|=\sqrt{p_{j}p_{j}}=\left|\boldsymbol{p}\right| ∣ σ ⋅ p ∣ = p j p j = ∣ p ∣
∣ ( σ ⋅ p ) u A ∣ = ∣ p ∣ ∣ u A ∣ \left|\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)u_{A}\right|=\left|\boldsymbol{p}\right|\left|u_{A}\right| ∣ ( σ ⋅ p ) u A ∣ = ∣ p ∣ ∣ u A ∣ Então sendo E ≈ m c 2 E\approx mc^{2} E ≈ m c 2 p = m v p=mv p = m v
∣ u B ∣ ≈ m v ( m c 2 c + m c ) ∣ u A ∣ \left|u_{B}\right|\approx\frac{mv}{\left(\frac{mc^{2}}{c}+mc\right)}\left|u_{A}\right| ∣ u B ∣ ≈ ( c m c 2 + m c ) m v ∣ u A ∣ Ou como queríamos demonstrar:
∣ u B ∣ ≈ 1 2 v c ∣ u A ∣ \left|u_{B}\right|\approx\frac{1}{2}\frac{v}{c}\left|u_{A}\right| ∣ u B ∣ ≈ 2 1 c v ∣ u A ∣ Alguma observações é que encontrei na literatura a razão entre a componentes após substituir p μ → p μ + e A μ p^{\mu}\rightarrow p^{\mu}+eA^{\mu} p μ → p μ + e A μ e A μ eA^{\mu} e A μ E E E 55 56 57 56 σ ⋅ p u B ≈ 0 \boldsymbol{\sigma}\cdot\boldsymbol{p}u_{B}\approx0 σ ⋅ p u B ≈ 0 σ ⋅ p ≠ 0 \boldsymbol{\sigma}\cdot\boldsymbol{p}\neq0 σ ⋅ p = 0 u B ≈ 0 u_{B}\approx0 u B ≈ 0 u B u_{B} u B v / c v/c v / c v / c ≈ 0 v/c\approx0 v / c ≈ 0
b)
Agora voltando a equação 99 da aula 35, temos que:
H = α ⋅ p + β m H=\boldsymbol{\alpha}\cdot\boldsymbol{p}+\beta m H = α ⋅ p + β m ( i γ μ ∂ μ − I m ) ψ = 0 \left(i\gamma^{\mu}\partial_{\mu}-\boldsymbol{I}m\right)\psi=0 ( i γ μ ∂ μ − I m ) ψ = 0 Lembrando que γ μ ≡ ( β , β α ) \gamma^{\mu}\equiv\left(\beta,\beta\boldsymbol{\alpha}\right) γ μ ≡ ( β , β α )
α i = ( 0 σ i σ i 0 ) , β = ( 1 0 0 − 1 ) \alpha_{i}=\left(\begin{array}{cc}
0 & \sigma_{i}\\
\sigma_{i} & 0
\end{array}\right),\qquad\beta=\left(\begin{array}{cc}
\boldsymbol{1} & 0\\
0 & -\boldsymbol{1}
\end{array}\right) α i = ( 0 σ i σ i 0 ) , β = ( 1 0 0 − 1 ) Reescrevendo:
( i γ μ p μ − I m ) ψ = 0 \left(i\gamma^{\mu}p_{\mu}-\boldsymbol{I}m\right)\psi=0 ( i γ μ p μ − I m ) ψ = 0 Fazendo a substituição p μ → p u + e A μ p^{\mu}\rightarrow p^{u}+eA^{\mu} p μ → p u + e A μ A μ = ( A 0 , A ) A^{\mu}=\left(A^{0},\boldsymbol{A}\right) A μ = ( A 0 , A )
( i γ μ ( ∂ μ + e A μ ) − I m ) ψ = 0 ( γ μ i ∂ μ + e γ μ A μ − I m ) ψ = 0 ( γ ν ( i ∂ ν + e A ν ) − I m ) ψ = − γ 0 ( i ∂ 0 + e A 0 ) ψ \begin{align*}
\left(i\gamma^{\mu}\left(\partial_{\mu}+eA_{\mu}\right)-\boldsymbol{I}m\right)\psi & =0\\
\left(\gamma^{\mu}i\partial_{\mu}+e\gamma^{\mu}A_{\mu}-\boldsymbol{I}m\right)\psi & =0\\
\left(\gamma^{\nu}\left(i\partial_{\nu}+eA_{\nu}\right)-\boldsymbol{I}m\right)\psi & =-\gamma^{0}\left(i\partial_{0}+eA_{0}\right)\psi
\end{align*} ( i γ μ ( ∂ μ + e A μ ) − I m ) ψ ( γ μ i ∂ μ + e γ μ A μ − I m ) ψ ( γ ν ( i ∂ ν + e A ν ) − I m ) ψ = 0 = 0 = − γ 0 ( i ∂ 0 + e A 0 ) ψ Onde μ = 0 , 1 , 2 , 3 \mu=0,1,2,3 μ = 0 , 1 , 2 , 3 ν = 1 , 2 , 3 \nu=1,2,3 ν = 1 , 2 , 3 p μ = i ∂ μ p_{\mu}=i\partial_{\mu} p μ = i ∂ μ E = i ∂ 0 E=i\partial_{0} E = i ∂ 0
( β α ⋅ ( p + e A ) − I m ) ψ = − ( E + e A 0 ) β ψ ( ( 0 σ i ⋅ ( p + e A ) − σ i ⋅ ( p + e A ) 0 ) − ( m 0 0 m ) ) ( u A u B ) = − ( E + e A 0 ) ( 1 0 0 − 1 ) ( u A u B ) ( − m σ i ⋅ ( p + e A ) − σ i ⋅ ( p + e A ) − m ) ( u A u B ) = ( E + e A 0 ) ( − u A u B ) ( ( E + e A 0 − m ) σ i ⋅ ( p + e A ) − σ i ⋅ ( p + e A ) − ( E + e A 0 + m ) ) ( u A u B ) = ( 0 0 ) \begin{align*}
\left(\beta\boldsymbol{\alpha}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)-\boldsymbol{I}m\right)\psi & =-\left(E+eA_{0}\right)\beta\psi\\
\left(\left(\begin{array}{cc}
0 & \boldsymbol{\sigma}_{i}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\\
-\boldsymbol{\sigma}_{i}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right) & 0
\end{array}\right)-\left(\begin{array}{cc}
m & 0\\
0 & m
\end{array}\right)\right)\left(\begin{array}{c}
u_{A}\\
u_{B}
\end{array}\right) & =-\left(E+eA_{0}\right)\left(\begin{array}{cc}
\boldsymbol{1} & 0\\
0 & -\boldsymbol{1}
\end{array}\right)\left(\begin{array}{c}
u_{A}\\
u_{B}
\end{array}\right)\\
\left(\begin{array}{cc}
-m & \boldsymbol{\sigma}_{i}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\\
-\boldsymbol{\sigma}_{i}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right) & -m
\end{array}\right)\left(\begin{array}{c}
u_{A}\\
u_{B}
\end{array}\right) & =\left(E+eA_{0}\right)\left(\begin{array}{c}
-u_{A}\\
u_{B}
\end{array}\right)\\
\left(\begin{array}{cc}
\left(E+eA_{0}-m\right) & \boldsymbol{\sigma}_{i}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\\
-\boldsymbol{\sigma}_{i}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right) & -\left(E+eA_{0}+m\right)
\end{array}\right)\left(\begin{array}{c}
u_{A}\\
u_{B}
\end{array}\right) & =\left(\begin{array}{c}
0\\
0
\end{array}\right)
\end{align*} ( β α ⋅ ( p + e A ) − I m ) ψ ( ( 0 − σ i ⋅ ( p + e A ) σ i ⋅ ( p + e A ) 0 ) − ( m 0 0 m ) ) ( u A u B ) ( − m − σ i ⋅ ( p + e A ) σ i ⋅ ( p + e A ) − m ) ( u A u B ) ( ( E + e A 0 − m ) − σ i ⋅ ( p + e A ) σ i ⋅ ( p + e A ) − ( E + e A 0 + m ) ) ( u A u B ) = − ( E + e A 0 ) β ψ = − ( E + e A 0 ) ( 1 0 0 − 1 ) ( u A u B ) = ( E + e A 0 ) ( − u A u B ) = ( 0 0 ) O que nos leva a um novo conjunto de equações:
σ ⋅ ( p + e A ) u B = ( E + e A 0 − m ) u A σ ⋅ ( p + e A ) u A = ( E + e A 0 + m ) u B \begin{align*}
\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)u_{B} & =\left(E+eA^{0}-m\right)u_{A}\\
\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)u_{A} & =\left(E+eA^{0}+m\right)u_{B}
\end{align*} σ ⋅ ( p + e A ) u B σ ⋅ ( p + e A ) u A = ( E + e A 0 − m ) u A = ( E + e A 0 + m ) u B Substituindo u B u_{B} u B
( E + e A 0 − m ) u A = [ σ ⋅ ( p + e A ) ] 2 ( E + e A 0 + m ) u A \begin{align*}
\left(E+eA^{0}-m\right)u_{A} & =\frac{\left[\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\right]^{2}}{\left(E+eA^{0}+m\right)}u_{A}
\end{align*} ( E + e A 0 − m ) u A = ( E + e A 0 + m ) [ σ ⋅ ( p + e A ) ] 2 u A Então reescrevendo E N R = E − m E_{NR}=E-m E NR = E − m
( E N R + e A 0 ) u A = [ σ ⋅ ( p + e A ) ] 2 ( E N R + e A 0 + 2 m ) u A \begin{align*}
\left(E_{NR}+eA^{0}\right)u_{A} & =\frac{\left[\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\right]^{2}}{\left(E_{NR}+eA^{0}+2m\right)}u_{A}
\end{align*} ( E NR + e A 0 ) u A = ( E NR + e A 0 + 2 m ) [ σ ⋅ ( p + e A ) ] 2 u A E usando e A 0 ≪ m eA^{0}\ll m e A 0 ≪ m
1 ( E N R + e A 0 + 2 m ) = 1 2 m [ 1 ( E N R 2 m + e A 0 2 m + 1 ) ] ≈ 1 2 m [ 1 ( E N R 2 m ) + 1 ] \begin{align*}
\frac{1}{\left(E_{NR}+eA^{0}+2m\right)} & =\frac{1}{2m}\left[\frac{1}{\left(\frac{E_{NR}}{2m}+\frac{eA^{0}}{2m}+1\right)}\right]\approx\frac{1}{2m}\left[\frac{1}{\left(\frac{E_{NR}}{2m}\right)+1}\right]
\end{align*} ( E NR + e A 0 + 2 m ) 1 = 2 m 1 ⎣ ⎡ ( 2 m E NR + 2 m e A 0 + 1 ) 1 ⎦ ⎤ ≈ 2 m 1 [ ( 2 m E NR ) + 1 1 ] E fazendo expansão em série de taylor com E N R 2 m = x \frac{E_{NR}}{2m}=x 2 m E NR = x E ≈ m E\approx m E ≈ m E N R ≈ 0 E_{NR}\approx0 E NR ≈ 0
1 ( E N R + e A 0 + 2 m ) ≈ 1 2 m [ 1 − E N R 2 m + ( E N R 2 m ) 2 + … ] \frac{1}{\left(E_{NR}+eA^{0}+2m\right)}\approx\frac{1}{2m}\left[1-\frac{E_{NR}}{2m}+\left(\frac{E_{NR}}{2m}\right)^{2}+\dots\right] ( E NR + e A 0 + 2 m ) 1 ≈ 2 m 1 [ 1 − 2 m E NR + ( 2 m E NR ) 2 + … ] Pegando só o primeiro termo da expansão:
:
1 ( E N R + e A 0 + 2 m ) ≈ 1 2 m \frac{1}{\left(E_{NR}+eA^{0}+2m\right)}\approx\frac{1}{2m} ( E NR + e A 0 + 2 m ) 1 ≈ 2 m 1 E voltando:
( E N R + e A 0 ) u A = 1 2 m [ σ ⋅ ( p + e A ) ] 2 u A \begin{align*}
\left(E_{NR}+eA^{0}\right)u_{A} & =\frac{1}{2m}\left[\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\right]^{2}u_{A}
\end{align*} ( E NR + e A 0 ) u A = 2 m 1 [ σ ⋅ ( p + e A ) ] 2 u A Abrindo então o termo ao quadrado, e lembrando que p μ = − i ℏ ∂ μ p_{\mu}=-i\hbar\partial_{\mu} p μ = − i ℏ ∂ μ σ μ \sigma_{\mu} σ μ A \boldsymbol{A} A
[ σ ⋅ ( p + e A ) ] 2 = [ σ i p i + e σ i A i ] 2 = [ σ j p j σ i p i + e σ j p j σ i A i + e σ j A j σ i p i + e 2 σ j A j σ i A i ] = σ j σ i [ p j p i + e p j A i + e A j p i + e 2 A j A i ] = σ j σ i [ p j p i + e 2 A j A i ] + σ j σ i [ e p j A i + e A j p i ] = 1 2 ( σ j σ i + σ i σ j ) [ p j p i + e 2 A j A i ] + σ j σ i e [ p j A i + A j p i ] \begin{align*}
\left[\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\right]^{2} & =\left[\sigma_{i}p_{i}+e\sigma_{i}A_{i}\right]^{2}\\
& =\left[\sigma_{j}p_{j}\sigma_{i}p_{i}+e\sigma_{j}p_{j}\sigma_{i}A_{i}+e\sigma_{j}A_{j}\sigma_{i}p_{i}+e^{2}\sigma_{j}A_{j}\sigma_{i}A_{i}\right]\\
& =\sigma_{j}\sigma_{i}\left[p_{j}p_{i}+ep_{j}A_{i}+eA_{j}p_{i}+e^{2}A_{j}A_{i}\right]\\
& =\sigma_{j}\sigma_{i}\left[p_{j}p_{i}+e^{2}A_{j}A_{i}\right]+\sigma_{j}\sigma_{i}\left[ep_{j}A_{i}+eA_{j}p_{i}\right]\\
& =\frac{1}{2}\left(\sigma_{j}\sigma_{i}+\sigma_{i}\sigma_{j}\right)\left[p_{j}p_{i}+e^{2}A_{j}A_{i}\right]+\sigma_{j}\sigma_{i}e\left[p_{j}A_{i}+A_{j}p_{i}\right]
\end{align*} [ σ ⋅ ( p + e A ) ] 2 = [ σ i p i + e σ i A i ] 2 = [ σ j p j σ i p i + e σ j p j σ i A i + e σ j A j σ i p i + e 2 σ j A j σ i A i ] = σ j σ i [ p j p i + e p j A i + e A j p i + e 2 A j A i ] = σ j σ i [ p j p i + e 2 A j A i ] + σ j σ i [ e p j A i + e A j p i ] = 2 1 ( σ j σ i + σ i σ j ) [ p j p i + e 2 A j A i ] + σ j σ i e [ p j A i + A j p i ] Como { σ i , σ j } = 2 δ i j \left\{ \sigma_{i},\sigma_{j}\right\} =2\delta_{ij} { σ i , σ j } = 2 δ ij σ i σ j = δ i j + i ε i j k σ k \sigma_{i}\sigma_{j}=\delta_{ij}+i\varepsilon_{ijk}\sigma_{k} σ i σ j = δ ij + i ε ijk σ k
[ σ ⋅ ( p + e A ) ] 2 = δ i j [ p j p i + e 2 A j A i ] + σ j σ i e [ p j A i + A j p i ] = p 2 + e 2 A 2 + ( δ i j + i ε i j k σ k ) e [ p j A i + A j p i ] \begin{align*}
\left[\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\right]^{2} & =\delta_{ij}\left[p_{j}p_{i}+e^{2}A_{j}A_{i}\right]+\sigma_{j}\sigma_{i}e\left[p_{j}A_{i}+A_{j}p_{i}\right]\\
& =\boldsymbol{p}^{2}+e^{2}\boldsymbol{A}^{2}+\left(\delta_{ij}+i\varepsilon_{ijk}\sigma_{k}\right)e\left[p_{j}A_{i}+A_{j}p_{i}\right]
\end{align*} [ σ ⋅ ( p + e A ) ] 2 = δ ij [ p j p i + e 2 A j A i ] + σ j σ i e [ p j A i + A j p i ] = p 2 + e 2 A 2 + ( δ ij + i ε ijk σ k ) e [ p j A i + A j p i ] Abrindo então último termo e sendo p = − i ∇ \boldsymbol{p}=-i\nabla p = − i ∇ u A u_{A} u A
( δ i j + i ε i j k σ k ) ( p j A i + A j p i ) u A = [ δ i j p j A i + δ i j A j p i + i ε j i k σ k p j A i + i ε j i k σ k A j p i ] u A = ( p ⋅ A + A ⋅ p ) u A + ε j i k σ k ∂ j ( A i u A ) + ε j i k σ k A j ∂ i ( u A ) = ( p ⋅ A + A ⋅ p ) u A + ε j i k σ k ∂ j ( A i ) u A + [ ε j i k σ k A i ∂ j ( u A ) + ε j i k σ k A j ∂ i ( u A ) ] \begin{align*}
\left(\delta_{ij}+i\varepsilon_{ijk}\sigma_{k}\right)\left(p_{j}A_{i}+A_{j}p_{i}\right)u_{A} & =\left[\delta_{ij}p_{j}A_{i}+\delta_{ij}A_{j}p_{i}+i\varepsilon_{jik}\sigma_{k}p_{j}A_{i}+i\varepsilon_{jik}\sigma_{k}A_{j}p_{i}\right]u_{A}\\
& =\left(\boldsymbol{p}\cdot\boldsymbol{A}+\boldsymbol{A}\cdot\boldsymbol{p}\right)u_{A}+\varepsilon_{jik}\sigma_{k}\partial_{j}\left(A_{i}u_{A}\right)+\varepsilon_{jik}\sigma_{k}A_{j}\partial_{i}\left(u_{A}\right)\\
& =\left(\boldsymbol{p}\cdot\boldsymbol{A}+\boldsymbol{A}\cdot\boldsymbol{p}\right)u_{A}+\varepsilon_{jik}\sigma_{k}\partial_{j}\left(A_{i}\right)u_{A}+\left[\varepsilon_{jik}\sigma_{k}A_{i}\partial_{j}\left(u_{A}\right)+\varepsilon_{jik}\sigma_{k}A_{j}\partial_{i}\left(u_{A}\right)\right]
\end{align*} ( δ ij + i ε ijk σ k ) ( p j A i + A j p i ) u A = [ δ ij p j A i + δ ij A j p i + i ε jik σ k p j A i + i ε jik σ k A j p i ] u A = ( p ⋅ A + A ⋅ p ) u A + ε jik σ k ∂ j ( A i u A ) + ε jik σ k A j ∂ i ( u A ) = ( p ⋅ A + A ⋅ p ) u A + ε jik σ k ∂ j ( A i ) u A + [ ε jik σ k A i ∂ j ( u A ) + ε jik σ k A j ∂ i ( u A ) ] Importante pestar atenção nos últimos dois termos. Por exemplo, mantendo 3 = 1 3=1 3 = 1
ε j i 3 A i ∂ j ( u A ) + ε j i 3 A j ∂ i ( u A ) = ε 123 A 2 ∂ 1 ( u A ) + ε 123 A 1 ∂ 2 ( u A ) + ε 213 A 1 ∂ 2 ( u A ) + ε 213 A 2 ∂ 1 ( u A ) = ( ε 123 + ε 213 ) A 2 ∂ 1 ( u A ) + ( ε 123 + ε 213 ) A 1 ∂ 2 ( u A ) = 0 \begin{align*}
\varepsilon_{ji3}A_{i}\partial_{j}\left(u_{A}\right)+\varepsilon_{ji3}A_{j}\partial_{i}\left(u_{A}\right) & =\varepsilon_{123}A_{2}\partial_{1}\left(u_{A}\right)+\varepsilon_{123}A_{1}\partial_{2}\left(u_{A}\right)+\varepsilon_{213}A_{1}\partial_{2}\left(u_{A}\right)+\varepsilon_{213}A_{2}\partial_{1}\left(u_{A}\right)\\
& =\left(\varepsilon_{123}+\varepsilon_{213}\right)A_{2}\partial_{1}\left(u_{A}\right)+\left(\varepsilon_{123}+\varepsilon_{213}\right)A_{1}\partial_{2}\left(u_{A}\right)\\
& =0
\end{align*} ε ji 3 A i ∂ j ( u A ) + ε ji 3 A j ∂ i ( u A ) = ε 123 A 2 ∂ 1 ( u A ) + ε 123 A 1 ∂ 2 ( u A ) + ε 213 A 1 ∂ 2 ( u A ) + ε 213 A 2 ∂ 1 ( u A ) = ( ε 123 + ε 213 ) A 2 ∂ 1 ( u A ) + ( ε 123 + ε 213 ) A 1 ∂ 2 ( u A ) = 0 É análogo para outras componentes. Então ficamos com:
( δ i j + i ε i j k σ k ) ( p j A i + A j p i ) u A = ( p ⋅ A + A ⋅ p ) u A + ε j i k σ k ∂ j ( A i ) u A = [ ( p ⋅ A + A ⋅ p ) + σ ⋅ ( ∇ × A ) ] u A = [ ( p ⋅ A + A ⋅ p ) + σ ⋅ B ] \begin{align*}
\left(\delta_{ij}+i\varepsilon_{ijk}\sigma_{k}\right)\left(p_{j}A_{i}+A_{j}p_{i}\right)u_{A} & =\left(\boldsymbol{p}\cdot\boldsymbol{A}+\boldsymbol{A}\cdot\boldsymbol{p}\right)u_{A}+\varepsilon_{jik}\sigma_{k}\partial_{j}\left(A_{i}\right)u_{A}\\
& =\left[\left(\boldsymbol{p}\cdot\boldsymbol{A}+\boldsymbol{A}\cdot\boldsymbol{p}\right)+\boldsymbol{\sigma}\cdot\left(\nabla\times\boldsymbol{A}\right)\right]u_{A}\\
& =\left[\left(\boldsymbol{p}\cdot\boldsymbol{A}+\boldsymbol{A}\cdot\boldsymbol{p}\right)+\boldsymbol{\sigma}\cdot\boldsymbol{B}\right]
\end{align*} ( δ ij + i ε ijk σ k ) ( p j A i + A j p i ) u A = ( p ⋅ A + A ⋅ p ) u A + ε jik σ k ∂ j ( A i ) u A = [ ( p ⋅ A + A ⋅ p ) + σ ⋅ ( ∇ × A ) ] u A = [ ( p ⋅ A + A ⋅ p ) + σ ⋅ B ] E então:
[ σ ⋅ ( p + e A ) ] 2 = p 2 + e 2 A 2 + ( δ i j + i ε i j k σ k ) e [ p j A i + A j p i ] = p 2 + e 2 A 2 + e p ⋅ A + e A ⋅ p + e σ ⋅ B = \begin{align*}
\left[\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\right]^{2} & =\boldsymbol{p}^{2}+e^{2}\boldsymbol{A}^{2}+\left(\delta_{ij}+i\varepsilon_{ijk}\sigma_{k}\right)e\left[p_{j}A_{i}+A_{j}p_{i}\right]\\
& =\boldsymbol{p}^{2}+e^{2}\boldsymbol{A}^{2}+e\boldsymbol{p}\cdot\boldsymbol{A}+e\boldsymbol{A}\cdot\boldsymbol{p}+e\boldsymbol{\sigma}\cdot\boldsymbol{B}\\
& =
\end{align*} [ σ ⋅ ( p + e A ) ] 2 = p 2 + e 2 A 2 + ( δ ij + i ε ijk σ k ) e [ p j A i + A j p i ] = p 2 + e 2 A 2 + e p ⋅ A + e A ⋅ p + e σ ⋅ B = E finalmente:
( E N R + e A 0 ) u A = 1 2 m [ σ ⋅ ( p + e A ) ] 2 u A ( E N R + e A 0 ) u A = ( p + e A ) 2 + e σ ⋅ B 2 m u A \begin{align*}
\left(E_{NR}+eA^{0}\right)u_{A} & =\frac{1}{2m}\left[\boldsymbol{\sigma}\cdot\left(\boldsymbol{p}+e\boldsymbol{A}\right)\right]^{2}u_{A}\\
\left(E_{NR}+eA^{0}\right)u_{A} & =\frac{\left(\boldsymbol{p}+e\boldsymbol{A}\right)^{2}+e\boldsymbol{\sigma}\cdot\boldsymbol{B}}{2m}u_{A}
\end{align*} ( E NR + e A 0 ) u A ( E NR + e A 0 ) u A = 2 m 1 [ σ ⋅ ( p + e A ) ] 2 u A = 2 m ( p + e A ) 2 + e σ ⋅ B u A Ou na formulação que o exercício pediu:
E N R u A = ( 1 2 m ( p + e A ) 2 + e 2 m σ ⋅ B − e A 0 ) u A E_{NR}u_{A}=\left(\frac{1}{2m}\left(\boldsymbol{p}+e\boldsymbol{A}\right)^{2}+\frac{e}{2m}\boldsymbol{\sigma}\cdot\boldsymbol{B}-eA^{0}\right)u_{A} E NR u A = ( 2 m 1 ( p + e A ) 2 + 2 m e σ ⋅ B − e A 0 ) u A Problem 5 ¶ Derive the completeness relations
∑ s = 1 , 2 u ( s ) ( p ) u ‾ ( s ) ( p ) = p + m ∑ s = 1 , 2 v ( s ) ( p ) v ‾ ( s ) ( p ) = p − m \begin{align*}
\sum_{s=1,2} & u^{\left(s\right)}\left(p\right)\overline{u}^{\left(s\right)}\left(p\right)=\cancel{p}+m\\
\sum_{s=1,2} & v^{\left(s\right)}\left(p\right)\overline{v}^{\left(s\right)}\left(p\right)=\cancel{p}-m
\end{align*} s = 1 , 2 ∑ s = 1 , 2 ∑ u ( s ) ( p ) u ( s ) ( p ) = p + m v ( s ) ( p ) v ( s ) ( p ) = p − m Queremos desenvolver:
∑ s = 1 , 2 u ( s ) ( p ) u ‾ ( s ) ( p ) = p + m ∑ s = 1 , 2 v ( s ) ( p ) v ‾ ( s ) ( p ) = p − m \begin{align*}
\sum_{s=1,2} & u^{\left(s\right)}\left(p\right)\overline{u}^{\left(s\right)}\left(p\right)=\cancel{p}+m\\
\sum_{s=1,2} & v^{\left(s\right)}\left(p\right)\overline{v}^{\left(s\right)}\left(p\right)=\cancel{p}-m
\end{align*} s = 1 , 2 ∑ s = 1 , 2 ∑ u ( s ) ( p ) u ( s ) ( p ) = p + m v ( s ) ( p ) v ( s ) ( p ) = p − m Temos então da equaçao 177 da aula 37:
χ ( 1 ) = ( 1 0 ) , χ ( 2 ) = ( 0 1 ) \chi^{\left(1\right)}=\left(\begin{array}{c}
1\\
0
\end{array}\right),\qquad\chi^{\left(2\right)}=\left(\begin{array}{c}
0\\
1
\end{array}\right) χ ( 1 ) = ( 1 0 ) , χ ( 2 ) = ( 0 1 ) E da 179 e 181, sendo u ( s ) = u ( s ) u^{\left(s\right)}=u^{\left(s\right)} u ( s ) = u ( s ) u ( s + 2 ) = v ( s ) u^{\left(s+2\right)}=v^{\left(s\right)} u ( s + 2 ) = v ( s )
u ( 1 ) = N ( χ ( 1 ) σ ⋅ p E + m χ ( 1 ) ) = N ( 1 0 p 3 E + m i p 2 + p 1 E + m ) \begin{align*}
u^{\left(1\right)} & =N\left(\begin{array}{c}
\chi^{\left(1\right)}\\
\frac{\boldsymbol{\sigma}\cdot\boldsymbol{p}}{E+m}\chi^{\left(1\right)}
\end{array}\right) & =N\left(\begin{array}{c}
\begin{array}{c}
1\\
0
\end{array}\\
\frac{p_{3}}{E+m}\\
\frac{ip_{2}+p_{1}}{E+m}
\end{array}\right)
\end{align*} u ( 1 ) = N ( χ ( 1 ) E + m σ ⋅ p χ ( 1 ) ) = N ⎝ ⎛ 1 0 E + m p 3 E + m i p 2 + p 1 ⎠ ⎞ Pois:
( σ ⋅ p ) χ ( 1 ) = ( σ 1 p 1 + σ 2 p 2 + σ 3 p 3 ) χ ( 1 ) = ( p 3 p 1 − i p 2 i p 2 + p 1 − p 3 ) ( 1 0 ) = ( p 3 i p 2 + p 1 ) \left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)\chi^{\left(1\right)}=\left(\sigma_{1}p_{1}+\sigma_{2}p_{2}+\sigma_{3}p_{3}\right)\chi^{\left(1\right)}=\left(\begin{array}{cc}
p_{3} & p_{1}-ip_{2}\\
ip_{2}+p_{1} & -p_{3}
\end{array}\right)\left(\begin{array}{c}
1\\
0
\end{array}\right)=\left(\begin{array}{c}
p_{3}\\
ip_{2}+p_{1}
\end{array}\right) ( σ ⋅ p ) χ ( 1 ) = ( σ 1 p 1 + σ 2 p 2 + σ 3 p 3 ) χ ( 1 ) = ( p 3 i p 2 + p 1 p 1 − i p 2 − p 3 ) ( 1 0 ) = ( p 3 i p 2 + p 1 ) De maneira análoga:
u ( 2 ) = N ( χ ( 2 ) σ ⋅ p E + m χ ( 2 ) ) = N ( 0 1 p 1 − i p 2 E + m − p 3 E + m ) v ( 1 ) = N ( − σ ⋅ p E + m χ ( 1 ) χ ( 1 ) ) = N ( − p 3 E + m − i p 2 − p 1 E + m 1 0 ) v ( 2 ) = N ( − σ ⋅ p E + m χ ( 2 ) χ ( 2 ) ) = N ( i p 2 − p 1 E + m p 3 E + m 0 1 ) \begin{align*}
u^{\left(2\right)} & =N\left(\begin{array}{c}
\chi^{\left(2\right)}\\
\frac{\boldsymbol{\sigma}\cdot\boldsymbol{p}}{E+m}\chi^{\left(2\right)}
\end{array}\right) & =N\left(\begin{array}{c}
\begin{array}{c}
0\\
1
\end{array}\\
\frac{p_{1}-ip_{2}}{E+m}\\
\frac{-p_{3}}{E+m}
\end{array}\right)\\
v^{\left(1\right)} & =N\left(\begin{array}{c}
-\frac{\boldsymbol{\sigma}\cdot\boldsymbol{p}}{E+m}\chi^{\left(1\right)}\\
\chi^{\left(1\right)}
\end{array}\right) & =N\left(\begin{array}{c}
\begin{array}{c}
\frac{-p_{3}}{E+m}\\
\frac{-ip_{2}-p_{1}}{E+m}
\end{array}\\
1\\
0
\end{array}\right)\\
v^{\left(2\right)} & =N\left(\begin{array}{c}
-\frac{\boldsymbol{\sigma}\cdot\boldsymbol{p}}{E+m}\chi^{\left(2\right)}\\
\chi^{\left(2\right)}
\end{array}\right) & =N\left(\begin{array}{c}
\begin{array}{c}
\frac{ip_{2}-p_{1}}{E+m}\\
\frac{p_{3}}{E+m}
\end{array}\\
0\\
1
\end{array}\right)
\end{align*} u ( 2 ) v ( 1 ) v ( 2 ) = N ( χ ( 2 ) E + m σ ⋅ p χ ( 2 ) ) = N ( − E + m σ ⋅ p χ ( 1 ) χ ( 1 ) ) = N ( − E + m σ ⋅ p χ ( 2 ) χ ( 2 ) ) = N ⎝ ⎛ 0 1 E + m p 1 − i p 2 E + m − p 3 ⎠ ⎞ = N ⎝ ⎛ E + m − p 3 E + m − i p 2 − p 1 1 0 ⎠ ⎞ = N ⎝ ⎛ E + m i p 2 − p 1 E + m p 3 0 1 ⎠ ⎞ Podemos definir no Wolfram Mathematica:
u1:=\{\{1\},\{0\},\{p3/(e+M)\},\{(I{*}p2+p1)/(e+M)\}\};~
u2:=\{\{0\},\{1\},\{(p1-I{*}p2)/(E+M)\},\{-p3/(e+M)\}\};~
v1:=\{\{-p3/(e+M)\},\{(-I{*}p2-p1)/(e+M)\},\{1\},\{0\}\};~
v2:=\{\{(I{*}p2-p1)/(e+M)\},\{p3/(e+M)\},\{0\},\{1\}\};
E agora u ‾ = u † γ 0 \overline{u}=u^{\dagger}\gamma^{0} u = u † γ 0 u ‾ ( 1 ) \overline{u}^{\left(1\right)} u ( 1 )
u ‾ ( 1 ) = N ( χ ( 1 ) ( σ ⋅ p ) † E + m χ ( 1 ) ) ( 1 0 0 − 1 ) = N ( χ ( 1 ) − ( σ ⋅ p ) † E + m χ ( 1 ) ) = ( 1 0 − p 3 E + m i p 2 − p 1 E + m ) \begin{align*}
\overline{u}^{\left(1\right)} & =N\left(\begin{array}{cc}
\chi^{\left(1\right)} & \frac{\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)^{\dagger}}{E+m}\chi^{\left(1\right)}\end{array}\right)\left(\begin{array}{cc}
\boldsymbol{1} & 0\\
0 & \boldsymbol{-1}
\end{array}\right) & =N\left(\begin{array}{cc}
\chi^{\left(1\right)} & -\frac{\left(\boldsymbol{\sigma}\cdot\boldsymbol{p}\right)^{\dagger}}{E+m}\chi^{\left(1\right)}\end{array}\right) & =\left(\begin{array}{cccc}
1 & 0 & \frac{-p_{3}}{E+m} & \frac{ip_{2}-p_{1}}{E+m}\end{array}\right)
\end{align*} u ( 1 ) = N ( χ ( 1 ) E + m ( σ ⋅ p ) † χ ( 1 ) ) ( 1 0 0 − 1 ) = N ( χ ( 1 ) − E + m ( σ ⋅ p ) † χ ( 1 ) ) = ( 1 0 E + m − p 3 E + m i p 2 − p 1 ) Checando no Wolfram Mathematica:
b:=\{\{1,0,0,0\},\{0,1,0,0\},\{0,0,-1,0\},\{0,0,0,-1\}\}
lu1:=Conjugate{[}Transpose{[}u1{]}{]}.b
MatrixForm{[}v2{]}Uma vez que está certo, declaramos o resto da mesma maneira:
u ‾ ( 2 ) = ( u ( 2 ) ) † γ 0 , v ‾ ( 1 ) = ( v ( 1 ) ) † γ 0 , v ‾ ( 2 ) = ( v ( 2 ) ) † γ 0 \overline{u}^{\left(2\right)}=\left(u^{\left(2\right)}\right)^{\dagger}\gamma^{0},\quad\overline{v}^{\left(1\right)}=\left(v^{\left(1\right)}\right)^{\dagger}\gamma^{0},\quad\overline{v}^{\left(2\right)}=\left(v^{\left(2\right)}\right)^{\dagger}\gamma^{0} u ( 2 ) = ( u ( 2 ) ) † γ 0 , v ( 1 ) = ( v ( 1 ) ) † γ 0 , v ( 2 ) = ( v ( 2 ) ) † γ 0 O que retorna:
u ‾ ( 2 ) = N ( 0 1 p 1 + i p 2 E + m p 3 E + m ) , v ‾ ( 1 ) = N ( − p 3 E + m − p 1 + i p 2 E + m − 1 0 ) , v ‾ ( 1 ) = N ( − p 1 − i p 2 E + m p 3 E + m − 1 0 ) \overline{u}^{\left(2\right)}=N\left(\begin{array}{cccc}
0 & 1 & \frac{p_{1}+ip_{2}}{E+m} & \frac{p_{3}}{E+m}\end{array}\right),\quad\overline{v}^{\left(1\right)}=N\left(\begin{array}{cccc}
\frac{-p_{3}}{E+m} & \frac{-p_{1}+ip_{2}}{E+m} & -1 & 0\end{array}\right),\quad\overline{v}^{\left(1\right)}=N\left(\begin{array}{cccc}
\frac{-p_{1}-ip_{2}}{E+m} & \frac{p_{3}}{E+m} & -1 & 0\end{array}\right) u ( 2 ) = N ( 0 1 E + m p 1 + i p 2 E + m p 3 ) , v ( 1 ) = N ( E + m − p 3 E + m − p 1 + i p 2 − 1 0 ) , v ( 1 ) = N ( E + m − p 1 − i p 2 E + m p 3 − 1 0 ) Fazendo então primeiro:
MatrixForm{[}u1.lu1+u2.lu2{]}∑ s = 1 , 2 u ( s ) ( p ) u ‾ ( s ) ( p ) = ( E + m 0 − p 3 − p 1 + p 2 0 E + m − p 1 − i p 2 p 3 p 3 p 1 − i p 2 − ( p 1 2 + p 2 2 + p 3 2 ) E + m 0 p 1 + i p 2 − p 3 0 − ( p 1 2 + p 2 2 + p 3 2 ) E + m ) \begin{align*}
\sum_{s=1,2} & u^{\left(s\right)}\left(p\right)\overline{u}^{\left(s\right)}\left(p\right)=\left(\begin{array}{cccc}
E+m & 0 & -p_{3} & -p_{1}+p_{2}\\
0 & E+m & -p_{1}-ip_{2} & p_{3}\\
p_{3} & p_{1}-ip_{2} & -\frac{\left(p^{2}_{1}+p^{2}_{2}+p^{2}_{3}\right)}{E+m} & 0\\
p_{1}+ip_{2} & -p_{3} & 0 & -\frac{\left(p^{2}_{1}+p^{2}_{2}+p^{2}_{3}\right)}{E+m}
\end{array}\right)
\end{align*} s = 1 , 2 ∑ u ( s ) ( p ) u ( s ) ( p ) = ⎝ ⎛ E + m 0 p 3 p 1 + i p 2 0 E + m p 1 − i p 2 − p 3 − p 3 − p 1 − i p 2 − E + m ( p 1 2 + p 2 2 + p 3 2 ) 0 − p 1 + p 2 p 3 0 − E + m ( p 1 2 + p 2 2 + p 3 2 ) ⎠ ⎞ Uma vez que N = E + M N=\sqrt{E+M} N = E + M E 2 = ( m c 2 ) 2 + ( p c ) 2 E^{2}=\left(mc^{2}\right)^{2}+\left(pc\right)^{2} E 2 = ( m c 2 ) 2 + ( p c ) 2 c = 1 c=1 c = 1 p 2 = E 2 − m 2 p^{2}=E^{2}-m^{2} p 2 = E 2 − m 2
E 2 − m 2 E + m = ( E − m ) ( E + m ) E + m = E − m \frac{E^{2}-m^{2}}{E+m}=\frac{\left(E-m\right)\left(E+m\right)}{E+m}=E-m E + m E 2 − m 2 = E + m ( E − m ) ( E + m ) = E − m Temos apenas:
∑ s = 1 , 2 u ( s ) ( p ) u ‾ ( s ) ( p ) = ( E + m 0 − p 3 − p 1 + p 2 0 E + m − p 1 − i p 2 p 3 p 3 p 1 − i p 2 − ( E − m ) 0 p 1 + i p 2 − p 3 0 − ( E − m ) ) \sum_{s=1,2}u^{\left(s\right)}\left(p\right)\overline{u}^{\left(s\right)}\left(p\right)=\left(\begin{array}{cccc}
E+m & 0 & -p_{3} & -p_{1}+p_{2}\\
0 & E+m & -p_{1}-ip_{2} & p_{3}\\
p_{3} & p_{1}-ip_{2} & -\left(E-m\right) & 0\\
p_{1}+ip_{2} & -p_{3} & 0 & -\left(E-m\right)
\end{array}\right) s = 1 , 2 ∑ u ( s ) ( p ) u ( s ) ( p ) = ⎝ ⎛ E + m 0 p 3 p 1 + i p 2 0 E + m p 1 − i p 2 − p 3 − p 3 − p 1 − i p 2 − ( E − m ) 0 − p 1 + p 2 p 3 0 − ( E − m ) ⎠ ⎞ Agora lembrando que:
γ 0 = ( 1 0 0 0 0 1 0 0 0 0 − 1 0 0 0 0 − 1 ) γ 1 = ( 0 0 0 1 0 0 1 0 0 − 1 0 0 − 1 0 0 0 ) γ 1 = ( 0 0 0 i 0 0 i 0 0 i 0 0 − i 0 0 0 ) γ 1 = ( 0 0 1 0 0 0 0 1 − 1 0 0 0 0 1 0 0 ) \begin{align*}
\gamma^{0}=\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{array}\right) & \gamma^{1}=\left(\begin{array}{cccc}
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0\\
0 & -1 & 0 & 0\\
-1 & 0 & 0 & 0
\end{array}\right)\\
\gamma^{1}=\left(\begin{array}{cccc}
0 & 0 & 0 & i\\
0 & 0 & i & 0\\
0 & i & 0 & 0\\
-i & 0 & 0 & 0
\end{array}\right) & \gamma^{1}=\left(\begin{array}{cccc}
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0
\end{array}\right)
\end{align*} γ 0 = ⎝ ⎛ 1 0 0 0 0 1 0 0 0 0 − 1 0 0 0 0 − 1 ⎠ ⎞ γ 1 = ⎝ ⎛ 0 0 0 − i 0 0 i 0 0 i 0 0 i 0 0 0 ⎠ ⎞ γ 1 = ⎝ ⎛ 0 0 0 − 1 0 0 − 1 0 0 1 0 0 1 0 0 0 ⎠ ⎞ γ 1 = ⎝ ⎛ 0 0 − 1 0 0 0 0 1 1 0 0 0 0 1 0 0 ⎠ ⎞ Neste ponto, uma observação é importante de ser eita. Até aqui estávamos usando p i p_{i} p i i i i p μ = ( E , p ) p^{\mu}=\left(E,\boldsymbol{p}\right) p μ = ( E , p ) p i p^{i} p i p i = ( E , − p ) p_{i}=\left(E,-\boldsymbol{p}\right) p i = ( E , − p ) γ μ p μ = p \gamma^{\mu}p_{\mu}=\cancel{p} γ μ p μ = p
∑ s = 1 , 2 u ( s ) ( p ) u ‾ ( s ) ( p ) = γ μ p μ + I m \sum_{s=1,2}u^{\left(s\right)}\left(p\right)\overline{u}^{\left(s\right)}\left(p\right)=\gamma^{\mu}p_{\mu}+\mathbb{I}m s = 1 , 2 ∑ u ( s ) ( p ) u ( s ) ( p ) = γ μ p μ + I m Como queríamos demonstrar. Fazendo um processo análogo:
MatrixForm{[}v1.lv1+v2.lv2{]}E já sabendo que ( p 1 2 + p 2 2 + p 3 2 ) / ( E + m ) = E − m \left(p^{2}_{1}+p^{2}_{2}+p^{2}_{3}\right)/(E+m)=E-m ( p 1 2 + p 2 2 + p 3 2 ) / ( E + m ) = E − m
∑ s = 1 , 2 v ( s ) ( p ) v ‾ ( s ) ( p ) = ( E − m 0 − p 3 − p 1 + p 2 0 E − m − p 1 − i p 2 p 3 p 3 p 1 − i p 2 − E − m 0 p 1 + i p 2 − p 3 0 − E − m ) \begin{align*}
\sum_{s=1,2} & v^{\left(s\right)}\left(p\right)\overline{v}^{\left(s\right)}\left(p\right)=\left(\begin{array}{cccc}
E-m & 0 & -p_{3} & -p_{1}+p_{2}\\
0 & E-m & -p_{1}-ip_{2} & p_{3}\\
p_{3} & p_{1}-ip_{2} & -E-m & 0\\
p_{1}+ip_{2} & -p_{3} & 0 & -E-m
\end{array}\right)
\end{align*} s = 1 , 2 ∑ v ( s ) ( p ) v ( s ) ( p ) = ⎝ ⎛ E − m 0 p 3 p 1 + i p 2 0 E − m p 1 − i p 2 − p 3 − p 3 − p 1 − i p 2 − E − m 0 − p 1 + p 2 p 3 0 − E − m ⎠ ⎞ Que de maneira análoga, nos leva a :
∑ s = 1 , 2 v ( s ) ( p ) v ‾ ( s ) ( p ) = γ μ p μ − I m \sum_{s=1,2}v^{\left(s\right)}\left(p\right)\overline{v}^{\left(s\right)}\left(p\right)=\gamma^{\mu}p_{\mu}-\mathbb{I}m s = 1 , 2 ∑ v ( s ) ( p ) v ( s ) ( p ) = γ μ p μ − I m
Lima, P. C. de. (2013). Equações Diferenciais Parciais I . http://150.164.25.15/ lima/apostilas/EDP1-aulas-22-8-13.pdf