A função de onda
Equação de Schrödinger ¶ A equação de Schrödinger desempenha um papel análogo na mecânica quântica ao da segunda lei de Newton na mecânica clássica:
i ℏ ∂ Ψ ∂ t = − ℏ 2 2 m ∂ 2 Ψ ∂ x 2 + V Ψ i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V\Psi i ℏ ∂ t ∂ Ψ = − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ + V Ψ Onde ℏ \hbar ℏ 2 π 2\pi 2 π m m m x x x t t t Ψ ( x , 0 ) \Psi\left(x,0\right) Ψ ( x , 0 ) 1 Ψ ( x , t ) \Psi\left(x,t\right) Ψ ( x , t ) x ( t ) x\left(t\right) x ( t )
Vale destacar que a função de onda não é algo que pode ser medido diretamente, o que fazemos é usá-la para obter grandezas que podem ser observadas diretamente. Por exemplo, a interpretação estatística de Born nos fornece a probabilidade de encontrar a partícula entre os pontos a a a b b b t t t
P = ∫ a b ∣ Ψ ( x , t ) ∣ 2 d x P=\int^{b}_{a}\left|\Psi\left(x,t\right)\right|^{2}dx P = ∫ a b ∣ Ψ ( x , t ) ∣ 2 d x Como ∣ Ψ ( x , t ) ∣ 2 \left|\Psi\left(x,t\right)\right|^{2} ∣ Ψ ( x , t ) ∣ 2 m − 1 m^{-1} m − 1 Ψ ( x , t ) \Psi\left(\boldsymbol{x},t\right) Ψ ( x , t ) m − 1 / 2 m^{-1/2} m − 1/2 n n n m − n / 2 m^{-n/2} m − n /2
∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 1 \int^{+\infty}_{-\infty}\left|\Psi\left(x,t\right)\right|^{2}dx=1 ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 1 Retornando à equação de Schrödinger: se Ψ \Psi Ψ A Ψ A\Psi A Ψ A A A A A A Ψ \Psi Ψ
d d t ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = ∫ − ∞ + ∞ ∂ ∂ t ∣ Ψ ( x , t ) ∣ 2 d x \frac{d}{dt}\int^{+\infty}_{-\infty}\left|\Psi\left(x,t\right)\right|^{2}dx=\int^{+\infty}_{-\infty}\frac{\partial}{\partial t}\left|\Psi\left(x,t\right)\right|^{2}dx d t d ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = ∫ − ∞ + ∞ ∂ t ∂ ∣ Ψ ( x , t ) ∣ 2 d x Aqui, fizemos a troca de derivada total por derivada parcial dentro da integral porque a integral resulta em uma função que depende apenas do tempo t t t ∣ Ψ ( x , t ) ∣ 2 |\Psi(x,t)|^{2} ∣Ψ ( x , t ) ∣ 2 x x x t t t
∂ ∂ t ∣ Ψ ∣ 2 = ∂ ∂ t ∣ Ψ ∣ 2 = ∂ ∂ t ( Ψ ∗ Ψ ) 2 = ∂ ∂ t ( Ψ ∗ Ψ ) = Ψ ∗ ∂ Ψ ∂ t + ∂ Ψ ∗ ∂ t Ψ \frac{\partial}{\partial t}\left|\Psi\right|^{2}=\frac{\partial}{\partial t}\left|\Psi\right|^{2}=\frac{\partial}{\partial t}\left(\sqrt{\Psi^{*}\Psi}\right)^{2}=\frac{\partial}{\partial t}\left(\Psi^{*}\Psi\right)=\Psi^{*}\frac{\partial\Psi}{\partial t}+\frac{\partial\Psi^{*}}{\partial t}\Psi ∂ t ∂ ∣ Ψ ∣ 2 = ∂ t ∂ ∣ Ψ ∣ 2 = ∂ t ∂ ( Ψ ∗ Ψ ) 2 = ∂ t ∂ ( Ψ ∗ Ψ ) = Ψ ∗ ∂ t ∂ Ψ + ∂ t ∂ Ψ ∗ Ψ A equação de Schrödinger é:
∂ Ψ ∂ t = i ℏ 2 m ∂ 2 Ψ ∂ x 2 − i ℏ V Ψ \frac{\partial\Psi}{\partial t}=\frac{i\hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{i}{\hbar}V\Psi ∂ t ∂ Ψ = 2 m i ℏ ∂ x 2 ∂ 2 Ψ − ℏ i V Ψ E tomando o complexo conjugado, assumindo V V V
∂ Ψ ∗ ∂ t = − i ℏ 2 m ∂ 2 Ψ ∗ ∂ x 2 + i ℏ V Ψ ∗ \frac{\partial\Psi^{*}}{\partial t}=-\frac{i\hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}+\frac{i}{\hbar}V\Psi^{*} ∂ t ∂ Ψ ∗ = − 2 m i ℏ ∂ x 2 ∂ 2 Ψ ∗ + ℏ i V Ψ ∗ Então:
∂ ∂ t ∣ Ψ ∣ 2 = Ψ ∗ ∂ Ψ ∂ t + ∂ Ψ ∗ ∂ t Ψ = Ψ ∗ ( i ℏ 2 m ∂ 2 Ψ ∂ x 2 − i ℏ V Ψ ) + ( − i ℏ 2 m ∂ 2 Ψ ∗ ∂ x 2 + i ℏ V Ψ ∗ ) Ψ = Ψ ∗ i ℏ 2 m ∂ 2 Ψ ∂ x 2 − i ℏ 2 m ∂ 2 Ψ ∗ ∂ x 2 Ψ = i ℏ 2 m ( Ψ ∗ ∂ 2 Ψ ∂ x 2 − ∂ 2 Ψ ∗ ∂ x 2 Ψ ) \begin{align}
\frac{\partial}{\partial t}\left|\Psi\right|^{2} & =\Psi^{*}\frac{\partial\Psi}{\partial t}+\frac{\partial\Psi^{*}}{\partial t}\Psi\\
& =\Psi^{*}\left(\frac{i\hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{i}{\hbar}V\Psi\right)+\left(-\frac{i\hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}+\frac{i}{\hbar}V\Psi^{*}\right)\Psi\\
& =\Psi^{*}\frac{i\hbar}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{i\hbar}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi\\
& =\frac{i\hbar}{2m}\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi\right)
\end{align} ∂ t ∂ ∣ Ψ ∣ 2 = Ψ ∗ ∂ t ∂ Ψ + ∂ t ∂ Ψ ∗ Ψ = Ψ ∗ ( 2 m i ℏ ∂ x 2 ∂ 2 Ψ − ℏ i V Ψ ) + ( − 2 m i ℏ ∂ x 2 ∂ 2 Ψ ∗ + ℏ i V Ψ ∗ ) Ψ = Ψ ∗ 2 m i ℏ ∂ x 2 ∂ 2 Ψ − 2 m i ℏ ∂ x 2 ∂ 2 Ψ ∗ Ψ = 2 m i ℏ ( Ψ ∗ ∂ x 2 ∂ 2 Ψ − ∂ x 2 ∂ 2 Ψ ∗ Ψ ) Então:
d d t ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = ∫ − ∞ + ∞ ∂ ∂ t ∣ Ψ ( x , t ) ∣ 2 d x = ∫ − ∞ + ∞ i ℏ 2 m ( Ψ ∗ ∂ 2 Ψ ∂ x 2 − ∂ 2 Ψ ∗ ∂ x 2 Ψ ) d x = i ℏ 2 m ∫ − ∞ + ∞ ( Ψ ∗ ∂ 2 Ψ ∂ x 2 − ∂ 2 Ψ ∗ ∂ x 2 Ψ ) d x \begin{align}
\frac{d}{dt}\int^{+\infty}_{-\infty}\left|\Psi\left(x,t\right)\right|^{2}dx & =\int^{+\infty}_{-\infty}\frac{\partial}{\partial t}\left|\Psi\left(x,t\right)\right|^{2}dx\\
& =\int^{+\infty}_{-\infty}\frac{i\hbar}{2m}\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi\right)dx\\
& =\frac{i\hbar}{2m}\int^{+\infty}_{-\infty}\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi\right)dx
\end{align} d t d ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = ∫ − ∞ + ∞ ∂ t ∂ ∣ Ψ ( x , t ) ∣ 2 d x = ∫ − ∞ + ∞ 2 m i ℏ ( Ψ ∗ ∂ x 2 ∂ 2 Ψ − ∂ x 2 ∂ 2 Ψ ∗ Ψ ) d x = 2 m i ℏ ∫ − ∞ + ∞ ( Ψ ∗ ∂ x 2 ∂ 2 Ψ − ∂ x 2 ∂ 2 Ψ ∗ Ψ ) d x Olhando o primeiro termo, vamos aplicar integral por partes (prova
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x \int^{b}_{a}f\left(x\right)g'\left(x\right)dx=\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx ∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x Onde também temos que u = f ( x ) u=f\left(x\right) u = f ( x ) d v = f ′ ( x ) d x dv=f'\left(x\right)dx d v = f ′ ( x ) d x f = Ψ ∗ f=\Psi^{*} f = Ψ ∗ g = Ψ ′ g=\Psi' g = Ψ ′
∫ − ∞ + ∞ Ψ ( x , t ) ∗ ∂ 2 Ψ ( x , t ) ∂ x 2 d x = Ψ ∗ Ψ ′ ∣ − ∞ + ∞ − ∫ − ∞ + ∞ Ψ ′ Ψ ∗ ′ d x \int^{+\infty}_{-\infty}\Psi\left(x,t\right)^{*}\frac{\partial^{2}\Psi\left(x,t\right)}{\partial x^{2}}dx=\left.\Psi^{*}\Psi'\right|^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\Psi'\Psi^{*}{}'dx ∫ − ∞ + ∞ Ψ ( x , t ) ∗ ∂ x 2 ∂ 2 Ψ ( x , t ) d x = Ψ ∗ Ψ ′ ∣ − ∞ + ∞ − ∫ − ∞ + ∞ Ψ ′ Ψ ∗ ′ d x De modo análogo temos pra segunda integral f = Ψ f=\Psi f = Ψ g = Ψ ∗ ′ g=\Psi^{*}{}' g = Ψ ∗ ′
∫ − ∞ + ∞ Ψ ∂ 2 Ψ ∗ ∂ x 2 d x = Ψ Ψ ∗ ′ ∣ − ∞ + ∞ − ∫ − ∞ + ∞ Ψ ∗ ′ Ψ ′ d x \int^{+\infty}_{-\infty}\Psi\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}dx=\left.\Psi\Psi^{*}{}'\right|^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\Psi^{*}{}'\Psi'dx ∫ − ∞ + ∞ Ψ ∂ x 2 ∂ 2 Ψ ∗ d x = Ψ Ψ ∗ ′ ∣ − ∞ + ∞ − ∫ − ∞ + ∞ Ψ ∗ ′ Ψ ′ d x Então:
d d t ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = i ℏ 2 m ∫ − ∞ + ∞ ( Ψ ∗ ∂ 2 Ψ ∂ x 2 − ∂ 2 Ψ ∗ ∂ x 2 Ψ ) d x = i ℏ 2 m [ Ψ ∗ Ψ ′ ∣ − ∞ + ∞ − ∫ − ∞ + ∞ Ψ ′ Ψ ∗ ′ d x − Ψ Ψ ∗ ′ ∣ − ∞ + ∞ + ∫ − ∞ + ∞ Ψ ∗ ′ Ψ ′ d x ] = i ℏ 2 m [ Ψ ∗ Ψ ′ ∣ − ∞ + ∞ − Ψ Ψ ∗ ′ ∣ − ∞ + ∞ ] = i ℏ 2 m ( Ψ ∗ Ψ ′ − Ψ Ψ ∗ ′ ) ∣ − ∞ + ∞ \begin{align}
\frac{d}{dt}\int^{+\infty}_{-\infty}\left|\Psi\left(x,t\right)\right|^{2}dx & =\frac{i\hbar}{2m}\int^{+\infty}_{-\infty}\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi\right)dx\\
& =\frac{i\hbar}{2m}\left[\left.\Psi^{*}\Psi'\right|^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\Psi'\Psi^{*}{}'dx-\left.\Psi\Psi^{*}{}'\right|^{+\infty}_{-\infty}+\int^{+\infty}_{-\infty}\Psi^{*}{}'\Psi'dx\right]\\
& =\frac{i\hbar}{2m}\left[\left.\Psi^{*}\Psi'\right|^{+\infty}_{-\infty}-\left.\Psi\Psi^{*}{}'\right|^{+\infty}_{-\infty}\right]\\
& =\frac{i\hbar}{2m}\left.\left(\Psi^{*}\Psi'-\Psi\Psi^{*}{}'\right)\right|^{+\infty}_{-\infty}
\end{align} d t d ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 2 m i ℏ ∫ − ∞ + ∞ ( Ψ ∗ ∂ x 2 ∂ 2 Ψ − ∂ x 2 ∂ 2 Ψ ∗ Ψ ) d x = 2 m i ℏ [ Ψ ∗ Ψ ′ ∣ − ∞ + ∞ − ∫ − ∞ + ∞ Ψ ′ Ψ ∗ ′ d x − Ψ Ψ ∗ ′ ∣ − ∞ + ∞ + ∫ − ∞ + ∞ Ψ ∗ ′ Ψ ′ d x ] = 2 m i ℏ [ Ψ ∗ Ψ ′ ∣ − ∞ + ∞ − Ψ Ψ ∗ ′ ∣ − ∞ + ∞ ] = 2 m i ℏ ( Ψ ∗ Ψ ′ − Ψ Ψ ∗ ′ ) ∣ − ∞ + ∞ Para que um resultado seja fisicamente válido, a função de onda e suas derivadas devem tender a zero no infinito.[1]
d d t ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 0 \begin{align}
\frac{d}{dt}\int^{+\infty}_{-\infty}\left|\Psi\left(x,t\right)\right|^{2}dx & =0
\end{align} d t d ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 0 Ou seja, como o valor da integral é constante ao longo do tempo, se a função de onda estiver normalizada no instante t = 0 t=0 t = 0
Exemplo: Considere a função de onda
Ψ ( x , t ) = A e − λ ∣ x ∣ e − i ω t \Psi\left(x,t\right)=Ae^{-\lambda\left|x\right|}e^{-i\omega t} Ψ ( x , t ) = A e − λ ∣ x ∣ e − iω t Onde A A A λ \lambda λ ω \omega ω
a) Normalização de Ψ \Psi Ψ
b) Determinar os valores esperados de x x x x 2 x^{2} x 2
c) Encontrar o desvio padrão de x x x ⟨ x ⟩ − σ \left\langle x\right\rangle -\sigma ⟨ x ⟩ − σ ⟨ x ⟩ + σ \left\langle x\right\rangle +\sigma ⟨ x ⟩ + σ
Sendo x x x
∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 1 ∫ − ∞ + ∞ Ψ Ψ ∗ d x = ∫ − ∞ + ∞ ( A e − λ ∣ x ∣ e − i ω t ) ( A e − λ ∣ x ∣ e i ω t ) d x = A 2 ∫ − ∞ + ∞ e − λ 2 ∣ x ∣ d x = A 2 ( ∫ − ∞ 0 e − λ 2 ( − x ) d x + ∫ 0 + ∞ e − λ 2 x d x ) = A 2 ( 1 2 λ ∫ − ∞ 0 e u d u + 1 − 2 λ ∫ 0 + ∞ e v d v ) = \begin{align}
\int^{+\infty}_{-\infty}\left|\Psi\left(x,t\right)\right|^{2}dx & =1\\
\int^{+\infty}_{-\infty}\Psi\Psi^{*}dx & =\\
\int^{+\infty}_{-\infty}\left(Ae^{-\lambda\left|x\right|}e^{-i\omega t}\right)\left(Ae^{-\lambda\left|x\right|}e^{i\omega t}\right)dx & =\\
A^{2}\int^{+\infty}_{-\infty}e^{-\lambda2\left|x\right|}dx & =\\
A^{2}\left(\int^{0}_{-\infty}e^{-\lambda2\left(-x\right)}dx+\int^{+\infty}_{0}e^{-\lambda2x}dx\right) & =\\
A^{2}\left(\frac{1}{2\lambda}\int^{0}_{-\infty}e^{u}du+\frac{1}{-2\lambda}\int^{+\infty}_{0}e^{v}dv\right) & =
\end{align} ∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x ∫ − ∞ + ∞ Ψ Ψ ∗ d x ∫ − ∞ + ∞ ( A e − λ ∣ x ∣ e − iω t ) ( A e − λ ∣ x ∣ e iω t ) d x A 2 ∫ − ∞ + ∞ e − λ 2 ∣ x ∣ d x A 2 ( ∫ − ∞ 0 e − λ 2 ( − x ) d x + ∫ 0 + ∞ e − λ 2 x d x ) A 2 ( 2 λ 1 ∫ − ∞ 0 e u d u + − 2 λ 1 ∫ 0 + ∞ e v d v ) = 1 = = = = = Chamo atenção para a 5ª linha onde dividimos a integral para respeitar o módulo, usando a propriedade de aditividade da integral[2] u = 2 λ x u=2\lambda x u = 2 λ x v = − 2 λ x v=-2\lambda x v = − 2 λ x
A 2 ( 1 2 λ e 2 λ x ∣ − ∞ 0 + 1 − 2 λ e − 2 λ x ∣ 0 + ∞ ) = 1 A 2 ( ( e 0 − 1 e ∞ ) 2 λ + ( 1 e ∞ − e 0 ) − 2 λ ) = A 2 ( 1 2 λ + 1 2 λ ) = A 2 λ = A = λ \begin{align}
A^{2}\left(\frac{1}{2\lambda}\left.e^{2\lambda x}\right|^{0}_{-\infty}+\frac{1}{-2\lambda}\left.e^{-2\lambda x}\right|^{+\infty}_{0}\right) & =1\\
A^{2}\left(\frac{\left(e^{0}-\frac{1}{e^{\infty}}\right)}{2\lambda}+\frac{\left(\frac{1}{e^{\infty}}-e^{0}\right)}{-2\lambda}\right) & =\\
A^{2}\left(\frac{1}{2\lambda}+\frac{1}{2\lambda}\right) & =\\
\frac{A^{2}}{\lambda} & =\\
A & =\sqrt{\lambda}
\end{align} A 2 ( 2 λ 1 e 2 λ x ∣ ∣ − ∞ 0 + − 2 λ 1 e − 2 λ x ∣ ∣ 0 + ∞ ) A 2 ( 2 λ ( e 0 − e ∞ 1 ) + − 2 λ ( e ∞ 1 − e 0 ) ) A 2 ( 2 λ 1 + 2 λ 1 ) λ A 2 A = 1 = = = = λ O valor esperado para x x x
∫ − ∞ + ∞ x ∣ Ψ ( x , t ) ∣ 2 d x = A 2 ( ∫ − ∞ 0 x e 2 λ x d x + ∫ 0 + ∞ x e − 2 λ x d x ) = λ ( ∫ − ∞ 0 1 2 λ u 2 λ e u d u + ∫ 0 + ∞ 1 2 λ u 2 λ e − u d u ) = 1 4 λ ( ∫ − ∞ 0 u e u d u + ∫ 0 + ∞ u e − u d u ) \begin{align}
\int^{+\infty}_{-\infty}x\left|\Psi\left(x,t\right)\right|^{2}dx & =A^{2}\left(\int^{0}_{-\infty}xe^{2\lambda x}dx+\int^{+\infty}_{0}xe^{-2\lambda x}dx\right)\\
& =\lambda\left(\int^{0}_{-\infty}\frac{1}{2\lambda}\frac{u}{2\lambda}e^{u}du+\int^{+\infty}_{0}\frac{1}{2\lambda}\frac{u}{2\lambda}e^{-u}du\right)\\
& =\frac{1}{4\lambda}\left(\int^{0}_{-\infty}ue^{u}du+\int^{+\infty}_{0}ue^{-u}du\right)
\end{align} ∫ − ∞ + ∞ x ∣ Ψ ( x , t ) ∣ 2 d x = A 2 ( ∫ − ∞ 0 x e 2 λ x d x + ∫ 0 + ∞ x e − 2 λ x d x ) = λ ( ∫ − ∞ 0 2 λ 1 2 λ u e u d u + ∫ 0 + ∞ 2 λ 1 2 λ u e − u d u ) = 4 λ 1 ( ∫ − ∞ 0 u e u d u + ∫ 0 + ∞ u e − u d u ) Onde usamos novamente que u = 2 λ x u=2\lambda x u = 2 λ x [3] f ( u ) = u f\left(u\right)=u f ( u ) = u g ( u ) = e u g\left(u\right)=e^{u} g ( u ) = e u
∫ a b f ( u ) g ′ ( u ) d u = f ( u ) g ( u ) ∣ a b − ∫ a b g ( u ) f ′ ( u ) d u ∫ − ∞ 0 u e u d u = u e u ∣ − ∞ 0 − ∫ − ∞ 9 e u d u = lim u → − ∞ ( u e u ) − ∫ − ∞ 9 e u d u = lim u → ∞ ( − u e u ) − ( e 0 − 1 e ∞ ) = lim u → ∞ ( − u e u ) − 1 \begin{align}
\int^{b}_{a}f\left(u\right)g'\left(u\right)du & =\left.f\left(u\right)g\left(u\right)\right|^{b}_{a}-\int^{b}_{a}g\left(u\right)f'\left(u\right)du\\
\int^{0}_{-\infty}ue^{u}du & =\left.ue^{u}\right|^{0}_{-\infty}-\int^{9}_{-\infty}e^{u}du\\
& =\lim_{u\rightarrow-\infty}\left(ue^{u}\right)-\int^{9}_{-\infty}e^{u}du\\
& =\lim_{u\rightarrow\infty}\left(-\frac{u}{e^{u}}\right)-\left(e^{0}-\frac{1}{e^{\infty}}\right)\\
& =\lim_{u\rightarrow\infty}\left(-\frac{u}{e^{u}}\right)-1
\end{align} ∫ a b f ( u ) g ′ ( u ) d u ∫ − ∞ 0 u e u d u = f ( u ) g ( u ) ∣ a b − ∫ a b g ( u ) f ′ ( u ) d u = u e u ∣ − ∞ 0 − ∫ − ∞ 9 e u d u = u → − ∞ lim ( u e u ) − ∫ − ∞ 9 e u d u = u → ∞ lim ( − e u u ) − ( e 0 − e ∞ 1 ) = u → ∞ lim ( − e u u ) − 1 Como temos uma indeterminação do tipo − ∞ / ∞ -\infty/\infty − ∞/∞ [4]
∫ − ∞ 0 u e u d u = lim u → ∞ ( − d u / d t d e u / d t ) − 1 = lim u → ∞ ( − d u / d t d e u / d t ) − 1 = lim u → ∞ ( − 1 e u ) − 1 = − 1 \begin{align}
\int^{0}_{-\infty}ue^{u}du & =\lim_{u\rightarrow\infty}\left(-\frac{du/dt}{de^{u}/dt}\right)-1\\
& =\lim_{u\rightarrow\infty}\left(-\frac{du/dt}{de^{u}/dt}\right)-1\\
& =\lim_{u\rightarrow\infty}\left(-\frac{1}{e^{u}}\right)-1\\
& =-1
\end{align} ∫ − ∞ 0 u e u d u = u → ∞ lim ( − d e u / d t d u / d t ) − 1 = u → ∞ lim ( − d e u / d t d u / d t ) − 1 = u → ∞ lim ( − e u 1 ) − 1 = − 1 E equivalente:
∫ 0 + ∞ u e − u d u = − u e − u ∣ 0 + ∞ + ∫ 0 + ∞ e − u d u = − lim u → ∞ ( u e − u ) − ( 1 e ∞ − 1 ) = − lim u → ∞ ( u e u ) + 1 = 1 \begin{align}
\int^{+\infty}_{0}ue^{-u}du & =\left.-ue^{-u}\right|^{+\infty}_{0}+\int^{+\infty}_{0}e^{-u}du\\
& =-\lim_{u\rightarrow\infty}\left(ue^{-u}\right)-\left(\frac{1}{e^{\infty}}-1\right)\\
& =-\lim_{u\rightarrow\infty}\left(\frac{u}{e^{u}}\right)+1\\
& =1
\end{align} ∫ 0 + ∞ u e − u d u = − u e − u ∣ ∣ 0 + ∞ + ∫ 0 + ∞ e − u d u = − u → ∞ lim ( u e − u ) − ( e ∞ 1 − 1 ) = − u → ∞ lim ( e u u ) + 1 = 1 Este é o mesmo limite que calculamos anteriormente aplicando a L’hopital e temos lim u → ∞ ( u / e u ) = 0 \lim_{u\rightarrow\infty}\left(u/e^{u}\right)=0 lim u → ∞ ( u / e u ) = 0 18
∫ − ∞ + ∞ x ∣ Ψ ( x , t ) ∣ 2 d x = 1 4 λ ( ∫ − ∞ 0 u e u d u + ∫ 0 + ∞ u e − u d u ) = 1 4 λ ( − 1 + 1 ) = 0 \begin{align}
\int^{+\infty}_{-\infty}x\left|\Psi\left(x,t\right)\right|^{2}dx & =\frac{1}{4\lambda}\left(\int^{0}_{-\infty}ue^{u}du+\int^{+\infty}_{0}ue^{-u}du\right)\\
& =\frac{1}{4\lambda}\left(-1+1\right)\\
& =0
\end{align} ∫ − ∞ + ∞ x ∣ Ψ ( x , t ) ∣ 2 d x = 4 λ 1 ( ∫ − ∞ 0 u e u d u + ∫ 0 + ∞ u e − u d u ) = 4 λ 1 ( − 1 + 1 ) = 0 Temos algo análogo para o valor de esperado de x 2 x^{2} x 2 u = 2 λ x u=2\lambda x u = 2 λ x
∫ − ∞ + ∞ x 2 ∣ Ψ ( x , t ) ∣ 2 d x = A 2 ( ∫ − ∞ 0 x 2 e 2 λ x d x + ∫ 0 + ∞ x 2 e − 2 λ x d x ) = λ ( ∫ − ∞ 0 1 2 λ ( u 2 λ ) 2 e u d u + ∫ 0 + ∞ 1 2 λ ( u 2 λ ) 2 e − u d u ) = 1 8 λ 2 ( ∫ − ∞ 0 u 2 e u d u + ∫ 0 + ∞ u 2 e − u d u ) \begin{align}
\int^{+\infty}_{-\infty}x^{2}\left|\Psi\left(x,t\right)\right|^{2}dx & =A^{2}\left(\int^{0}_{-\infty}x^{2}e^{2\lambda x}dx+\int^{+\infty}_{0}x^{2}e^{-2\lambda x}dx\right)\\
& =\lambda\left(\int^{0}_{-\infty}\frac{1}{2\lambda}\left(\frac{u}{2\lambda}\right)^{2}e^{u}du+\int^{+\infty}_{0}\frac{1}{2\lambda}\left(\frac{u}{2\lambda}\right)^{2}e^{-u}du\right)\\
& =\frac{1}{8\lambda^{2}}\left(\int^{0}_{-\infty}u^{2}e^{u}du+\int^{+\infty}_{0}u^{2}e^{-u}du\right)
\end{align} ∫ − ∞ + ∞ x 2 ∣ Ψ ( x , t ) ∣ 2 d x = A 2 ( ∫ − ∞ 0 x 2 e 2 λ x d x + ∫ 0 + ∞ x 2 e − 2 λ x d x ) = λ ( ∫ − ∞ 0 2 λ 1 ( 2 λ u ) 2 e u d u + ∫ 0 + ∞ 2 λ 1 ( 2 λ u ) 2 e − u d u ) = 8 λ 2 1 ( ∫ − ∞ 0 u 2 e u d u + ∫ 0 + ∞ u 2 e − u d u ) Aplicando novamente a integração por partes mas sendo agora f ( u ) = u 2 f\left(u\right)=u^{2} f ( u ) = u 2 g ( u ) = e u g\left(u\right)=e^{u} g ( u ) = e u
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ − ∞ 0 u 2 e u d u = u 2 e u ∣ − ∞ 0 − 2 ∫ − ∞ 0 u e u d u = lim u → ∞ u 2 e u − 2 ∫ − ∞ 0 u e u d u e u \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int^{0}_{-\infty}u^{2}e^{u}du & =\left.u^{2}e^{u}\right|^{0}_{-\infty}-2\int^{0}_{-\infty}ue^{u}du\\
& =\lim_{u\rightarrow\infty}\frac{u^{2}}{e^{u}}-2\int^{0}_{-\infty}ue^{u}due^{u}
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ − ∞ 0 u 2 e u d u = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = u 2 e u ∣ ∣ − ∞ 0 − 2 ∫ − ∞ 0 u e u d u = u → ∞ lim e u u 2 − 2 ∫ − ∞ 0 u e u d u e u O resultado desta integral já foi encontrado na equação 20
∫ − ∞ 0 u 2 e u d u = 2 \int^{0}_{-\infty}u^{2}e^{u}du=2 ∫ − ∞ 0 u 2 e u d u = 2 Para a segunda integral temos:
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ 0 + ∞ u 2 e − u d u = − u 2 e − u ∣ 0 ∞ + 2 ∫ 0 + ∞ u e − u d u = − lim u → ∞ u 2 e u + 2 ∫ − ∞ 0 u e − u d u \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int^{+\infty}_{0}u^{2}e^{-u}du & =-\left.u^{2}e^{-u}\right|^{\infty}_{0}+2\int^{+\infty}_{0}ue^{-u}du\\
& =-\lim_{u\rightarrow\infty}\frac{u^{2}}{e^{u}}+2\int^{0}_{-\infty}ue^{-u}du
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ 0 + ∞ u 2 e − u d u = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = − u 2 e − u ∣ ∣ 0 ∞ + 2 ∫ 0 + ∞ u e − u d u = − u → ∞ lim e u u 2 + 2 ∫ − ∞ 0 u e − u d u De forma análoga podemos aplicar duas vezes l’hopital para resolver o limite e o resultado dessa integral foi obtido em 20
∫ 0 + ∞ u 2 e − u d u = 2 \int^{+\infty}_{0}u^{2}e^{-u}du=2 ∫ 0 + ∞ u 2 e − u d u = 2 Agora combinando os resultados 25 27 23
∫ − ∞ + ∞ x 2 ∣ Ψ ( x , t ) ∣ 2 d x = 1 8 λ 2 ( ∫ − ∞ 0 u 2 e u d u + ∫ 0 + ∞ u 2 e − u d u ) = 1 8 λ 2 ( 2 + 2 ) = 1 2 λ 2 \begin{align}
\int^{+\infty}_{-\infty}x^{2}\left|\Psi\left(x,t\right)\right|^{2}dx & =\frac{1}{8\lambda^{2}}\left(\int^{0}_{-\infty}u^{2}e^{u}du+\int^{+\infty}_{0}u^{2}e^{-u}du\right)\\
& =\frac{1}{8\lambda^{2}}\left(2+2\right)\\
& =\frac{1}{2\lambda^{2}}
\end{align} ∫ − ∞ + ∞ x 2 ∣ Ψ ( x , t ) ∣ 2 d x = 8 λ 2 1 ( ∫ − ∞ 0 u 2 e u d u + ∫ 0 + ∞ u 2 e − u d u ) = 8 λ 2 1 ( 2 + 2 ) = 2 λ 2 1 O desvio padrão de x x x
σ = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 \begin{align}
\sigma & =\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}
\end{align} σ = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 Como ⟨ x ⟩ = ∫ − ∞ + ∞ x ∣ Ψ ( x , t ) ∣ 2 = 0 \left\langle x\right\rangle =\int^{+\infty}_{-\infty}x\left|\Psi\left(x,t\right)\right|^{2}=0 ⟨ x ⟩ = ∫ − ∞ + ∞ x ∣ Ψ ( x , t ) ∣ 2 = 0 ⟨ x 2 ⟩ = ∫ − ∞ + ∞ x 2 ∣ Ψ ( x , t ) ∣ 2 = 1 / 2 λ 2 \left\langle x^{2}\right\rangle =\int^{+\infty}_{-\infty}x^{2}\left|\Psi\left(x,t\right)\right|^{2}=1/2\lambda^{2} ⟨ x 2 ⟩ = ∫ − ∞ + ∞ x 2 ∣ Ψ ( x , t ) ∣ 2 = 1/2 λ 2
σ = 1 2 λ 2 − 0 = 1 2 λ \sigma=\sqrt{\frac{1}{2\lambda^{2}}-0}=\frac{1}{\sqrt{2}\lambda} σ = 2 λ 2 1 − 0 = 2 λ 1 E por fim a probabilidade de não estar no intervalo entre − σ -\sigma − σ + σ +\sigma + σ
∫ − ∞ − − σ ∣ Ψ ∣ 2 d x + ∫ + σ + ∞ ∣ Ψ ∣ 2 d x = − ∫ − σ − − ∞ ∣ Ψ ( x , t ) ∣ 2 d x + ∫ + σ + ∞ Ψ ( x , t ) 2 d x = ∫ + σ + ∞ Ψ ( − u , t ) 2 d u + ∫ + σ + ∞ Ψ ( x , t ) 2 d x \begin{align}
\int^{--\sigma}_{-\infty}\left|\Psi\right|^{2}dx+\int^{+\infty}_{+\sigma}\left|\Psi\right|^{2}dx & =-\int^{--\infty}_{-\sigma}\left|\Psi\left(x,t\right)\right|^{2}dx+\int^{+\infty}_{+\sigma}\Psi\left(x,t\right)^{2}dx\\
& =\int^{+\infty}_{+\sigma}\Psi\left(-u,t\right)^{2}du+\int^{+\infty}_{+\sigma}\Psi\left(x,t\right)^{2}dx
\end{align} ∫ − ∞ −− σ ∣ Ψ ∣ 2 d x + ∫ + σ + ∞ ∣ Ψ ∣ 2 d x = − ∫ − σ −− ∞ ∣ Ψ ( x , t ) ∣ 2 d x + ∫ + σ + ∞ Ψ ( x , t ) 2 d x = ∫ + σ + ∞ Ψ ( − u , t ) 2 d u + ∫ + σ + ∞ Ψ ( x , t ) 2 d x Onde usamos a propriedade sobre inverter os limites da integral[5] u = − x , u=-x, u = − x , d u / d x = − 1 du/dx=-1 d u / d x = − 1 d u = − d x du=-dx d u = − d x
lim x → − v u = lim x → − v − x = v \lim_{x\rightarrow-v}u=\lim_{x\rightarrow-v}-x=v x → − v lim u = x → − v lim − x = v No limite inferior v = σ v=\sigma v = σ v = − ∞ v=-\infty v = − ∞ Ψ ( x , t ) \Psi\left(x,t\right) Ψ ( x , t ) Ψ ( − x , t ) = Ψ ( x , t ) \Psi\left(-x,t\right)=\Psi\left(x,t\right) Ψ ( − x , t ) = Ψ ( x , t ) x x x ∣ x ∣ \left|x\right| ∣ x ∣
∫ − ∞ − − σ ∣ Ψ ∣ 2 d x + ∫ + σ + ∞ ∣ Ψ ∣ 2 d x = ∫ + σ + ∞ Ψ ( u , t ) 2 d u + ∫ + σ + ∞ Ψ ( x , t ) 2 d x = 2 ∫ + σ + ∞ Ψ ( x , t ) 2 d x \begin{align}
\int^{--\sigma}_{-\infty}\left|\Psi\right|^{2}dx+\int^{+\infty}_{+\sigma}\left|\Psi\right|^{2}dx & =\int^{+\infty}_{+\sigma}\Psi\left(u,t\right)^{2}du+\int^{+\infty}_{+\sigma}\Psi\left(x,t\right)^{2}dx\\
& =2\int^{+\infty}_{+\sigma}\Psi\left(x,t\right)^{2}dx
\end{align} ∫ − ∞ −− σ ∣ Ψ ∣ 2 d x + ∫ + σ + ∞ ∣ Ψ ∣ 2 d x = ∫ + σ + ∞ Ψ ( u , t ) 2 d u + ∫ + σ + ∞ Ψ ( x , t ) 2 d x = 2 ∫ + σ + ∞ Ψ ( x , t ) 2 d x Onde temos o segundo resultado pois x x x v v v Ψ \Psi Ψ 0 < σ < ∞ 0<\sigma<\infty 0 < σ < ∞ ∣ x ∣ = x \left|x\right|=x ∣ x ∣ = x
∫ − ∞ − − σ ∣ Ψ ∣ 2 d x + ∫ + σ + ∞ ∣ Ψ ∣ 2 d x = 2 ∫ + σ + ∞ Ψ ( x , t ) 2 d x = 2 A 2 ∫ + σ + ∞ e − λ 2 ∣ x ∣ d x = 2 λ ∫ + σ + ∞ e − λ 2 x d x = 2 λ − 2 λ [ 1 e λ 2 x ] + σ + ∞ = − [ 1 e ∞ − 1 e λ 2 σ ] + σ + ∞ = exp [ − λ 2 σ ] = exp [ − 2 λ 2 λ ] \begin{align}
\int^{--\sigma}_{-\infty}\left|\Psi\right|^{2}dx+\int^{+\infty}_{+\sigma}\left|\Psi\right|^{2}dx & =2\int^{+\infty}_{+\sigma}\Psi\left(x,t\right)^{2}dx\\
& =2A^{2}\int^{+\infty}_{+\sigma}e^{-\lambda2\left|x\right|}dx\\
& =2\lambda\int^{+\infty}_{+\sigma}e^{-\lambda2x}dx\\
& =\frac{2\lambda}{-2\lambda}\left[\frac{1}{e^{\lambda2x}}\right]^{+\infty}_{+\sigma}\\
& =-\left[\frac{1}{e^{\infty}}-\frac{1}{e^{\lambda2\sigma}}\right]^{+\infty}_{+\sigma}\\
& =\exp\left[-\lambda2\sigma\right]\\
& =\exp\left[-\frac{2\lambda}{\sqrt{2}\lambda}\right]
\end{align} ∫ − ∞ −− σ ∣ Ψ ∣ 2 d x + ∫ + σ + ∞ ∣ Ψ ∣ 2 d x = 2 ∫ + σ + ∞ Ψ ( x , t ) 2 d x = 2 A 2 ∫ + σ + ∞ e − λ 2 ∣ x ∣ d x = 2 λ ∫ + σ + ∞ e − λ 2 x d x = − 2 λ 2 λ [ e λ 2 x 1 ] + σ + ∞ = − [ e ∞ 1 − e λ 2 σ 1 ] + σ + ∞ = exp [ − λ 2 σ ] = exp [ − 2 λ 2 λ ] Onde usamos o fato de que calculamos já A A A σ . \sigma. σ . − 2 ^{-\sqrt{2}} − 2 24 , 31 % 24,31\% 24 , 31%
Momentum ¶ O valor esperado ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩ ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩ x ¨ \ddot{x} x ¨
d ⟨ x ⟩ d t = ∫ x ∂ ∂ t ∣ Ψ ( x , t ) ∣ 2 \begin{align}
\frac{d\left\langle x\right\rangle }{dt} & =\int x\frac{\partial}{\partial t}\left|\Psi\left(x,t\right)\right|^{2}
\end{align} d t d ⟨ x ⟩ = ∫ x ∂ t ∂ ∣ Ψ ( x , t ) ∣ 2 Onde estamos derivando apenas em relação ao que depende explicitamente do tempo, e aqui há algo que talvez nos pareça estranho. Diferentemente da mecânica clássica, em que temos x ( t ) x(t) x ( t ) x x x x x x x x x Ψ ( x , 0 ) \Psi(x,0) Ψ ( x , 0 ) Ψ ( x , 1 ) \Psi(x,1) Ψ ( x , 1 ) x x x Ψ ( x , 0 ) \Psi(x,0) Ψ ( x , 0 ) x = 1 x=1 x = 1 x = 1 x=1 x = 1 x ( t ) x(t) x ( t ) x ( 0 ) = 0 x(0)=0 x ( 0 ) = 0 x ( 1 ) = 1 x(1)=1 x ( 1 ) = 1
O que ocorre é que Ψ ( 0 , 0 ) \Psi(0,0) Ψ ( 0 , 0 ) Ψ ( 1 , 0 ) \Psi(1,0) Ψ ( 1 , 0 ) Ψ ( 0 , 1 ) \Psi(0,1) Ψ ( 0 , 1 ) Ψ ( 1 , 1 ) \Psi(1,1) Ψ ( 1 , 1 ) Ψ ( x , t ) \Psi(x,t) Ψ ( x , t ) x = 0 x=0 x = 0 x = 1 x=1 x = 1 t = 0 t=0 t = 0 t = 1 t=1 t = 1 8
∂ ∂ t ∣ Ψ ∣ 2 = i ℏ 2 m ( Ψ ∗ ∂ 2 Ψ ∂ x 2 − ∂ 2 Ψ ∗ ∂ x 2 Ψ ) \frac{\partial}{\partial t}\left|\Psi\right|^{2}=\frac{i\hbar}{2m}\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi\right) ∂ t ∂ ∣ Ψ ∣ 2 = 2 m i ℏ ( Ψ ∗ ∂ x 2 ∂ 2 Ψ − ∂ x 2 ∂ 2 Ψ ∗ Ψ ) Temos:
d ⟨ x ⟩ d t = ∫ x ∂ ∂ t ∣ Ψ ( x , t ) ∣ 2 = i ℏ 2 m ∫ x ( Ψ ∗ ∂ 2 Ψ ∂ x 2 − ∂ 2 Ψ ∗ ∂ x 2 Ψ ) d x \begin{align}
\frac{d\left\langle x\right\rangle }{dt} & =\int x\frac{\partial}{\partial t}\left|\Psi\left(x,t\right)\right|^{2}\\
& =\frac{i\hbar}{2m}\int x\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi\right)dx
\end{align} d t d ⟨ x ⟩ = ∫ x ∂ t ∂ ∣ Ψ ( x , t ) ∣ 2 = 2 m i ℏ ∫ x ( Ψ ∗ ∂ x 2 ∂ 2 Ψ − ∂ x 2 ∂ 2 Ψ ∗ Ψ ) d x Onde usando a regra do produto:
d d x ( f g ) = f d g d x + d f d x g \frac{d}{dx}\left(fg\right)=f\frac{dg}{dx}+\frac{df}{dx}g d x d ( f g ) = f d x d g + d x df g Então:
∂ ∂ x ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) = ∂ ∂ x ( Ψ ∗ ∂ Ψ ∂ x ) − ∂ ∂ x ( ∂ Ψ ∗ ∂ x Ψ ) = [ ∂ Ψ ∗ ∂ x ∂ Ψ ∂ x + Ψ ∗ ∂ 2 Ψ ∂ x 2 ] − [ ∂ 2 Ψ ∗ ∂ x 2 Ψ + ∂ Ψ ∗ ∂ x ∂ Ψ ∂ x ] ∂ Ψ ∗ ∂ x ∂ Ψ ∂ x + Ψ ∗ ∂ 2 Ψ ∂ x 2 − ∂ 2 Ψ ∗ ∂ x 2 Ψ \begin{align}
\frac{\partial}{\partial x}\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right) & =\frac{\partial}{\partial x}\left(\Psi^{*}\frac{\partial\Psi}{\partial x}\right)-\frac{\partial}{\partial x}\left(\frac{\partial\Psi^{*}}{\partial x}\Psi\right)\\
& =\left[\frac{\partial\Psi^{*}}{\partial x}\frac{\partial\Psi}{\partial x}+\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}\right]-\left[\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi+\frac{\partial\Psi^{*}}{\partial x}\frac{\partial\Psi}{\partial x}\right]\\
& \frac{\partial\Psi^{*}}{\partial x}\frac{\partial\Psi}{\partial x}+\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\Psi
\end{align} ∂ x ∂ ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ ) = ∂ x ∂ ( Ψ ∗ ∂ x ∂ Ψ ) − ∂ x ∂ ( ∂ x ∂ Ψ ∗ Ψ ) = [ ∂ x ∂ Ψ ∗ ∂ x ∂ Ψ + Ψ ∗ ∂ x 2 ∂ 2 Ψ ] − [ ∂ x 2 ∂ 2 Ψ ∗ Ψ + ∂ x ∂ Ψ ∗ ∂ x ∂ Ψ ] ∂ x ∂ Ψ ∗ ∂ x ∂ Ψ + Ψ ∗ ∂ x 2 ∂ 2 Ψ − ∂ x 2 ∂ 2 Ψ ∗ Ψ Então podemos reescrever a integral como:
d ⟨ x ⟩ d t = i ℏ 2 m ∫ x [ ∂ ∂ x ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) ] d x \frac{d\left\langle x\right\rangle }{dt}=\frac{i\hbar}{2m}\int x\left[\frac{\partial}{\partial x}\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right)\right]dx d t d ⟨ x ⟩ = 2 m i ℏ ∫ x [ ∂ x ∂ ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ ) ] d x Por conveniência não estamos escrevendo os limites da integral ∫ − ∞ + ∞ → ∫ \int^{+\infty}_{-\infty}\rightarrow\int ∫ − ∞ + ∞ → ∫ Ψ ( x , t ) = Ψ \Psi\left(x,t\right)=\Psi Ψ ( x , t ) = Ψ f ( x ) = x f\left(x\right)=x f ( x ) = x g ′ ( x ) = ∂ ∂ x ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) g'\left(x\right)=\frac{\partial}{\partial x}\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right) g ′ ( x ) = ∂ x ∂ ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ )
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ x ( Ψ ∗ ∂ 2 Ψ ∂ x 2 ) = [ x ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) ] − ∞ + ∞ − ∫ ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) d x \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int x\left(\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}\right) & =\left[x\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right)\right]^{+\infty}_{-\infty}-\int\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right)dx
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ x ( Ψ ∗ ∂ x 2 ∂ 2 Ψ ) = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = [ x ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ ) ] − ∞ + ∞ − ∫ ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ ) d x Uma demonstração mais robusta matematicamente pode ser necessária futuramente, mas exigimos que Ψ ( x ) \Psi(x) Ψ ( x ) x → ∞ x\to\infty x → ∞ x x x 1 / ∣ x ∣ → 0 1/|x|\to0 1/∣ x ∣ → 0 x → ∞ x\to\infty x → ∞ 1 / ∣ x ∣ 1/|x| 1/∣ x ∣ Ψ \Psi Ψ
Podemos dizer que essas exigências são para funções de onda normalizadas no sentido mais tradicional. É possível a existência de sistemas quânticos cuja função de onda não seja normalizável no sentido que estamos discutindo aqui, e que podem ser tratados por outras abordagens. Contudo, essas são exceções e exigem um conhecimento mais avançado. Neste momento, devemos nos ater aos casos que obedecem a essas condições, por serem tanto mais comuns quanto mais simples de trabalhar.
Mas uma vez que o termo entre parênteses vai então mais rápido a 0 quando x → ∞ x\rightarrow\infty x → ∞
d ⟨ x ⟩ d t = − i ℏ 2 m ∫ ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) d x \frac{d\left\langle x\right\rangle }{dt}=-\frac{i\hbar}{2m}\int\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right)dx d t d ⟨ x ⟩ = − 2 m i ℏ ∫ ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ ) d x Ainda vale comentar que Griffiths define que os estados fisicamente possíveis correspondem a soluções quadrado-integráveis da equação de Schrödinger [6] ∣ f ( x ) ∣ 2 |f(x)|^{2} ∣ f ( x ) ∣ 2 ∣ x ∣ − p |x|^{-p} ∣ x ∣ − p
∫ − ∞ + ∞ d x ∣ x ∣ p = ∫ − ∞ + ∞ d x ∣ x ∣ p + = ∫ − ∞ 0 d x ( − x ) p + ∫ 0 + ∞ d x x p = − ∫ + ∞ 0 d u u p + ∫ 0 + ∞ d x x p = ∫ 0 + ∞ d u u p + ∫ 0 + ∞ d x x p = 2 ∫ 0 + ∞ d x x p = − ∫ 0 + ∞ x − p d x = x − p + 1 p − 1 ∣ 0 + ∞ = ( p − 1 ) lim x → ∞ 1 x p − 1 − 0 = ( p − 1 ) lim x → ∞ 1 x p − 1 \begin{align}
\int^{+\infty}_{-\infty}\frac{dx}{\left|x\right|^{p}} & =\int^{+\infty}_{-\infty}\frac{dx}{\left|x\right|^{p}}+=\int^{0}_{-\infty}\frac{dx}{\left(-x\right)^{p}}+\int^{+\infty}_{0}\frac{dx}{x^{p}}\\
& =-\int^{0}_{+\infty}\frac{du}{u^{p}}+\int^{+\infty}_{0}\frac{dx}{x^{p}}\\
& =\int^{+\infty}_{0}\frac{du}{u^{p}}+\int^{+\infty}_{0}\frac{dx}{x^{p}}\\
& =2\int^{+\infty}_{0}\frac{dx}{x^{p}}\\
& =-\int^{+\infty}_{0}x^{-p}dx\\
& =\left.\frac{x^{-p+1}}{p-1}\right|^{+\infty}_{0}\\
& =\left(p-1\right)\lim_{x\rightarrow\infty}\frac{1}{x^{p-1}}-0\\
& =\left(p-1\right)\lim_{x\rightarrow\infty}\frac{1}{x^{p-1}}
\end{align} ∫ − ∞ + ∞ ∣ x ∣ p d x = ∫ − ∞ + ∞ ∣ x ∣ p d x + = ∫ − ∞ 0 ( − x ) p d x + ∫ 0 + ∞ x p d x = − ∫ + ∞ 0 u p d u + ∫ 0 + ∞ x p d x = ∫ 0 + ∞ u p d u + ∫ 0 + ∞ x p d x = 2 ∫ 0 + ∞ x p d x = − ∫ 0 + ∞ x − p d x = p − 1 x − p + 1 ∣ ∣ 0 + ∞ = ( p − 1 ) x → ∞ lim x p − 1 1 − 0 = ( p − 1 ) x → ∞ lim x p − 1 1 Para que a integral não diverja, precisamos quep − 1 ≥ 0 p-1\geq0 p − 1 ≥ 0 p ≥ 1 , p\geq1, p ≥ 1 , p p p p = 1 p=1 p = 1 ∣ f ( x ) ∣ 2 ∼ 1 ∣ x ∣ p |f(x)|^{2}\sim\frac{1}{|x|^{p}} ∣ f ( x ) ∣ 2 ∼ ∣ x ∣ p 1 ∣ f ( x ) ∣ ∼ 1 ∣ x ∣ p |f(x)|\sim\frac{1}{\sqrt{\left|x\right|^{p}}} ∣ f ( x ) ∣ ∼ ∣ x ∣ p 1 [7]
∣ f ( x ) ∣ 2 = ∣ 1 x p ∣ 2 = 1 ∣ x p ∣ \left|f\left(x\right)\right|^{2}=\left|\frac{1}{\sqrt{x^{p}}}\right|^{2}=\frac{1}{\left|x^{p}\right|} ∣ f ( x ) ∣ 2 = ∣ ∣ x p 1 ∣ ∣ 2 = ∣ x p ∣ 1 Como p p p f ( x ) f(x) f ( x ) f ( x ) ∼ 1 ∣ x ∣ . f(x)\sim\frac{1}{\sqrt{|x|}}. f ( x ) ∼ ∣ x ∣ 1 . p p p f ( x ) f(x) f ( x ) x → ∞ x\to\infty x → ∞ Ψ ( x , t ) \Psi(x,t) Ψ ( x , t ) 1 ∣ x ∣ \frac{1}{\sqrt{|x|}} ∣ x ∣ 1 ∣ x ∣ → ∞ |x|\to\infty ∣ x ∣ → ∞
Mas este caso, apesar de ser uma aproximação útil e cobrir a maioria dos casos fisicamente válidos, como foi discutido, não cobre necessariamente todos. É possível que uma função de onda fisicamente válida possua outra aproximação assintótica ou mesmo não seja quadrado-integrável. Neste último caso, é necessário usar alguma generalização da condição de normalização. Para a função de onda ser válida fisicamente, é necessário principalmente que ela seja uma solução da equação de Schrödinger e que possua uma interpretação probabilística (que pode ser feita através de uma generalização da normalização), além, evidentemente, de obedecer às condições impostas pelo problema físico.
Por enquanto, trabalharemos com os casos em que essas condições a respeito da normalização da função de onda são válidas, isto é, que a função seja quadrado-integrável e, consequentemente, apresente uma normalização do tipo∫ − ∞ + ∞ ∣ Ψ ( x , t ) ∣ 2 d x = 1 , \int^{+\infty}_{-\infty}|\Psi(x,t)|^{2}\,dx=1, ∫ − ∞ + ∞ ∣Ψ ( x , t ) ∣ 2 d x = 1 , 1 ∣ x ∣ p , \frac{1}{|x|^{p}}, ∣ x ∣ p 1 , p ≥ 1 2 p\geq\frac{1}{2} p ≥ 2 1 [8]
Faremos isso da mesma forma que é assumido por Griffiths, pois assim podemos trabalhar com os casos mais gerais e deixar as exceções para um estudo mais especializado no futuro. Retomando, então, a discussão do limite fazendo uso dessas condições, ou seja, considerando que o comportamento assintótico da função de onda no infinito é similar a 1 x \frac{1}{\sqrt{x}} x 1
Ψ ∗ ∂ Ψ ∂ x ∼ 1 x ( − 1 2 x 3 2 ) = − 1 2 x 2 \Psi^{*}\frac{\partial\Psi}{\partial x}\sim\frac{1}{\sqrt{x}}\left(-\frac{1}{2x^{\frac{3}{2}}}\right)=-\frac{1}{2x^{2}} Ψ ∗ ∂ x ∂ Ψ ∼ x 1 ( − 2 x 2 3 1 ) = − 2 x 2 1 E consequentemente
lim x → ∞ x ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) ∼ lim x → ∞ ( − x 2 x 2 ) = lim x → ∞ ( − 1 2 x ) = 0 \lim_{x\rightarrow\infty}x\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right)\sim\lim_{x\rightarrow\infty}\left(-\frac{x}{2x^{2}}\right)=\lim_{x\rightarrow\infty}\left(-\frac{1}{2x}\right)=0 x → ∞ lim x ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ ) ∼ x → ∞ lim ( − 2 x 2 x ) = x → ∞ lim ( − 2 x 1 ) = 0 Então ficamos com:
d ⟨ x ⟩ d t = − i ℏ 2 m ∫ ( Ψ ∗ ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x Ψ ) d x \frac{d\left\langle x\right\rangle }{dt}=-\frac{i\hbar}{2m}\int\left(\Psi^{*}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\Psi\right)dx d t d ⟨ x ⟩ = − 2 m i ℏ ∫ ( Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ Ψ ) d x Usando novamente a integral por substituição, para a segunda integral fazendo f = Ψ f=\Psi f = Ψ g = Ψ ∗ g=\Psi^{*} g = Ψ ∗
∫ a b f d g d x d x = [ f g ] a b − ∫ a b d f d x g d x ∫ − ∞ ∞ ∂ Ψ ∗ ∂ x Ψ d x = [ Ψ Ψ ∗ ] − ∞ + ∞ − ∫ − ∞ ∞ Ψ ∗ ∂ Ψ ∂ x = − ∫ − ∞ ∞ Ψ ∗ ∂ Ψ ∂ x \begin{align}
\int^{b}_{a}f\frac{dg}{dx}dx & =\left[fg\right]^{b}_{a}-\int^{b}_{a}\frac{df}{dx}gdx\\
\int^{\infty}_{-\infty}\frac{\partial\Psi^{*}}{\partial x}\Psi dx & =\left[\Psi\Psi^{*}\right]^{+\infty}_{-\infty}-\int^{\infty}_{-\infty}\Psi^{*}\frac{\partial\Psi}{\partial x}\\
& =-\int^{\infty}_{-\infty}\Psi^{*}\frac{\partial\Psi}{\partial x}
\end{align} ∫ a b f d x d g d x ∫ − ∞ ∞ ∂ x ∂ Ψ ∗ Ψ d x = [ f g ] a b − ∫ a b d x df g d x = [ Ψ Ψ ∗ ] − ∞ + ∞ − ∫ − ∞ ∞ Ψ ∗ ∂ x ∂ Ψ = − ∫ − ∞ ∞ Ψ ∗ ∂ x ∂ Ψ Onde usamos o fato de que a função de onda vai a 0 no infinito . Temos então:
d ⟨ x ⟩ d t = − i ℏ 2 m ∫ ( Ψ ∗ ∂ Ψ ∂ x + Ψ ∗ ∂ Ψ ∂ x ) d x = − i ℏ m ∫ Ψ ∗ ∂ Ψ ∂ x d x \frac{d\left\langle x\right\rangle }{dt}=-\frac{i\hbar}{2m}\int\left(\Psi^{*}\frac{\partial\Psi}{\partial x}+\Psi^{*}\frac{\partial\Psi}{\partial x}\right)dx=-\frac{i\hbar}{m}\int\Psi^{*}\frac{\partial\Psi}{\partial x}dx d t d ⟨ x ⟩ = − 2 m i ℏ ∫ ( Ψ ∗ ∂ x ∂ Ψ + Ψ ∗ ∂ x ∂ Ψ ) d x = − m i ℏ ∫ Ψ ∗ ∂ x ∂ Ψ d x Vamos postular[9]
⟨ v ⟩ = d ⟨ x ⟩ d t \left\langle v\right\rangle =\frac{d\left\langle x\right\rangle }{dt} ⟨ v ⟩ = d t d ⟨ x ⟩ Temos então o momento:
⟨ p ⟩ = m d ⟨ x ⟩ d t = − i ℏ ∫ Ψ ∗ ∂ Ψ ∂ x d x \left\langle p\right\rangle =m\frac{d\left\langle x\right\rangle }{dt}=-i\hbar\int\Psi^{*}\frac{\partial\Psi}{\partial x}dx ⟨ p ⟩ = m d t d ⟨ x ⟩ = − i ℏ ∫ Ψ ∗ ∂ x ∂ Ψ d x Podemos reescrever as expressões como:
⟨ x ⟩ = ∫ Ψ ∗ [ x ] Ψ d x ⟨ p ⟩ = ∫ Ψ ∗ [ − i ℏ ( ∂ ∂ x ) ] Ψ d x \begin{align}
\left\langle x\right\rangle & =\int\Psi^{*}\left[x\right]\Psi dx\\
\left\langle p\right\rangle & =\int\Psi^{*}\left[-i\hbar\left(\frac{\partial}{\partial x}\right)\right]\Psi dx
\end{align} ⟨ x ⟩ ⟨ p ⟩ = ∫ Ψ ∗ [ x ] Ψ d x = ∫ Ψ ∗ [ − i ℏ ( ∂ x ∂ ) ] Ψ d x Dizemos que o operador [10] x x x − i ℏ ∂ ∂ x -i\hbar\frac{\partial}{\partial x} − i ℏ ∂ x ∂ Ψ ∗ \Psi^{*} Ψ ∗ Ψ \Psi Ψ
T = 1 2 m v 2 = p 2 m T=\frac{1}{2}mv^{2}=\frac{p^{2}}{m} T = 2 1 m v 2 = m p 2 O momento angular é dado por:
L = r × m v = r × p \boldsymbol{L}=\boldsymbol{r}\times m\boldsymbol{v}=\boldsymbol{r}\times\boldsymbol{p} L = r × m v = r × p Então, para calcular o valor esperado de qualquer quantidade Q ( x , p ) Q(x,p) Q ( x , p ) p p p − i ℏ ∂ ∂ x -i\hbar\frac{\partial}{\partial x} − i ℏ ∂ x ∂ Ψ ∗ \Psi^{*} Ψ ∗ Ψ \Psi Ψ
⟨ Q ( x , p ) ⟩ = ∫ Ψ ∗ [ Q ( x , − i ℏ ∂ ∂ x ) ] Ψ d x \left\langle Q\left(x,p\right)\right\rangle =\int\Psi^{*}\left[Q\left(x,-i\hbar\frac{\partial}{\partial x}\right)\right]\Psi dx ⟨ Q ( x , p ) ⟩ = ∫ Ψ ∗ [ Q ( x , − i ℏ ∂ x ∂ ) ] Ψ d x Griffiths promete fundamentar essa equação em uma base teórica sólida futuramente, mas por enquanto podemos considerá-la um axioma[11]
⟨ T ⟩ = − ℏ 2 m ∫ Ψ ∗ ∂ 2 Ψ ∂ x 2 d x \left\langle T\right\rangle =-\frac{\hbar^{2}}{m}\int\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}dx ⟨ T ⟩ = − m ℏ 2 ∫ Ψ ∗ ∂ x 2 ∂ 2 Ψ d x Exemplo: Calcular d ⟨ p ⟩ / d t d\left\langle p\right\rangle /dt d ⟨ p ⟩ / d t
Queremos obter:
d ⟨ p ⟩ d t = ⟨ − ∂ V ∂ x ⟩ \frac{d\left\langle p\right\rangle }{dt}=\left\langle -\frac{\partial V}{\partial x}\right\rangle d t d ⟨ p ⟩ = ⟨ − ∂ x ∂ V ⟩ Vamos retomar a definição de ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩
d ⟨ p ⟩ d t = d d t ( − i ℏ ∫ Ψ ∗ ∂ Ψ ∂ x d x ) = − i ℏ d d t ( ∫ Ψ ∗ ∂ Ψ ∂ x d x ) = − [ ∫ ( i ℏ ∂ Ψ ∗ ∂ t ∂ Ψ ∂ x + i ℏ Ψ ∗ ∂ ∂ t ( ∂ Ψ ∂ x ) ) d x ] = [ ∫ ( ( − i ℏ ∂ Ψ ∗ ∂ t ) ∂ Ψ ∂ x − Ψ ∗ ∂ ∂ x ( i ℏ ∂ Ψ ∂ t ) ) d x ] \begin{align}
\frac{d\left\langle p\right\rangle }{dt} & =\frac{d}{dt}\left(-i\hbar\int\Psi^{*}\frac{\partial\Psi}{\partial x}dx\right)\\
& =-i\hbar\frac{d}{dt}\left(\int\Psi^{*}\frac{\partial\Psi}{\partial x}dx\right)\\
& =-\left[\int\left(i\hbar\frac{\partial\Psi^{*}}{\partial t}\frac{\partial\Psi}{\partial x}+i\hbar\Psi^{*}\frac{\partial}{\partial t}\left(\frac{\partial\Psi}{\partial x}\right)\right)dx\right]\\
& =\left[\int\left(\left(-i\hbar\frac{\partial\Psi^{*}}{\partial t}\right)\frac{\partial\Psi}{\partial x}-\Psi^{*}\frac{\partial}{\partial x}\left(i\hbar\frac{\partial\Psi}{\partial t}\right)\right)dx\right]
\end{align} d t d ⟨ p ⟩ = d t d ( − i ℏ ∫ Ψ ∗ ∂ x ∂ Ψ d x ) = − i ℏ d t d ( ∫ Ψ ∗ ∂ x ∂ Ψ d x ) = − [ ∫ ( i ℏ ∂ t ∂ Ψ ∗ ∂ x ∂ Ψ + i ℏ Ψ ∗ ∂ t ∂ ( ∂ x ∂ Ψ ) ) d x ] = [ ∫ ( ( − i ℏ ∂ t ∂ Ψ ∗ ) ∂ x ∂ Ψ − Ψ ∗ ∂ x ∂ ( i ℏ ∂ t ∂ Ψ ) ) d x ] Onde usamos a simetria das segundas derivadas[12]
i ℏ ∂ Ψ ∂ t = − ℏ 2 2 m ∂ 2 Ψ ∂ x 2 + V Ψ i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V\Psi i ℏ ∂ t ∂ Ψ = − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ + V Ψ Temos:
d ⟨ p ⟩ d t = − [ ∫ ( i ℏ ∂ Ψ ∗ ∂ t ∂ Ψ ∂ x + i ℏ Ψ ∗ ∂ ∂ x ( ∂ Ψ ∂ t ) ) d x ] = ∫ ( ( − ℏ 2 2 m ∂ 2 Ψ ∗ ∂ x 2 + V Ψ ∗ ) ∂ Ψ ∂ x − Ψ ∗ ∂ ∂ x ( − ℏ 2 2 m ∂ 2 Ψ ∂ x 2 + V Ψ ) ) d x = ∫ ( − ℏ 2 2 m ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x + V Ψ ∗ ∂ Ψ ∂ x + ℏ 2 2 m Ψ ∗ ∂ 3 Ψ ∂ x 3 − Ψ ∗ ∂ ∂ x ( V Ψ ) ) d x = ∫ [ ℏ 2 2 m ( − ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x + Ψ ∗ ∂ 3 Ψ ∂ x 3 ) + ( V Ψ ∗ ∂ Ψ ∂ x − Ψ ∗ ∂ ∂ x ( V Ψ ) ) ] d x = ∫ [ ℏ 2 2 m ( − ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x + Ψ ∗ ∂ 3 Ψ ∂ x 3 ) + ( V Ψ ∗ ∂ Ψ ∂ x − Ψ ∗ ∂ V ∂ x Ψ − Ψ ∗ V ∂ ∂ x Ψ ) ] d x = ℏ 2 2 m ∫ ( − ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x + Ψ ∗ ∂ 3 Ψ ∂ x 3 ) d x − ∫ Ψ ∗ ∂ V ∂ x Ψ d x \begin{align}
\frac{d\left\langle p\right\rangle }{dt} & =-\left[\int\left(i\hbar\frac{\partial\Psi^{*}}{\partial t}\frac{\partial\Psi}{\partial x}+i\hbar\Psi^{*}\frac{\partial}{\partial x}\left(\frac{\partial\Psi}{\partial t}\right)\right)dx\right]\\
& =\int\left(\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}+V\Psi^{*}\right)\frac{\partial\Psi}{\partial x}-\Psi^{*}\frac{\partial}{\partial x}\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V\Psi\right)\right)dx\\
& =\int\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}+V\Psi^{*}\frac{\partial\Psi}{\partial x}+\frac{\hbar^{2}}{2m}\Psi^{*}\frac{\partial^{3}\Psi}{\partial x^{3}}-\Psi^{*}\frac{\partial}{\partial x}\left(V\Psi\right)\right)dx\\
& =\int\left[\frac{\hbar^{2}}{2m}\left(-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}+\Psi^{*}\frac{\partial^{3}\Psi}{\partial x^{3}}\right)+\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\Psi^{*}\frac{\partial}{\partial x}\left(V\Psi\right)\right)\right]dx\\
& =\int\left[\frac{\hbar^{2}}{2m}\left(-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}+\Psi^{*}\frac{\partial^{3}\Psi}{\partial x^{3}}\right)+\left(V\Psi^{*}\frac{\partial\Psi}{\partial x}-\Psi^{*}\frac{\partial V}{\partial x}\Psi-\Psi^{*}V\frac{\partial}{\partial x}\Psi\right)\right]dx\\
& =\frac{\hbar^{2}}{2m}\int\left(-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}+\Psi^{*}\frac{\partial^{3}\Psi}{\partial x^{3}}\right)dx-\int\Psi^{*}\frac{\partial V}{\partial x}\Psi dx
\end{align} d t d ⟨ p ⟩ = − [ ∫ ( i ℏ ∂ t ∂ Ψ ∗ ∂ x ∂ Ψ + i ℏ Ψ ∗ ∂ x ∂ ( ∂ t ∂ Ψ ) ) d x ] = ∫ ( ( − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ ∗ + V Ψ ∗ ) ∂ x ∂ Ψ − Ψ ∗ ∂ x ∂ ( − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ + V Ψ ) ) d x = ∫ ( − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ + V Ψ ∗ ∂ x ∂ Ψ + 2 m ℏ 2 Ψ ∗ ∂ x 3 ∂ 3 Ψ − Ψ ∗ ∂ x ∂ ( V Ψ ) ) d x = ∫ [ 2 m ℏ 2 ( − ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ + Ψ ∗ ∂ x 3 ∂ 3 Ψ ) + ( V Ψ ∗ ∂ x ∂ Ψ − Ψ ∗ ∂ x ∂ ( V Ψ ) ) ] d x = ∫ [ 2 m ℏ 2 ( − ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ + Ψ ∗ ∂ x 3 ∂ 3 Ψ ) + ( V Ψ ∗ ∂ x ∂ Ψ − Ψ ∗ ∂ x ∂ V Ψ − Ψ ∗ V ∂ x ∂ Ψ ) ] d x = 2 m ℏ 2 ∫ ( − ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ + Ψ ∗ ∂ x 3 ∂ 3 Ψ ) d x − ∫ Ψ ∗ ∂ x ∂ V Ψ d x Na primeira integral precisamos integrar por partes. Pegando a segunda integral fazendo f = Ψ ∗ f=\Psi^{*} f = Ψ ∗ g = ∂ 2 Ψ / ∂ x 2 g=\partial^{2}\Psi/\partial x^{2} g = ∂ 2 Ψ/ ∂ x 2
∫ a b f d g d x d x = [ f g ] a b − ∫ a b d f d x g d x ∫ Ψ ∗ ∂ 3 Ψ ∂ x 3 d x = [ Ψ ∗ ∂ 2 Ψ ∂ x 2 ] − ∞ + ∞ − ∫ ∂ Ψ ∗ ∂ x ∂ 2 Ψ ∂ x 2 d x = − ∫ ∂ Ψ ∗ ∂ x ∂ 2 Ψ ∂ x 2 d x \begin{align}
\int^{b}_{a}f\frac{dg}{dx}dx & =\left[fg\right]^{b}_{a}-\int^{b}_{a}\frac{df}{dx}gdx\\
\int\Psi^{*}\frac{\partial^{3}\Psi}{\partial x^{3}}dx & =\left[\Psi^{*}\frac{\partial^{2}\Psi}{\partial x^{2}}\right]^{+\infty}_{-\infty}-\int\frac{\partial\Psi^{*}}{\partial x}\frac{\partial^{2}\Psi}{\partial x^{2}}dx\\
& =-\int\frac{\partial\Psi^{*}}{\partial x}\frac{\partial^{2}\Psi}{\partial x^{2}}dx
\end{align} ∫ a b f d x d g d x ∫ Ψ ∗ ∂ x 3 ∂ 3 Ψ d x = [ f g ] a b − ∫ a b d x df g d x = [ Ψ ∗ ∂ x 2 ∂ 2 Ψ ] − ∞ + ∞ − ∫ ∂ x ∂ Ψ ∗ ∂ x 2 ∂ 2 Ψ d x = − ∫ ∂ x ∂ Ψ ∗ ∂ x 2 ∂ 2 Ψ d x Então:
d ⟨ p ⟩ d t = ℏ 2 2 m ∫ ( − ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x − ∂ Ψ ∗ ∂ x ∂ 2 Ψ ∂ x 2 ) d x − ∫ Ψ ∗ ∂ V ∂ x Ψ d x \begin{align}
\frac{d\left\langle p\right\rangle }{dt} & =\frac{\hbar^{2}}{2m}\int\left(-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}-\frac{\partial\Psi^{*}}{\partial x}\frac{\partial^{2}\Psi}{\partial x^{2}}\right)dx-\int\Psi^{*}\frac{\partial V}{\partial x}\Psi dx
\end{align} d t d ⟨ p ⟩ = 2 m ℏ 2 ∫ ( − ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ − ∂ x ∂ Ψ ∗ ∂ x 2 ∂ 2 Ψ ) d x − ∫ Ψ ∗ ∂ x ∂ V Ψ d x Vamos aplicar a integral por partes novamente no segundo termo da primeira integral sendo agora f = ∂ Ψ ∗ / ∂ x f=\partial\Psi^{*}/\partial x f = ∂ Ψ ∗ / ∂ x g = ∂ Ψ / ∂ x g=\partial\Psi/\partial x g = ∂ Ψ/ ∂ x
∫ a b f d g d x d x = [ f g ] a b − ∫ a b d f d x g d x ∫ a b ∂ Ψ ∗ ∂ x ∂ Ψ 2 ∂ x 2 d x = [ ∂ Ψ ∗ ∂ x ∂ Ψ ∂ x ] a b − ∫ a b ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x d x = − ∫ a b ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x d x \begin{align}
\int^{b}_{a}f\frac{dg}{dx}dx & =\left[fg\right]^{b}_{a}-\int^{b}_{a}\frac{df}{dx}gdx\\
\int^{b}_{a}\frac{\partial\Psi^{*}}{\partial x}\frac{\partial\Psi^{2}}{\partial x^{2}}dx & =\left[\frac{\partial\Psi^{*}}{\partial x}\frac{\partial\Psi}{\partial x}\right]^{b}_{a}-\int^{b}_{a}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}dx\\
& =-\int^{b}_{a}\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}dx
\end{align} ∫ a b f d x d g d x ∫ a b ∂ x ∂ Ψ ∗ ∂ x 2 ∂ Ψ 2 d x = [ f g ] a b − ∫ a b d x df g d x = [ ∂ x ∂ Ψ ∗ ∂ x ∂ Ψ ] a b − ∫ a b ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ d x = − ∫ a b ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ d x Onde como no caso anterior, usamos a condição de que Ψ \Psi Ψ
d ⟨ p ⟩ d t = ℏ 2 2 m ∫ ( − ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x + ∂ 2 Ψ ∗ ∂ x 2 ∂ Ψ ∂ x d x ) d x − ∫ Ψ ∗ ∂ V ∂ x Ψ d x = − ∫ Ψ ∗ ∂ V ∂ x Ψ d x d x \begin{align}
\frac{d\left\langle p\right\rangle }{dt} & =\frac{\hbar^{2}}{2m}\int\left(-\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}+\frac{\partial^{2}\Psi^{*}}{\partial x^{2}}\frac{\partial\Psi}{\partial x}dx\right)dx-\int\Psi^{*}\frac{\partial V}{\partial x}\Psi dx\\
& =-\int\Psi^{*}\frac{\partial V}{\partial x}\Psi dxdx
\end{align} d t d ⟨ p ⟩ = 2 m ℏ 2 ∫ ( − ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ + ∂ x 2 ∂ 2 Ψ ∗ ∂ x ∂ Ψ d x ) d x − ∫ Ψ ∗ ∂ x ∂ V Ψ d x = − ∫ Ψ ∗ ∂ x ∂ V Ψ d x d x E pela equação 55
⟨ Q ( x , p ) ⟩ = ∫ Ψ ∗ [ Q ( x , − i ℏ ∂ ∂ x ) ] Ψ d x ⟨ − ∂ V ∂ x ⟩ = ∫ Ψ ∗ [ − ∂ V ∂ x ] Ψ d x \begin{align}
\left\langle Q\left(x,p\right)\right\rangle & =\int\Psi^{*}\left[Q\left(x,-i\hbar\frac{\partial}{\partial x}\right)\right]\Psi dx\\
\left\langle -\frac{\partial V}{\partial x}\right\rangle & =\int\Psi^{*}\left[-\frac{\partial V}{\partial x}\right]\Psi dx
\end{align} ⟨ Q ( x , p ) ⟩ ⟨ − ∂ x ∂ V ⟩ = ∫ Ψ ∗ [ Q ( x , − i ℏ ∂ x ∂ ) ] Ψ d x = ∫ Ψ ∗ [ − ∂ x ∂ V ] Ψ d x Logo:
d ⟨ p ⟩ d t = ⟨ − ∂ V ∂ x ⟩ \frac{d\left\langle p\right\rangle }{dt}=\left\langle -\frac{\partial V}{\partial x}\right\rangle d t d ⟨ p ⟩ = ⟨ − ∂ x ∂ V ⟩ Como queríamos demonstrar.
A relação da incerteza ¶ Se pegarmos uma corda, temos duas situações extremas:
Se a oscilamos periodicamente, temos claramente um comprimento de onda, mas não podemos determinar onde ela está localizada.
Se enviamos apenas um pulso, podemos localizar o pulso, mas não podemos falar em comprimento de onda.
Casos intermediários apresentam uma situação de troca entre uma condição e outra. Isso se aplica a qualquer fenômeno ondulatório, inclusive à função de onda da mecânica quântica. Embora exista um teorema que pode ser demonstrado rigorosamente por meio da análise de Fourier, vamos nos concentrar aqui no argumento qualitativo. A fórmula de de Broglie[13]
p = h λ = 2 π ℏ λ p=\frac{h}{\lambda}=\frac{2\pi\hbar}{\lambda} p = λ h = λ 2 π ℏ Temos, portanto, uma relação entre o momento e o comprimento de onda. Como essas grandezas são inversamente proporcionais, pode surgir a tendência de pensar que suas dispersões[14] y = f ( x ) y=f(x) y = f ( x ) prova
σ y ≈ ∣ d f d x ∣ σ x \sigma_{y}\approx\left|\frac{df}{dx}\right|\sigma_{x} σ y ≈ ∣ ∣ d x df ∣ ∣ σ x Ou seja:
σ p ≈ ∣ − h λ 2 ∣ σ λ = h λ 2 σ λ \sigma_{p}\approx\left|-\frac{h}{\lambda^{2}}\right|\sigma_{\lambda}=\frac{h}{\lambda^{2}}\sigma_{\lambda} σ p ≈ ∣ ∣ − λ 2 h ∣ ∣ σ λ = λ 2 h σ λ Ou ainda:
σ p p ≈ σ λ λ \frac{\sigma_{p}}{p}\approx\frac{\sigma_{\lambda}}{\lambda} p σ p ≈ λ σ λ Essa aproximação é suficiente para notarmos que chegamos a uma relação que expressa que a dispersão relativa de p p p λ \lambda λ λ \lambda λ p = h / λ p=h/\lambda p = h / λ σ λ \sigma_{\lambda} σ λ σ p \sigma_{p} σ p p p p λ \lambda λ
σ x σ p ≥ ℏ 2 \sigma_{x}\sigma_{p}\geq\frac{\hbar}{2} σ x σ p ≥ 2 ℏ Observação: A dispersão aqui se refere aos resultados obtidos em medições realizadas sobre sistemas identicamente preparados. Mesmo com a mesma preparação, os resultados não são necessariamente idênticos, mas se distribuem em torno de um valor médio.
Exemplo: Uma partícula de massa m m m
Ψ ( x , t ) = A exp [ − a ( m x 2 ℏ + i t ) ] \Psi\left(x,t\right)=A\exp\left[-a\left(\frac{mx^{2}}{\hbar}+it\right)\right] Ψ ( x , t ) = A exp [ − a ( ℏ m x 2 + i t ) ]
Onde A A A a a a
a) Queremos encontrar A A A
b) Encontrar qual função de energia potencial, V ( x ) V\left(x\right) V ( x )
c) Calcular os valores esperados x x x x 2 x^{2} x 2 p p p p 2 p^{2} p 2
d) Encontrar σ x \sigma_{x} σ x σ p \sigma_{p} σ p
Vamos começar calculando A A A
∣ Ψ ( x , t ) ∣ 2 = Ψ ∗ ( x , t ) Ψ ( x , t ) = ( A exp [ − a ( m x 2 ℏ + i t ) ] ) ( A exp [ − a ( m x 2 ℏ + i t ) ] ) = A 2 exp [ − 2 a m ℏ x 2 ] \begin{align}
\left|\Psi\left(x,t\right)\right|^{2} & =\Psi^{*}\left(x,t\right)\Psi\left(x,t\right)\\
& =\left(A\exp\left[-a\left(\frac{mx^{2}}{\hbar}+it\right)\right]\right)\left(A\exp\left[-a\left(\frac{mx^{2}}{\hbar}+it\right)\right]\right)\\
& =A^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right]
\end{align} ∣ Ψ ( x , t ) ∣ 2 = Ψ ∗ ( x , t ) Ψ ( x , t ) = ( A exp [ − a ( ℏ m x 2 + i t ) ] ) ( A exp [ − a ( ℏ m x 2 + i t ) ] ) = A 2 exp [ − ℏ 2 am x 2 ] Vamos substituir u = 2 a m ℏ x u=\sqrt{\frac{2am}{\hbar}}x u = ℏ 2 am x d u = 2 a m ℏ d x du=\sqrt{\frac{2am}{\hbar}}dx d u = ℏ 2 am d x
∫ ∣ Ψ ( x , t ) ∣ 2 d x = ∫ A 2 exp [ − 2 a m x 2 ℏ ] d x 1 = A 2 ℏ 2 a m ∫ e − u 2 d u \begin{align}
\int\left|\Psi\left(x,t\right)\right|^{2}dx & =\int A^{2}\exp\left[-\frac{2amx^{2}}{\hbar}\right]dx\\
1 & =A^{2}\sqrt{\frac{\hbar}{2am}}\int e^{-u^{2}}du
\end{align} ∫ ∣ Ψ ( x , t ) ∣ 2 d x 1 = ∫ A 2 exp [ − ℏ 2 am x 2 ] d x = A 2 2 am ℏ ∫ e − u 2 d u Essa é uma integral gaussiana de resultado bastante conhecido, mas vamos resolver uma vez (solução ) pelo menos. Definimos então uma integral:
G = ∫ − ∞ + ∞ e − a x 2 d x G=\int^{+\infty}_{-\infty}e^{-ax^{2}}dx G = ∫ − ∞ + ∞ e − a x 2 d x Então:
G 2 = ( ∫ − ∞ + ∞ e − a x 2 d x ) 2 = ( ∫ − ∞ + ∞ e − a x 2 d x ) ( ∫ − ∞ + ∞ e − a x 2 d x ) G^{2}=\left(\int^{+\infty}_{-\infty}e^{-ax^{2}}dx\right)^{2}=\left(\int^{+\infty}_{-\infty}e^{-ax^{2}}dx\right)\left(\int^{+\infty}_{-\infty}e^{-ax^{2}}dx\right) G 2 = ( ∫ − ∞ + ∞ e − a x 2 d x ) 2 = ( ∫ − ∞ + ∞ e − a x 2 d x ) ( ∫ − ∞ + ∞ e − a x 2 d x ) Ou como x x x x → y x\rightarrow y x → y
G 2 = ( ∫ − ∞ + ∞ e − a x 2 d x ) ( ∫ − ∞ + ∞ e − a y 2 d x ) = ( ∫ − ∞ + ∞ ∫ − ∞ + ∞ e − a ( x 2 + y 2 ) d x d y ) G^{2}=\left(\int^{+\infty}_{-\infty}e^{-ax^{2}}dx\right)\left(\int^{+\infty}_{-\infty}e^{-ay^{2}}dx\right)=\left(\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}e^{-a\left(x^{2}+y^{2}\right)}dxdy\right) G 2 = ( ∫ − ∞ + ∞ e − a x 2 d x ) ( ∫ − ∞ + ∞ e − a y 2 d x ) = ( ∫ − ∞ + ∞ ∫ − ∞ + ∞ e − a ( x 2 + y 2 ) d x d y ) Agora podemos fazer uma mudança no sistema de coordenadas, obviamente, mudando também o elemento de área. Usando o sistema de coordenadas polares (definição ):
x = r cos θ y = r sin θ d a = r d r d θ \begin{align}
x & =r\cos\theta\\
y & =r\sin\theta\\
da & =rdrd\theta
\end{align} x y d a = r cos θ = r sin θ = r d r d θ Para cobrir todo o espaço, nossos limites são 0 ≤ r < ∞ 0\leq r<\infty 0 ≤ r < ∞ 0 ≤ θ < 2 π 0\leq\theta<2\pi 0 ≤ θ < 2 π
G 2 = ( ∫ 0 2 π ∫ 0 ∞ e − a r 2 r d r d θ ) = ( ∫ 0 2 π d θ ) ( ∫ 0 ∞ e − a r 2 r d r ) = 2 π ( ∫ 0 ∞ e − a r 2 r d r ) \begin{align}
G^{2} & =\left(\int^{2\pi}_{0}\int^{\infty}_{0}e^{-ar^{2}}rdrd\theta\right)\\
& =\left(\int^{2\pi}_{0}d\theta\right)\left(\int^{\infty}_{0}e^{-ar^{2}}rdr\right)\\
& =2\pi\left(\int^{\infty}_{0}e^{-ar^{2}}rdr\right)
\end{align} G 2 = ( ∫ 0 2 π ∫ 0 ∞ e − a r 2 r d r d θ ) = ( ∫ 0 2 π d θ ) ( ∫ 0 ∞ e − a r 2 r d r ) = 2 π ( ∫ 0 ∞ e − a r 2 r d r ) Substituindo u = − a r 2 u=-ar^{2} u = − a r 2 d u d r = − a 2 r → − d u a = 2 r d r \frac{du}{dr}=-a2r\rightarrow-\frac{du}{a}=2rdr d r d u = − a 2 r → − a d u = 2 r d r u = − a 0 2 = 0 u=-a0^{2}=0 u = − a 0 2 = 0 u = − a r ∞ 2 = − ∞ u=-ar\infty^{2}=-\infty u = − a r ∞ 2 = − ∞
G 2 = π ( ∫ 0 − ∞ e − a r 2 2 r d r ) = − π a ( ∫ 0 − ∞ e u d u ) = − π a [ e u ] 0 − ∞ = − π a [ 1 e u − e 0 ] 0 − ∞ = π a \begin{align}
G^{2} & =\pi\left(\int^{-\infty}_{0}e^{-ar^{2}}2rdr\right)\\
& =-\frac{\pi}{a}\left(\int^{-\infty}_{0}e^{u}du\right)\\
& =-\frac{\pi}{a}\left[e^{u}\right]^{-\infty}_{0}\\
& =-\frac{\pi}{a}\left[\frac{1}{e^{u}}-e^{0}\right]^{-\infty}_{0}\\
& =\frac{\pi}{a}
\end{align} G 2 = π ( ∫ 0 − ∞ e − a r 2 2 r d r ) = − a π ( ∫ 0 − ∞ e u d u ) = − a π [ e u ] 0 − ∞ = − a π [ e u 1 − e 0 ] 0 − ∞ = a π Logo se G 2 = π / a G^{2}=\pi/a G 2 = π / a G = π / a G=\sqrt{\pi/a} G = π / a
∫ − ∞ + ∞ e − a x 2 d x = G = π a \int^{+\infty}_{-\infty}e^{-ax^{2}}dx=G=\sqrt{\frac{\pi}{a}} ∫ − ∞ + ∞ e − a x 2 d x = G = a π Com isso podemos retomar e resolver nossa integral:
1 = A 2 ℏ 2 a m ∫ e − u 2 d u 1 = A 2 ℏ 2 a m π 2 a m ℏ π = A 2 \begin{align}
1 & =A^{2}\sqrt{\frac{\hbar}{2am}}\int e^{-u^{2}}du\\
1 & =A^{2}\sqrt{\frac{\hbar}{2am}}\sqrt{\pi}\\
\sqrt{\frac{2am}{\hbar\pi}} & =A^{2}
\end{align} 1 1 ℏ π 2 am = A 2 2 am ℏ ∫ e − u 2 d u = A 2 2 am ℏ π = A 2 Ou então:
A = ( 2 a m ℏ π ) 1 4 A=\left(\frac{2am}{\hbar\pi}\right)^{\frac{1}{4}} A = ( ℏ π 2 am ) 4 1 Agora queremos saber para qual potencial de energia V ( x ) V\left(x\right) V ( x )
i ℏ ∂ Ψ ∂ t = − ℏ 2 2 m ∂ 2 Ψ ∂ x 2 + V Ψ i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V\Psi i ℏ ∂ t ∂ Ψ = − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ + V Ψ E derivando a função de onda:
Ψ = A exp [ − a ( m x 2 ℏ + i t ) ] ∂ Ψ ∂ t = A exp [ − a m x 2 ℏ ] ( − a i ) exp [ − a i t ] = − a i Ψ ∂ Ψ ∂ x = A [ ( − 2 a m x ℏ ) ] exp [ − a m x 2 ℏ ] exp [ − a i t ] = − 2 a m x ℏ Ψ ∂ 2 Ψ ∂ x 2 = ∂ ∂ x ( − 2 a m x ℏ Ψ ) = − 2 a m ℏ Ψ + ( 2 a m x ℏ ) 2 Ψ \begin{align}
\Psi & =A\exp\left[-a\left(\frac{mx^{2}}{\hbar}+it\right)\right]\\
\frac{\partial\Psi}{\partial t} & =A\exp\left[-\frac{amx^{2}}{\hbar}\right]\left(-ai\right)\exp\left[-ait\right]=-ai\Psi\\
\frac{\partial\Psi}{\partial x} & =A\left[\left(-\frac{2amx}{\hbar}\right)\right]\exp\left[-\frac{amx^{2}}{\hbar}\right]\exp\left[-ait\right]=-\frac{2amx}{\hbar}\Psi\\
\frac{\partial^{2}\Psi}{\partial x^{2}} & =\frac{\partial}{\partial x}\left(-\frac{2amx}{\hbar}\Psi\right)=-\frac{2am}{\hbar}\Psi+\left(\frac{2amx}{\hbar}\right)^{2}\Psi
\end{align} Ψ ∂ t ∂ Ψ ∂ x ∂ Ψ ∂ x 2 ∂ 2 Ψ = A exp [ − a ( ℏ m x 2 + i t ) ] = A exp [ − ℏ am x 2 ] ( − ai ) exp [ − ai t ] = − ai Ψ = A [ ( − ℏ 2 am x ) ] exp [ − ℏ am x 2 ] exp [ − ai t ] = − ℏ 2 am x Ψ = ∂ x ∂ ( − ℏ 2 am x Ψ ) = − ℏ 2 am Ψ + ( ℏ 2 am x ) 2 Ψ Então:
i ℏ ∂ Ψ ∂ t = − ℏ 2 2 m ∂ 2 Ψ ∂ x 2 + V Ψ V Ψ = ℏ 2 2 m ∂ 2 Ψ ∂ x 2 + i ℏ ∂ Ψ ∂ t = ℏ 2 2 m [ − 2 a m ℏ Ψ + ( 2 a m x ℏ ) 2 Ψ ] + i ℏ [ − a i Ψ ] = − a ℏ Ψ + 2 m ( a x ) 2 Ψ + ℏ a Ψ = [ − a ℏ + 2 m ( a x ) 2 + ℏ a ] Ψ [ V ] Ψ = [ 2 m ( a x ) 2 ] Ψ \begin{align}
i\hbar\frac{\partial\Psi}{\partial t} & =-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+V\Psi\\
V\Psi & =\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}}+i\hbar\frac{\partial\Psi}{\partial t}\\
& =\frac{\hbar^{2}}{2m}\left[-\frac{2am}{\hbar}\Psi+\left(\frac{2amx}{\hbar}\right)^{2}\Psi\right]+i\hbar\left[-ai\Psi\right]\\
& =-a\hbar\Psi+2m\left(ax\right)^{2}\Psi+\hbar a\Psi\\
& =\left[-a\hbar+2m\left(ax\right)^{2}+\hbar a\right]\Psi\\
\left[V\right]\Psi & =\left[2m\left(ax\right)^{2}\right]\Psi
\end{align} i ℏ ∂ t ∂ Ψ V Ψ [ V ] Ψ = − 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ + V Ψ = 2 m ℏ 2 ∂ x 2 ∂ 2 Ψ + i ℏ ∂ t ∂ Ψ = 2 m ℏ 2 [ − ℏ 2 am Ψ + ( ℏ 2 am x ) 2 Ψ ] + i ℏ [ − ai Ψ ] = − a ℏΨ + 2 m ( a x ) 2 Ψ + ℏ a Ψ = [ − a ℏ + 2 m ( a x ) 2 + ℏ a ] Ψ = [ 2 m ( a x ) 2 ] Ψ Então:
V = 2 m a 2 x 2 V=2ma^{2}x^{2} V = 2 m a 2 x 2 Agora queremos calcular diversos valores esperados, vamos relembrar que já calculamos ∣ Ψ ( x , t ) ∣ 2 = A 2 exp [ − 2 a m ℏ x 2 ] \left|\Psi\left(x,t\right)\right|^{2}=A^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right] ∣ Ψ ( x , t ) ∣ 2 = A 2 exp [ − ℏ 2 am x 2 ] ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩
⟨ x ⟩ = ∫ x A 2 exp [ − 2 a m ℏ x 2 ] d x \left\langle x\right\rangle =\int xA^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right]dx ⟨ x ⟩ = ∫ x A 2 exp [ − ℏ 2 am x 2 ] d x Nesse caso, como o integrando é impar, isso é, se G ( x ) = x A 2 exp [ − 2 a m ℏ x 2 ] G\left(x\right)=xA^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right] G ( x ) = x A 2 exp [ − ℏ 2 am x 2 ]
G ( − x ) = − G ( x ) ( − x ) A 2 exp [ − 2 a m ℏ ( − x ) 2 ] = − ( x A 2 exp [ − 2 a m ℏ x 2 ] ) − x A 2 exp [ − 2 a m ℏ x 2 ] = − x A 2 exp [ − 2 a m ℏ x 2 ] \begin{align}
G\left(-x\right) & =-G\left(x\right)\\
\left(-x\right)A^{2}\exp\left[-\frac{2am}{\hbar}\left(-x\right)^{2}\right] & =-\left(xA^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right]\right)\\
-xA^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right] & =-xA^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right]
\end{align} G ( − x ) ( − x ) A 2 exp [ − ℏ 2 am ( − x ) 2 ] − x A 2 exp [ − ℏ 2 am x 2 ] = − G ( x ) = − ( x A 2 exp [ − ℏ 2 am x 2 ] ) = − x A 2 exp [ − ℏ 2 am x 2 ] Então simplesmente a integral é zero e temos ⟨ x ⟩ = 0 \left\langle x\right\rangle =0 ⟨ x ⟩ = 0 ⟨ p ⟩ = m d ⟨ x ⟩ d t \left\langle p\right\rangle =m\frac{d\left\langle x\right\rangle }{dt} ⟨ p ⟩ = m d t d ⟨ x ⟩ ⟨ p ⟩ = 0 \left\langle p\right\rangle =0 ⟨ p ⟩ = 0 ⟨ x 2 ⟩ \left\langle x^{2}\right\rangle ⟨ x 2 ⟩
⟨ x 2 ⟩ = A 2 ∫ x 2 exp [ − 2 a m ℏ x 2 ] d x \left\langle x^{2}\right\rangle =A^{2}\int x^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right]dx ⟨ x 2 ⟩ = A 2 ∫ x 2 exp [ − ℏ 2 am x 2 ] d x Escrevendo α = 2 a m ℏ \alpha=\frac{2am}{\hbar} α = ℏ 2 am
I = ∫ x 2 e − α x 2 d x I=\int x^{2}e^{-\alpha x^{2}}dx I = ∫ x 2 e − α x 2 d x Fazendo integral por partes sendo f = x f=x f = x g ′ = x e − α x 2 g'=xe^{-\alpha x^{2}} g ′ = x e − α x 2
g = ∫ x e − α x 2 d x g=\int xe^{-\alpha x^{2}}dx g = ∫ x e − α x 2 d x Substituindo u = − α x 2 u=-\alpha x^{2} u = − α x 2 − d u 2 α = − x d x -\frac{du}{2\alpha}=-xdx − 2 α d u = − x d x
g = − 1 2 α ∫ e u d u = − e u 2 α = − e − α x 2 2 α g=-\frac{1}{2\alpha}\int e^{u}du=-\frac{e^{u}}{2\alpha}=-\frac{e^{-\alpha x^{2}}}{2\alpha} g = − 2 α 1 ∫ e u d u = − 2 α e u = − 2 α e − α x 2 Evidentemente temos f ′ = 1 f'=1 f ′ = 1
∫ a b f d g d x d x = [ f g ] a b − ∫ a b d f d x g d x ∫ x ( x e − α x 2 ) d x = [ − x e − α x 2 2 α ] − ∞ + ∞ + 1 2 α ∫ e − α x 2 d x \begin{align}
\int^{b}_{a}f\frac{dg}{dx}dx & =\left[fg\right]^{b}_{a}-\int^{b}_{a}\frac{df}{dx}gdx\\
\int x\left(xe^{-\alpha x^{2}}\right)dx & =\left[-\frac{xe^{-\alpha x^{2}}}{2\alpha}\right]^{+\infty}_{-\infty}+\frac{1}{2\alpha}\int e^{-\alpha x^{2}}dx
\end{align} ∫ a b f d x d g d x ∫ x ( x e − α x 2 ) d x = [ f g ] a b − ∫ a b d x df g d x = [ − 2 α x e − α x 2 ] − ∞ + ∞ + 2 α 1 ∫ e − α x 2 d x O primeiro termo vai a zero pois e − x 2 e^{-x^{2}} e − x 2 1 ∣ x ∣ \frac{1}{\left|x\right|} ∣ x ∣ 1 α = 2 a m ℏ \alpha=\frac{2am}{\hbar} α = ℏ 2 am
∫ x ( x e − α x 2 ) d x = 1 2 α π α = π 2 ( ℏ 2 a m ) 3 2 \begin{align}
\int x\left(xe^{-\alpha x^{2}}\right)dx & =\frac{1}{2\alpha}\sqrt{\frac{\pi}{\alpha}}=\frac{\sqrt{\pi}}{2}\left(\frac{\hbar}{2am}\right)^{\frac{3}{2}}
\end{align} ∫ x ( x e − α x 2 ) d x = 2 α 1 α π = 2 π ( 2 am ℏ ) 2 3 E sendo A = ( 2 a m ℏ π ) 1 4 A=\left(\frac{2am}{\hbar\pi}\right)^{\frac{1}{4}} A = ( ℏ π 2 am ) 4 1
⟨ x 2 ⟩ = A 2 ∫ x 2 exp [ − 2 a m ℏ x 2 ] d x = ( 2 a m ℏ π ) 2 4 π 2 ( ℏ 2 a m ) 3 2 \left\langle x^{2}\right\rangle =A^{2}\int x^{2}\exp\left[-\frac{2am}{\hbar}x^{2}\right]dx=\left(\frac{2am}{\hbar\pi}\right)^{\frac{2}{4}}\frac{\sqrt{\pi}}{2}\left(\frac{\hbar}{2am}\right)^{\frac{3}{2}} ⟨ x 2 ⟩ = A 2 ∫ x 2 exp [ − ℏ 2 am x 2 ] d x = ( ℏ π 2 am ) 4 2 2 π ( 2 am ℏ ) 2 3 Ou apenas:
⟨ x 2 ⟩ = ℏ 4 a m \left\langle x^{2}\right\rangle =\frac{\hbar}{4am} ⟨ x 2 ⟩ = 4 am ℏ E por fim para calcular ⟨ p 2 ⟩ \left\langle p^{2}\right\rangle ⟨ p 2 ⟩ p p p − i ℏ ∂ ∂ x -i\hbar\frac{\partial}{\partial x} − i ℏ ∂ x ∂
⟨ p 2 ⟩ = ∫ Ψ ∗ ( − i ℏ ∂ ∂ x ) 2 Ψ d x \begin{align}
\left\langle p^{2}\right\rangle & =\int\Psi^{*}\left(-i\hbar\frac{\partial}{\partial x}\right)^{2}\Psi dx
\end{align} ⟨ p 2 ⟩ = ∫ Ψ ∗ ( − i ℏ ∂ x ∂ ) 2 Ψ d x Já calculamos anteriormente:
∂ 2 Ψ ∂ x 2 = − 2 a m ℏ Ψ + ( 2 a m x ℏ ) 2 Ψ \frac{\partial^{2}\Psi}{\partial x^{2}}=-\frac{2am}{\hbar}\Psi+\left(\frac{2amx}{\hbar}\right)^{2}\Psi ∂ x 2 ∂ 2 Ψ = − ℏ 2 am Ψ + ( ℏ 2 am x ) 2 Ψ Então:
⟨ p 2 ⟩ = ∫ Ψ ∗ ( − i ℏ ) 2 [ − 2 a m ℏ Ψ + ( 2 a m x ℏ ) 2 Ψ ] Ψ d x = ∫ Ψ ∗ ( − ℏ 2 ) [ − 2 a m ℏ Ψ + ( 2 a m x ℏ ) 2 Ψ ] Ψ d x = ( − ℏ 2 ) ( − 2 a m ℏ ∫ Ψ ∗ Ψ d x + ( 2 a m ℏ ) 2 ∫ x 2 Ψ 2 Ψ d x ) = ℏ 2 a m ∫ Ψ ∗ Ψ d x − ( 2 a m ) 2 ∫ x 2 Ψ 2 Ψ d x \begin{align}
\left\langle p^{2}\right\rangle & =\int\Psi^{*}\left(-i\hbar\right)^{2}\left[-\frac{2am}{\hbar}\Psi+\left(\frac{2amx}{\hbar}\right)^{2}\Psi\right]\Psi dx\\
& =\int\Psi^{*}\left(-\hbar^{2}\right)\left[-\frac{2am}{\hbar}\Psi+\left(\frac{2amx}{\hbar}\right)^{2}\Psi\right]\Psi dx\\
& =\left(-\hbar^{2}\right)\left(-\frac{2am}{\hbar}\int\Psi^{*}\Psi dx+\left(\frac{2am}{\hbar}\right)^{2}\int x^{2}\Psi^{2}\Psi dx\right)\\
& =\hbar2am\int\Psi^{*}\Psi dx-\left(2am\right)^{2}\int x^{2}\Psi^{2}\Psi dx
\end{align} ⟨ p 2 ⟩ = ∫ Ψ ∗ ( − i ℏ ) 2 [ − ℏ 2 am Ψ + ( ℏ 2 am x ) 2 Ψ ] Ψ d x = ∫ Ψ ∗ ( − ℏ 2 ) [ − ℏ 2 am Ψ + ( ℏ 2 am x ) 2 Ψ ] Ψ d x = ( − ℏ 2 ) ( − ℏ 2 am ∫ Ψ ∗ Ψ d x + ( ℏ 2 am ) 2 ∫ x 2 Ψ 2 Ψ d x ) = ℏ2 am ∫ Ψ ∗ Ψ d x − ( 2 am ) 2 ∫ x 2 Ψ 2 Ψ d x Pela normalização sabemos que a primeira integral vale 1 e a segunda ⟨ x 2 ⟩ = ℏ 4 a m \left\langle x^{2}\right\rangle =\frac{\hbar}{4am} ⟨ x 2 ⟩ = 4 am ℏ
⟨ p 2 ⟩ = 2 ℏ a m − ( 2 a m ) 2 ℏ 4 a m = 2 ℏ a m − ℏ a m \begin{align}
\left\langle p^{2}\right\rangle & =2\hbar am-\left(2am\right)^{2}\frac{\hbar}{4am}\\
& =2\hbar am-\hbar am
\end{align} ⟨ p 2 ⟩ = 2ℏ am − ( 2 am ) 2 4 am ℏ = 2ℏ am − ℏ am Então ⟨ p 2 ⟩ = ℏ a m \left\langle p^{2}\right\rangle =\hbar am ⟨ p 2 ⟩ = ℏ am
σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 σ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ℏ 4 a m − 0 = ℏ a m − 0 = ℏ 4 a m = ℏ a m \begin{align}
\sigma_{x} & =\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}} & \sigma_{p} & =\sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}\\
& =\sqrt{\frac{\hbar}{4am}-0} & & =\sqrt{\hbar am-0}\\
& =\sqrt{\frac{\hbar}{4am}} & & =\sqrt{\hbar am}
\end{align} σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = 4 am ℏ − 0 = 4 am ℏ σ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ℏ am − 0 = ℏ am Então:
σ x σ p = ℏ 4 a m ℏ a m = ℏ 2 \sigma_{x}\sigma_{p}=\sqrt{\frac{\hbar}{4am}}\sqrt{\hbar am}=\frac{\hbar}{2} σ x σ p = 4 am ℏ ℏ am = 2 ℏ Este resultado é consistente com a relação de incerteza:
σ x σ p ≥ ℏ 2 \sigma_{x}\sigma_{p}\geq\frac{\hbar}{2} σ x σ p ≥ 2 ℏ