Oscilador harmônico
A ideia do oscilador harmônico clássico é uma massa m m m k k k
F = − k x = m d 2 x d t 2 F=-kx=m\frac{d^{2}x}{dt^{2}} F = − k x = m d t 2 d 2 x Então:
k x = − m k d 2 x d t 2 = − ω d 2 x d t 2 kx=-\frac{m}{k}\frac{d^{2}x}{dt^{2}}=-\omega\frac{d^{2}x}{dt^{2}} k x = − k m d t 2 d 2 x = − ω d t 2 d 2 x A solução então é:
x ( t ) = A sin ( ω t ) + B cos ( ω t ) x\left(t\right)=A\sin\left(\omega t\right)+B\cos\left(\omega t\right) x ( t ) = A sin ( ω t ) + B cos ( ω t ) onde:
ω ≡ m k \omega\equiv\sqrt{\frac{m}{k}} ω ≡ k m Para uma força conservativa ( ∇ × F = 0 ) \left(\nabla\times F=0\right) ( ∇ × F = 0 ) V ( x ) V\left(x\right) V ( x ) F = − d V d x F=-\frac{dV}{dx} F = − d x d V
V ( x ) = 1 2 k x 2 V\left(x\right)=\frac{1}{2}kx^{2} V ( x ) = 2 1 k x 2 Se expandirmos o potencial em uma série de Taylor, lembrando que podemos expandir uma função f ( x ) f\left(x\right) f ( x ) a a a
f ( x ) = ∑ n = 0 ∞ a n ( x − a ) n , a n = f ( n ) ( a ) n ! f\left(x\right)=\sum^{\infty}_{n=0}a_{n}\left(x-a\right)^{n},\qquad a_{n}=\frac{f^{\left(n\right)}\left(a\right)}{n!} f ( x ) = n = 0 ∑ ∞ a n ( x − a ) n , a n = n ! f ( n ) ( a ) Expandindo então V ( x ) V\left(x\right) V ( x ) x 0 x_{0} x 0
V ( x ) = V ( 0 ) ( x 0 ) 0 ! ( x − x 0 ) 0 + V ( 1 ) ( x 0 ) 1 ! ( x − x 0 ) 1 + V 2 ( x 0 ) 2 ! ( x − x 0 ) 2 + … = V ( x 0 ) + V ′ ( x 0 ) ( x − x 0 ) + 1 2 V ′ ′ ( x 0 ) ( x − a ) 2 + … \begin{align}
V\left(x\right) & =\frac{V^{\left(0\right)}\left(x_{0}\right)}{0!}\left(x-x_{0}\right)^{0}+\frac{V^{\left(1\right)}\left(x_{0}\right)}{1!}\left(x-x_{0}\right)^{1}+\frac{V^{2}\left(x_{0}\right)}{2!}\left(x-x_{0}\right)^{2}+\dots\\
& =V\left(x_{0}\right)+V^{'}\left(x_{0}\right)\left(x-x_{0}\right)+\frac{1}{2}V^{''}\left(x_{0}\right)\left(x-a\right)^{2}+\dots
\end{align} V ( x ) = 0 ! V ( 0 ) ( x 0 ) ( x − x 0 ) 0 + 1 ! V ( 1 ) ( x 0 ) ( x − x 0 ) 1 + 2 ! V 2 ( x 0 ) ( x − x 0 ) 2 + … = V ( x 0 ) + V ′ ( x 0 ) ( x − x 0 ) + 2 1 V ′′ ( x 0 ) ( x − a ) 2 + … Podemos adicionar uma constante a V ( x ) V\left(x\right) V ( x ) F F F − V ( x 0 ) -V\left(x_{0}\right) − V ( x 0 ) V ′ ( x 0 ) = 0 V'\left(x_{0}\right)=0 V ′ ( x 0 ) = 0 x 0 x_{0} x 0 ( x − x 0 ) \left(x-x_{0}\right) ( x − x 0 ) n > 2 n>2 n > 2
V ( x ) ≈ 1 2 V ′ ′ ( x 0 ) ( x − x 0 ) 2 V\left(x\right)\approx\frac{1}{2}V''\left(x_{0}\right)\left(x-x_{0}\right)^{2} V ( x ) ≈ 2 1 V ′′ ( x 0 ) ( x − x 0 ) 2 Essa aproximação descreve um oscilador harmônico simples em torno do ponto x 0 x_{0} x 0 x 0 = 0 x_{0}=0 x 0 = 0
V ( x ) ≈ 1 2 V ′ ′ ( 0 ) x 2 V\left(x\right)\approx\frac{1}{2}V''\left(0\right)x^{2} V ( x ) ≈ 2 1 V ′′ ( 0 ) x 2 Podemos calcular a força:
F = − d V d x = − d d x ( 1 2 V ′ ′ ( 0 ) x 2 ) = − V ′ ′ ( 0 ) x F=-\frac{dV}{dx}=-\frac{d}{dx}\left(\frac{1}{2}V''\left(0\right)x^{2}\right)=-V''\left(0\right)x F = − d x d V = − d x d ( 2 1 V ′′ ( 0 ) x 2 ) = − V ′′ ( 0 ) x Lembrando que começamos o capítulo discutindo a lei de Hooke F = − k x F=-kx F = − k x k = V ′ ′ ( x 0 ) k=V''\left(x_{0}\right) k = V ′′ ( x 0 ) V ′ ′ ( x 0 ) = 0 V''\left(x_{0}\right)=0 V ′′ ( x 0 ) = 0
O problema quântico é resolver a equação de Schrödinger pro seguinte potencial:
V ( x ) = 1 2 m ω 2 x 2 V\left(x\right)=\frac{1}{2}m\omega^{2}x^{2} V ( x ) = 2 1 m ω 2 x 2 Onde podemos lembrar que:
ω = k m → k = m ω 2 \omega=\sqrt{\frac{k}{m}}\rightarrow k=m\omega^{2} ω = m k → k = m ω 2 A equação de Schrödinger independente do tempo, com a separação de variáveis é então:
− ℏ 2 2 m ∂ 2 ψ ∂ x 2 + V ψ = E ψ − ℏ 2 2 m ∂ 2 ψ ∂ x 2 + 1 2 m ω 2 x 2 ψ = E ψ \begin{align}
-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+V\psi & =E\psi\\
-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\frac{1}{2}m\omega^{2}x^{2}\psi & =E\psi
\end{align} − 2 m ℏ 2 ∂ x 2 ∂ 2 ψ + V ψ − 2 m ℏ 2 ∂ x 2 ∂ 2 ψ + 2 1 m ω 2 x 2 ψ = E ψ = E ψ Há duas formas de resolver. Uma mais geral e mais difícil e outra mais específica e mais simples, começaremos pela mais simples.
Método algébrico ¶ Vamos reescrever:
( − ℏ 2 2 m ∂ 2 ∂ x 2 + 1 2 m ω 2 x 2 ) ψ = E ψ 1 2 m ( − ℏ 2 ∂ 2 ∂ x 2 + m 2 ω 2 x 2 ) ψ = 1 2 m [ ( − i ℏ ∂ ∂ x ) 2 + ( m ω x ) 2 ] ψ = 1 2 m [ p ^ 2 + ( m ω x ) 2 ] ψ \begin{align}
\left(-\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+\frac{1}{2}m\omega^{2}x^{2}\right)\psi & =E\psi\\
\frac{1}{2m}\left(-\hbar^{2}\frac{\partial^{2}}{\partial x^{2}}+m^{2}\omega^{2}x^{2}\right)\psi & =\\
\frac{1}{2m}\left[\left(-i\hbar\frac{\partial}{\partial x}\right)^{2}+\left(m\omega x\right)^{2}\right]\psi & =\\
\frac{1}{2m}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right]\psi
\end{align} ( − 2 m ℏ 2 ∂ x 2 ∂ 2 + 2 1 m ω 2 x 2 ) ψ 2 m 1 ( − ℏ 2 ∂ x 2 ∂ 2 + m 2 ω 2 x 2 ) ψ 2 m 1 [ ( − i ℏ ∂ x ∂ ) 2 + ( mω x ) 2 ] ψ 2 m 1 [ p 2 + ( mω x ) 2 ] ψ = E ψ = = Onde p ^ = − i ℏ ∂ ∂ x \widehat{p}=-i\hbar\frac{\partial}{\partial x} p = − i ℏ ∂ x ∂
u 2 + v 2 = ( i u + v ) ( − i u + v ) u^{2}+v^{2}=\left(iu+v\right)\left(-iu+v\right) u 2 + v 2 = ( i u + v ) ( − i u + v ) Isso só é possível porque u u u v v v [ u , v ] = u v − v u = 0 \left[u,v\right]=uv-vu=0 [ u , v ] = uv − vu = 0 p ^ \widehat{p} p x x x
( − i p ^ + m ω x ) , ( + i p ^ + m ω x ) \left(-i\widehat{p}+m\omega x\right),\qquad\left(+i\widehat{p}+m\omega x\right) ( − i p + mω x ) , ( + i p + mω x ) Ou então escrevendo de forma mais sucinta e colocando um fator a frente (que seja justificado posteriormente, apenas para o resultado ficar melhor) e prestando atenção nos sinais:
a ^ ± ≡ 1 2 ℏ m ω ( ∓ i p ^ + m ω x ) \widehat{a}_{\pm}\equiv\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\widehat{p}+m\omega x\right) a ± ≡ 2ℏ mω 1 ( ∓ i p + mω x ) Então:
a ^ − a ^ + = 1 2 ℏ m ω ( i p ^ + m ω x ) 1 2 ℏ m ω ( − i p ^ + m ω x ) = 1 2 ℏ m ω [ p ^ 2 + i m ω p ^ x − i m ω x p ^ + ( m ω x ) 2 ] = 1 2 ℏ m ω [ p ^ 2 − i m ω ( x p ^ − p ^ x ) + ( m ω x ) 2 ] \begin{align}
\widehat{a}_{-}\widehat{a}_{+} & =\frac{1}{\sqrt{2\hbar m\omega}}\left(i\widehat{p}+m\omega x\right)\frac{1}{\sqrt{2\hbar m\omega}}\left(-i\widehat{p}+m\omega x\right)\\
& =\frac{1}{2\hbar m\omega}\left[\widehat{p}^{2}+im\omega\widehat{p}x-im\omega x\widehat{p}+\left(m\omega x\right)^{2}\right]\\
& =\frac{1}{2\hbar m\omega}\left[\widehat{p}^{2}-im\omega\left(x\widehat{p}-\widehat{p}x\right)+\left(m\omega x\right)^{2}\right]
\end{align} a − a + = 2ℏ mω 1 ( i p + mω x ) 2ℏ mω 1 ( − i p + mω x ) = 2ℏ mω 1 [ p 2 + imω p x − imω x p + ( mω x ) 2 ] = 2ℏ mω 1 [ p 2 − imω ( x p − p x ) + ( mω x ) 2 ] Se x x x p ^ \widehat{p} p [ x , p ^ ] = x p ^ − p ^ x = 0 \left[x,\widehat{p}\right]=x\widehat{p}-\widehat{p}x=0 [ x , p ] = x p − p x = 0 [ x , p ^ ] ≠ 0 \left[x,\widehat{p}\right]\neq0 [ x , p ] = 0
a ^ − a ^ + = 1 2 ℏ m ω [ p ^ 2 − i m ω [ x , p ^ ] + ( m ω x ) 2 ] = 1 2 ℏ m ω [ p ^ 2 + ( m ω x ) 2 ] − i m ω 2 ℏ m ω [ x , p ^ ] = 1 2 ℏ m ω [ p ^ 2 + ( m ω x ) 2 ] − i 2 ℏ [ x , p ^ ] \begin{align}
\widehat{a}_{-}\widehat{a}_{+} & =\frac{1}{2\hbar m\omega}\left[\widehat{p}^{2}-im\omega\left[x,\widehat{p}\right]+\left(m\omega x\right)^{2}\right]\\
& =\frac{1}{2\hbar m\omega}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right]-\frac{im\omega}{2\hbar m\omega}\left[x,\widehat{p}\right]\\
& =\frac{1}{2\hbar m\omega}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right]-\frac{i}{2\hbar}\left[x,\widehat{p}\right]
\end{align} a − a + = 2ℏ mω 1 [ p 2 − imω [ x , p ] + ( mω x ) 2 ] = 2ℏ mω 1 [ p 2 + ( mω x ) 2 ] − 2ℏ mω imω [ x , p ] = 2ℏ mω 1 [ p 2 + ( mω x ) 2 ] − 2ℏ i [ x , p ] Agora queremos então calcular o comutador de x x x [1] p ^ \widehat{p} p f ( x ) f\left(x\right) f ( x )
[ x , p ^ ] f ( x ) = ( x p ^ − p ^ x ) f ( x ) = ( − x i ℏ ∂ ∂ x + i ℏ ∂ ∂ x x ) f ( x ) = − x i ℏ ∂ f ∂ x + i ℏ ∂ ( x f ) ∂ x = i ℏ ( − x ∂ f ∂ x + x ∂ f ∂ x + f ∂ x ∂ x ) = i ℏ f ( x ) [ x , p ^ ] = i ℏ \begin{align}
\left[x,\widehat{p}\right]f\left(x\right) & =\left(x\widehat{p}-\widehat{p}x\right)f\left(x\right)\\
& =\left(-xi\hbar\frac{\partial}{\partial x}+i\hbar\frac{\partial}{\partial x}x\right)f\left(x\right)\\
& =-xi\hbar\frac{\partial f}{\partial x}+i\hbar\frac{\partial\left(xf\right)}{\partial x}\\
& =i\hbar\left(-x\frac{\partial f}{\partial x}+x\frac{\partial f}{\partial x}+f\frac{\partial x}{\partial x}\right)\\
& =i\hbar f\left(x\right)\\
\left[x,\widehat{p}\right] & =i\hbar
\end{align} [ x , p ] f ( x ) [ x , p ] = ( x p − p x ) f ( x ) = ( − x i ℏ ∂ x ∂ + i ℏ ∂ x ∂ x ) f ( x ) = − x i ℏ ∂ x ∂ f + i ℏ ∂ x ∂ ( x f ) = i ℏ ( − x ∂ x ∂ f + x ∂ x ∂ f + f ∂ x ∂ x ) = i ℏ f ( x ) = i ℏ Essa fórmula é conhecida como relação de comutação canônica. Segundo o livro, todos os “mistérios” da mecânica quântica podem ser traçados no fato de que posição e momento não comutam.Alguns autores tratam este fato como um axioma da teoria e derivam a partir dele p ^ = − i ℏ d / d x \widehat{p}=-i\hbar d/dx p = − i ℏ d / d x
a ^ − a ^ + = 1 2 ℏ m ω [ p ^ 2 + ( m ω x ) 2 ] − i 2 ℏ [ x , p ^ ] = 1 2 ℏ m ω [ p ^ 2 + ( m ω x ) 2 ] − i 2 ℏ ( i ℏ ) = 1 ℏ ω ( 1 2 m [ p ^ 2 + ( m ω x ) 2 ] ) + 1 2 \begin{align}
\widehat{a}_{-}\widehat{a}_{+} & =\frac{1}{2\hbar m\omega}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right]-\frac{i}{2\hbar}\left[x,\widehat{p}\right]\\
& =\frac{1}{2\hbar m\omega}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right]-\frac{i}{2\hbar}\left(i\hbar\right)\\
& =\frac{1}{\hbar\omega}\left(\frac{1}{2m}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right]\right)+\frac{1}{2}
\end{align} a − a + = 2ℏ mω 1 [ p 2 + ( mω x ) 2 ] − 2ℏ i [ x , p ] = 2ℏ mω 1 [ p 2 + ( mω x ) 2 ] − 2ℏ i ( i ℏ ) = ℏ ω 1 ( 2 m 1 [ p 2 + ( mω x ) 2 ] ) + 2 1 Lembrando:
1 2 m [ p ^ 2 + ( m ω x ) 2 ] ψ = E ψ \begin{align}
\frac{1}{2m}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right]\psi & =E\psi
\end{align} 2 m 1 [ p 2 + ( mω x ) 2 ] ψ = E ψ E como vimos que a equação de Schrödinger independente do tempo pode ser escrito como H ^ ψ = E ψ \widehat{H}\psi=E\psi H ψ = E ψ
H ^ = 1 2 m [ p ^ 2 + ( m ω x ) 2 ] \widehat{H}=\frac{1}{2m}\left[\widehat{p}^{2}+\left(m\omega x\right)^{2}\right] H = 2 m 1 [ p 2 + ( mω x ) 2 ] Ou seja:
a ^ − a ^ + = H ^ ℏ ω + 1 2 \begin{align}
\widehat{a}_{-}\widehat{a}_{+} & =\frac{\widehat{H}}{\hbar\omega}+\frac{1}{2}
\end{align} a − a + = ℏ ω H + 2 1 Um cálculo análogo nos leva a:
a ^ + a ^ − = H ^ ℏ ω − 1 2 \widehat{a}_{+}\widehat{a}_{-}=\frac{\widehat{H}}{\hbar\omega}-\frac{1}{2} a + a − = ℏ ω H − 2 1 Então:
[ a ^ − , a ^ + ] = a ^ − a ^ + − a ^ + a ^ − = ( H ^ ℏ ω + 1 2 ) − ( H ^ ℏ ω − 1 2 ) = 1 \left[\widehat{a}_{-},\widehat{a}_{+}\right]=\widehat{a}_{-}\widehat{a}_{+}-\widehat{a}_{+}\widehat{a}_{-}=\left(\frac{\widehat{H}}{\hbar\omega}+\frac{1}{2}\right)-\left(\frac{\widehat{H}}{\hbar\omega}-\frac{1}{2}\right)=1 [ a − , a + ] = a − a + − a + a − = ( ℏ ω H + 2 1 ) − ( ℏ ω H − 2 1 ) = 1 E o Hamiltoniano pode ser escrito:
H ^ = ( a ^ − a ^ + − 1 2 ) ℏ ω H ^ = ( a ^ + a ^ − + 1 2 ) ℏ ω \begin{align}
\widehat{H} & =\left(\widehat{a}_{-}\widehat{a}_{+}-\frac{1}{2}\right)\hbar\omega\\
\widehat{H} & =\left(\widehat{a}_{+}\widehat{a}_{-}+\frac{1}{2}\right)\hbar\omega
\end{align} H H = ( a − a + − 2 1 ) ℏ ω = ( a + a − + 2 1 ) ℏ ω Então a equação de Schrödinger se torna[2]
H ^ ψ = E ψ ℏ ω ( a ^ − a ^ + − 1 2 ) ψ = E ψ \begin{align}
\widehat{H}\psi & =E\psi\\
\hbar\omega\left(\widehat{a}_{-}\widehat{a}_{+}-\frac{1}{2}\right)\psi & =E\psi
\end{align} H ψ ℏ ω ( a − a + − 2 1 ) ψ = E ψ = E ψ E:
H ^ ψ = E ψ ℏ ω ( a ^ + a ^ − + 1 2 ) ψ = E ψ \begin{align}
\widehat{H}\psi & =E\psi\\
\hbar\omega\left(\widehat{a}_{+}\widehat{a}_{-}+\frac{1}{2}\right)\psi & =E\psi
\end{align} H ψ ℏ ω ( a + a − + 2 1 ) ψ = E ψ = E ψ Então de forma mais sucinta podemos escrever então a equação de Schrödinger para o oscilador harmônico como:
ℏ ω ( a ^ ± a ^ ∓ ± 1 2 ) ψ = E ψ \hbar\omega\left(\widehat{a}_{\pm}\widehat{a}_{\mp}\pm\frac{1}{2}\right)\psi=E\psi ℏ ω ( a ± a ∓ ± 2 1 ) ψ = E ψ Então chegamos em uma das primeiras afirmações que precisamos provar: Se a equação de Schrödinger H ^ ψ = E ψ \widehat{H}\psi=E\psi H ψ = E ψ
H ^ ( a ^ + ψ ) = ( E + ℏ ω ) ( a ^ + ψ ) \widehat{H}\left(\widehat{a}_{+}\psi\right)=\left(E+\hbar\omega\right)\left(\widehat{a}_{+}\psi\right) H ( a + ψ ) = ( E + ℏ ω ) ( a + ψ ) Exemplo: Prove que H ^ ( a ^ + ψ ) = ( E + ℏ ω ) ( a ^ + ψ ) \widehat{H}\left(\widehat{a}_{+}\psi\right)=\left(E+\hbar\omega\right)\left(\widehat{a}_{+}\psi\right) H ( a + ψ ) = ( E + ℏ ω ) ( a + ψ )
Vamos provar usando o resultado prévio:
H ^ ( a ^ + ψ ) = ℏ ω ( a ^ + a ^ − + 1 2 ) ( a ^ + ψ ) = ℏ ω ( a ^ + a ^ − a ^ + + a ^ + 2 ) ψ = a ^ + [ ℏ ω ( a ^ − a ^ + + 1 2 ) ] ψ \begin{align}
\widehat{H}\left(\widehat{a}_{+}\psi\right) & =\hbar\omega\left(\widehat{a}_{+}\widehat{a}_{-}+\frac{1}{2}\right)\left(\widehat{a}_{+}\psi\right)\\
& =\hbar\omega\left(\widehat{a}_{+}\widehat{a}_{-}\widehat{a}_{+}+\frac{\widehat{a}_{+}}{2}\right)\psi\\
& =\widehat{a}_{+}\left[\hbar\omega\left(\widehat{a}_{-}\widehat{a}_{+}+\frac{1}{2}\right)\right]\psi
\end{align} H ( a + ψ ) = ℏ ω ( a + a − + 2 1 ) ( a + ψ ) = ℏ ω ( a + a − a + + 2 a + ) ψ = a + [ ℏ ω ( a − a + + 2 1 ) ] ψ E sendo:
[ a ^ − , a ^ + ] = a ^ − a ^ + − a ^ + a ^ − = 1 → a ^ − a ^ + = a ^ + a ^ − + 1 \left[\widehat{a}_{-},\widehat{a}_{+}\right]=\widehat{a}_{-}\widehat{a}_{+}-\widehat{a}_{+}\widehat{a}_{-}=1\rightarrow\quad\widehat{a}_{-}\widehat{a}_{+}=\widehat{a}_{+}\widehat{a}_{-}+1 [ a − , a + ] = a − a + − a + a − = 1 → a − a + = a + a − + 1 Então:
H ^ ( a ^ + ψ ) = a ^ + [ ℏ ω ( a ^ − a ^ + + 1 2 ) ] ψ = a ^ + [ ℏ ω ( a ^ + a ^ − + 1 + 1 2 ) ] ψ = a ^ + [ ℏ ω ( a ^ + a ^ − + 1 2 ) + ℏ ω ] ψ = a ^ + [ H ^ + ℏ ω ] ψ = a ^ + [ H ^ ψ + ℏ ω ψ ] = a ^ + [ E ψ + ℏ ω ψ ] \begin{align}
\widehat{H}\left(\widehat{a}_{+}\psi\right) & =\widehat{a}_{+}\left[\hbar\omega\left(\widehat{a}_{-}\widehat{a}_{+}+\frac{1}{2}\right)\right]\psi\\
& =\widehat{a}_{+}\left[\hbar\omega\left(\widehat{a}_{+}\widehat{a}_{-}+1+\frac{1}{2}\right)\right]\psi\\
& =\widehat{a}_{+}\left[\hbar\omega\left(\widehat{a}_{+}\widehat{a}_{-}+\frac{1}{2}\right)+\hbar\omega\right]\psi\\
& =\widehat{a}_{+}\left[\widehat{H}+\hbar\omega\right]\psi\\
& =\widehat{a}_{+}\left[\widehat{H}\psi+\hbar\omega\psi\right]\\
& =\widehat{a}_{+}\left[E\psi+\hbar\omega\psi\right]
\end{align} H ( a + ψ ) = a + [ ℏ ω ( a − a + + 2 1 ) ] ψ = a + [ ℏ ω ( a + a − + 1 + 2 1 ) ] ψ = a + [ ℏ ω ( a + a − + 2 1 ) + ℏ ω ] ψ = a + [ H + ℏ ω ] ψ = a + [ H ψ + ℏ ω ψ ] = a + [ E ψ + ℏ ω ψ ] Então:
H ^ ( a ^ + ψ ) = ( E + ℏ ω ) ( a ^ + ψ ) \widehat{H}\left(\widehat{a}_{+}\psi\right)=\left(E+\hbar\omega\right)\left(\widehat{a}_{+}\psi\right) H ( a + ψ ) = ( E + ℏ ω ) ( a + ψ ) Uma vez que E E E ℏ ω \hbar\omega ℏ ω
H ^ ( a ^ − ψ ) = ( E − ℏ ω ) ( a ^ − ψ ) \widehat{H}\left(\widehat{a}_{-}\psi\right)=\left(E-\hbar\omega\right)\left(\widehat{a}_{-}\psi\right) H ( a − ψ ) = ( E − ℏ ω ) ( a − ψ ) Ou seja:
H ^ ( a ^ ± ψ ) = ( E ± ℏ ω ) ( a ^ ± ψ ) \widehat{H}\left(\widehat{a}_{\pm}\psi\right)=\left(E\pm\hbar\omega\right)\left(\widehat{a}_{\pm}\psi\right) H ( a ± ψ ) = ( E ± ℏ ω ) ( a ± ψ ) Ou seja, podemos gerar novas soluções se podemos encontrar apenas uma solução para começar. a ^ ± \widehat{a}_{\pm} a ± a ^ + \widehat{a}_{+} a + a ^ − \widehat{a}_{-} a − ψ 0 \psi_{0} ψ 0
a ^ − ψ 0 = 0 \widehat{a}_{-}\psi_{0}=0 a − ψ 0 = 0 Podemos determinar qual é esta função, usando a definição de a ^ − \widehat{a}_{-} a −
a ^ − = 1 2 ℏ m ω ( i p ^ + m ω x ) \begin{align}
\widehat{a}_{-} & =\frac{1}{\sqrt{2\hbar m\omega}}\left(i\widehat{p}+m\omega x\right)
\end{align} a − = 2ℏ mω 1 ( i p + mω x ) Então:
0 = a ^ − ψ 0 = 1 2 ℏ m ω ( i p ^ + m ω x ) ψ 0 = i ( − i ℏ d ψ 0 d x ) + m ω x ψ 0 = i ℏ d ψ 0 d x + m ω x ψ 0 d ψ 0 d x = − m ω ℏ x ψ 0 d ψ 0 ψ 0 = − m ω ℏ x d x ∫ d ψ 0 ψ 0 = − m ω ℏ ∫ x d x ln ψ 0 = − m ω ℏ x 2 2 + c exp ( ln ψ 0 ) = exp ( − m ω ℏ x 2 2 + c ) ψ 0 ( x ) = e − m ω 2 ℏ e c \begin{align}
0 & =\widehat{a}_{-}\psi_{0}\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(i\widehat{p}+m\omega x\right)\psi_{0}\\
& =i\left(-i\hbar\frac{d\psi_{0}}{dx}\right)+m\omega x\psi_{0}\\
& =i\hbar\frac{d\psi_{0}}{dx}+m\omega x\psi_{0}\\
\frac{d\psi_{0}}{dx} & =-\frac{m\omega}{\hbar}x\psi_{0}\\
\frac{d\psi_{0}}{\psi_{0}} & =-\frac{m\omega}{\hbar}xdx\\
\int\frac{d\psi_{0}}{\psi_{0}} & =-\frac{m\omega}{\hbar}\int xdx\\
\ln\psi_{0} & =-\frac{m\omega}{\hbar}\frac{x^{2}}{2}+c\\
\exp\left(\ln\psi_{0}\right) & =\exp\left(-\frac{m\omega}{\hbar}\frac{x^{2}}{2}+c\right)\\
\psi_{0}\left(x\right) & =e^{-\frac{m\omega}{2\hbar}}e^{c}
\end{align} 0 d x d ψ 0 ψ 0 d ψ 0 ∫ ψ 0 d ψ 0 ln ψ 0 exp ( ln ψ 0 ) ψ 0 ( x ) = a − ψ 0 = 2ℏ mω 1 ( i p + mω x ) ψ 0 = i ( − i ℏ d x d ψ 0 ) + mω x ψ 0 = i ℏ d x d ψ 0 + mω x ψ 0 = − ℏ mω x ψ 0 = − ℏ mω x d x = − ℏ mω ∫ x d x = − ℏ mω 2 x 2 + c = exp ( − ℏ mω 2 x 2 + c ) = e − 2ℏ mω e c Ou então como e c = A e^{c}=A e c = A
ψ 0 ( x ) = A e − m ω 2 ℏ x 2 \psi_{0}\left(x\right)=Ae^{-\frac{m\omega}{2\hbar}x^{2}} ψ 0 ( x ) = A e − 2ℏ mω x 2 Podemos determinar a constante normalizando:
1 = ∫ − ∞ + ∞ ∣ ψ 0 ∣ 2 d x = ∫ − ∞ + ∞ ∣ A e − m ω 2 ℏ x 2 ∣ 2 d x = ∣ A ∣ 2 ∫ − ∞ + ∞ e − m ω ℏ x 2 d x 1=\int^{+\infty}_{-\infty}\left|\psi_{0}\right|^{2}dx=\int^{+\infty}_{-\infty}\left|Ae^{-\frac{m\omega}{2\hbar}x^{2}}\right|^{2}dx=\left|A\right|^{2}\int^{+\infty}_{-\infty}e^{-\frac{m\omega}{\hbar}x^{2}}dx 1 = ∫ − ∞ + ∞ ∣ ψ 0 ∣ 2 d x = ∫ − ∞ + ∞ ∣ ∣ A e − 2ℏ mω x 2 ∣ ∣ 2 d x = ∣ A ∣ 2 ∫ − ∞ + ∞ e − ℏ mω x 2 d x Lembrando que essa é a integral gaussiana∫ − ∞ + ∞ e − a x 2 d x = π a \int^{+\infty}_{-\infty}e^{-ax^{2}}dx=\sqrt{\frac{\pi}{a}} ∫ − ∞ + ∞ e − a x 2 d x = a π
1 = ∣ A ∣ 2 π m ω ℏ = ∣ A ∣ 2 ℏ π m ω 1=\left|A\right|^{2}\sqrt{\frac{\pi}{\frac{m\omega}{\hbar}}}=\left|A\right|^{2}\sqrt{\frac{\hbar\pi}{m\omega}} 1 = ∣ A ∣ 2 ℏ mω π = ∣ A ∣ 2 mω ℏ π Então:
A = ( m ω ℏ π ) 1 / 4 A=\left(\frac{m\omega}{\hbar\pi}\right)^{1/4} A = ( ℏ π mω ) 1/4 Ou seja:
ψ 0 ( x ) = ( m ω ℏ π ) 1 / 4 e − m ω 2 ℏ x 2 \psi_{0}\left(x\right)=\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^{2}} ψ 0 ( x ) = ( ℏ π mω ) 1/4 e − 2ℏ mω x 2 Agora para determinar a energia vamos retomar a equação de Schrödinger:
ℏ ω ( a ^ + a ^ − + 1 2 ) ψ 0 = E 0 ψ 0 ℏ ω ( a ^ + a ^ − ψ 0 + 1 2 ψ 0 ) = E 0 ψ 0 ℏ ω 2 ψ 0 = E 0 ψ 0 ℏ ω 2 = E 0 \begin{align}
\hbar\omega\left(\widehat{a}_{+}\widehat{a}_{-}+\frac{1}{2}\right)\psi_{0} & =E_{0}\psi_{0}\\
\hbar\omega\left(\widehat{a}_{+}\widehat{a}_{-}\psi_{0}+\frac{1}{2}\psi_{0}\right) & =E_{0}\psi_{0}\\
\frac{\hbar\omega}{2}\psi_{0} & =E_{0}\psi_{0}\\
\frac{\hbar\omega}{2} & =E_{0}
\end{align} ℏ ω ( a + a − + 2 1 ) ψ 0 ℏ ω ( a + a − ψ 0 + 2 1 ψ 0 ) 2 ℏ ω ψ 0 2 ℏ ω = E 0 ψ 0 = E 0 ψ 0 = E 0 ψ 0 = E 0 Onde usamos o fato de que a ^ − ψ 0 = 0 \widehat{a}_{-}\psi_{0}=0 a − ψ 0 = 0 ℏ ω \hbar\omega ℏ ω ψ n ( x ) \psi_{n}\left(x\right) ψ n ( x ) a + a_{+} a + n n n ψ 0 ( x ) \psi_{0}\left(x\right) ψ 0 ( x )
ψ n ( x ) = A n ( a + ) n ψ 0 ( x ) \psi_{n}\left(x\right)=A_{n}\left(a_{+}\right)^{n}\psi_{0}\left(x\right) ψ n ( x ) = A n ( a + ) n ψ 0 ( x ) E a energia desse estado vai sofrer um incremento de ℏ ω \hbar\omega ℏ ω n n n E 0 = ℏ ω / 2 E_{0}=\hbar\omega/2 E 0 = ℏ ω /2
E n = ( n + 1 2 ) ℏ ω E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega E n = ( n + 2 1 ) ℏ ω Com esse método podemos construir todos estados estacionários do oscilador harmônico.
Exemplo: Calcule o primeiro estado excitado do oscilador harmônico.
Usando a equação 17 p ^ = − i ℏ d / d x \widehat{p}=-i\hbar d/dx p = − i ℏ d / d x
ψ 1 ( x ) = A 1 ( a + ) 1 ψ 0 ( x ) = A 1 a + ψ 0 ( x ) = A 1 ( 1 2 ℏ m ω ( − i p ^ + m ω x ) ) ( ( m ω ℏ π ) 1 / 4 e − m ω 2 ℏ x 2 ) = A 1 ( m ω ℏ π ) 1 / 4 1 2 ℏ m ω ( − i p ^ e − m ω 2 ℏ x 2 + m ω x e − m ω 2 ℏ x 2 ) = A 1 ( m ω ℏ π ) 1 / 4 ( − i [ − i ℏ d d x ] e − m ω 2 ℏ x 2 + m ω x e − m ω 2 ℏ x 2 ) = A 1 ( m ω ℏ π ) 1 / 4 1 2 ℏ m ω ( − ℏ d d x e − m ω 2 ℏ x 2 + m ω x e − m ω 2 ℏ x 2 ) = A 1 ( m ω ℏ π ) 1 / 4 e − m ω 2 ℏ x 2 2 ℏ m ω ( − ℏ d d x ( − m ω 2 ℏ x 2 ) + m ω x ) = A 1 ( m ω ℏ π ) 1 / 4 e − m ω 2 ℏ x 2 2 ℏ m ω ( ℏ m ω 2 ℏ 2 x + m ω x ) = A 1 ( m ω ℏ π ) 1 / 4 e − m ω 2 ℏ x 2 2 ℏ m ω ( m ω x + m ω x ) = A 1 ( m ω ℏ π ) 1 / 4 2 m ω x 2 ℏ m ω e − m ω 2 ℏ x 2 = A 1 ( m ω ℏ π ) 1 / 4 2 m ω ℏ x e − m ω 2 ℏ x 2 \begin{align}
\psi_{1}\left(x\right) & =A_{1}\left(a_{+}\right)^{1}\psi_{0}\left(x\right)\\
& =A_{1}a_{+}\psi_{0}\left(x\right)\\
& =A_{1}\left(\frac{1}{\sqrt{2\hbar m\omega}}\left(-i\widehat{p}+m\omega x\right)\right)\left(\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^{2}}\right)\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\frac{1}{\sqrt{2\hbar m\omega}}\left(-i\widehat{p}e^{-\frac{m\omega}{2\hbar}x^{2}}+m\omega xe^{-\frac{m\omega}{2\hbar}x^{2}}\right)\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\left(-i\left[-i\hbar\frac{d}{dx}\right]e^{-\frac{m\omega}{2\hbar}x^{2}}+m\omega xe^{-\frac{m\omega}{2\hbar}x^{2}}\right)\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\frac{1}{\sqrt{2\hbar m\omega}}\left(-\hbar\frac{d}{dx}e^{-\frac{m\omega}{2\hbar}x^{2}}+m\omega xe^{-\frac{m\omega}{2\hbar}x^{2}}\right)\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\frac{e^{-\frac{m\omega}{2\hbar}x^{2}}}{\sqrt{2\hbar m\omega}}\left(-\hbar\frac{d}{dx}\left(-\frac{m\omega}{2\hbar}x^{2}\right)+m\omega x\right)\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\frac{e^{-\frac{m\omega}{2\hbar}x^{2}}}{\sqrt{2\hbar m\omega}}\left(\hbar\frac{m\omega}{2\hbar}2x+m\omega x\right)\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\frac{e^{-\frac{m\omega}{2\hbar}x^{2}}}{\sqrt{2\hbar m\omega}}\left(m\omega x+m\omega x\right)\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\frac{2m\omega x}{\sqrt{2\hbar m\omega}}e^{-\frac{m\omega}{2\hbar}x^{2}}\\
& =A_{1}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\sqrt{\frac{2m\omega}{\hbar}x}e^{-\frac{m\omega}{2\hbar}x^{2}}
\end{align} ψ 1 ( x ) = A 1 ( a + ) 1 ψ 0 ( x ) = A 1 a + ψ 0 ( x ) = A 1 ( 2ℏ mω 1 ( − i p + mω x ) ) ( ( ℏ π mω ) 1/4 e − 2ℏ mω x 2 ) = A 1 ( ℏ π mω ) 1/4 2ℏ mω 1 ( − i p e − 2ℏ mω x 2 + mω x e − 2ℏ mω x 2 ) = A 1 ( ℏ π mω ) 1/4 ( − i [ − i ℏ d x d ] e − 2ℏ mω x 2 + mω x e − 2ℏ mω x 2 ) = A 1 ( ℏ π mω ) 1/4 2ℏ mω 1 ( − ℏ d x d e − 2ℏ mω x 2 + mω x e − 2ℏ mω x 2 ) = A 1 ( ℏ π mω ) 1/4 2ℏ mω e − 2ℏ mω x 2 ( − ℏ d x d ( − 2ℏ mω x 2 ) + mω x ) = A 1 ( ℏ π mω ) 1/4 2ℏ mω e − 2ℏ mω x 2 ( ℏ 2ℏ mω 2 x + mω x ) = A 1 ( ℏ π mω ) 1/4 2ℏ mω e − 2ℏ mω x 2 ( mω x + mω x ) = A 1 ( ℏ π mω ) 1/4 2ℏ mω 2 mω x e − 2ℏ mω x 2 = A 1 ( ℏ π mω ) 1/4 ℏ 2 mω x e − 2ℏ mω x 2 Agora só precisamos normalizar.
∫ − ∞ + ∞ ∣ ψ ∣ 2 d x = ∣ A 1 ∣ 2 ( m ω ℏ π ) 1 / 2 ( 2 m ω ℏ ) ∫ − ∞ + ∞ x 2 e − m ω ℏ x 2 d x \int^{+\infty}_{-\infty}\left|\psi\right|^{2}dx=\left|A_{1}\right|^{2}\left(\frac{m\omega}{\hbar\pi}\right)^{1/2}\left(\frac{2m\omega}{\hbar}\right)\int^{+\infty}_{-\infty}x^{2}e^{-\frac{m\omega}{\hbar}x^{2}}dx ∫ − ∞ + ∞ ∣ ψ ∣ 2 d x = ∣ A 1 ∣ 2 ( ℏ π mω ) 1/2 ( ℏ 2 mω ) ∫ − ∞ + ∞ x 2 e − ℏ mω x 2 d x Para resolver essa integral, vamos seguir esta Solução , onde vamos usar então o seguinte truque:
∂ ∂ a e − a x 2 = e − a x 2 ∂ ∂ a ( − a x 2 ) = − x 2 e − a x 2 \frac{\partial}{\partial a}e^{-ax^{2}}=e^{-ax^{2}}\frac{\partial}{\partial a}\left(-ax^{2}\right)=-x^{2}e^{-ax^{2}} ∂ a ∂ e − a x 2 = e − a x 2 ∂ a ∂ ( − a x 2 ) = − x 2 e − a x 2 Então:
I = ∫ − ∞ + ∞ x 2 e − m ω ℏ x 2 d x = − ∫ − ∞ + ∞ ∂ ∂ a e − a x 2 d x = − ∂ ∂ a ∫ − ∞ + ∞ e − a x 2 d x I=\int^{+\infty}_{-\infty}x^{2}e^{-\frac{m\omega}{\hbar}x^{2}}dx=-\int^{+\infty}_{-\infty}\frac{\partial}{\partial a}e^{-ax^{2}}dx=-\frac{\partial}{\partial a}\int^{+\infty}_{-\infty}e^{-ax^{2}}dx I = ∫ − ∞ + ∞ x 2 e − ℏ mω x 2 d x = − ∫ − ∞ + ∞ ∂ a ∂ e − a x 2 d x = − ∂ a ∂ ∫ − ∞ + ∞ e − a x 2 d x E como sabemos o resultado esta integral Gaussiana:
I = − ∂ ∂ a π a = − π ∂ a − 1 / 2 ∂ a = − π ( − 1 2 ) a − 3 / 2 = π 4 a 3 I=-\frac{\partial}{\partial a}\sqrt{\frac{\pi}{a}}=-\sqrt{\pi}\frac{\partial a^{-1/2}}{\partial a}=-\sqrt{\pi}\left(-\frac{1}{2}\right)a^{-3/2}=\sqrt{\frac{\pi}{4a^{3}}} I = − ∂ a ∂ a π = − π ∂ a ∂ a − 1/2 = − π ( − 2 1 ) a − 3/2 = 4 a 3 π Então:
1 = ∣ A 1 ∣ 2 ( m ω ℏ π ) 1 / 2 ( 2 m ω ℏ ) 1 2 π ( m ω ℏ ) 3 = ∣ A 1 ∣ 2 ( m ω ℏ ) 3 / 2 ( ℏ m ω ) 3 / 2 = ∣ A 1 ∣ 2 \begin{align}
1 & =\left|A_{1}\right|^{2}\left(\frac{m\omega}{\hbar\pi}\right)^{1/2}\left(\frac{2m\omega}{\hbar}\right)\frac{1}{2}\sqrt{\frac{\pi}{\left(\frac{m\omega}{\hbar}\right)^{3}}}\\
& =\left|A_{1}\right|^{2}\left(\frac{m\omega}{\hbar}\right)^{3/2}\left(\frac{\hbar}{m\omega}\right)^{3/2}\\
& =\left|A_{1}\right|^{2}
\end{align} 1 = ∣ A 1 ∣ 2 ( ℏ π mω ) 1/2 ( ℏ 2 mω ) 2 1 ( ℏ mω ) 3 π = ∣ A 1 ∣ 2 ( ℏ mω ) 3/2 ( mω ℏ ) 3/2 = ∣ A 1 ∣ 2 Então A 1 = 1 A_{1}=1 A 1 = 1 a ^ + \widehat{a}_{+} a + ψ n ± 1 \psi_{n\pm1} ψ n ± 1
a ^ + ψ n = c n ψ n + 1 , a ^ − ψ n = d n ψ n − 1 \widehat{a}_{+}\psi_{n}=c_{n}\psi_{n+1},\quad\widehat{a}_{-}\psi_{n}=d_{n}\psi_{n-1} a + ψ n = c n ψ n + 1 , a − ψ n = d n ψ n − 1 Agora queremos obter as constantes de proporcionalidades. Vamos começar provando que para quaisquer funções f ( x ) f\left(x\right) f ( x ) g ( x ) g\left(x\right) g ( x )
Exemplo: Prove que
∫ − ∞ + ∞ f ∗ ( a ^ ± g ) d x = ∫ − ∞ + ∞ ( a ^ ∓ f ) ∗ g d x \int^{+\infty}_{-\infty}f^{*}\left(\widehat{a}_{\pm}g\right)dx=\int^{+\infty}_{-\infty}\left(\widehat{a}_{\mp}f\right)^{*}gdx ∫ − ∞ + ∞ f ∗ ( a ± g ) d x = ∫ − ∞ + ∞ ( a ∓ f ) ∗ g d x Então lembrando que a ^ ± ≡ 1 2 ℏ m ω ( ∓ i p ^ + m ω x ) \widehat{a}_{\pm}\equiv\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\widehat{p}+m\omega x\right) a ± ≡ 2ℏ mω 1 ( ∓ i p + mω x ) p ^ = − i ℏ d / d x \widehat{p}=-i\hbar d/dx p = − i ℏ d / d x
∫ − ∞ + ∞ f ∗ ( a ^ ± g ) d x = ∫ − ∞ + ∞ f ∗ ( 1 2 ℏ m ω ( ∓ i p ^ + m ω x ) ) g d x = 1 2 ℏ m ω ∫ − ∞ + ∞ f ∗ ( ( ∓ i [ − i ℏ d d x ] + m ω x ) g ) d x = 1 2 ℏ m ω ∫ − ∞ + ∞ f ∗ ( ∓ ℏ d d x + m ω x ) g d x = 1 2 ℏ m ω ( ∓ ℏ ∫ − ∞ + ∞ f ∗ d g d x d x + m ω ∫ − ∞ + ∞ f ∗ x g d x ) \begin{align}
\int^{+\infty}_{-\infty}f^{*}\left(\widehat{a}_{\pm}g\right)dx & =\int^{+\infty}_{-\infty}f^{*}\left(\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\widehat{p}+m\omega x\right)\right)gdx\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\int^{+\infty}_{-\infty}f^{*}\left(\left(\mp i\left[-i\hbar\frac{d}{dx}\right]+m\omega x\right)g\right)dx\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\int^{+\infty}_{-\infty}f^{*}\left(\mp\hbar\frac{d}{dx}+m\omega x\right)gdx\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp\hbar\int^{+\infty}_{-\infty}f^{*}\frac{dg}{dx}dx+m\omega\int^{+\infty}_{-\infty}f^{*}xgdx\right)
\end{align} ∫ − ∞ + ∞ f ∗ ( a ± g ) d x = ∫ − ∞ + ∞ f ∗ ( 2ℏ mω 1 ( ∓ i p + mω x ) ) g d x = 2ℏ mω 1 ∫ − ∞ + ∞ f ∗ ( ( ∓ i [ − i ℏ d x d ] + mω x ) g ) d x = 2ℏ mω 1 ∫ − ∞ + ∞ f ∗ ( ∓ ℏ d x d + mω x ) g d x = 2ℏ mω 1 ( ∓ ℏ ∫ − ∞ + ∞ f ∗ d x d g d x + mω ∫ − ∞ + ∞ f ∗ xg d x ) Aplicando então integral por partes na primeira integral sendo f ( x ) = f ∗ f\left(x\right)=f^{*} f ( x ) = f ∗
∫ a b f ( x ) g ′ ( x ) d x = [ f ( x ) g ( x ) ] a b − ∫ a b f ′ ( x ) g ( x ) d x ∫ − ∞ + ∞ f ∗ d g d x d x = [ f ∗ g ] − ∞ + ∞ − ∫ − ∞ + ∞ d f ∗ d x g 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}f'\left(x\right)g\left(x\right)dx\\
\int^{+\infty}_{-\infty}f^{*}\frac{dg}{dx}dx & =\left[f^{*}g\right]^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\frac{df^{*}}{dx}gdx
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ − ∞ + ∞ f ∗ d x d g d x = [ f ( x ) g ( x ) ] a b − ∫ a b f ′ ( x ) g ( x ) d x = [ f ∗ g ] − ∞ + ∞ − ∫ − ∞ + ∞ d x d f ∗ g d x Na verdade, f ( x ) f\left(x\right) f ( x ) g ( x ) g\left(x\right) g ( x ) ± ∞ \pm\infty ± ∞ [3]
∫ − ∞ + ∞ f ∗ d g d x d x = − ∫ − ∞ + ∞ d f ∗ d x g d x \begin{align}
\int^{+\infty}_{-\infty}f^{*}\frac{dg}{dx}dx & =-\int^{+\infty}_{-\infty}\frac{df^{*}}{dx}gdx
\end{align} ∫ − ∞ + ∞ f ∗ d x d g d x = − ∫ − ∞ + ∞ d x d f ∗ g d x Então retomando:
∫ − ∞ + ∞ f ∗ ( a ^ ± g ) d x = 1 2 ℏ m ω ( ∓ ℏ ∫ − ∞ + ∞ f ∗ d g d x d x + m ω ∫ − ∞ + ∞ f ∗ x g d x ) = 1 2 ℏ m ω ( ± ℏ ∫ − ∞ + ∞ d f ∗ d x g d x + m ω ∫ − ∞ + ∞ f ∗ x g d x ) = 1 2 ℏ m ω ∫ − ∞ + ∞ [ ( ± ℏ d d x + m ω x ) f ∗ ] g d x \begin{align}
\int^{+\infty}_{-\infty}f^{*}\left(\widehat{a}_{\pm}g\right)dx & =\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp\hbar\int^{+\infty}_{-\infty}f^{*}\frac{dg}{dx}dx+m\omega\int^{+\infty}_{-\infty}f^{*}xgdx\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(\pm\hbar\int^{+\infty}_{-\infty}\frac{df^{*}}{dx}gdx+m\omega\int^{+\infty}_{-\infty}f^{*}xgdx\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\int^{+\infty}_{-\infty}\left[\left(\pm\hbar\frac{d}{dx}+m\omega x\right)f^{*}\right]gdx
\end{align} ∫ − ∞ + ∞ f ∗ ( a ± g ) d x = 2ℏ mω 1 ( ∓ ℏ ∫ − ∞ + ∞ f ∗ d x d g d x + mω ∫ − ∞ + ∞ f ∗ xg d x ) = 2ℏ mω 1 ( ± ℏ ∫ − ∞ + ∞ d x d f ∗ g d x + mω ∫ − ∞ + ∞ f ∗ xg d x ) = 2ℏ mω 1 ∫ − ∞ + ∞ [ ( ± ℏ d x d + mω x ) f ∗ ] g d x E como:
a ^ ± = 1 2 ℏ m ω ( ∓ i p ^ + m ω x ) = 1 2 ℏ m ω ( ∓ i ( − i ℏ d d x ) + m ω x ) 1 2 ℏ m ω ( ∓ ℏ d d x + m ω x ) \begin{align}
\widehat{a}_{\pm} & =\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\widehat{p}+m\omega x\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\left(-i\hbar\frac{d}{dx}\right)+m\omega x\right)\\
& \frac{1}{\sqrt{2\hbar m\omega}}\left(\mp\hbar\frac{d}{dx}+m\omega x\right)
\end{align} a ± = 2ℏ mω 1 ( ∓ i p + mω x ) = 2ℏ mω 1 ( ∓ i ( − i ℏ d x d ) + mω x ) 2ℏ mω 1 ( ∓ ℏ d x d + mω x ) Evidentemente então a ^ ± = a ^ ± ∗ \widehat{a}_{\pm}=\widehat{a}^{*}_{\pm} a ± = a ± ∗
∫ − ∞ + ∞ f ∗ ( a ^ ± g ) d x = ∫ − ∞ + ∞ [ 1 2 ℏ m ω ( ± ℏ d d x + m ω x ) f ∗ ] g d x = ∫ − ∞ + ∞ a ^ ∓ f ∗ g d x = ∫ − ∞ + ∞ a ^ ∓ ∗ f ∗ g d x ∫ − ∞ + ∞ ( a ^ ∓ f ) ∗ g d x \begin{align}
\int^{+\infty}_{-\infty}f^{*}\left(\widehat{a}_{\pm}g\right)dx & =\int^{+\infty}_{-\infty}\left[\frac{1}{\sqrt{2\hbar m\omega}}\left(\pm\hbar\frac{d}{dx}+m\omega x\right)f^{*}\right]gdx\\
& =\int^{+\infty}_{-\infty}\widehat{a}_{\mp}f^{*}gdx\\
& =\int^{+\infty}_{-\infty}\widehat{a}^{*}_{\mp}f^{*}gdx\\
& \int^{+\infty}_{-\infty}\left(\widehat{a}_{\mp}f\right)^{*}gdx
\end{align} ∫ − ∞ + ∞ f ∗ ( a ± g ) d x = ∫ − ∞ + ∞ [ 2ℏ mω 1 ( ± ℏ d x d + mω x ) f ∗ ] g d x = ∫ − ∞ + ∞ a ∓ f ∗ g d x = ∫ − ∞ + ∞ a ∓ ∗ f ∗ g d x ∫ − ∞ + ∞ ( a ∓ f ) ∗ g d x Ou seja:
∫ − ∞ + ∞ f ∗ ( a ^ ± g ) d x = ∫ − ∞ + ∞ ( a ^ ∓ f ) ∗ g d x \int^{+\infty}_{-\infty}f^{*}\left(\widehat{a}_{\pm}g\right)dx=\int^{+\infty}_{-\infty}\left(\widehat{a}_{\mp}f\right)^{*}gdx ∫ − ∞ + ∞ f ∗ ( a ± g ) d x = ∫ − ∞ + ∞ ( a ∓ f ) ∗ g d x Veremos melhor depois, mas isso significa na linguagem da álgebra linear que a ^ ∓ \widehat{a}_{\mp} a ∓ a ^ ± \widehat{a}_{\pm} a ± ψ ± n = a ^ ± ψ n \psi_{\pm n}=\widehat{a}_{\pm}\psi_{n} ψ ± n = a ± ψ n
I = ∫ − ∞ + ∞ ∣ ψ ± n ∣ 2 = ∫ − ∞ + ∞ ∣ a ^ ± ψ n ∣ 2 d x = ∫ − ∞ + ∞ ( a ^ ± ψ n ) ∗ ( a ^ ± ψ n ) d x = ∫ − ∞ + ∞ ( a ^ ∓ a ^ ± ψ n ) ∗ ( ψ n ) I=\int^{+\infty}_{-\infty}\left|\psi_{\pm n}\right|^{2}=\int^{+\infty}_{-\infty}\left|\widehat{a}_{\pm}\psi_{n}\right|^{2}dx=\int^{+\infty}_{-\infty}\left(\widehat{a}_{\pm}\psi_{n}\right)^{*}\left(\widehat{a}_{\pm}\psi_{n}\right)dx=\int^{+\infty}_{-\infty}\left(\widehat{a}_{\mp}\widehat{a}_{\pm}\psi_{n}\right)^{*}\left(\psi_{n}\right) I = ∫ − ∞ + ∞ ∣ ψ ± n ∣ 2 = ∫ − ∞ + ∞ ∣ a ± ψ n ∣ 2 d x = ∫ − ∞ + ∞ ( a ± ψ n ) ∗ ( a ± ψ n ) d x = ∫ − ∞ + ∞ ( a ∓ a ± ψ n ) ∗ ( ψ n ) Agora somos apresentado aos resultados:
a ^ + a ^ − ψ n = n ψ n , a ^ − a ^ + ψ n = ( n + 1 ) ψ n \widehat{a}_{+}\widehat{a}_{-}\psi_{n}=n\psi_{n},\qquad\widehat{a}_{-}\widehat{a}_{+}\psi_{n}=\left(n+1\right)\psi_{n} a + a − ψ n = n ψ n , a − a + ψ n = ( n + 1 ) ψ n Onde é alegado que usamos, ℏ ω ( a ^ ± a ^ ∓ ± 1 2 ) ψ = E ψ \hbar\omega\left(\widehat{a}_{\pm}\widehat{a}_{\mp}\pm\frac{1}{2}\right)\psi=E\psi ℏ ω ( a ± a ∓ ± 2 1 ) ψ = E ψ ψ n ( x ) = A n ( a ^ + ) n ψ 0 ( x ) \psi_{n}\left(x\right)=A_{n}\left(\widehat{a}_{+}\right)^{n}\psi{}_{0}\left(x\right) ψ n ( x ) = A n ( a + ) n ψ 0 ( x ) E n = ( n + 1 2 ) ℏ ω E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega E n = ( n + 2 1 ) ℏ ω
ℏ ω ( a ^ ± a ^ ∓ ± 1 2 ) ψ n = E n ψ n ℏ ω ( a ^ ± a ^ ∓ ± 1 2 ) ψ n = ℏ ω ( n + 1 2 ) ψ n ( a ^ ± a ^ ∓ ± 1 2 ) ψ n = ( n + 1 2 ) ψ n \begin{align}
\hbar\omega\left(\widehat{a}_{\pm}\widehat{a}_{\mp}\pm\frac{1}{2}\right)\psi_{n} & =E_{n}\psi_{n}\\
\hbar\omega\left(\widehat{a}_{\pm}\widehat{a}_{\mp}\pm\frac{1}{2}\right)\psi_{n} & =\hbar\omega\left(n+\frac{1}{2}\right)\psi_{n}\\
\left(\widehat{a}_{\pm}\widehat{a}_{\mp}\pm\frac{1}{2}\right)\psi_{n} & =\left(n+\frac{1}{2}\right)\psi_{n}
\end{align} ℏ ω ( a ± a ∓ ± 2 1 ) ψ n ℏ ω ( a ± a ∓ ± 2 1 ) ψ n ( a ± a ∓ ± 2 1 ) ψ n = E n ψ n = ℏ ω ( n + 2 1 ) ψ n = ( n + 2 1 ) ψ n Então:
( a ^ + a ^ − + 1 2 ) ψ n = ( n + 1 2 ) ψ n → a ^ + a ^ − ψ n = n ψ n ( a ^ − a ^ + − 1 2 ) ψ n = ( n + 1 2 ) ψ n → a ^ − a ^ + ψ n = ( n + 1 ) ψ n \begin{align}
\left(\widehat{a}_{+}\widehat{a}_{-}+\frac{1}{2}\right)\psi_{n} & =\left(n+\frac{1}{2}\right)\psi_{n} & \rightarrow\widehat{a}_{+}\widehat{a}_{-}\psi_{n} & =n\psi_{n}\\
\left(\widehat{a}_{-}\widehat{a}_{+}-\frac{1}{2}\right)\psi_{n} & =\left(n+\frac{1}{2}\right)\psi_{n} & \rightarrow\widehat{a}_{-}\widehat{a}_{+}\psi_{n} & =\left(n+1\right)\psi_{n}
\end{align} ( a + a − + 2 1 ) ψ n ( a − a + − 2 1 ) ψ n = ( n + 2 1 ) ψ n = ( n + 2 1 ) ψ n → a + a − ψ n → a − a + ψ n = n ψ n = ( n + 1 ) ψ n Então retomando a integral:
I = { ∫ − ∞ + ∞ ( a ^ − a ^ + ψ n ) ∗ ( ψ n ) d x = ∫ − ∞ + ∞ ( [ n + 1 ] ψ n ) ∗ ( ψ n ) d x ∫ − ∞ + ∞ ( a ^ + a ^ − ψ n ) ∗ ( ψ n ) d x = ∫ − ∞ + ∞ ( n ψ n ) ∗ ( ψ n ) d x I=\begin{cases}
\int^{+\infty}_{-\infty}\left(\widehat{a}_{-}\widehat{a}_{+}\psi_{n}\right)^{*}\left(\psi_{n}\right)dx & =\int^{+\infty}_{-\infty}\left(\left[n+1\right]\psi_{n}\right)^{*}\left(\psi_{n}\right)dx\\
\int^{+\infty}_{-\infty}\left(\widehat{a}_{+}\widehat{a}_{-}\psi_{n}\right)^{*}\left(\psi_{n}\right)dx & =\int^{+\infty}_{-\infty}\left(n\psi_{n}\right)^{*}\left(\psi_{n}\right)dx
\end{cases} I = { ∫ − ∞ + ∞ ( a − a + ψ n ) ∗ ( ψ n ) d x ∫ − ∞ + ∞ ( a + a − ψ n ) ∗ ( ψ n ) d x = ∫ − ∞ + ∞ ( [ n + 1 ] ψ n ) ∗ ( ψ n ) d x = ∫ − ∞ + ∞ ( n ψ n ) ∗ ( ψ n ) d x Ou então apenas:
∫ − ∞ + ∞ ∣ ψ + n ∣ 2 d x = ( n + 1 ) ∫ − ∞ + ∞ ( ψ n ) ∗ ( ψ n ) d x = ( n + 1 ) ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x ∫ − ∞ + ∞ ∣ ψ − n ∣ 2 d x = n ∫ − ∞ + ∞ ( ψ n ) ∗ ( ψ n ) d x = n ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x \begin{align}
\int^{+\infty}_{-\infty}\left|\psi_{+n}\right|^{2}dx & =\left(n+1\right)\int^{+\infty}_{-\infty}\left(\psi_{n}\right)^{*}\left(\psi_{n}\right)dx & = & \left(n+1\right)\int^{+\infty}_{-\infty}\left|\psi_{n}\right|^{2}dx\\
\int^{+\infty}_{-\infty}\left|\psi_{-n}\right|^{2}dx & =n\int^{+\infty}_{-\infty}\left(\psi_{n}\right)^{*}\left(\psi_{n}\right)dx & = & n\int^{+\infty}_{-\infty}\left|\psi_{n}\right|^{2}dx
\end{align} ∫ − ∞ + ∞ ∣ ψ + n ∣ 2 d x ∫ − ∞ + ∞ ∣ ψ − n ∣ 2 d x = ( n + 1 ) ∫ − ∞ + ∞ ( ψ n ) ∗ ( ψ n ) d x = n ∫ − ∞ + ∞ ( ψ n ) ∗ ( ψ n ) d x = = ( n + 1 ) ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x n ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x Evidentemente, outra forma de vermos isso é usando nossas definições iniciais a ^ + ψ n = c n ψ n + 1 \widehat{a}_{+}\psi_{n}=c_{n}\psi_{n+1} a + ψ n = c n ψ n + 1 a ^ − ψ n = d n ψ n − 1 \widehat{a}_{-}\psi_{n}=d_{n}\psi_{n-1} a − ψ n = d n ψ n − 1
∫ − ∞ + ∞ ∣ ψ + n ∣ 2 d x = ∫ − ∞ + ∞ ∣ c n ψ n + 1 ∣ 2 d x = ∣ c n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x ∫ − ∞ + ∞ ∣ ψ − n ∣ 2 d x = ∫ − ∞ + ∞ ∣ d n ψ n − 1 ∣ 2 d x = ∣ d n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n − 1 ∣ 2 d x \begin{align}
\int^{+\infty}_{-\infty}\left|\psi_{+n}\right|^{2} & dx=\int^{+\infty}_{-\infty}\left|c_{n}\psi_{n+1}\right|^{2}dx=\left|c_{n}\right|^{2}\int^{+\infty}_{-\infty}\left|\psi_{n+1}\right|^{2}dx\\
\int^{+\infty}_{-\infty}\left|\psi_{-n}\right|^{2} & dx=\int^{+\infty}_{-\infty}\left|d_{n}\psi_{n-1}\right|^{2}dx=\left|d_{n}\right|^{2}\int^{+\infty}_{-\infty}\left|\psi_{n-1}\right|^{2}dx
\end{align} ∫ − ∞ + ∞ ∣ ψ + n ∣ 2 ∫ − ∞ + ∞ ∣ ψ − n ∣ 2 d x = ∫ − ∞ + ∞ ∣ c n ψ n + 1 ∣ 2 d x = ∣ c n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x d x = ∫ − ∞ + ∞ ∣ d n ψ n − 1 ∣ 2 d x = ∣ d n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n − 1 ∣ 2 d x Ou seja:
( n + 1 ) ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x = ∣ c n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x n ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x = ∣ d n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x \begin{align}
\left(n+1\right)\int^{+\infty}_{-\infty}\left|\psi_{n}\right|^{2}dx & =\left|c_{n}\right|^{2}\int^{+\infty}_{-\infty}\left|\psi_{n+1}\right|^{2}dx\\
n\int^{+\infty}_{-\infty}\left|\psi_{n}\right|^{2} & dx=\left|d_{n}\right|^{2}\int^{+\infty}_{-\infty}\left|\psi_{n+1}\right|^{2}dx
\end{align} ( n + 1 ) ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x n ∫ − ∞ + ∞ ∣ ψ n ∣ 2 = ∣ c n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x d x = ∣ d n ∣ 2 ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x Mas se todas as funções de onda (ψ n \psi_{n} ψ n ψ n + 1 \psi_{n+1} ψ n + 1 ψ n − 1 \psi_{n-1} ψ n − 1
∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x = ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x = ∫ − ∞ + ∞ ∣ ψ n − 1 ∣ 2 d x = 1 \int^{+\infty}_{-\infty}\left|\psi_{n}\right|^{2}dx=\int^{+\infty}_{-\infty}\left|\psi_{n+1}\right|^{2}dx=\int^{+\infty}_{-\infty}\left|\psi_{n-1}\right|^{2}dx=1 ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x = ∫ − ∞ + ∞ ∣ ψ n + 1 ∣ 2 d x = ∫ − ∞ + ∞ ∣ ψ n − 1 ∣ 2 d x = 1 Ou seja, escolhendo coeficientes de normalização reais:
( n + 1 ) = ∣ c n ∣ 2 n = ∣ d n ∣ 2 n + 1 = c n n = d n \begin{align}
\left(n+1\right) & =\left|c_{n}\right|^{2} & n & =\left|d_{n}\right|^{2}\\
\sqrt{n+1} & =c_{n} & \sqrt{n} & =d_{n}
\end{align} ( n + 1 ) n + 1 = ∣ c n ∣ 2 = c n n n = ∣ d n ∣ 2 = d n Então, as nossas definições iniciais, temos:
a ^ + ψ n = c n ψ n + 1 a ^ − ψ n = d n ψ n − 1 = n + 1 ψ n + 1 = n ψ n − 1 \begin{align}
\widehat{a}_{+}\psi_{n} & =c_{n}\psi_{n+1} & \widehat{a}_{-}\psi_{n} & =d_{n}\psi_{n-1}\\
& =\sqrt{n+1}\psi_{n+1} & & =\sqrt{n}\psi_{n-1}
\end{align} a + ψ n = c n ψ n + 1 = n + 1 ψ n + 1 a − ψ n = d n ψ n − 1 = n ψ n − 1 Podemos ainda ir além, isso ficará claro depois de analisarmos os primeiros valores de n n n ψ 0 \psi_{0} ψ 0 ψ n + 1 \psi_{n+1} ψ n + 1 a ^ + \widehat{a}_{+} a +
ψ n + 1 = 1 n + 1 ( a ^ + ψ n ) \psi_{n+1}=\frac{1}{\sqrt{n+1}}\left(\widehat{a}_{+}\psi_{n}\right) ψ n + 1 = n + 1 1 ( a + ψ n ) Então partindo de ψ 0 \psi_{0} ψ 0
ψ 0 = ψ 0 ψ 1 = 1 0 + 1 ( a ^ + ψ 0 ) = a ^ + ψ 0 ψ 2 = 1 1 + 1 ( a ^ + ψ 1 ) = 1 2 a ^ + ψ 1 = 1 2 a ^ + ( a ^ + ψ 0 ) = 1 2 ⋅ 1 ( a ^ + ) 2 ψ 0 = 1 2 ! ( a ^ + ) 2 ψ 0 ψ 3 = 1 2 + 1 ( a ^ + ψ 2 ) = 1 3 a ^ + ψ 2 = 1 3 a ^ + ( 1 2 ! ( a ^ + ) 2 ψ 0 ) = 1 3 ! ( a ^ + ) 3 ψ 0 \begin{align}
\psi_{0} & =\psi_{0}\\
\psi_{1} & =\frac{1}{\sqrt{0+1}}\left(\widehat{a}_{+}\psi_{0}\right)=\widehat{a}_{+}\psi_{0}\\
\psi_{2} & =\frac{1}{\sqrt{1+1}}\left(\widehat{a}_{+}\psi_{1}\right)=\frac{1}{\sqrt{2}}\widehat{a}_{+}\psi_{1}=\frac{1}{\sqrt{2}}\widehat{a}_{+}\left(\widehat{a}_{+}\psi_{0}\right)=\frac{1}{\sqrt{2\cdot1}}\left(\widehat{a}_{+}\right)^{2}\psi_{0}=\frac{1}{\sqrt{2!}}\left(\widehat{a}_{+}\right)^{2}\psi_{0}\\
\psi_{3} & =\frac{1}{\sqrt{2+1}}\left(\widehat{a}_{+}\psi_{2}\right)=\frac{1}{\sqrt{3}}\widehat{a}_{+}\psi_{2}=\frac{1}{\sqrt{3}}\widehat{a}_{+}\left(\frac{1}{\sqrt{2!}}\left(\widehat{a}_{+}\right)^{2}\psi_{0}\right)=\frac{1}{\sqrt{3!}}\left(\widehat{a}_{+}\right)^{3}\psi_{0}
\end{align} ψ 0 ψ 1 ψ 2 ψ 3 = ψ 0 = 0 + 1 1 ( a + ψ 0 ) = a + ψ 0 = 1 + 1 1 ( a + ψ 1 ) = 2 1 a + ψ 1 = 2 1 a + ( a + ψ 0 ) = 2 ⋅ 1 1 ( a + ) 2 ψ 0 = 2 ! 1 ( a + ) 2 ψ 0 = 2 + 1 1 ( a + ψ 2 ) = 3 1 a + ψ 2 = 3 1 a + ( 2 ! 1 ( a + ) 2 ψ 0 ) = 3 ! 1 ( a + ) 3 ψ 0 Então:
ψ n = 1 n ! ( a ^ + ) n ψ 0 \psi_{n}=\frac{1}{\sqrt{n!}}\left(\widehat{a}_{+}\right)^{n}\psi_{0} ψ n = n ! 1 ( a + ) n ψ 0 Lembrando que ψ 0 ( x ) = ( m ω ℏ π ) 1 / 4 e − m ω 2 ℏ x 2 \psi_{0}\left(x\right)=\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^{2}} ψ 0 ( x ) = ( ℏ π mω ) 1/4 e − 2ℏ mω x 2
ψ n ( x ) = 1 n ! ( m ω ℏ π ) 1 / 4 ( a ^ + ) n e − m ω 2 ℏ x 2 \psi_{n}\left(x\right)=\frac{1}{\sqrt{n!}}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\left(\widehat{a}_{+}\right)^{n}e^{-\frac{m\omega}{2\hbar}x^{2}} ψ n ( x ) = n ! 1 ( ℏ π mω ) 1/4 ( a + ) n e − 2ℏ mω x 2 Ou ainda:
ψ n ( x ) = 1 n ! ( m ω ℏ π ) 1 / 4 ( 1 2 ℏ m ω ( − ℏ d d x + m ω x ) ) n e − m ω 2 ℏ x 2 \psi_{n}\left(x\right)=\frac{1}{\sqrt{n!}}\left(\frac{m\omega}{\hbar\pi}\right)^{1/4}\left(\frac{1}{\sqrt{2\hbar m\omega}}\left(-\hbar\frac{d}{dx}+m\omega x\right)\right)^{n}e^{-\frac{m\omega}{2\hbar}x^{2}} ψ n ( x ) = n ! 1 ( ℏ π mω ) 1/4 ( 2ℏ mω 1 ( − ℏ d x d + mω x ) ) n e − 2ℏ mω x 2 Temos novamente um caso em que os estados são ortogonais, ou seja:
∫ − ∞ + ∞ ψ m ∗ ψ n d x = δ m n \int^{+\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx=\delta_{mn} ∫ − ∞ + ∞ ψ m ∗ ψ n d x = δ mn Primeiro usando a ^ + a ^ − ψ n = n ψ n \widehat{a}_{+}\widehat{a}_{-}\psi_{n}=n\psi_{n} a + a − ψ n = n ψ n
∫ − ∞ ∞ ψ m ∗ ( a ^ + a ^ − ) ψ n d x = ∫ − ∞ ∞ ψ m ∗ ( a ^ a ^ − ) ψ n d x = ∫ − ∞ ∞ ψ m ∗ n ψ n d x = n ∫ − ∞ ∞ ψ m ∗ ψ n d x \int^{\infty}_{-\infty}\psi^{*}_{m}\left(\widehat{a}_{+}\widehat{a}_{-}\right)\psi_{n}dx=\int^{\infty}_{-\infty}\psi^{*}_{m}\left(\widehat{a}\widehat{a}_{-}\right)\psi_{n}dx=\int^{\infty}_{-\infty}\psi^{*}_{m}n\psi_{n}dx=n\int^{\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx ∫ − ∞ ∞ ψ m ∗ ( a + a − ) ψ n d x = ∫ − ∞ ∞ ψ m ∗ ( a a − ) ψ n d x = ∫ − ∞ ∞ ψ m ∗ n ψ n d x = n ∫ − ∞ ∞ ψ m ∗ ψ n d x Mas usando :∫ − ∞ + ∞ f ∗ ( a ^ ± g ) d x = ∫ − ∞ + ∞ ( a ^ ∓ f ) ∗ g d x \int^{+\infty}_{-\infty}f^{*}\left(\widehat{a}_{\pm}g\right)dx=\int^{+\infty}_{-\infty}\left(\widehat{a}_{\mp}f\right)^{*}gdx ∫ − ∞ + ∞ f ∗ ( a ± g ) d x = ∫ − ∞ + ∞ ( a ∓ f ) ∗ g d x
∫ − ∞ ∞ ψ m ∗ ( a ^ + a ^ − ) ψ n d x = ∫ − ∞ ∞ ψ m ∗ ( a ^ + [ a ^ − ψ n ] ) d x = ∫ − ∞ ∞ ( a ^ − ψ m ) ∗ ( a ^ − ψ n ) d x = ∫ − ∞ ∞ ( a ^ + a ^ − ψ m ) ∗ ψ n d x \begin{align}
\int^{\infty}_{-\infty}\psi^{*}_{m}\left(\widehat{a}_{+}\widehat{a}_{-}\right)\psi_{n}dx= & \int^{\infty}_{-\infty}\psi^{*}_{m}\left(\widehat{a}_{+}\left[\widehat{a}_{-}\psi_{n}\right]\right)dx\\
= & \int^{\infty}_{-\infty}\left(\widehat{a}_{-}\psi_{m}\right)^{*}\left(\widehat{a}_{-}\psi_{n}\right)dx\\
= & \int^{\infty}_{-\infty}\left(\widehat{a}_{+}\widehat{a}_{-}\psi_{m}\right)^{*}\psi_{n}dx
\end{align} ∫ − ∞ ∞ ψ m ∗ ( a + a − ) ψ n d x = = = ∫ − ∞ ∞ ψ m ∗ ( a + [ a − ψ n ] ) d x ∫ − ∞ ∞ ( a − ψ m ) ∗ ( a − ψ n ) d x ∫ − ∞ ∞ ( a + a − ψ m ) ∗ ψ n d x Ou seja:
∫ − ∞ ∞ ψ m ∗ ( a ^ + a ^ − ) ψ n d x = ∫ − ∞ ∞ ( m ψ m ) ∗ ψ n d x = m ∫ − ∞ ∞ ψ m ∗ ψ n d x \int^{\infty}_{-\infty}\psi^{*}_{m}\left(\widehat{a}_{+}\widehat{a}_{-}\right)\psi_{n}dx=\int^{\infty}_{-\infty}\left(m\psi_{m}\right)^{*}\psi_{n}dx=m\int^{\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx ∫ − ∞ ∞ ψ m ∗ ( a + a − ) ψ n d x = ∫ − ∞ ∞ ( m ψ m ) ∗ ψ n d x = m ∫ − ∞ ∞ ψ m ∗ ψ n d x Evidentemente como 81 83
n ∫ − ∞ ∞ ψ m ∗ ψ n d x = m ∫ − ∞ ∞ ψ m ∗ ψ n d x 0 = ( m − n ) ∫ − ∞ ∞ ψ m ∗ ψ n d x \begin{align}
n\int^{\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx & =m\int^{\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx\\
0 & =\left(m-n\right)\int^{\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx
\end{align} n ∫ − ∞ ∞ ψ m ∗ ψ n d x 0 = m ∫ − ∞ ∞ ψ m ∗ ψ n d x = ( m − n ) ∫ − ∞ ∞ ψ m ∗ ψ n d x Então se m ≠ n m\neq n m = n ∫ − ∞ ∞ ψ m ∗ ψ n d x = 0 \int^{\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx=0 ∫ − ∞ ∞ ψ m ∗ ψ n d x = 0 m = n , m=n, m = n ,
∫ − ∞ + ∞ ψ m ∗ ψ n d x = ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x = 1 \int^{+\infty}_{-\infty}\psi^{*}_{m}\psi_{n}dx=\int^{+\infty}_{-\infty}\left|\psi_{n}\right|^{2}dx=1 ∫ − ∞ + ∞ ψ m ∗ ψ n d x = ∫ − ∞ + ∞ ∣ ψ n ∣ 2 d x = 1 Que é essencialmente o critério de normalização da função de onda já discutido. A ortonormalidade nos permite usar novamente o truque de Fourier para avaliar os coeficientes c n c_{n} c n ∣ c n ∣ 2 \left|c_{n}\right|^{2} ∣ c n ∣ 2 E n E_{n} E n
Exemplo: Encontre o valor esperado da energia potencial no enésimo estado do oscilador harmônico.
A anergia potencial[4]
⟨ V ⟩ = ⟨ 1 2 m ω 2 x 2 ⟩ = ∫ − ∞ + ∞ ψ n ∗ ( 1 2 m ω 2 x 2 ) ψ n d x \left\langle V\right\rangle =\left\langle \frac{1}{2}m\omega^{2}x^{2}\right\rangle =\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\frac{1}{2}m\omega^{2}x^{2}\right)\psi_{n}dx ⟨ V ⟩ = ⟨ 2 1 m ω 2 x 2 ⟩ = ∫ − ∞ + ∞ ψ n ∗ ( 2 1 m ω 2 x 2 ) ψ n d x Ou apenas:
⟨ V ⟩ = 1 2 m ω 2 ∫ − ∞ + ∞ ψ n ∗ x 2 ψ n d x \left\langle V\right\rangle =\frac{1}{2}m\omega^{2}\int^{+\infty}_{-\infty}\psi^{*}_{n}x^{2}\psi_{n}dx ⟨ V ⟩ = 2 1 m ω 2 ∫ − ∞ + ∞ ψ n ∗ x 2 ψ n d x Mas agora podemos usar um truque lembrando que na equação 17 a ^ ± ≡ 1 2 ℏ m ω ( ∓ i p ^ + m ω x ) \widehat{a}_{\pm}\equiv\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\widehat{p}+m\omega x\right) a ± ≡ 2ℏ mω 1 ( ∓ i p + mω x )
Então:
a ^ + + a ^ − = 1 2 ℏ m ω ( − i p ^ + m ω x ) + 1 2 ℏ m ω ( + i p ^ + m ω x ) = 2 m ω 2 ℏ m ω x = 2 m ω ℏ x ℏ 2 m ω ( a ^ + + a ^ − ) = x \begin{align}
\widehat{a}_{+}+\widehat{a}_{-} & =\frac{1}{\sqrt{2\hbar m\omega}}\left(-i\widehat{p}+m\omega x\right)+\frac{1}{\sqrt{2\hbar m\omega}}\left(+i\widehat{p}+m\omega x\right)\\
& =\frac{2m\omega}{\sqrt{2\hbar m\omega}}x\\
& =\sqrt{\frac{2m\omega}{\hbar}}x\\
\sqrt{\frac{\hbar}{2m\omega}}\left(\widehat{a}_{+}+\widehat{a}_{-}\right) & =x
\end{align} a + + a − 2 mω ℏ ( a + + a − ) = 2ℏ mω 1 ( − i p + mω x ) + 2ℏ mω 1 ( + i p + mω x ) = 2ℏ mω 2 mω x = ℏ 2 mω x = x De modo análogo poderíamos calcular que:
p ^ = i ℏ m ω 2 ( a ^ + − a ^ − ) \widehat{p}=i\sqrt{\frac{\hbar m\omega}{2}}\left(\widehat{a}_{+}-\widehat{a}_{-}\right) p = i 2 ℏ mω ( a + − a − ) Fazendo então:
x 2 = ( ℏ 2 m ω [ a ^ + + a ^ − ] ) ( ℏ 2 m ω [ a ^ + + a ^ − ] ) = ℏ 2 m ω ( a ^ + 2 + a ^ + a ^ − + a ^ − a ^ + + a ^ − 2 ) \begin{align}
x^{2} & =\left(\sqrt{\frac{\hbar}{2m\omega}}\left[\widehat{a}_{+}+\widehat{a}_{-}\right]\right)\left(\sqrt{\frac{\hbar}{2m\omega}}\left[\widehat{a}_{+}+\widehat{a}_{-}\right]\right)\\
& =\frac{\hbar}{2m\omega}\left(\widehat{a}^{2}_{+}+\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{-}\widehat{a}_{+}+\widehat{a}^{2}_{-}\right)
\end{align} x 2 = ( 2 mω ℏ [ a + + a − ] ) ( 2 mω ℏ [ a + + a − ] ) = 2 mω ℏ ( a + 2 + a + a − + a − a + + a − 2 ) E como:
( a ^ + 2 + a ^ − 2 ) ψ n = a ^ + a ^ + ψ n + a ^ − a ^ − ψ n = a ^ + c n ψ n + 1 + a ^ − d n ψ n − 1 = c n a ^ + ψ n + 1 + d n a ^ − ψ n − 1 = c n c n + 1 ψ n + 2 + d n d n − 1 ψ n − 2 = A ψ n + 2 + B ψ n − 2 \begin{align}
\left(\widehat{a}^{2}_{+}+\widehat{a}^{2}_{-}\right)\psi_{n} & =\widehat{a}_{+}\widehat{a}_{+}\psi_{n}+\widehat{a}_{-}\widehat{a}_{-}\psi_{n}\\
& =\widehat{a}_{+}c_{n}\psi_{n+1}+\widehat{a}_{-}d_{n}\psi_{n-1}\\
& =c_{n}\widehat{a}_{+}\psi_{n+1}+d_{n}\widehat{a}_{-}\psi_{n-1}\\
& =c_{n}c_{n+1}\psi_{n+2}+d_{n}d_{n-1}\psi_{n-2}\\
& =A\psi_{n+2}+B\psi_{n-2}
\end{align} ( a + 2 + a − 2 ) ψ n = a + a + ψ n + a − a − ψ n = a + c n ψ n + 1 + a − d n ψ n − 1 = c n a + ψ n + 1 + d n a − ψ n − 1 = c n c n + 1 ψ n + 2 + d n d n − 1 ψ n − 2 = A ψ n + 2 + B ψ n − 2 Pela ortogonalidade temos:
∫ − ∞ + ∞ ψ n ∗ ( a ^ + 2 + a ^ − 2 ) ψ n d x = ∫ − ∞ + ∞ ψ n ∗ ( A ψ n + 2 + B ψ n − 2 ) d x = A ∫ − ∞ + ∞ ψ n ∗ ψ n + 2 d x + B A ∫ − ∞ + ∞ ψ n ∗ ψ n − 2 d x = A δ n , n + 1 + B δ n , n − 2 = 0 \begin{align}
\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}^{2}_{+}+\widehat{a}^{2}_{-}\right)\psi_{n}dx & =\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(A\psi_{n+2}+B\psi_{n-2}\right)dx\\
& =A\int^{+\infty}_{-\infty}\psi^{*}_{n}\psi_{n+2}dx+BA\int^{+\infty}_{-\infty}\psi^{*}_{n}\psi_{n-2}dx\\
& =A\delta_{n,n+1}+B\delta_{n,n-2}\\
& =0
\end{align} ∫ − ∞ + ∞ ψ n ∗ ( a + 2 + a − 2 ) ψ n d x = ∫ − ∞ + ∞ ψ n ∗ ( A ψ n + 2 + B ψ n − 2 ) d x = A ∫ − ∞ + ∞ ψ n ∗ ψ n + 2 d x + B A ∫ − ∞ + ∞ ψ n ∗ ψ n − 2 d x = A δ n , n + 1 + B δ n , n − 2 = 0 Dessa forma ficamos apenas com:
⟨ V ⟩ = 1 2 m ω 2 ∫ − ∞ + ∞ ψ n ∗ x 2 ψ n d x = 1 2 m ω 2 ∫ − ∞ + ∞ ψ n ∗ [ ℏ 2 m ω ( a ^ + 2 + a ^ + a ^ − + a ^ − a ^ + + a ^ − 2 ) ] ψ n d x = ℏ ω 4 [ ∫ − ∞ + ∞ ψ n ∗ ( a ^ + 2 + a ^ − 2 ) ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( a ^ + a ^ − + a ^ − a ^ + ) ψ n d x ] = ℏ ω 4 ∫ − ∞ + ∞ ψ n ∗ ( a ^ + a ^ − + a ^ − a ^ + ) ψ n d x = ℏ ω 4 [ ∫ − ∞ + ∞ ψ n ∗ ( a ^ + a ^ − ) ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( a ^ − a ^ + ) ψ n d x ] \begin{align}
\left\langle V\right\rangle & =\frac{1}{2}m\omega^{2}\int^{+\infty}_{-\infty}\psi^{*}_{n}x^{2}\psi_{n}dx\\
& =\frac{1}{2}m\omega^{2}\int^{+\infty}_{-\infty}\psi^{*}_{n}\left[\frac{\hbar}{2m\omega}\left(\widehat{a}^{2}_{+}+\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{-}\widehat{a}_{+}+\widehat{a}^{2}_{-}\right)\right]\psi_{n}dx\\
& =\frac{\hbar\omega}{4}\left[\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}^{2}_{+}+\widehat{a}^{2}_{-}\right)\psi_{n}dx+\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{-}\widehat{a}_{+}\right)\psi_{n}dx\right]\\
& =\frac{\hbar\omega}{4}\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{-}\widehat{a}_{+}\right)\psi_{n}dx\\
& =\frac{\hbar\omega}{4}\left[\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}_{+}\widehat{a}_{-}\right)\psi_{n}dx+\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}_{-}\widehat{a}_{+}\right)\psi_{n}dx\right]
\end{align} ⟨ V ⟩ = 2 1 m ω 2 ∫ − ∞ + ∞ ψ n ∗ x 2 ψ n d x = 2 1 m ω 2 ∫ − ∞ + ∞ ψ n ∗ [ 2 mω ℏ ( a + 2 + a + a − + a − a + + a − 2 ) ] ψ n d x = 4 ℏ ω [ ∫ − ∞ + ∞ ψ n ∗ ( a + 2 + a − 2 ) ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( a + a − + a − a + ) ψ n d x ] = 4 ℏ ω ∫ − ∞ + ∞ ψ n ∗ ( a + a − + a − a + ) ψ n d x = 4 ℏ ω [ ∫ − ∞ + ∞ ψ n ∗ ( a + a − ) ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( a − a + ) ψ n d x ] E como em 65 a ^ + a ^ − ψ n = n ψ n \widehat{a}_{+}\widehat{a}_{-}\psi_{n}=n\psi_{n} a + a − ψ n = n ψ n a ^ − a ^ + ψ n = ( n + 1 ) ψ n \widehat{a}_{-}\widehat{a}_{+}\psi_{n}=\left(n+1\right)\psi_{n} a − a + ψ n = ( n + 1 ) ψ n
⟨ V ⟩ = ℏ ω 4 [ ∫ − ∞ + ∞ ψ n ∗ ( a ^ + a ^ − ) ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( a ^ − a ^ + ) ψ n d x ] = ℏ ω 4 [ ∫ − ∞ + ∞ ψ n ∗ n ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( n + 1 ) ψ n d x ] = ℏ ω 4 [ n ∫ − ∞ + ∞ ψ n ∗ ψ n d x + ( n + 1 ) ∫ − ∞ + ∞ ψ n ∗ ψ n d x ] = ℏ ω 4 [ ( n + n + 1 ) ∫ − ∞ + ∞ ψ n ∗ ψ n d x ] = ℏ ω 4 ( 2 n + 1 ) ∫ − ∞ + ∞ ψ n ∗ ψ n d x \begin{align}
\left\langle V\right\rangle & =\frac{\hbar\omega}{4}\left[\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}_{+}\widehat{a}_{-}\right)\psi_{n}dx+\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(\widehat{a}_{-}\widehat{a}_{+}\right)\psi_{n}dx\right]\\
& =\frac{\hbar\omega}{4}\left[\int^{+\infty}_{-\infty}\psi^{*}_{n}n\psi_{n}dx+\int^{+\infty}_{-\infty}\psi^{*}_{n}\left(n+1\right)\psi_{n}dx\right]\\
& =\frac{\hbar\omega}{4}\left[n\int^{+\infty}_{-\infty}\psi^{*}_{n}\psi_{n}dx+\left(n+1\right)\int^{+\infty}_{-\infty}\psi^{*}_{n}\psi_{n}dx\right]\\
& =\frac{\hbar\omega}{4}\left[\left(n+n+1\right)\int^{+\infty}_{-\infty}\psi^{*}_{n}\psi_{n}dx\right]\\
& =\frac{\hbar\omega}{4}\left(2n+1\right)\int^{+\infty}_{-\infty}\psi^{*}_{n}\psi_{n}dx
\end{align} ⟨ V ⟩ = 4 ℏ ω [ ∫ − ∞ + ∞ ψ n ∗ ( a + a − ) ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( a − a + ) ψ n d x ] = 4 ℏ ω [ ∫ − ∞ + ∞ ψ n ∗ n ψ n d x + ∫ − ∞ + ∞ ψ n ∗ ( n + 1 ) ψ n d x ] = 4 ℏ ω [ n ∫ − ∞ + ∞ ψ n ∗ ψ n d x + ( n + 1 ) ∫ − ∞ + ∞ ψ n ∗ ψ n d x ] = 4 ℏ ω [ ( n + n + 1 ) ∫ − ∞ + ∞ ψ n ∗ ψ n d x ] = 4 ℏ ω ( 2 n + 1 ) ∫ − ∞ + ∞ ψ n ∗ ψ n d x E pela normalização da função de onda, ficamos então com:
⟨ V ⟩ = ℏ ω 4 ( 2 n + 1 ) = ℏ ω 2 ( n + 1 2 ) \left\langle V\right\rangle =\frac{\hbar\omega}{4}\left(2n+1\right)=\frac{\hbar\omega}{2}\left(n+\frac{1}{2}\right) ⟨ V ⟩ = 4 ℏ ω ( 2 n + 1 ) = 2 ℏ ω ( n + 2 1 ) Ou seja, o valor esperado da energia potencial é metade da energia total (a outra metade é energia cinética).
Exemplo:
a) Construa ψ 2 ( x ) \psi_{2}\left(x\right) ψ 2 ( x )
b) Verifique a ortogonalidade de ψ 0 \psi_{0} ψ 0 ψ 1 \psi_{1} ψ 1 ψ 2 \psi_{2} ψ 2
Partindo de:
a ^ ± = 1 2 ℏ m ω ( ∓ i p ^ + m ω x ) ψ 0 ( x ) = ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) \widehat{a}_{\pm}=\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp i\widehat{p}+m\omega x\right)\quad\psi_{0}\left(x\right)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right) a ± = 2ℏ mω 1 ( ∓ i p + mω x ) ψ 0 ( x ) = ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) E sendo p ^ = − i ℏ d d x \widehat{p}=-i\hbar\frac{d}{dx} p = − i ℏ d x d
a ^ + ψ 0 = 1 2 ℏ m ω ( − i p ^ + m ω x ) ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) = 1 2 ℏ m ω ( − i [ − i ℏ d d x ] + m ω x ) ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) = 1 2 ℏ m ω ( m ω π ℏ ) 1 / 4 ( − ℏ d d x ( exp ( − m ω 2 ℏ x 2 ) ) + m ω x exp ( − m ω 2 ℏ x 2 ) ) = 1 2 ℏ m ω ( m ω π ℏ ) 1 / 4 ( − ℏ exp ( − m ω 2 ℏ x 2 ) ( − 2 m ω x 2 ℏ ) + m ω x exp ( − m ω 2 ℏ x 2 ) ) = 1 2 ℏ m ω ( m ω π ℏ ) 1 / 4 ( exp ( − m ω 2 ℏ x 2 ) m ω x + m ω x exp ( − m ω 2 ℏ x 2 ) ) = 2 m ω 2 ℏ m ω ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) x = 2 m ω ℏ ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) x \begin{alignat}{1}
\widehat{a}_{+}\psi_{0} & =\frac{1}{\sqrt{2\hbar m\omega}}\left(-i\widehat{p}+m\omega x\right)\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(-i\left[-i\hbar\frac{d}{dx}\right]+m\omega x\right)\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\hbar\frac{d}{dx}\left(\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\right)+m\omega x\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\hbar\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\left(-\frac{2m\omega x}{2\hbar}\right)+m\omega x\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)m\omega x+m\omega x\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\right)\\
& =\frac{2m\omega}{\sqrt{2\hbar m\omega}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)x\\
& =\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)x
\end{alignat} a + ψ 0 = 2ℏ mω 1 ( − i p + mω x ) ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) = 2ℏ mω 1 ( − i [ − i ℏ d x d ] + mω x ) ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) = 2ℏ mω 1 ( π ℏ mω ) 1/4 ( − ℏ d x d ( exp ( − 2ℏ mω x 2 ) ) + mω x exp ( − 2ℏ mω x 2 ) ) = 2ℏ mω 1 ( π ℏ mω ) 1/4 ( − ℏ exp ( − 2ℏ mω x 2 ) ( − 2ℏ 2 mω x ) + mω x exp ( − 2ℏ mω x 2 ) ) = 2ℏ mω 1 ( π ℏ mω ) 1/4 ( exp ( − 2ℏ mω x 2 ) mω x + mω x exp ( − 2ℏ mω x 2 ) ) = 2ℏ mω 2 mω ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) x = ℏ 2 mω ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) x Aplicando o operador escada novamente:
( a ^ + ) 2 ψ 0 = a ^ + ( a ^ + ψ 0 ) = a ^ + ( 2 m ω ℏ ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) x ) = 1 2 ℏ m ω ( − ℏ d d x + m ω x ) ( 2 m ω ℏ ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) x ) = 2 m ω 2 ℏ 2 m ω ( m ω π ℏ ) 1 / 4 ( − ℏ d d x [ e − m ω 2 ℏ x 2 x ] + m ω x e − m ω 2 ℏ x 2 x ) = 1 ℏ ( m ω π ℏ ) 1 / 4 ( − ℏ [ e − m ω 2 π ℏ x 2 d d x x + x d d x e − m ω 2 ℏ x 2 ] + m ω x 2 e − m ω 2 π ℏ x 2 ) = 1 ℏ ( m ω π ℏ ) 1 / 4 ( − ℏ [ e − m ω 2 ℏ x 2 + x e − m ω 2 ℏ x 2 ( − 2 x m ω 2 ℏ x ) ] + m ω x 2 e − m ω 2 ℏ x 2 ) = 1 ℏ ( m ω π ℏ ) 1 / 4 ( − ℏ e − m ω 2 ℏ x 2 + e − m ω 2 ℏ x 2 ( x 2 m ω ) + m ω x 2 e − m ω 2 ℏ x 2 ) = 1 ℏ ( m ω π ℏ ) 1 / 4 ( − ℏ + x 2 m ω + m ω x 2 ) e − m ω 2 ℏ x 2 = 1 ℏ ( m ω π ℏ ) 1 / 4 ( 2 m ω x 2 − ℏ ) e − m ω 2 ℏ x 2 = ( m ω π ℏ ) 1 / 4 ( 2 m ω ℏ x 2 − 1 ) e − m ω 2 ℏ x 2 \begin{alignat}{1}
\left(\widehat{a}_{+}\right)^{2}\psi_{0} & =\widehat{a}_{+}\left(\widehat{a}_{+}\psi_{0}\right)\\
& =\widehat{a}_{+}\left(\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)x\right)\\
& =\frac{1}{\sqrt{2\hbar m\omega}}\left(-\hbar\frac{d}{dx}+m\omega x\right)\left(\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)x\right)\\
& =\sqrt{\frac{2m\omega}{2\hbar^{2}m\omega}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\hbar\frac{d}{dx}\left[e^{-\frac{m\omega}{2\hbar}x^{2}}x\right]+m\omega xe^{-\frac{m\omega}{2\hbar}x^{2}}x\right)\\
& =\frac{1}{\hbar}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\hbar\left[e^{-\frac{m\omega}{2\pi\hbar}x^{2}}\frac{d}{dx}x+x\frac{d}{dx}e^{-\frac{m\omega}{2\hbar}x^{2}}\right]+m\omega x^{2}e^{-\frac{m\omega}{2\pi\hbar}x^{2}}\right)\\
& =\frac{1}{\hbar}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\hbar\left[e^{-\frac{m\omega}{2\hbar}x^{2}}+xe^{-\frac{m\omega}{2\hbar}x^{2}}\left(-\frac{2xm\omega}{2\hbar}x\right)\right]+m\omega x^{2}e^{-\frac{m\omega}{2\hbar}x^{2}}\right)\\
& =\frac{1}{\hbar}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\hbar e^{-\frac{m\omega}{2\hbar}x^{2}}+e^{-\frac{m\omega}{2\hbar}x^{2}}\left(x^{2}m\omega\right)+m\omega x^{2}e^{-\frac{m\omega}{2\hbar}x^{2}}\right)\\
& =\frac{1}{\hbar}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(-\hbar+x^{2}m\omega+m\omega x^{2}\right)e^{-\frac{m\omega}{2\hbar}x^{2}}\\
& =\frac{1}{\hbar}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(2m\omega x^{2}-\hbar\right)e^{-\frac{m\omega}{2\hbar}x^{2}}\\
& =\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{2\hbar}x^{2}}
\end{alignat} ( a + ) 2 ψ 0 = a + ( a + ψ 0 ) = a + ( ℏ 2 mω ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) x ) = 2ℏ mω 1 ( − ℏ d x d + mω x ) ( ℏ 2 mω ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) x ) = 2 ℏ 2 mω 2 mω ( π ℏ mω ) 1/4 ( − ℏ d x d [ e − 2ℏ mω x 2 x ] + mω x e − 2ℏ mω x 2 x ) = ℏ 1 ( π ℏ mω ) 1/4 ( − ℏ [ e − 2 π ℏ mω x 2 d x d x + x d x d e − 2ℏ mω x 2 ] + mω x 2 e − 2 π ℏ mω x 2 ) = ℏ 1 ( π ℏ mω ) 1/4 ( − ℏ [ e − 2ℏ mω x 2 + x e − 2ℏ mω x 2 ( − 2ℏ 2 x mω x ) ] + mω x 2 e − 2ℏ mω x 2 ) = ℏ 1 ( π ℏ mω ) 1/4 ( − ℏ e − 2ℏ mω x 2 + e − 2ℏ mω x 2 ( x 2 mω ) + mω x 2 e − 2ℏ mω x 2 ) = ℏ 1 ( π ℏ mω ) 1/4 ( − ℏ + x 2 mω + mω x 2 ) e − 2ℏ mω x 2 = ℏ 1 ( π ℏ mω ) 1/4 ( 2 mω x 2 − ℏ ) e − 2ℏ mω x 2 = ( π ℏ mω ) 1/4 ( ℏ 2 mω x 2 − 1 ) e − 2ℏ mω x 2 E como vimos que:
ψ n = 1 n ! ( a ^ + ) n ψ 0 ψ 2 = 1 2 ! ( a ^ + ) 2 ψ 0 = 1 2 ( m ω π ℏ ) 1 / 4 ( 2 m ω ℏ x 2 − 1 ) e − m ω 2 ℏ x 2 \begin{align}
\psi_{n} & =\frac{1}{\sqrt{n!}}\left(\widehat{a}_{+}\right)^{n}\psi_{0}\\
\psi_{2} & =\frac{1}{\sqrt{2!}}\left(\widehat{a}_{+}\right)^{2}\psi_{0}\\
& =\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{2\hbar}x^{2}}
\end{align} ψ n ψ 2 = n ! 1 ( a + ) n ψ 0 = 2 ! 1 ( a + ) 2 ψ 0 = 2 1 ( π ℏ mω ) 1/4 ( ℏ 2 mω x 2 − 1 ) e − 2ℏ mω x 2 Também:
ψ 1 = 1 1 ! ( a ^ + ) 1 ψ 0 = 2 m ω ℏ ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) x \begin{align}
\psi_{1} & =\frac{1}{\sqrt{1!}}\left(\widehat{a}_{+}\right)^{1}\psi_{0}\\
& =\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)x
\end{align} ψ 1 = 1 ! 1 ( a + ) 1 ψ 0 = ℏ 2 mω ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) x Então agora queremos verificar a ortogonalidade. Para f ( x ) = ψ 0 ∗ ψ 1 f\left(x\right)=\psi^{*}_{0}\psi_{1} f ( x ) = ψ 0 ∗ ψ 1
ψ 0 ∗ ψ 1 = ( ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) ) ( 2 m ω ℏ ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) x ) f ( x ) = [ ( m ω π ℏ ) 1 / 4 2 m ω ℏ ( m ω π ℏ ) 1 / 4 ] exp ( − m ω 2 ℏ x 2 − m ω 2 ℏ x 2 ) x = A exp ( − m ω ℏ x 2 ) x \begin{align}
\psi^{*}_{0}\psi_{1} & =\left(\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)\right)\left(\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega}{2\hbar}x^{2}\right)x\right)\\
f\left(x\right) & =\left[\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\right]\exp\left(-\frac{m\omega}{2\hbar}x^{2}-\frac{m\omega}{2\hbar}x^{2}\right)x\\
& =A\exp\left(-\frac{m\omega}{\hbar}x^{2}\right)x
\end{align} ψ 0 ∗ ψ 1 f ( x ) = ( ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) ) ( ℏ 2 mω ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 ) x ) = [ ( π ℏ mω ) 1/4 ℏ 2 mω ( π ℏ mω ) 1/4 ] exp ( − 2ℏ mω x 2 − 2ℏ mω x 2 ) x = A exp ( − ℏ mω x 2 ) x Podemos ver que é uma função ímpar:
f ( − x ) = A e ( − m ω ℏ ( − x ) 2 ) ( − x ) = − A e ( − m ω ℏ x 2 ) x = − f ( x ) f\left(-x\right)=Ae^{\left(-\frac{m\omega}{\hbar}\left(-x\right)^{2}\right)}\left(-x\right)=-Ae^{\left(-\frac{m\omega}{\hbar}x^{2}\right)}x=-f\left(x\right) f ( − x ) = A e ( − ℏ mω ( − x ) 2 ) ( − x ) = − A e ( − ℏ mω x 2 ) x = − f ( x ) Logo:
∫ − ∞ + ∞ ψ 0 ∗ ψ 1 d x = ∫ − ∞ + ∞ f ( x ) d x = 0 \int^{+\infty}_{-\infty}\psi^{*}_{0}\psi_{1}dx=\int^{+\infty}_{-\infty}f\left(x\right)dx=0 ∫ − ∞ + ∞ ψ 0 ∗ ψ 1 d x = ∫ − ∞ + ∞ f ( x ) d x = 0 Vamos ter um resultado análogo para ψ 2 ∗ ψ 1 ∗ \psi^{*}_{2}\psi^{*}_{1} ψ 2 ∗ ψ 1 ∗
ψ 2 ∗ ψ 1 = ( 1 2 ( m ω π ℏ ) 1 / 4 ( 2 m ω ℏ x 2 − 1 ) e − m ω 2 ℏ x 2 ) ( 2 m ω ℏ ( m ω π ℏ ) 1 / 4 e − m ω 2 ℏ x 2 x ) f ( x ) = [ 1 2 ( m ω π ℏ ) 1 / 4 2 m ω ℏ ( m ω π ℏ ) 1 / 4 ] ( 2 m ω ℏ x 2 − 1 ) e − m ω 2 ℏ x 2 − m ω 2 ℏ x 2 x = A ( 2 m ω ℏ x 2 − 1 ) e − m ω ℏ x 2 x \begin{align}
\psi^{*}_{2}\psi_{1} & =\left(\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{2\hbar}x^{2}}\right)\left(\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^{2}}x\right)\\
f\left(x\right) & =\left[\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\right]\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{2\hbar}x^{2}-\frac{m\omega}{2\hbar}x^{2}}x\\
& =A\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{\hbar}x^{2}}x
\end{align} ψ 2 ∗ ψ 1 f ( x ) = ( 2 1 ( π ℏ mω ) 1/4 ( ℏ 2 mω x 2 − 1 ) e − 2ℏ mω x 2 ) ( ℏ 2 mω ( π ℏ mω ) 1/4 e − 2ℏ mω x 2 x ) = [ 2 1 ( π ℏ mω ) 1/4 ℏ 2 mω ( π ℏ mω ) 1/4 ] ( ℏ 2 mω x 2 − 1 ) e − 2ℏ mω x 2 − 2ℏ mω x 2 x = A ( ℏ 2 mω x 2 − 1 ) e − ℏ mω x 2 x Temos novamente uma função ímpar:
f ( − x ) = A ( 2 m ω ℏ ( − x ) 2 − 1 ) e − m ω ℏ ( − x ) 2 ( − x ) = − A ( 2 m ω ℏ x 2 − 1 ) e − m ω ℏ x 2 x = − f ( x ) f\left(-x\right)=A\left(\frac{2m\omega}{\hbar}\left(-x\right)^{2}-1\right)e^{-\frac{m\omega}{\hbar}\left(-x\right)^{2}}\left(-x\right)=-A\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{\hbar}x^{2}}x=-f\left(x\right) f ( − x ) = A ( ℏ 2 mω ( − x ) 2 − 1 ) e − ℏ mω ( − x ) 2 ( − x ) = − A ( ℏ 2 mω x 2 − 1 ) e − ℏ mω x 2 x = − f ( x ) Logo novamente:
∫ − ∞ + ∞ ψ 2 ∗ ψ 1 d x = ∫ − ∞ + ∞ f ( x ) d x = 0 \int^{+\infty}_{-\infty}\psi^{*}_{2}\psi_{1}dx=\int^{+\infty}_{-\infty}f\left(x\right)dx=0 ∫ − ∞ + ∞ ψ 2 ∗ ψ 1 d x = ∫ − ∞ + ∞ f ( x ) d x = 0 Isso muda para ψ 2 ∗ ψ 0 \psi^{*}_{2}\psi_{0} ψ 2 ∗ ψ 0
ψ 2 ∗ ψ 0 = ( 1 2 ( m ω π ℏ ) 1 / 4 ( 2 m ω ℏ x 2 − 1 ) e − m ω 2 ℏ x 2 ) ( ( m ω π ℏ ) 1 / 4 e − m ω 2 ℏ x 2 ) = A ( 2 m ω ℏ x 2 − 1 ) e − m ω ℏ x 2 = A 2 m ω ℏ x 2 e − m ω ℏ x 2 − A e − m ω ℏ x 2 = B x 2 e − a x 2 − A e − a x 2 \begin{align}
\psi^{*}_{2}\psi_{0} & =\left(\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{2\hbar}x^{2}}\right)\left(\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega}{2\hbar}x^{2}}\right)\\
& =A\left(\frac{2m\omega}{\hbar}x^{2}-1\right)e^{-\frac{m\omega}{\hbar}x^{2}}\\
& =A\frac{2m\omega}{\hbar}x^{2}e^{-\frac{m\omega}{\hbar}x^{2}}-Ae^{-\frac{m\omega}{\hbar}x^{2}}\\
& =Bx^{2}e^{-ax^{2}}-Ae^{-ax^{2}}
\end{align} ψ 2 ∗ ψ 0 = ( 2 1 ( π ℏ mω ) 1/4 ( ℏ 2 mω x 2 − 1 ) e − 2ℏ mω x 2 ) ( ( π ℏ mω ) 1/4 e − 2ℏ mω x 2 ) = A ( ℏ 2 mω x 2 − 1 ) e − ℏ mω x 2 = A ℏ 2 mω x 2 e − ℏ mω x 2 − A e − ℏ mω x 2 = B x 2 e − a x 2 − A e − a x 2 Integrando temos então:
∫ − ∞ + ∞ ψ 2 ∗ ψ 0 d x = B ∫ − ∞ + ∞ x 2 e − a x 2 d x − A ∫ − ∞ + ∞ e − a x 2 d x \int^{+\infty}_{-\infty}\psi^{*}_{2}\psi_{0}dx=B\int^{+\infty}_{-\infty}x^{2}e^{-ax^{2}}dx-A\int^{+\infty}_{-\infty}e^{-ax^{2}}dx ∫ − ∞ + ∞ ψ 2 ∗ ψ 0 d x = B ∫ − ∞ + ∞ x 2 e − a x 2 d x − A ∫ − ∞ + ∞ e − a x 2 d x Estas duas integrais já foram resolvidas. ∫ − ∞ + ∞ e − a x 2 d x = π a \int^{+\infty}_{-\infty}e^{-ax^{2}}dx=\sqrt{\frac{\pi}{a}} ∫ − ∞ + ∞ e − a x 2 d x = a π ∫ − ∞ + ∞ x 2 e − a x 2 d x = π 4 a 3 \int^{+\infty}_{-\infty}x^{2}e^{-ax^{2}}dx=\sqrt{\frac{\pi}{4a^{3}}} ∫ − ∞ + ∞ x 2 e − a x 2 d x = 4 a 3 π A = 1 2 ( m ω π ℏ ) 1 / 2 A=\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2} A = 2 1 ( π ℏ mω ) 1/2 B = A 2 m ω ℏ B=A\frac{2m\omega}{\hbar} B = A ℏ 2 mω a = m ω ℏ a=\frac{m\omega}{\hbar} a = ℏ mω
∫ − ∞ + ∞ ψ 2 ∗ ψ 0 d x = B ∫ − ∞ + ∞ x 2 e − α x 2 d x − A ∫ − ∞ + ∞ e − α x 2 d x = B ( π 4 a 3 ) − A π a = A 2 m ω ℏ ( π 4 ( m ω ℏ ) 3 ) − 1 2 ( m ω π ℏ ) 1 / 2 π m ω ℏ = 1 2 ( m ω π ℏ ) 1 / 2 2 m ω ℏ ( ℏ 3 π 4 ( m ω ) 3 ) − 1 2 ( m ω π ℏ ) 1 / 2 ℏ π m ω = 1 2 ( m ω π ℏ ) 1 / 2 ( 2 2 m 2 ω 2 ℏ 2 ( ℏ 3 π 4 ( m ω ) 3 ) − ℏ π m ω ) = 1 2 ( m ω π ℏ ) 1 / 2 ( ℏ π m ω − ℏ π m ω ) = 0 \begin{align}
\int^{+\infty}_{-\infty}\psi^{*}_{2}\psi_{0}dx & =B\int^{+\infty}_{-\infty}x^{2}e^{-\alpha x^{2}}dx-A\int^{+\infty}_{-\infty}e^{-\alpha x^{2}}dx\\
& =B\left(\sqrt{\frac{\pi}{4a^{3}}}\right)-A\sqrt{\frac{\pi}{a}}\\
& =A\frac{2m\omega}{\hbar}\left(\sqrt{\frac{\pi}{4\left(\frac{m\omega}{\hbar}\right)^{3}}}\right)-\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\sqrt{\frac{\pi}{\frac{m\omega}{\hbar}}}\\
& =\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\frac{2m\omega}{\hbar}\left(\sqrt{\frac{\hbar^{3}\pi}{4\left(m\omega\right)^{3}}}\right)-\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\sqrt{\frac{\hbar\pi}{m\omega}}\\
& =\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\left(\sqrt{\frac{2^{2}m^{2}\omega^{2}}{\hbar^{2}}}\left(\sqrt{\frac{\hbar^{3}\pi}{4\left(m\omega\right)^{3}}}\right)-\sqrt{\frac{\hbar\pi}{m\omega}}\right)\\
& =\frac{1}{\sqrt{2}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\left(\sqrt{\frac{\hbar\pi}{m\omega}}-\sqrt{\frac{\hbar\pi}{m\omega}}\right)\\
& =0
\end{align} ∫ − ∞ + ∞ ψ 2 ∗ ψ 0 d x = B ∫ − ∞ + ∞ x 2 e − α x 2 d x − A ∫ − ∞ + ∞ e − α x 2 d x = B ( 4 a 3 π ) − A a π = A ℏ 2 mω ⎝ ⎛ 4 ( ℏ mω ) 3 π ⎠ ⎞ − 2 1 ( π ℏ mω ) 1/2 ℏ mω π = 2 1 ( π ℏ mω ) 1/2 ℏ 2 mω ( 4 ( mω ) 3 ℏ 3 π ) − 2 1 ( π ℏ mω ) 1/2 mω ℏ π = 2 1 ( π ℏ mω ) 1/2 ( ℏ 2 2 2 m 2 ω 2 ( 4 ( mω ) 3 ℏ 3 π ) − mω ℏ π ) = 2 1 ( π ℏ mω ) 1/2 ( mω ℏ π − mω ℏ π ) = 0 Exemplo:
a) Calcule ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩ ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩ ⟨ x 2 ⟩ \left\langle x^{2}\right\rangle ⟨ x 2 ⟩ ⟨ p 2 ⟩ \left\langle p^{2}\right\rangle ⟨ p 2 ⟩ ψ 0 \psi_{0} ψ 0 ψ 1 \psi_{1} ψ 1
b) Verifique o princípio da incerteza para estes estados.
c) Calcule ⟨ T ⟩ \left\langle T\right\rangle ⟨ T ⟩ ⟨ V ⟩ \left\langle V\right\rangle ⟨ V ⟩
Vamos analisar algo primeiro. O produto entre uma função par g ( x ) g\left(x\right) g ( x ) f ( x ) f\left(x\right) f ( x )
g ( x ) = f ( x ) g ( x ) g\left(x\right)=f\left(x\right)g\left(x\right) g ( x ) = f ( x ) g ( x ) Então:
g ( − x ) = f ( − x ) g ( − x ) = − f ( x ) g ( x ) = − g ( x ) g\left(-x\right)=f\left(-x\right)g\left(-x\right)=-f\left(x\right)g\left(x\right)=-g\left(x\right) g ( − x ) = f ( − x ) g ( − x ) = − f ( x ) g ( x ) = − g ( x ) Agora queremos analisar ∣ f ( x ) ∣ 2 \left|f\left(x\right)\right|^{2} ∣ f ( x ) ∣ 2 f ∗ ( x ) f^{*}\left(x\right) f ∗ ( x ) f ( x ) = u ( x ) + i v ( x ) f\left(x\right)=u\left(x\right)+iv\left(x\right) f ( x ) = u ( x ) + i v ( x )
f ( − x ) = u ( − x ) + i v ( − x ) = u ( x ) + i v ( x ) = f ( x ) f\left(-x\right)=u\left(-x\right)+iv\left(-x\right)=u\left(x\right)+iv\left(x\right)=f\left(x\right) f ( − x ) = u ( − x ) + i v ( − x ) = u ( x ) + i v ( x ) = f ( x ) O conjugado então:
f ∗ ( − x ) = u ( − x ) − i v ( − x ) = u ( x ) − i v ( x ) = f ( x ) f^{*}\left(-x\right)=u\left(-x\right)-iv\left(-x\right)=u\left(x\right)-iv\left(x\right)=f\left(x\right) f ∗ ( − x ) = u ( − x ) − i v ( − x ) = u ( x ) − i v ( x ) = f ( x ) Acontece algo análogo se for uma função ímpar, como podemos i i i ∣ f ( x ) ∣ 2 \left|f\left(x\right)\right|^{2} ∣ f ( x ) ∣ 2 f ( x ) f\left(x\right) f ( x )
∣ f ( − x ) ∣ 2 = f ∗ ( − x ) f ( − x ) = [ − f ∗ ( x ) ] [ − f ( x ) ] = f ∗ ( x ) f ( x ) = ∣ f ( x ) ∣ 2 \begin{align}
\left|f\left(-x\right)\right|^{2} & =f^{*}\left(-x\right)f\left(-x\right)=\left[-f^{*}\left(x\right)\right]\left[-f\left(x\right)\right]=f^{*}\left(x\right)f\left(x\right)=\left|f\left(x\right)\right|^{2}
\end{align} ∣ f ( − x ) ∣ 2 = f ∗ ( − x ) f ( − x ) = [ − f ∗ ( x ) ] [ − f ( x ) ] = f ∗ ( x ) f ( x ) = ∣ f ( x ) ∣ 2 Usando a propriedade que discutimos de que o conjugado da função conserva a paridade.
Ou seja se temos f ( x ) ∣ g ( x ) ∣ 2 f\left(x\right)\left|g\left(x\right)\right|^{2} f ( x ) ∣ g ( x ) ∣ 2 g ( x ) g\left(x\right) g ( x ) f ( x ) f\left(x\right) f ( x )
Vamos usar isso para calcular ⟨ x ⟩ = ∫ − ∞ + ∞ x ∣ ψ n ∣ d x \left\langle x\right\rangle =\int^{+\infty}_{-\infty}x\left|\psi_{n}\right|dx ⟨ x ⟩ = ∫ − ∞ + ∞ x ∣ ψ n ∣ d x n = 0 n=0 n = 0 n = 1 n=1 n = 1 ⟨ x ⟩ = 0 \left\langle x\right\rangle =0 ⟨ x ⟩ = 0
Consequentemente ⟨ p ⟩ = m d ⟨ x ⟩ d t = 0 \left\langle p\right\rangle =m\frac{d\left\langle x\right\rangle }{dt}=0 ⟨ p ⟩ = m d t d ⟨ x ⟩ = 0 ⟨ x 2 ⟩ \left\langle x^{2}\right\rangle ⟨ x 2 ⟩ ⟨ p 2 ⟩ \left\langle p^{2}\right\rangle ⟨ p 2 ⟩ ψ 0 \psi_{0} ψ 0 ψ 0 ( x ) = a e − b 2 x 2 \psi_{0}\left(x\right)=ae^{-\frac{b}{2}x^{2}} ψ 0 ( x ) = a e − 2 b x 2 a = ( m ω π ℏ ) 1 / 4 a=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} a = ( π ℏ mω ) 1/4 b = m ω ℏ b=\frac{m\omega}{\hbar} b = ℏ mω
⟨ x 2 ⟩ = ∫ − ∞ + ∞ ψ 0 ∗ x 2 ψ 0 d x = ∫ − ∞ + ∞ a e − b 2 x 2 x 2 a e − b 2 x 2 d x = a 2 ∫ − ∞ + ∞ x 2 e − b x 2 d x = a 2 π 4 b 3 \begin{align}
\left\langle x^{2}\right\rangle & =\int^{+\infty}_{-\infty}\psi^{*}_{0}x^{2}\psi_{0}dx\\
& =\int^{+\infty}_{-\infty}ae^{-\frac{b}{2}x^{2}}x^{2}ae^{-\frac{b}{2}x^{2}}dx\\
& =a^{2}\int^{+\infty}_{-\infty}x^{2}e^{-bx^{2}}dx\\
& =a^{2}\sqrt{\frac{\pi}{4b^{3}}}
\end{align} ⟨ x 2 ⟩ = ∫ − ∞ + ∞ ψ 0 ∗ x 2 ψ 0 d x = ∫ − ∞ + ∞ a e − 2 b x 2 x 2 a e − 2 b x 2 d x = a 2 ∫ − ∞ + ∞ x 2 e − b x 2 d x = a 2 4 b 3 π Onde usamos novamente o resultado já conhecido que ∫ − ∞ + ∞ x 2 e − a x 2 d x = π 4 a 3 \int^{+\infty}_{-\infty}x^{2}e^{-ax^{2}}dx=\sqrt{\frac{\pi}{4a^{3}}} ∫ − ∞ + ∞ x 2 e − a x 2 d x = 4 a 3 π
⟨ x 2 ⟩ = [ ( m ω π ℏ ) 1 / 4 ] 2 π 4 ( m ω ℏ ) 3 = m ω π ℏ π ℏ 3 4 ( m ω ) 3 = ( ℏ 4 m ω ) 2 = ℏ 2 m ω \left\langle x^{2}\right\rangle =\left[\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\right]^{2}\sqrt{\frac{\pi}{4\left(\frac{m\omega}{\hbar}\right)^{3}}}=\sqrt{\frac{m\omega}{\pi\hbar}}\sqrt{\frac{\pi\hbar^{3}}{4\left(m\omega\right)^{3}}}=\sqrt{\left(\frac{\hbar}{4m\omega}\right)^{2}}=\frac{\hbar}{2m\omega} ⟨ x 2 ⟩ = [ ( π ℏ mω ) 1/4 ] 2 4 ( ℏ mω ) 3 π = π ℏ mω 4 ( mω ) 3 π ℏ 3 = ( 4 mω ℏ ) 2 = 2 mω ℏ E para ⟨ p 2 ⟩ \left\langle p^{2}\right\rangle ⟨ p 2 ⟩
⟨ p 2 ⟩ = ∫ − ∞ + ∞ ψ 0 ∗ ( − i ℏ d d x ) 2 ψ 0 d x = ∫ − ∞ + ∞ a e − b 2 x 2 ( − ℏ 2 d 2 d x 2 ) a e − b 2 x 2 d x = − ( a ℏ ) 2 ∫ − ∞ + ∞ e − b 2 x 2 d d x ( − b x e − b 2 x 2 ) d x = ( a ℏ ) 2 b ∫ − ∞ + ∞ e − b 2 x 2 ( e − b 2 x 2 − x e − b 2 x 2 b x ) d x = ( a ℏ ) 2 b ( ∫ − ∞ + ∞ e − b x 2 d x − b ∫ − ∞ + ∞ x 2 e − b x 2 d x ) \begin{align}
\left\langle p^{2}\right\rangle & =\int^{+\infty}_{-\infty}\psi^{*}_{0}\left(-i\hbar\frac{d}{dx}\right)^{2}\psi_{0}dx\\
& =\int^{+\infty}_{-\infty}ae^{-\frac{b}{2}x^{2}}\left(-\hbar^{2}\frac{d^{2}}{dx^{2}}\right)ae^{-\frac{b}{2}x^{2}}dx\\
& =-\left(a\hbar\right)^{2}\int^{+\infty}_{-\infty}e^{-\frac{b}{2}x^{2}}\frac{d}{dx}\left(-bxe^{-\frac{b}{2}x^{2}}\right)dx\\
& =\left(a\hbar\right)^{2}b\int^{+\infty}_{-\infty}e^{-\frac{b}{2}x^{2}}\left(e^{-\frac{b}{2}x^{2}}-xe^{-\frac{b}{2}x^{2}}bx\right)dx\\
& =\left(a\hbar\right)^{2}b\left(\int^{+\infty}_{-\infty}e^{-bx^{2}}dx-b\int^{+\infty}_{-\infty}x^{2}e^{-bx^{2}}dx\right)
\end{align} ⟨ p 2 ⟩ = ∫ − ∞ + ∞ ψ 0 ∗ ( − i ℏ d x d ) 2 ψ 0 d x = ∫ − ∞ + ∞ a e − 2 b x 2 ( − ℏ 2 d x 2 d 2 ) a e − 2 b x 2 d x = − ( a ℏ ) 2 ∫ − ∞ + ∞ e − 2 b x 2 d x d ( − b x e − 2 b x 2 ) d x = ( a ℏ ) 2 b ∫ − ∞ + ∞ e − 2 b x 2 ( e − 2 b x 2 − x e − 2 b x 2 b x ) d x = ( a ℏ ) 2 b ( ∫ − ∞ + ∞ e − b x 2 d x − b ∫ − ∞ + ∞ x 2 e − b x 2 d x ) Novamente utilizand os resultados já conhecidos ∫ − ∞ + ∞ x 2 e − a x 2 d x = π 4 a 3 \int^{+\infty}_{-\infty}x^{2}e^{-ax^{2}}dx=\sqrt{\frac{\pi}{4a^{3}}} ∫ − ∞ + ∞ x 2 e − a x 2 d x = 4 a 3 π ∫ − ∞ + ∞ e − a x 2 d x = π a \int^{+\infty}_{-\infty}e^{-ax^{2}}dx=\sqrt{\frac{\pi}{a}} ∫ − ∞ + ∞ e − a x 2 d x = a π
⟨ p 2 ⟩ = ( a ℏ ) 2 b ( π b − b π 4 b 3 ) = ( ( m ω π ℏ ) 1 / 4 ℏ ) 2 m ω ℏ ( π m ω ℏ − m ω ℏ π 4 ( m ω ℏ ) 3 ) = ( m ω π ℏ ) 1 / 2 m ℏ ω ( ℏ π m ω − 1 2 ( 2 m ω ℏ ) 2 ℏ 3 π 4 ( m ω ) 3 ) = 1 2 m ω π ℏ m ℏ ω ℏ π m ω = m ℏ ω 2 \begin{align}
\left\langle p^{2}\right\rangle & =\left(a\hbar\right)^{2}b\left(\sqrt{\frac{\pi}{b}}-b\sqrt{\frac{\pi}{4b^{3}}}\right)\\
& =\left(\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\hbar\right)^{2}\frac{m\omega}{\hbar}\left(\sqrt{\frac{\pi}{\frac{m\omega}{\hbar}}}-\frac{m\omega}{\hbar}\sqrt{\frac{\pi}{4\left(\frac{m\omega}{\hbar}\right)^{3}}}\right)\\
& =\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}m\hbar\omega\left(\sqrt{\frac{\hbar\pi}{m\omega}}-\frac{1}{2}\sqrt{\left(\frac{2m\omega}{\hbar}\right)^{2}}\sqrt{\frac{\hbar^{3}\pi}{4\left(m\omega\right)^{3}}}\right)\\
& =\frac{1}{2}\sqrt{\frac{m\omega}{\pi\hbar}}m\hbar\omega\sqrt{\frac{\hbar\pi}{m\omega}}\\
& =\frac{m\hbar\omega}{2}
\end{align} ⟨ p 2 ⟩ = ( a ℏ ) 2 b ( b π − b 4 b 3 π ) = ( ( π ℏ mω ) 1/4 ℏ ) 2 ℏ mω ⎝ ⎛ ℏ mω π − ℏ mω 4 ( ℏ mω ) 3 π ⎠ ⎞ = ( π ℏ mω ) 1/2 m ℏ ω ⎝ ⎛ mω ℏ π − 2 1 ( ℏ 2 mω ) 2 4 ( mω ) 3 ℏ 3 π ⎠ ⎞ = 2 1 π ℏ mω m ℏ ω mω ℏ π = 2 m ℏ ω Onde a = ( m ω π ℏ ) 1 / 4 a=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} a = ( π ℏ mω ) 1/4 b = m ω ℏ b=\frac{m\omega}{\hbar} b = ℏ mω n = 1 n=1 n = 1
ψ 1 = a e − b 2 x 2 x \psi_{1}=ae^{-\frac{b}{2}x^{2}}x ψ 1 = a e − 2 b x 2 x Onde agora a = 2 m ω ℏ ( m ω π ℏ ) 1 / 4 a=\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} a = ℏ 2 mω ( π ℏ mω ) 1/4 b = m ω ℏ b=\frac{m\omega}{\hbar} b = ℏ mω
⟨ x 2 ⟩ = ∫ − ∞ + ∞ ψ 1 ∗ x 2 ψ 1 d x = ∫ − ∞ + ∞ ( a e − b 2 x 2 x ) x 2 ( a e − b 2 x 2 x ) d x = a 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x \begin{align}
\left\langle x^{2}\right\rangle & =\int^{+\infty}_{-\infty}\psi^{*}_{1}x^{2}\psi_{1}dx\\
& =\int^{+\infty}_{-\infty}\left(ae^{-\frac{b}{2}x^{2}}x\right)x^{2}\left(ae^{-\frac{b}{2}x^{2}}x\right)dx\\
& =a^{2}\int^{+\infty}_{-\infty}x^{4}e^{-bx^{2}}dx
\end{align} ⟨ x 2 ⟩ = ∫ − ∞ + ∞ ψ 1 ∗ x 2 ψ 1 d x = ∫ − ∞ + ∞ ( a e − 2 b x 2 x ) x 2 ( a e − 2 b x 2 x ) d x = a 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x Usando integral por partes sendo u = x 3 u=x^{3} u = x 3 d u d x = 3 x 2 \frac{du}{dx}=3x^{2} d x d u = 3 x 2 d v d u = x e − a x 2 \frac{dv}{du}=xe^{-ax^{2}} d u d v = x e − a x 2 w = − a x 2 w=-ax^{2} w = − a x 2 d w d x = − 2 a x \frac{dw}{dx}=-2ax d x d w = − 2 a x
v = ∫ x e − a x 2 d x = − 1 2 a ∫ e w d w = − e w 2 a + C = − e − a x 2 2 a + C v=\int xe^{-ax^{2}}dx=-\frac{1}{2a}\int e^{w}dw=-\frac{e^{w}}{2a}+C=-\frac{e^{-ax^{2}}}{2a}+C v = ∫ x e − a x 2 d x = − 2 a 1 ∫ e w d w = − 2 a e w + C = − 2 a e − a x 2 + C Então:
∫ u d v = u v − ∫ v d u ∫ − ∞ + ∞ x 4 e − b x 2 d x = [ − x 3 e − b x 2 2 b ] − ∞ + ∞ + ∫ − ∞ + ∞ e − b x 2 2 b 3 x 2 d x = 3 2 b ∫ − ∞ + ∞ x 2 e − b x 2 d x \begin{align}
\int udv & =uv-\int vdu\\
\int^{+\infty}_{-\infty}x^{4}e^{-bx^{2}}dx & =\left[-x^{3}\frac{e^{-bx^{2}}}{2b}\right]^{+\infty}_{-\infty}+\int^{+\infty}_{-\infty}\frac{e^{-bx^{2}}}{2b}3x^{2}dx\\
& =\frac{3}{2b}\int^{+\infty}_{-\infty}x^{2}e^{-bx^{2}}dx
\end{align} ∫ u d v ∫ − ∞ + ∞ x 4 e − b x 2 d x = uv − ∫ v d u = [ − x 3 2 b e − b x 2 ] − ∞ + ∞ + ∫ − ∞ + ∞ 2 b e − b x 2 3 x 2 d x = 2 b 3 ∫ − ∞ + ∞ x 2 e − b x 2 d x Como temos o resultado dessa integral, então:
∫ − ∞ + ∞ x 4 e − b x 2 d x = 3 2 b ∫ − ∞ + ∞ x 2 e − b x 2 d x = 3 2 b π 4 b 3 = 3 4 π b 5 \begin{align}
\int^{+\infty}_{-\infty}x^{4}e^{-bx^{2}}dx & =\frac{3}{2b}\int^{+\infty}_{-\infty}x^{2}e^{-bx^{2}}dx\\
& =\frac{3}{2b}\sqrt{\frac{\pi}{4b^{3}}}\\
& =\frac{3}{4}\sqrt{\frac{\pi}{b^{5}}}
\end{align} ∫ − ∞ + ∞ x 4 e − b x 2 d x = 2 b 3 ∫ − ∞ + ∞ x 2 e − b x 2 d x = 2 b 3 4 b 3 π = 4 3 b 5 π Retomando então o valor esperando de x 2 x^{2} x 2
⟨ x 2 ⟩ = a 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x = a 2 3 4 π b 5 = [ 2 m ω ℏ ( m ω π ℏ ) 1 / 4 ] 2 3 4 π ( m ω ℏ ) 5 = 2 m ω ℏ m ω π ℏ 3 4 ℏ 5 π ( m ω ) 5 = m ω π ℏ 3 2 ℏ 3 π ( m ω ) 3 = 3 2 ( ℏ m ω ) 2 = 3 2 ℏ m ω \begin{align}
\left\langle x^{2}\right\rangle & =a^{2}\int^{+\infty}_{-\infty}x^{4}e^{-bx^{2}}dx\\
& =a^{2}\frac{3}{4}\sqrt{\frac{\pi}{b^{5}}}\\
& =\left[\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\right]^{2}\frac{3}{4}\sqrt{\frac{\pi}{\left(\frac{m\omega}{\hbar}\right)^{5}}}\\
& =\frac{2m\omega}{\hbar}\sqrt{\frac{m\omega}{\pi\hbar}}\frac{3}{4}\sqrt{\frac{\hbar^{5}\pi}{\left(m\omega\right)^{5}}}\\
& =\sqrt{\frac{m\omega}{\pi\hbar}}\frac{3}{2}\sqrt{\frac{\hbar^{3}\pi}{\left(m\omega\right)^{3}}}\\
& =\frac{3}{2}\sqrt{\left(\frac{\hbar}{m\omega}\right)^{2}}\\
& =\frac{3}{2}\frac{\hbar}{m\omega}
\end{align} ⟨ x 2 ⟩ = a 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x = a 2 4 3 b 5 π = [ ℏ 2 mω ( π ℏ mω ) 1/4 ] 2 4 3 ( ℏ mω ) 5 π = ℏ 2 mω π ℏ mω 4 3 ( mω ) 5 ℏ 5 π = π ℏ mω 2 3 ( mω ) 3 ℏ 3 π = 2 3 ( mω ℏ ) 2 = 2 3 mω ℏ E para ⟨ p 2 ⟩ \left\langle p^{2}\right\rangle ⟨ p 2 ⟩
⟨ p 2 ⟩ = ∫ − ∞ + ∞ ψ 1 ∗ ( − i ℏ d d x ) 2 ψ 1 d x = ∫ − ∞ + ∞ ( a e − b 2 x 2 x ) ( − ℏ 2 d 2 d x 2 ) ( a e − b 2 x 2 x ) d x = − ℏ 2 a 2 ∫ − ∞ + ∞ e − b 2 x 2 x ( d d x ( e − b 2 x 2 − b x e − b 2 x 2 x ) ) d x = − ℏ 2 a 2 ∫ − ∞ + ∞ e − b 2 x 2 x ( − b x e − b 2 x 2 − b ( 2 x e − b 2 x 2 − b x 3 e − b 2 x 2 ) ) d x = − ℏ 2 a 2 ∫ − ∞ + ∞ ( − b x 2 e − b x 2 − b 2 x 2 e − b x 2 + b 2 x 4 e − b x 2 ) d x = − ℏ 2 a 2 ( − 3 b ∫ − ∞ + ∞ x 2 e − b x 2 d x + b 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x ) = − ℏ 2 a 2 ( − 3 b ∫ − ∞ + ∞ x 2 e − b x 2 d x + b 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x ) \begin{align}
\left\langle p^{2}\right\rangle & =\int^{+\infty}_{-\infty}\psi^{*}_{1}\left(-i\hbar\frac{d}{dx}\right)^{2}\psi_{1}dx\\
& =\int^{+\infty}_{-\infty}\left(ae^{-\frac{b}{2}x^{2}}x\right)\left(-\hbar^{2}\frac{d^{2}}{dx^{2}}\right)\left(ae^{-\frac{b}{2}x^{2}}x\right)dx\\
& =-\hbar^{2}a^{2}\int^{+\infty}_{-\infty}e^{-\frac{b}{2}x^{2}}x\left(\frac{d}{dx}\left(e^{-\frac{b}{2}x^{2}}-bxe^{-\frac{b}{2}x^{2}}x\right)\right)dx\\
& =-\hbar^{2}a^{2}\int^{+\infty}_{-\infty}e^{-\frac{b}{2}x^{2}}x\left(-bxe^{-\frac{b}{2}x^{2}}-b\left(2xe^{-\frac{b}{2}x^{2}}-bx^{3}e^{-\frac{b}{2}x^{2}}\right)\right)dx\\
& =-\hbar^{2}a^{2}\int^{+\infty}_{-\infty}\left(-bx^{2}e^{-bx^{2}}-b2x^{2}e^{-bx^{2}}+b^{2}x^{4}e^{-bx^{2}}\right)dx\\
& =-\hbar^{2}a^{2}\left(-3b\int^{+\infty}_{-\infty}x^{2}e^{-bx^{2}}dx+b^{2}\int^{+\infty}_{-\infty}x^{4}e^{-bx^{2}}dx\right)\\
& =-\hbar^{2}a^{2}\left(-3b\int^{+\infty}_{-\infty}x^{2}e^{-bx^{2}}dx+b^{2}\int^{+\infty}_{-\infty}x^{4}e^{-bx^{2}}dx\right)
\end{align} ⟨ p 2 ⟩ = ∫ − ∞ + ∞ ψ 1 ∗ ( − i ℏ d x d ) 2 ψ 1 d x = ∫ − ∞ + ∞ ( a e − 2 b x 2 x ) ( − ℏ 2 d x 2 d 2 ) ( a e − 2 b x 2 x ) d x = − ℏ 2 a 2 ∫ − ∞ + ∞ e − 2 b x 2 x ( d x d ( e − 2 b x 2 − b x e − 2 b x 2 x ) ) d x = − ℏ 2 a 2 ∫ − ∞ + ∞ e − 2 b x 2 x ( − b x e − 2 b x 2 − b ( 2 x e − 2 b x 2 − b x 3 e − 2 b x 2 ) ) d x = − ℏ 2 a 2 ∫ − ∞ + ∞ ( − b x 2 e − b x 2 − b 2 x 2 e − b x 2 + b 2 x 4 e − b x 2 ) d x = − ℏ 2 a 2 ( − 3 b ∫ − ∞ + ∞ x 2 e − b x 2 d x + b 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x ) = − ℏ 2 a 2 ( − 3 b ∫ − ∞ + ∞ x 2 e − b x 2 d x + b 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x ) Como já resolvemos essas contas:
⟨ p 2 ⟩ = − ℏ 2 a 2 ( − 3 b ∫ − ∞ + ∞ x 2 e − b x 2 d x + b 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x ) = − ℏ 2 a 2 ( − 3 b [ π 4 b 3 ] + b 2 [ 3 4 π b 5 ] ) = − 3 ℏ 2 a 2 ( 1 2 π b + 1 4 π b ) = 3 ℏ 2 a 2 4 π b = 3 ℏ 2 4 ( 2 m ω ℏ ( m ω π ℏ ) 1 / 4 ) 2 π m ω ℏ = 3 ℏ 2 4 2 m ω ℏ m ω π ℏ ℏ π m ω = 3 m ω ℏ 2 \begin{align}
\left\langle p^{2}\right\rangle & =-\hbar^{2}a^{2}\left(-3b\int^{+\infty}_{-\infty}x^{2}e^{-bx^{2}}dx+b^{2}\int^{+\infty}_{-\infty}x^{4}e^{-bx^{2}}dx\right)\\
& =-\hbar^{2}a^{2}\left(-3b\left[\sqrt{\frac{\pi}{4b^{3}}}\right]+b^{2}\left[\frac{3}{4}\sqrt{\frac{\pi}{b^{5}}}\right]\right)\\
& =-3\hbar^{2}a^{2}\left(\frac{1}{2}\sqrt{\frac{\pi}{b}}+\frac{1}{4}\sqrt{\frac{\pi}{b}}\right)\\
& =\frac{3\hbar^{2}a^{2}}{4}\sqrt{\frac{\pi}{b}}\\
& =\frac{3\hbar^{2}}{4}\left(\sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\right)^{2}\sqrt{\frac{\pi}{\frac{m\omega}{\hbar}}}\\
& =\frac{3\hbar^{2}}{4}\frac{2m\omega}{\hbar}\sqrt{\frac{m\omega}{\pi\hbar}}\sqrt{\frac{\hbar\pi}{m\omega}}\\
& =\frac{3m\omega\hbar}{2}
\end{align} ⟨ p 2 ⟩ = − ℏ 2 a 2 ( − 3 b ∫ − ∞ + ∞ x 2 e − b x 2 d x + b 2 ∫ − ∞ + ∞ x 4 e − b x 2 d x ) = − ℏ 2 a 2 ( − 3 b [ 4 b 3 π ] + b 2 [ 4 3 b 5 π ] ) = − 3 ℏ 2 a 2 ( 2 1 b π + 4 1 b π ) = 4 3 ℏ 2 a 2 b π = 4 3 ℏ 2 ( ℏ 2 mω ( π ℏ mω ) 1/4 ) 2 ℏ mω π = 4 3 ℏ 2 ℏ 2 mω π ℏ mω mω ℏ π = 2 3 mω ℏ Agora queremos verificar o princípio de incerteza (σ x σ p ≥ ℏ 2 \sigma_{x}\sigma_{p}\geq\frac{\hbar}{2} σ x σ p ≥ 2 ℏ n = 0 n=0 n = 0
σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = ℏ 2 m ω − 0 2 = ℏ 2 m ω σ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = m ℏ ω 2 − 0 2 = m ℏ ω 2 \begin{align}
\sigma_{x} & =\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}=\sqrt{\frac{\hbar}{2m\omega}-0^{2}}=\sqrt{\frac{\hbar}{2m\omega}}\\
\sigma_{p} & =\sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}=\sqrt{\frac{m\hbar\omega}{2}-0^{2}}=\sqrt{\frac{m\hbar\omega}{2}}
\end{align} σ x σ p = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = 2 mω ℏ − 0 2 = 2 mω ℏ = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = 2 m ℏ ω − 0 2 = 2 m ℏ ω Então:
σ x σ p = ℏ 2 m ω m ℏ ω 2 = ℏ 2 \sigma_{x}\sigma_{p}=\sqrt{\frac{\hbar}{2m\omega}}\sqrt{\frac{m\hbar\omega}{2}}=\frac{\hbar}{2} σ x σ p = 2 mω ℏ 2 m ℏ ω = 2 ℏ Para n = 1 : n=1: n = 1 :
σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = 3 2 ℏ m ω − 0 2 = 3 2 ℏ m ω σ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = 3 m ω ℏ 2 − 0 2 = 3 m ω ℏ 2 \begin{align}
\sigma_{x} & =\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}=\sqrt{\frac{3}{2}\frac{\hbar}{m\omega}-0^{2}}=\sqrt{\frac{3}{2}\frac{\hbar}{m\omega}}\\
\sigma_{p} & =\sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}=\sqrt{\frac{3m\omega\hbar}{2}-0^{2}}=\sqrt{\frac{3m\omega\hbar}{2}}
\end{align} σ x σ p = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = 2 3 mω ℏ − 0 2 = 2 3 mω ℏ = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = 2 3 mω ℏ − 0 2 = 2 3 mω ℏ Logo:
σ x σ p = 3 2 ℏ m ω 3 m ω ℏ 2 = 3 ℏ 2 \sigma_{x}\sigma_{p}=\sqrt{\frac{3}{2}\frac{\hbar}{m\omega}}\sqrt{\frac{3m\omega\hbar}{2}}=3\frac{\hbar}{2} σ x σ p = 2 3 mω ℏ 2 3 mω ℏ = 3 2 ℏ Ambos os resultados são válidos e obedecem o princípio. Por último vamos calcular a energia cinética ⟨ T ⟩ \left\langle T\right\rangle ⟨ T ⟩ ⟨ V ⟩ \left\langle V\right\rangle ⟨ V ⟩ ⟨ H ⟩ = ⟨ T ⟩ + ⟨ V ⟩ \left\langle H\right\rangle =\left\langle T\right\rangle +\left\langle V\right\rangle ⟨ H ⟩ = ⟨ T ⟩ + ⟨ V ⟩ E n = ( n + 1 2 ) ℏ ω E_{n}=\left(n+\frac{1}{2}\right)\hbar\omega E n = ( n + 2 1 ) ℏ ω
Então para a energia cinética, que temos a fórmula clássica:
⟨ T 0 ⟩ = ⟨ p 2 ⟩ 2 m = m ℏ ω 2 2 m = ℏ ω 4 ⟨ T 1 ⟩ = ⟨ p 2 ⟩ 2 m = 3 m ω ℏ 2 2 m = 3 ω ℏ 4 \begin{align}
\left\langle T_{0}\right\rangle = & \frac{\left\langle p^{2}\right\rangle }{2m}=\frac{\frac{m\hbar\omega}{2}}{2m}=\frac{\hbar\omega}{4}\\
\left\langle T_{1}\right\rangle = & \frac{\left\langle p^{2}\right\rangle }{2m}=\frac{\frac{3m\omega\hbar}{2}}{2m}=\frac{3\omega\hbar}{4}
\end{align} ⟨ T 0 ⟩ = ⟨ T 1 ⟩ = 2 m ⟨ p 2 ⟩ = 2 m 2 m ℏ ω = 4 ℏ ω 2 m ⟨ p 2 ⟩ = 2 m 2 3 mω ℏ = 4 3 ω ℏ E para a potencial esperada, conforme discutido no início do capítulo, para um oscilador simples:
⟨ V 0 ⟩ = m ω 2 2 ⟨ x 2 ⟩ = m ω 2 2 ℏ 2 m ω = ℏ ω 4 ⟨ V 1 ⟩ = m ω 2 2 ⟨ x 2 ⟩ = m ω 2 2 3 2 ℏ m ω = 3 ω ℏ 4 \begin{align}
\left\langle V_{0}\right\rangle = & \frac{m\omega^{2}}{2}\left\langle x^{2}\right\rangle =\frac{m\omega^{2}}{2}\frac{\hbar}{2m\omega}=\frac{\hbar\omega}{4}\\
\left\langle V_{1}\right\rangle = & \frac{m\omega^{2}}{2}\left\langle x^{2}\right\rangle =\frac{m\omega^{2}}{2}\frac{3}{2}\frac{\hbar}{m\omega}=\frac{3\omega\hbar}{4}
\end{align} ⟨ V 0 ⟩ = ⟨ V 1 ⟩ = 2 m ω 2 ⟨ x 2 ⟩ = 2 m ω 2 2 mω ℏ = 4 ℏ ω 2 m ω 2 ⟨ x 2 ⟩ = 2 m ω 2 2 3 mω ℏ = 4 3 ω ℏ Então a energia total é então:
⟨ H 0 ⟩ = ℏ ω 4 + ℏ ω 4 = 1 2 ℏ ω = E 0 E 1 = 3 ω ℏ 4 + 3 ω ℏ 4 = 6 4 ℏ ω = 3 2 ℏ ω = ( 1 + 1 2 ) ℏ ω = E 1 \begin{align}
\left\langle H_{0}\right\rangle = & \frac{\hbar\omega}{4}+\frac{\hbar\omega}{4}=\frac{1}{2}\hbar\omega=E_{0}\\
E_{1}= & \frac{3\omega\hbar}{4}+\frac{3\omega\hbar}{4}=\frac{6}{4}\hbar\omega=\frac{3}{2}\hbar\omega=\left(1+\frac{1}{2}\right)\hbar\omega=E_{1}
\end{align} ⟨ H 0 ⟩ = E 1 = 4 ℏ ω + 4 ℏ ω = 2 1 ℏ ω = E 0 4 3 ω ℏ + 4 3 ω ℏ = 4 6 ℏ ω = 2 3 ℏ ω = ( 1 + 2 1 ) ℏ ω = E 1 Exemplo: Encontre ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩ ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩ ⟨ x 2 ⟩ \left\langle x^{2}\right\rangle ⟨ x 2 ⟩ ⟨ p 2 ⟩ \left\langle p^{2}\right\rangle ⟨ p 2 ⟩
De um dos exemplos que fizemos temos:
x = ℏ 2 m ω ( a ^ + + a ^ − ) p ^ = i ℏ m ω 2 ( a ^ + − a ^ − ) x 2 = ℏ 2 m ω ( a ^ + 2 + a ^ + a ^ − + a ^ + a ^ − + a ^ − 2 ) \begin{align}
x & =\sqrt{\frac{\hbar}{2m\omega}}\left(\widehat{a}_{+}+\widehat{a}_{-}\right)\\
\widehat{p} & =i\sqrt{\frac{\hbar m\omega}{2}}\left(\widehat{a}_{+}-\widehat{a}_{-}\right)\\
x^{2} & =\frac{\hbar}{2m\omega}\left(\widehat{a}^{2}_{+}+\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}^{2}_{-}\right)
\end{align} x p x 2 = 2 mω ℏ ( a + + a − ) = i 2 ℏ mω ( a + − a − ) = 2 mω ℏ ( a + 2 + a + a − + a + a − + a − 2 ) Vamos calcular p 2 p^{2} p 2 ⟨ x ⟩ , ⟨ p ⟩ , ⟨ x 2 ⟩ \left\langle x\right\rangle ,\left\langle p\right\rangle ,\left\langle x^{2}\right\rangle ⟨ x ⟩ , ⟨ p ⟩ , ⟨ x 2 ⟩ ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩ a ^ + ψ n = n + 1 ψ n + 1 \widehat{a}_{+}\psi_{n}=\sqrt{n+1}\psi_{n+1} a + ψ n = n + 1 ψ n + 1 a ^ − ψ n = n ψ n − 1 \widehat{a}_{-}\psi_{n}=\sqrt{n}\psi_{n-1} a − ψ n = n ψ n − 1
⟨ x ⟩ = ∫ ψ n ∗ x ψ n d x = ∫ ψ n ∗ ( ℏ 2 m ω ( a ^ + + a ^ − ) ) ψ n d x = ℏ 2 m ω ∫ ψ n ∗ ( a ^ + ψ n + a ^ − ψ n ) d x = ℏ 2 m ω ∫ ψ n ∗ ( n + 1 ψ n + 1 + n ψ n − 1 ) d x = ℏ 2 m ω ( n + 1 ∫ ψ n ∗ ψ n + 1 d x + n ∫ ψ n ∗ ψ n − 1 d x ) = ℏ 2 m ω ( 0 + 0 ) = 0 \begin{align}
\left\langle x\right\rangle & =\int\psi^{*}_{n}x\psi_{n}dx\\
& =\int\psi^{*}_{n}\left(\sqrt{\frac{\hbar}{2m\omega}}\left(\widehat{a}_{+}+\widehat{a}_{-}\right)\right)\psi_{n}dx\\
& =\sqrt{\frac{\hbar}{2m\omega}}\int\psi^{*}_{n}\left(\widehat{a}_{+}\psi_{n}+\widehat{a}_{-}\psi_{n}\right)dx\\
& =\sqrt{\frac{\hbar}{2m\omega}}\int\psi^{*}_{n}\left(\sqrt{n+1}\psi_{n+1}+\sqrt{n}\psi_{n-1}\right)dx\\
& =\sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n+1}\int\psi^{*}_{n}\psi_{n+1}dx+\sqrt{n}\int\psi^{*}_{n}\psi_{n-1}dx\right)\\
& =\sqrt{\frac{\hbar}{2m\omega}}\left(0+0\right)\\
& =0
\end{align} ⟨ x ⟩ = ∫ ψ n ∗ x ψ n d x = ∫ ψ n ∗ ( 2 mω ℏ ( a + + a − ) ) ψ n d x = 2 mω ℏ ∫ ψ n ∗ ( a + ψ n + a − ψ n ) d x = 2 mω ℏ ∫ ψ n ∗ ( n + 1 ψ n + 1 + n ψ n − 1 ) d x = 2 mω ℏ ( n + 1 ∫ ψ n ∗ ψ n + 1 d x + n ∫ ψ n ∗ ψ n − 1 d x ) = 2 mω ℏ ( 0 + 0 ) = 0 Onde usamos os fatos da ortogonalidade para sabermos que o resultado da integral é 0. De forma geral v vimos que devido a ortogonalidade:
⟨ a ^ ± ⟩ = ∫ ψ n ∗ a ^ ± ψ n d x = 0 \left\langle \widehat{a}_{\pm}\right\rangle =\int\psi^{*}_{n}\widehat{a}_{\pm}\psi_{n}dx=0 ⟨ a ± ⟩ = ∫ ψ n ∗ a ± ψ n d x = 0 Vamos usar estes resultados agora para calcular ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩
⟨ p ⟩ = ∫ ψ n ∗ p ψ n d x = ∫ ψ n ∗ ( i ℏ m ω 2 ( a ^ + − a ^ − ) ) ψ n d x = i ℏ m ω 2 ( ∫ ψ n ∗ a ^ + ψ n d x − ∫ ψ n ∗ a ^ − ψ n d x ) = 0 \begin{align}
\left\langle p\right\rangle & =\int\psi^{*}_{n}p\psi_{n}dx\\
& =\int\psi^{*}_{n}\left(i\sqrt{\frac{\hbar m\omega}{2}}\left(\widehat{a}_{+}-\widehat{a}_{-}\right)\right)\psi_{n}dx\\
& =i\sqrt{\frac{\hbar m\omega}{2}}\left(\int\psi^{*}_{n}\widehat{a}_{+}\psi_{n}dx-\int\psi^{*}_{n}\widehat{a}_{-}\psi_{n}dx\right)\\
& =0
\end{align} ⟨ p ⟩ = ∫ ψ n ∗ p ψ n d x = ∫ ψ n ∗ ( i 2 ℏ mω ( a + − a − ) ) ψ n d x = i 2 ℏ mω ( ∫ ψ n ∗ a + ψ n d x − ∫ ψ n ∗ a − ψ n d x ) = 0 Agora para ⟨ x 2 ⟩ \left\langle x^{2}\right\rangle ⟨ x 2 ⟩
⟨ x 2 ⟩ = ∫ ψ n ∗ x 2 ψ n d x = ∫ ψ n ∗ ( ℏ 2 m ω ( a ^ + 2 + a ^ + a ^ − + a ^ − a ^ + + a ^ − 2 ) ) ψ n d x = ℏ 2 m ω ∫ ψ n ∗ ( a ^ + 2 ψ n + a ^ + a ^ − ψ n + a ^ − a ^ + ψ n + a ^ − 2 ψ n ) d x \begin{align}
\left\langle x^{2}\right\rangle & =\int\psi^{*}_{n}x^{2}\psi_{n}dx\\
& =\int\psi^{*}_{n}\left(\frac{\hbar}{2m\omega}\left(\widehat{a}^{2}_{+}+\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{-}\widehat{a}_{+}+\widehat{a}^{2}_{-}\right)\right)\psi_{n}dx\\
& =\frac{\hbar}{2m\omega}\int\psi^{*}_{n}\left(\widehat{a}^{2}_{+}\psi_{n}+\widehat{a}_{+}\widehat{a}_{-}\psi_{n}+\widehat{a}_{-}\widehat{a}_{+}\psi_{n}+\widehat{a}^{2}_{-}\psi_{n}\right)dx
\end{align} ⟨ x 2 ⟩ = ∫ ψ n ∗ x 2 ψ n d x = ∫ ψ n ∗ ( 2 mω ℏ ( a + 2 + a + a − + a − a + + a − 2 ) ) ψ n d x = 2 mω ℏ ∫ ψ n ∗ ( a + 2 ψ n + a + a − ψ n + a − a + ψ n + a − 2 ψ n ) d x Precisamos analisar qual a funaçõ de onda depois de cada operação, já tínhamos conferidos para a ^ − a ^ + ψ n \widehat{a}_{-}\widehat{a}_{+}\psi_{n} a − a + ψ n a ^ + a ^ − \widehat{a}_{+}\widehat{a}_{-} a + a −
a ^ + a ^ + ψ n = a ^ + ( n + 1 ψ n + 1 ) = n + 1 a ^ + ψ n + 1 = ( n + 1 ) ( n + 2 ) ψ n + 2 a ^ − a ^ + ψ n = a ^ − ( n + 1 ψ n + 1 ) = n + 1 a ^ − ψ n + 1 = ( n + 1 ) ( n + 1 ) ψ n = ( n + 1 ) ψ n a ^ + a ^ − ψ n = a ^ + ( n ψ n − 1 ) = n a ^ + ψ n − 1 = n n ψ n = n ψ n a ^ − a ^ − ψ n = a ^ − ( n ψ n − 1 ) = n a ^ − ψ n − 1 = ( n − 1 ) n ψ n − 2 \begin{align*}
\widehat{a}_{+}\widehat{a}_{+}\psi_{n} & =\widehat{a}_{+}\left(\sqrt{n+1}\psi_{n+1}\right) & =\sqrt{n+1}\widehat{a}_{+}\psi_{n+1} & =\sqrt{\left(n+1\right)\left(n+2\right)}\psi_{n+2}\\
\widehat{a}_{-}\widehat{a}_{+}\psi_{n} & =\widehat{a}_{-}\left(\sqrt{n+1}\psi_{n+1}\right) & =\sqrt{n+1}\widehat{a}_{-}\psi_{n+1} & =\sqrt{\left(n+1\right)\left(n+1\right)}\psi_{n}=\left(n+1\right)\psi_{n}\\
\widehat{a}_{+}\widehat{a}_{-}\psi_{n} & =\widehat{a}_{+}\left(\sqrt{n}\psi_{n-1}\right) & =\sqrt{n}\widehat{a}_{+}\psi_{n-1} & =\sqrt{nn}\psi_{n}=n\psi_{n}\\
\widehat{a}_{-}\widehat{a}_{-}\psi_{n} & =\widehat{a}_{-}\left(\sqrt{n}\psi_{n-1}\right) & =\sqrt{n}\widehat{a}_{-}\psi_{n-1} & =\sqrt{\left(n-1\right)n}\psi_{n-2}
\end{align*} a + a + ψ n a − a + ψ n a + a − ψ n a − a − ψ n = a + ( n + 1 ψ n + 1 ) = a − ( n + 1 ψ n + 1 ) = a + ( n ψ n − 1 ) = a − ( n ψ n − 1 ) = n + 1 a + ψ n + 1 = n + 1 a − ψ n + 1 = n a + ψ n − 1 = n a − ψ n − 1 = ( n + 1 ) ( n + 2 ) ψ n + 2 = ( n + 1 ) ( n + 1 ) ψ n = ( n + 1 ) ψ n = nn ψ n = n ψ n = ( n − 1 ) n ψ n − 2 Então podemos ver que as integrais correspondendo a a ^ ± \widehat{a}_{\pm} a ± a ^ ± ψ n = c a ± ψ n ± 2 \widehat{a}_{\pm}\psi_{n}=c_{a_{\pm}}\psi_{n\pm2} a ± ψ n = c a ± ψ n ± 2
∫ ψ n ∗ a ^ ± ψ n d x = c a ± ∫ ψ n ∗ ψ n ± 2 d x = 0 \int\psi^{*}_{n}\widehat{a}_{\pm}\psi_{n}dx=c_{a_{\pm}}\int\psi^{*}_{n}\psi_{n\pm2}dx=0 ∫ ψ n ∗ a ± ψ n d x = c a ± ∫ ψ n ∗ ψ n ± 2 d x = 0 Restando então apenas as integrais com a ^ ∓ a ^ ± \widehat{a}_{\mp}\widehat{a}_{\pm} a ∓ a ±
⟨ x 2 ⟩ = ℏ 2 m ω ∫ ψ n ∗ ( a ^ + 2 ψ n + a ^ + a ^ − ψ n + a ^ − a ^ + ψ n + a ^ − 2 ψ n ) d x = ℏ 2 m ω ∫ ψ n ∗ ( a ^ + a ^ − ψ n + a ^ − a ^ + ψ n ) d x = ℏ 2 m ω ∫ ψ n ∗ ( n ψ n + ( n + 1 ) ψ n ) d x = ℏ 2 m ω ∫ ψ n ∗ ( n + n + 1 ) ψ n d x = ℏ ( 2 n + 1 ) 2 m ω ∫ ψ n ∗ ψ n d x = ℏ ( 2 n + 1 ) 2 m ω \begin{align}
\left\langle x^{2}\right\rangle & =\frac{\hbar}{2m\omega}\int\psi^{*}_{n}\left(\widehat{a}^{2}_{+}\psi_{n}+\widehat{a}_{+}\widehat{a}_{-}\psi_{n}+\widehat{a}_{-}\widehat{a}_{+}\psi_{n}+\widehat{a}^{2}_{-}\psi_{n}\right)dx\\
& =\frac{\hbar}{2m\omega}\int\psi^{*}_{n}\left(\widehat{a}_{+}\widehat{a}_{-}\psi_{n}+\widehat{a}_{-}\widehat{a}_{+}\psi_{n}\right)dx\\
& =\frac{\hbar}{2m\omega}\int\psi^{*}_{n}\left(n\psi_{n}+\left(n+1\right)\psi_{n}\right)dx\\
& =\frac{\hbar}{2m\omega}\int\psi^{*}_{n}\left(n+n+1\right)\psi_{n}dx\\
& =\frac{\hbar\left(2n+1\right)}{2m\omega}\int\psi^{*}_{n}\psi_{n}dx\\
& =\frac{\hbar\left(2n+1\right)}{2m\omega}
\end{align} ⟨ x 2 ⟩ = 2 mω ℏ ∫ ψ n ∗ ( a + 2 ψ n + a + a − ψ n + a − a + ψ n + a − 2 ψ n ) d x = 2 mω ℏ ∫ ψ n ∗ ( a + a − ψ n + a − a + ψ n ) d x = 2 mω ℏ ∫ ψ n ∗ ( n ψ n + ( n + 1 ) ψ n ) d x = 2 mω ℏ ∫ ψ n ∗ ( n + n + 1 ) ψ n d x = 2 mω ℏ ( 2 n + 1 ) ∫ ψ n ∗ ψ n d x = 2 mω ℏ ( 2 n + 1 ) Ou então:
⟨ x 2 ⟩ = ( n + 1 2 ) ℏ m ω \left\langle x^{2}\right\rangle =\left(n+\frac{1}{2}\right)\frac{\hbar}{m\omega} ⟨ x 2 ⟩ = ( n + 2 1 ) mω ℏ Agora para ⟨ p 2 ⟩ \left\langle p^{2}\right\rangle ⟨ p 2 ⟩
⟨ p 2 ⟩ = ∫ ψ n ∗ p 2 ψ n d x = ∫ ψ n ∗ ( i ℏ m ω 2 ( a ^ + − a ^ − ) ) 2 ψ n d x = − ℏ m ω 2 ∫ ψ n ∗ ( a ^ + 2 − a ^ + a ^ − − a ^ − a ^ + + a ^ − 2 ) ψ n d x \begin{align}
\left\langle p^{2}\right\rangle & =\int\psi^{*}_{n}p^{2}\psi_{n}dx\\
& =\int\psi^{*}_{n}\left(i\sqrt{\frac{\hbar m\omega}{2}}\left(\widehat{a}_{+}-\widehat{a}_{-}\right)\right)^{2}\psi_{n}dx\\
& =-\frac{\hbar m\omega}{2}\int\psi^{*}_{n}\left(\widehat{a}^{2}_{+}-\widehat{a}_{+}\widehat{a}_{-}-\widehat{a}_{-}\widehat{a}_{+}+\widehat{a}^{2}_{-}\right)\psi_{n}dx
\end{align} ⟨ p 2 ⟩ = ∫ ψ n ∗ p 2 ψ n d x = ∫ ψ n ∗ ( i 2 ℏ mω ( a + − a − ) ) 2 ψ n d x = − 2 ℏ mω ∫ ψ n ∗ ( a + 2 − a + a − − a − a + + a − 2 ) ψ n d x Já vimos que as integrais envolvendo a ^ ± \widehat{a}_{\pm} a ±
∫ ψ n ∗ ( a ^ + a ^ − + a ^ − a ^ + ) ψ n d x = ( 2 n + 1 ) \int\psi^{*}_{n}\left(\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{-}\widehat{a}_{+}\right)\psi_{n}dx=\left(2n+1\right) ∫ ψ n ∗ ( a + a − + a − a + ) ψ n d x = ( 2 n + 1 ) Então:
⟨ p 2 ⟩ = ℏ m ω 2 ∫ ψ n ∗ ( a ^ + a ^ − + a ^ − a ^ + ) ψ n d x = ℏ m ω 2 ( 2 n + 1 ) \begin{align}
\left\langle p^{2}\right\rangle & =\frac{\hbar m\omega}{2}\int\psi^{*}_{n}\left(\widehat{a}_{+}\widehat{a}_{-}+\widehat{a}_{-}\widehat{a}_{+}\right)\psi_{n}dx\\
& =\frac{\hbar m\omega}{2}\left(2n+1\right)
\end{align} ⟨ p 2 ⟩ = 2 ℏ mω ∫ ψ n ∗ ( a + a − + a − a + ) ψ n d x = 2 ℏ mω ( 2 n + 1 ) Ou simplesmente:
⟨ p 2 ⟩ = ℏ m ω ( n + 1 2 ) \left\langle p^{2}\right\rangle =\hbar m\omega\left(n+\frac{1}{2}\right) ⟨ p 2 ⟩ = ℏ mω ( n + 2 1 ) Agora é fácil calcularmos ⟨ T ⟩ \left\langle T\right\rangle ⟨ T ⟩
⟨ T ⟩ = ⟨ p 2 ⟩ 2 m = 1 2 ( n + 1 2 ) ℏ ω \left\langle T\right\rangle =\frac{\left\langle p^{2}\right\rangle }{2m}=\frac{1}{2}\left(n+\frac{1}{2}\right)\hbar\omega ⟨ T ⟩ = 2 m ⟨ p 2 ⟩ = 2 1 ( n + 2 1 ) ℏ ω Assim como podemos calcular facilmente também a relação de incerteza:
σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = ( n + 1 2 ) ℏ m ω = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ℏ m ω ( n + 1 2 ) \begin{align}
\sigma_{x} & =\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}=\sqrt{\left(n+\frac{1}{2}\right)\frac{\hbar}{m\omega}}\\
& =\sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}=\sqrt{\hbar m\omega\left(n+\frac{1}{2}\right)}
\end{align} σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = ( n + 2 1 ) mω ℏ = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ℏ mω ( n + 2 1 ) Ou seja:
σ x σ p = ( n + 1 2 ) ℏ m ω ℏ m ω ( n + 1 2 ) = ( n + 1 2 ) ℏ \sigma_{x}\sigma_{p}=\sqrt{\left(n+\frac{1}{2}\right)\frac{\hbar}{m\omega}}\sqrt{\hbar m\omega\left(n+\frac{1}{2}\right)}=\left(n+\frac{1}{2}\right)\hbar σ x σ p = ( n + 2 1 ) mω ℏ ℏ mω ( n + 2 1 ) = ( n + 2 1 ) ℏ Como n ≥ 0 n\geq0 n ≥ 0
σ x σ p ≥ ℏ 2 \sigma_{x}\sigma_{p}\geq\frac{\hbar}{2} σ x σ p ≥ 2 ℏ Método analítico ¶ Agora podemos retomar a equação de schroedinger para o oscilador harmônico e resolver usando o método das séries de potências[5]
− ℏ 2 2 m d 2 d x 2 ψ + 1 2 m ω 2 x 2 ψ = E ψ -\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi+\frac{1}{2}m\omega^{2}x^{2}\psi=E\psi − 2 m ℏ 2 d x 2 d 2 ψ + 2 1 m ω 2 x 2 ψ = E ψ Introduzindo a variável ε ≡ m ω ℏ x \varepsilon\equiv\sqrt{\frac{m\omega}{\hbar}}x ε ≡ ℏ mω x K ≖ 2 E ℏ ω K\eqcirc\frac{2E}{\hbar\omega} K ≖ ℏ ω 2 E
d f d x = d f d ε d ε d x \frac{df}{dx}=\frac{df}{d\varepsilon}\frac{d\varepsilon}{dx} d x df = d ε df d x d ε Então a segunda derivada é:
d 2 f d x 2 = d d x ( d f d x ) = d d ε ( d f d ε d ε d x ) d ε d x = d 2 f d ε 2 ( d ε d x ) 2 = d 2 f d ε 2 ( m ω ℏ ) 2 = m ω ℏ d 2 f d ε 2 \begin{align}
\frac{d^{2}f}{dx^{2}} & =\frac{d}{dx}\left(\frac{df}{dx}\right)=\frac{d}{d\varepsilon}\left(\frac{df}{d\varepsilon}\frac{d\varepsilon}{dx}\right)\frac{d\varepsilon}{dx}=\frac{d^{2}f}{d\varepsilon^{2}}\left(\frac{d\varepsilon}{dx}\right)^{2}=\frac{d^{2}f}{d\varepsilon^{2}}\left(\sqrt{\frac{m\omega}{\hbar}}\right)^{2}=\frac{m\omega}{\hbar}\frac{d^{2}f}{d\varepsilon^{2}}
\end{align} d x 2 d 2 f = d x d ( d x df ) = d ε d ( d ε df d x d ε ) d x d ε = d ε 2 d 2 f ( d x d ε ) 2 = d ε 2 d 2 f ( ℏ mω ) 2 = ℏ mω d ε 2 d 2 f sendo
m ω 2 x 2 = m ω 2 ( ℏ m ω ε ) 2 = m ω 2 ( ℏ m ω ) ε 2 = ω ℏ ε 2 m\omega^{2}x^{2}=m\omega^{2}\left(\sqrt{\frac{\hbar}{m\omega}}\varepsilon\right)^{2}=m\omega^{2}\left(\frac{\hbar}{m\omega}\right)\varepsilon^{2}=\omega\hbar\varepsilon^{2} m ω 2 x 2 = m ω 2 ( mω ℏ ε ) 2 = m ω 2 ( mω ℏ ) ε 2 = ω ℏ ε 2 − ℏ 2 2 m d 2 d x 2 ψ + 1 2 m ω 2 x 2 ψ = E ψ − ℏ 2 2 m m ω ℏ d 2 ψ d ε 2 + 1 2 ℏ ω ε 2 ψ = E ψ − m ω 2 d 2 ψ d ε 2 = ( E − 1 2 ℏ ω ε 2 ) ψ d 2 ψ d ε 2 = ( ε 2 − 2 E ℏ ω ) ψ d 2 ψ d ε 2 = ( ε 2 − K ) ψ \begin{align}
-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2\psi}}+\frac{1}{2}m\omega^{2}x^{2}\psi & =E\psi\\
-\frac{\hbar^{2}}{2m}\frac{m\omega}{\hbar}\frac{d^{2}\psi}{d\varepsilon^{2}}+\frac{1}{2}\hbar\omega\varepsilon^{2}\psi & =E\psi\\
-\frac{m\omega}{2}\frac{d^{2}\psi}{d\varepsilon^{2}} & =\left(E-\frac{1}{2}\hbar\omega\varepsilon^{2}\right)\psi\\
\frac{d^{2}\psi}{d\varepsilon^{2}} & =\left(\varepsilon^{2}-\frac{2E}{\hbar\omega}\right)\psi\\
\frac{d^{2}\psi}{d\varepsilon^{2}} & =\left(\varepsilon^{2}-K\right)\psi
\end{align} − 2 m ℏ 2 d x 2 ψ d 2 + 2 1 m ω 2 x 2 ψ − 2 m ℏ 2 ℏ mω d ε 2 d 2 ψ + 2 1 ℏ ω ε 2 ψ − 2 mω d ε 2 d 2 ψ d ε 2 d 2 ψ d ε 2 d 2 ψ = E ψ = E ψ = ( E − 2 1 ℏ ω ε 2 ) ψ = ( ε 2 − ℏ ω 2 E ) ψ = ( ε 2 − K ) ψ Nosso objetivo é resolver esta equação e obter os valores permitidos de K K K x x x ε \varepsilon ε ε 2 ≫ K \varepsilon^{2}\gg K ε 2 ≫ K
d 2 ψ d ε 2 ≈ ε 2 ψ \frac{d^{2}\psi}{d\varepsilon^{2}}\approx\varepsilon^{2}\psi d ε 2 d 2 ψ ≈ ε 2 ψ Onde uma solução é:
ψ ( ε ) ≈ A e − ε 2 / 2 + B e + ε 2 / 2 \psi\left(\varepsilon\right)\approx Ae^{-\varepsilon^{2}/2}+Be^{+\varepsilon^{2}/2} ψ ( ε ) ≈ A e − ε 2 /2 + B e + ε 2 /2 Derivando duas vezes a função de onda proposta:
d ψ d ε = − ε A e − ε 2 / 2 + ε B e + ε 2 / 2 d 2 ψ d ε 2 = − ( A e − ε 2 / 2 − ε 2 A e − ε 2 / 2 ) + ( B e + ε 2 / 2 + ε 2 B e + ε 2 / 2 ) d 2 ψ d ε 2 = − ( 1 − ε 2 ) A e − ε 2 / 2 + ( 1 + ε 2 ) B e + ε 2 / 2 \begin{align}
\frac{d\psi}{d\varepsilon} & =-\varepsilon Ae^{-\varepsilon^{2}/2}+\varepsilon Be^{+\varepsilon^{2}/2}\\
\frac{d^{2}\psi}{d\varepsilon^{2}} & =-\left(Ae^{-\varepsilon^{2}/2}-\varepsilon^{2}Ae^{-\varepsilon^{2}/2}\right)+\left(Be^{+\varepsilon^{2}/2}+\varepsilon^{2}Be^{+\varepsilon^{2}/2}\right)\\
\frac{d^{2}\psi}{d\varepsilon^{2}} & =-\left(1-\varepsilon^{2}\right)Ae^{-\varepsilon^{2}/2}+\left(1+\varepsilon^{2}\right)Be^{+\varepsilon^{2}/2}
\end{align} d ε d ψ d ε 2 d 2 ψ d ε 2 d 2 ψ = − ε A e − ε 2 /2 + εB e + ε 2 /2 = − ( A e − ε 2 /2 − ε 2 A e − ε 2 /2 ) + ( B e + ε 2 /2 + ε 2 B e + ε 2 /2 ) = − ( 1 − ε 2 ) A e − ε 2 /2 + ( 1 + ε 2 ) B e + ε 2 /2 Agora se considerarmos a aproximação onde ε → ∞ \varepsilon\rightarrow\infty ε → ∞ 1 ≪ ε 2 1\ll\varepsilon^{2} 1 ≪ ε 2
d 2 ψ d ε 2 ≈ A ε 2 e − ε 2 / 2 + ε 2 B e + ε 2 / 2 ≈ ε 2 ( A e − ε 2 / 2 + B e + ε 2 / 2 ) ≈ ε 2 ψ \begin{align}
\frac{d^{2}\psi}{d\varepsilon^{2}} & \approx A\varepsilon^{2}e^{-\varepsilon^{2}/2}+\varepsilon^{2}Be^{+\varepsilon^{2}/2}\\
& \approx\varepsilon^{2}\left(Ae^{-\varepsilon^{2}/2}+Be^{+\varepsilon^{2}/2}\right)\\
& \approx\varepsilon^{2}\psi
\end{align} d ε 2 d 2 ψ ≈ A ε 2 e − ε 2 /2 + ε 2 B e + ε 2 /2 ≈ ε 2 ( A e − ε 2 /2 + B e + ε 2 /2 ) ≈ ε 2 ψ Então a equação 157 156 B B B ∣ x ∣ → ∞ \left|x\right|\rightarrow\infty ∣ x ∣ → ∞ [6] ε \varepsilon ε
ψ ( ε ) → ( ) e − ε 2 / 2 \psi\left(\varepsilon\right)\rightarrow\left(\quad\right)e^{-\varepsilon^{2}/2} ψ ( ε ) → ( ) e − ε 2 /2 Isso nos motiva a propor uma solução do tipo:
ψ ( ε ) = h ( ε ) e − ε 2 / 2 \psi\left(\varepsilon\right)=h\left(\varepsilon\right)e^{-\varepsilon^{2}/2} ψ ( ε ) = h ( ε ) e − ε 2 /2 Com a esperança de que h ( ε ) h\left(\varepsilon\right) h ( ε ) ψ ( ε ) \psi\left(\varepsilon\right) ψ ( ε ) [7]
d ψ d ε = d h d ε e − ε 2 / 2 + h d d ε e − ε 2 / 2 = d h d ε e − ε 2 / 2 − ε h e − ε 2 / 2 = ( d h d ε − ε h ) e − ε 2 / 2 \frac{d\psi}{d\varepsilon}=\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}+h\frac{d}{d\varepsilon}e^{-\varepsilon^{2}/2}=\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}-\varepsilon he^{-\varepsilon^{2}/2}=\left(\frac{dh}{d\varepsilon}-\varepsilon h\right)e^{-\varepsilon^{2}/2} d ε d ψ = d ε d h e − ε 2 /2 + h d ε d e − ε 2 /2 = d ε d h e − ε 2 /2 − ε h e − ε 2 /2 = ( d ε d h − ε h ) e − ε 2 /2 Derivando novamente:
d 2 ψ d ε 2 = d d ε [ ( d h d ε − ε h ) e − ε 2 / 2 ] = d d ε [ d h d ε e − ε 2 / 2 ] − d d ε [ ( ε ) ( h e − ε 2 / 2 ) ] = [ d 2 h d ε 2 e − ε 2 / 2 − ε d h d ε e − ε 2 / 2 ] − [ h e − ε 2 / 2 + ε d d ε ( h e − ε 2 / 2 ) ] = [ d 2 h d ε 2 e − ε 2 / 2 − ε d h d ε e − ε 2 / 2 ] − [ h e − ε 2 / 2 + ε ( d h d ε e − ε 2 / 2 − ε h e − ε 2 / 2 ) ] = d 2 h d ε 2 e − ε 2 / 2 − ε d h d ε e − ε 2 / 2 − h e − ε 2 / 2 − ε d h d ε e − ε 2 / 2 + ε 2 h e − ε 2 / 2 = ( d 2 h d ε 2 − 2 ε d h d ε + ε 2 h − h ) e − ε 2 / 2 = ( d 2 h d ε 2 − 2 ε d h d ε + ( ε 2 − 1 ) h ) e − ε 2 / 2 \begin{align}
\frac{d^{2}\psi}{d\varepsilon^{2}} & =\frac{d}{d\varepsilon}\left[\left(\frac{dh}{d\varepsilon}-\varepsilon h\right)e^{-\varepsilon^{2}/2}\right]\\
& =\frac{d}{d\varepsilon}\left[\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}\right]-\frac{d}{d\varepsilon}\left[\left(\varepsilon\right)\left(he^{-\varepsilon^{2}/2}\right)\right]\\
& =\left[\frac{d^{2}h}{d\varepsilon^{2}}e^{-\varepsilon^{2}/2}-\varepsilon\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}\right]-\left[he^{-\varepsilon^{2}/2}+\varepsilon\frac{d}{d\varepsilon}\left(he^{-\varepsilon^{2}/2}\right)\right]\\
& =\left[\frac{d^{2}h}{d\varepsilon^{2}}e^{-\varepsilon^{2}/2}-\varepsilon\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}\right]-\left[he^{-\varepsilon^{2}/2}+\varepsilon\left(\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}-\varepsilon he^{-\varepsilon^{2}/2}\right)\right]\\
& =\frac{d^{2}h}{d\varepsilon^{2}}e^{-\varepsilon^{2}/2}-\varepsilon\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}-he^{-\varepsilon^{2}/2}-\varepsilon\frac{dh}{d\varepsilon}e^{-\varepsilon^{2}/2}+\varepsilon^{2}he^{-\varepsilon^{2}/2}\\
& =\left(\frac{d^{2}h}{d\varepsilon^{2}}-2\varepsilon\frac{dh}{d\varepsilon}+\varepsilon^{2}h-h\right)e^{-\varepsilon^{2}/2}\\
& =\left(\frac{d^{2}h}{d\varepsilon^{2}}-2\varepsilon\frac{dh}{d\varepsilon}+\left(\varepsilon^{2}-1\right)h\right)e^{-\varepsilon^{2}/2}
\end{align} d ε 2 d 2 ψ = d ε d [ ( d ε d h − ε h ) e − ε 2 /2 ] = d ε d [ d ε d h e − ε 2 /2 ] − d ε d [ ( ε ) ( h e − ε 2 /2 ) ] = [ d ε 2 d 2 h e − ε 2 /2 − ε d ε d h e − ε 2 /2 ] − [ h e − ε 2 /2 + ε d ε d ( h e − ε 2 /2 ) ] = [ d ε 2 d 2 h e − ε 2 /2 − ε d ε d h e − ε 2 /2 ] − [ h e − ε 2 /2 + ε ( d ε d h e − ε 2 /2 − ε h e − ε 2 /2 ) ] = d ε 2 d 2 h e − ε 2 /2 − ε d ε d h e − ε 2 /2 − h e − ε 2 /2 − ε d ε d h e − ε 2 /2 + ε 2 h e − ε 2 /2 = ( d ε 2 d 2 h − 2 ε d ε d h + ε 2 h − h ) e − ε 2 /2 = ( d ε 2 d 2 h − 2 ε d ε d h + ( ε 2 − 1 ) h ) e − ε 2 /2 Lembrando que h = h ( ε ) h=h\left(\varepsilon\right) h = h ( ε )
d 2 ψ d ε 2 = ( ε 2 − K ) ψ ( d 2 h d ε 2 − 2 ε d h d ε + ( ε 2 − 1 ) h ) e − ε 2 / 2 = ( ε 2 − K ) h e − ε 2 / 2 d 2 h d ε 2 − 2 ε d h d ε − 1 h = − K h d 2 h d ε 2 − 2 ε d h d ε + ( K − 1 ) h = 0 \begin{align}
\frac{d^{2}\psi}{d\varepsilon^{2}} & =\left(\varepsilon^{2}-K\right)\psi\\
\left(\frac{d^{2}h}{d\varepsilon^{2}}-2\varepsilon\frac{dh}{d\varepsilon}+\left(\varepsilon^{2}-1\right)h\right)e^{-\varepsilon^{2}/2} & =\left(\varepsilon^{2}-K\right)he^{-\varepsilon^{2}/2}\\
\frac{d^{2}h}{d\varepsilon^{2}}-2\varepsilon\frac{dh}{d\varepsilon}-1h & =-Kh\\
\frac{d^{2}h}{d\varepsilon^{2}}-2\varepsilon\frac{dh}{d\varepsilon}+\left(K-1\right)h & =0
\end{align} d ε 2 d 2 ψ ( d ε 2 d 2 h − 2 ε d ε d h + ( ε 2 − 1 ) h ) e − ε 2 /2 d ε 2 d 2 h − 2 ε d ε d h − 1 h d ε 2 d 2 h − 2 ε d ε d h + ( K − 1 ) h = ( ε 2 − K ) ψ = ( ε 2 − K ) h e − ε 2 /2 = − K h = 0 Seguindo então nosso livro base, vamos propor uma solução na forma da série de potências em ε \varepsilon ε [8]
h ( ε ) = a 0 + a 1 ε + a 2 ε 2 + ⋯ = ∑ j = 0 ∞ a j ε j h\left(\varepsilon\right)=a_{0}+a_{1}\varepsilon+a_{2}\varepsilon^{2}+\dots=\sum^{\infty}_{j=0}a_{j}\varepsilon^{j} h ( ε ) = a 0 + a 1 ε + a 2 ε 2 + ⋯ = j = 0 ∑ ∞ a j ε j Logo:
d h d ε = 0 + ∑ j = 1 ∞ a j d ε j d ε = ∑ j = 1 ∞ a j j ε j − 1 = ∑ j = 0 ∞ a j j ε j − 1 d 2 h d ε 2 = 0 + ∑ j = 2 ∞ a j j d ε j − 1 d ε = ∑ j = 0 ∞ a j + 2 ( j + 2 ) ( j + 1 ) ε j \begin{align}
\frac{dh}{d\varepsilon} & =0+\sum^{\infty}_{j=1}a_{j}\frac{d\varepsilon^{j}}{d\varepsilon}=\sum^{\infty}_{j=1}a_{j}j\varepsilon^{j-1}=\sum^{\infty}_{j=0}a_{j}j\varepsilon^{j-1}\\
\frac{d^{2}h}{d\varepsilon^{2}} & =0+\sum^{\infty}_{j=2}a_{j}j\frac{d\varepsilon^{j-1}}{d\varepsilon}=\sum^{\infty}_{j=0}a_{j+2}\left(j+2\right)\left(j+1\right)\varepsilon^{j}
\end{align} d ε d h d ε 2 d 2 h = 0 + j = 1 ∑ ∞ a j d ε d ε j = j = 1 ∑ ∞ a j j ε j − 1 = j = 0 ∑ ∞ a j j ε j − 1 = 0 + j = 2 ∑ ∞ a j j d ε d ε j − 1 = j = 0 ∑ ∞ a j + 2 ( j + 2 ) ( j + 1 ) ε j Onde decorre que quando derivamos, sempre zeramos o primeiro termo por ser constante, começando então a derivada do somatório com o índice seguinte. Na primeira derivada podemos começar em 0 pois todos termos multiplicam j j j j = 0 j=0 j = 0 j j j j = 0 j=0 j = 0
Substituindo então em 164
d 2 h d ε 2 − 2 ε d h d ε + ( K − 1 ) h = 0 ∑ j = 0 ∞ a j + 2 ( j + 2 ) ( j + 1 ) ε j − 2 ε ∑ j = 0 ∞ a j j ε j − 1 + ( K − 1 ) ∑ j = 0 ∞ a j ε j = 0 ∑ j = 0 ∞ a j + 2 ( j + 2 ) ( j + 1 ) ε j − 2 ∑ j = 0 ∞ a j j ε j + ( K − 1 ) ∑ j = 0 ∞ a j ε j = 0 ∑ j = 0 ∞ ( a j + 2 ( j + 2 ) ( j + 1 ) − 2 a j j + ( K − 1 ) a j ) ε j = 0 \begin{align}
\frac{d^{2}h}{d\varepsilon^{2}}-2\varepsilon\frac{dh}{d\varepsilon}+\left(K-1\right)h & =0\\
\sum^{\infty}_{j=0}a_{j+2}\left(j+2\right)\left(j+1\right)\varepsilon^{j}-2\varepsilon\sum^{\infty}_{j=0}a_{j}j\varepsilon^{j-1}+\left(K-1\right)\sum^{\infty}_{j=0}a_{j}\varepsilon^{j} & =0\\
\sum^{\infty}_{j=0}a_{j+2}\left(j+2\right)\left(j+1\right)\varepsilon^{j}-2\sum^{\infty}_{j=0}a_{j}j\varepsilon^{j}+\left(K-1\right)\sum^{\infty}_{j=0}a_{j}\varepsilon^{j} & =0\\
\sum^{\infty}_{j=0}\left(a_{j+2}\left(j+2\right)\left(j+1\right)-2a_{j}j+\left(K-1\right)a_{j}\right)\varepsilon^{j} & =0
\end{align} d ε 2 d 2 h − 2 ε d ε d h + ( K − 1 ) h j = 0 ∑ ∞ a j + 2 ( j + 2 ) ( j + 1 ) ε j − 2 ε j = 0 ∑ ∞ a j j ε j − 1 + ( K − 1 ) j = 0 ∑ ∞ a j ε j j = 0 ∑ ∞ a j + 2 ( j + 2 ) ( j + 1 ) ε j − 2 j = 0 ∑ ∞ a j j ε j + ( K − 1 ) j = 0 ∑ ∞ a j ε j j = 0 ∑ ∞ ( a j + 2 ( j + 2 ) ( j + 1 ) − 2 a j j + ( K − 1 ) a j ) ε j = 0 = 0 = 0 = 0 A única forma disso ser verdade, isto é, deste soma ser zero é se coeficiente de cada potência seja zero[9] f ( x ) = a x 0 + b x 1 = 0 f\left(x\right)=ax^{0}+bx^{1}=0 f ( x ) = a x 0 + b x 1 = 0 a = − b x a=-bx a = − b x x x x a = b = 0 a=b=0 a = b = 0
a j + 2 ( j + 2 ) ( j + 1 ) − 2 a j j + ( K − 1 ) a j = 0 a j + 2 ( j + 2 ) ( j + 1 ) − [ 2 j − K + 1 ] a j = 0 a j + 2 ( j + 2 ) ( j + 1 ) = [ 2 j − K + 1 ] a j a j + 2 = ( 2 j − K + 1 ) ( j + 2 ) ( j + 1 ) a j \begin{align}
a_{j+2}\left(j+2\right)\left(j+1\right)-2a_{j}j+\left(K-1\right)a_{j} & =0\\
a_{j+2}\left(j+2\right)\left(j+1\right)-\left[2j-K+1\right]a_{j} & =0\\
a_{j+2}\left(j+2\right)\left(j+1\right) & =\left[2j-K+1\right]a_{j}\\
a_{j+2} & =\frac{\left(2j-K+1\right)}{\left(j+2\right)\left(j+1\right)}a_{j}
\end{align} a j + 2 ( j + 2 ) ( j + 1 ) − 2 a j j + ( K − 1 ) a j a j + 2 ( j + 2 ) ( j + 1 ) − [ 2 j − K + 1 ] a j a j + 2 ( j + 2 ) ( j + 1 ) a j + 2 = 0 = 0 = [ 2 j − K + 1 ] a j = ( j + 2 ) ( j + 1 ) ( 2 j − K + 1 ) a j A fórmula de recorrência é equivalente à solução da equação de Schrödinger, no sentido de que qualquer solução obtida da recorrência satisfaz a equação de Schrödinger original. Começando com a 0 a_{0} a 0
a 2 = 1 − K 2 a 0 , a 4 = ( 5 − K ) 12 a 2 , … a_{2}=\frac{1-K}{2}a_{0},\quad a_{4}=\frac{\left(5-K\right)}{12}a_{2},\quad\dots a 2 = 2 1 − K a 0 , a 4 = 12 ( 5 − K ) a 2 , … E começando com a 1 a_{1} a 1
a 3 = ( 3 − K ) 6 a 1 , a 5 = ( 7 − K ) 20 a 3 … a_{3}=\frac{\left(3-K\right)}{6}a_{1},\quad a_{5}=\frac{\left(7-K\right)}{20}a_{3}\quad\dots a 3 = 6 ( 3 − K ) a 1 , a 5 = 20 ( 7 − K ) a 3 … Então podemos escrever a solução completa como:
h ( ε ) = h p a r ( ε ) + h i m p a r ( ε ) h\left(\varepsilon\right)=h_{par}\left(\varepsilon\right)+h_{impar}\left(\varepsilon\right) h ( ε ) = h p a r ( ε ) + h im p a r ( ε ) Onde:
h p a r ( ε ) = a 0 + a 2 ε 2 + a 4 ε 4 + … h i m p a r ( ε ) = a 1 + a 3 ε 3 + a 5 ε 5 + … \begin{align*}
h_{par}\left(\varepsilon\right) & =a_{0}+a_{2}\varepsilon^{2}+a_{4}\varepsilon^{4}+\dots\\
h_{impar}\left(\varepsilon\right) & =a_{1}+a_{3}\varepsilon^{3}+a_{5}\varepsilon^{5}+\dots
\end{align*} h p a r ( ε ) h im p a r ( ε ) = a 0 + a 2 ε 2 + a 4 ε 4 + … = a 1 + a 3 ε 3 + a 5 ε 5 + … Então nossa solução a partir da fórmula de recorrência determina h ( ε ) h\left(\varepsilon\right) h ( ε ) a 0 a_{0} a 0 a 1 a_{1} a 1 j j j
a j + 2 ≈ lim j → ∞ ( 2 j − K + 1 ) ( j + 2 ) ( j + 1 ) a j ≈ lim j → ∞ 2 j − K + 1 j 2 + 3 j + 2 a j ≈ a j j \begin{align}
a_{j+2} & \approx\lim_{j\rightarrow\infty}\frac{\left(2j-K+1\right)}{\left(j+2\right)\left(j+1\right)}a_{j}\\
& \approx\lim_{j\rightarrow\infty}\frac{2j-K+1}{j^{2}+3j+2}a_{j}\\
& \approx\frac{a_{j}}{j}
\end{align} a j + 2 ≈ j → ∞ lim ( j + 2 ) ( j + 1 ) ( 2 j − K + 1 ) a j ≈ j → ∞ lim j 2 + 3 j + 2 2 j − K + 1 a j ≈ j a j Uma solução aproximada proposta é[10]
a j ≈ C ( j / 2 ) ! a_{j}\approx\frac{C}{\left(j/2\right)!} a j ≈ ( j /2 ) ! C Podemos testar essa solução fazendo:
a j + 2 a j ≈ C ( [ j + 2 ] / 2 ) ! C ( j / 2 ) ! = C ( j + 2 2 ) ! ( j 2 ) ! C = ( j 2 ) ! ( j 2 + 1 ) ! = 1 j 2 + 1 = 2 j + 2 \frac{a_{j+2}}{a_{j}}\approx\frac{\frac{C}{\left(\left[j+2\right]/2\right)!}}{\frac{C}{\left(j/2\right)!}}=\frac{C}{\left(\frac{j+2}{2}\right)!}\frac{\left(\frac{j}{2}\right)!}{C}=\frac{\left(\frac{j}{2}\right)!}{\left(\frac{j}{2}+1\right)!}=\frac{1}{\frac{j}{2}+1}=\frac{2}{j+2} a j a j + 2 ≈ ( j /2 ) ! C ( [ j + 2 ] /2 ) ! C = ( 2 j + 2 ) ! C C ( 2 j ) ! = ( 2 j + 1 ) ! ( 2 j ) ! = 2 j + 1 1 = j + 2 2 Ou apenas:
a j + 2 ≈ 2 j + 2 a j a_{j+2}\approx\frac{2}{j+2}a_{j} a j + 2 ≈ j + 2 2 a j E como estamos interessados em um regime de grandes valores de j j j
a j + 2 ≈ a j j a_{j+2}\approx\frac{a_{j}}{j} a j + 2 ≈ j a j Que era nosso resultado anterior, demonstrando a validade da solução aproximada. Utilizando então nossa aproximação a j ≈ C ( j / 2 ) ! a_{j}\approx\frac{C}{\left(j/2\right)!} a j ≈ ( j /2 ) ! C h ( ε ) h\left(\varepsilon\right) h ( ε ) [11]
h p a r ( ε ) ≈ C ∑ j = 0 , 2 , 4 ∞ 1 ( j / 2 ) ! ε j h_{par}\left(\varepsilon\right)\approx C\sum^{\infty}_{j=0,2,4}\frac{1}{\left(j/2\right)!}\varepsilon^{j} h p a r ( ε ) ≈ C j = 0 , 2 , 4 ∑ ∞ ( j /2 ) ! 1 ε j Fazendo j = 2 n j=2n j = 2 n
h p a r ( ε ) ≈ C ∑ n ∞ 1 ( 2 n / 2 ) ! ε 2 n = C ∑ n ∞ ε 2 n n ! h_{par}\left(\varepsilon\right)\approx C\sum^{\infty}_{n}\frac{1}{\left(2n/2\right)!}\varepsilon^{2n}=C\sum^{\infty}_{n}\frac{\varepsilon^{2n}}{n!} h p a r ( ε ) ≈ C n ∑ ∞ ( 2 n /2 ) ! 1 ε 2 n = C n ∑ ∞ n ! ε 2 n A expansão em série de taylor de e x 2 e^{x^{2}} e x 2
f ( x ) = e x 2 f ( 0 ) = 1 f ′ ( x ) = 2 x e x 2 f ′ ( 0 ) = 0 f ′ ′ ( x ) = 2 e x 2 + 4 x 2 e x 2 f ′ ′ ( 0 ) = 2 f ′ ′ ′ ( x ) = 12 x e x 2 + 8 x 3 e x 2 f ′ ′ ′ ( 0 ) = 0 f ′ ′ ′ ′ ( x ) = 12 e x 2 + 48 x 2 e x 2 + 16 x 4 e x 2 f ′ ′ ′ ′ ( 0 ) = 12 \begin{align}
f\left(x\right) & =e^{x^{2}} & f\left(0\right) & =1\\
f'\left(x\right) & =2xe^{x^{2}} & f'\left(0\right) & =0\\
f''\left(x\right) & =2e^{x^{2}}+4x^{2}e^{x^{2}} & f''\left(0\right) & =2\\
f'''\left(x\right) & =12xe^{x^{2}}+8x^{3}e^{x^{2}} & f'''\left(0\right) & =0\\
f''''\left(x\right) & =12e^{x^{2}}+48x^{2}e^{x^{2}}+16x^{4}e^{x^{2}} & f''''\left(0\right) & =12
\end{align} f ( x ) f ′ ( x ) f ′′ ( x ) f ′′′ ( x ) f ′′′′ ( x ) = e x 2 = 2 x e x 2 = 2 e x 2 + 4 x 2 e x 2 = 12 x e x 2 + 8 x 3 e x 2 = 12 e x 2 + 48 x 2 e x 2 + 16 x 4 e x 2 f ( 0 ) f ′ ( 0 ) f ′′ ( 0 ) f ′′′ ( 0 ) f ′′′′ ( 0 ) = 1 = 0 = 2 = 0 = 12 Temos então:
f ( x ) = f ( 0 ) + f ′ ( 0 ) x + f ′ ′ ( 0 ) x 2 2 ! + f ′ ′ ′ ( 0 ) x 3 3 ! + … e x 2 = 1 + 2 x 2 2 ! + 12 x 4 4 ! + … = x 2 ⋅ 0 0 ! + x 2 ⋅ 1 1 ! + x 2 ⋅ 2 2 ! + … = ∑ n x 2 n n ! \begin{alignat}{1}
f\left(x\right) & =f\left(0\right)+f'\left(0\right)x+f''\left(0\right)\frac{x^{2}}{2!}+f'''\left(0\right)\frac{x^{3}}{3!}+\dots\\
e^{x^{2}} & =1+\frac{2x^{2}}{2!}+\frac{12x^{4}}{4!}+\dots\\
& =\frac{x^{2\cdot0}}{0!}+\frac{x^{2\cdot1}}{1!}+\frac{x^{2\cdot2}}{2!}+\dots\\
& =\sum_{n}\frac{x^{2n}}{n!}
\end{alignat} f ( x ) e x 2 = f ( 0 ) + f ′ ( 0 ) x + f ′′ ( 0 ) 2 ! x 2 + f ′′′ ( 0 ) 3 ! x 3 + … = 1 + 2 ! 2 x 2 + 4 ! 12 x 4 + … = 0 ! x 2 ⋅ 0 + 1 ! x 2 ⋅ 1 + 2 ! x 2 ⋅ 2 + … = n ∑ n ! x 2 n Logo:
h p a r ( ε ) ≈ C e ε 2 h_{par}\left(\varepsilon\right)\approx Ce^{\varepsilon^{2}} h p a r ( ε ) ≈ C e ε 2 Porém substituindo isto em nossa função de onda teríamos:
ψ p a r ( ε ) = C e ε 2 ⋅ e − ε 2 / 2 \psi_{par}\left(\varepsilon\right)=Ce^{\varepsilon^{2}}\cdot e^{-\varepsilon^{2}/2} ψ p a r ( ε ) = C e ε 2 ⋅ e − ε 2 /2 Porém este é o comportamento assintótico que estávamos tentando evitar, afinal, supomos que o comportamento assintótico fosse dado pelo exponencial e − ε 2 / 2 e^{-\varepsilon^{2}/2} e − ε 2 /2 h ( ε ) h\left(\varepsilon\right) h ( ε )
Ou seja, toda esta seção de cálculos é para mostrar que não queremos que o somatório se estenda para j → ∞ j\rightarrow\infty j → ∞ h ( ε ) h\left(\varepsilon\right) h ( ε ) ψ \psi ψ ( ε ) \left(\varepsilon\right) ( ε ) j j j n n n a n + 2 = 0 a_{n+2}=0 a n + 2 = 0 h p a r h_{par} h p a r h i m p a r h_{impar} h im p a r a 1 = 0 a_{1}=0 a 1 = 0 a 0 = 0 a_{0}=0 a 0 = 0 [12]
Então para j = n j=n j = n
a n + 2 = ( 2 n − K + 1 ) ( n + 2 ) ( n + 1 ) a n 0 = ( 2 n − K + 1 ) ( n + 2 ) ( n + 1 ) a n 0 = 2 n − K + 1 \begin{align}
a_{n+2} & =\frac{\left(2n-K+1\right)}{\left(n+2\right)\left(n+1\right)}a_{n}\\
0 & =\frac{\left(2n-K+1\right)}{\left(n+2\right)\left(n+1\right)}a_{n}\\
0 & =2n-K+1
\end{align} a n + 2 0 0 = ( n + 2 ) ( n + 1 ) ( 2 n − K + 1 ) a n = ( n + 2 ) ( n + 1 ) ( 2 n − K + 1 ) a n = 2 n − K + 1 Ou então:
Recuperando adefinição de K ≖ 2 E ℏ ω , K\eqcirc\frac{2E}{\hbar\omega}, K ≖ ℏ ω 2 E ,
2 E ℏ ω = 2 n + 1 → E = ( n + 1 2 ) ℏ ω \frac{2E}{\hbar\omega}=2n+1\rightarrow E=\left(n+\frac{1}{2}\right)\hbar\omega ℏ ω 2 E = 2 n + 1 → E = ( n + 2 1 ) ℏ ω Ou seja, recuperamos a condição de quantização fundamental[13] K K K
a j + 2 = ( 2 j − K + 1 ) ( j + 2 ) ( j + 1 ) a j = ( 2 j − ( 2 n + 1 ) + 1 ) ( j + 2 ) ( j + 1 ) a j = 2 ( j − n ) ( j + 2 ) ( j + 1 ) a j a_{j+2}=\frac{\left(2j-K+1\right)}{\left(j+2\right)\left(j+1\right)}a_{j}=\frac{\left(2j-\left(2n+1\right)+1\right)}{\left(j+2\right)\left(j+1\right)}a_{j}=\frac{2\left(j-n\right)}{\left(j+2\right)\left(j+1\right)}a_{j} a j + 2 = ( j + 2 ) ( j + 1 ) ( 2 j − K + 1 ) a j = ( j + 2 ) ( j + 1 ) ( 2 j − ( 2 n + 1 ) + 1 ) a j = ( j + 2 ) ( j + 1 ) 2 ( j − n ) a j Se n = 0 n=0 n = 0 a 1 = 0 a_{1}=0 a 1 = 0 a 0 a_{0} a 0 a 2 a_{2} a 2 j = n = 0 j=n=0 j = n = 0
a 2 = 2 ( 0 − 0 ) ( 0 + 2 ) ( 0 + 1 ) a 0 = 0 a_{2}=\frac{2\left(0-0\right)}{\left(0+2\right)\left(0+1\right)}a_{0}=0 a 2 = ( 0 + 2 ) ( 0 + 1 ) 2 ( 0 − 0 ) a 0 = 0 Consequente todos os termos pares futuros, assim os ímpares serão 0, ou seja para qualquer j > 0 j>0 j > 0 a j = 0 a_{j}=0 a j = 0 h n ( ε ) h_{n}\left(\varepsilon\right) h n ( ε ) n n n h 0 ( ε ) = a 0 h_{0}\left(\varepsilon\right)=a_{0} h 0 ( ε ) = a 0
ψ 0 ( ε ) = h 0 ( ε ) e − ε 2 / 2 = a 0 e − ε 2 / 2 \psi_{0}\left(\varepsilon\right)=h_{0}\left(\varepsilon\right)e^{-\varepsilon^{2}/2}=a_{0}e^{-\varepsilon^{2}/2} ψ 0 ( ε ) = h 0 ( ε ) e − ε 2 /2 = a 0 e − ε 2 /2 De forma análoga se escolhermos n = 1 n=1 n = 1 a 0 = 0 a_{0}=0 a 0 = 0 h 1 ( ε ) = a 1 ε h_{1}\left(\varepsilon\right)=a_{1}\varepsilon h 1 ( ε ) = a 1 ε a j = 0 a_{j}=0 a j = 0 j > 1 j>1 j > 1 ψ 1 ( ε ) = a 1 ε e − ε 2 / 2 \psi_{1}\left(\varepsilon\right)=a_{1}\varepsilon e^{-\varepsilon^{2}/2} ψ 1 ( ε ) = a 1 ε e − ε 2 /2 h n ( ε ) h_{n}\left(\varepsilon\right) h n ( ε ) n n n ε \varepsilon ε n n n n n n
Estes polinômios, ignorando o fator geral (a 0 a_{0} a 0 a 1 a_{1} a 1 H n ( ε ) H_{n}\left(\varepsilon\right) H n ( ε ) ε \varepsilon ε 2 n 2^{n} 2 n [14]
ψ n ( x ) = ( m ω π ℏ ) 1 / 4 1 2 n n ! H n ( ε ) e − ε 2 / 2 \psi_{n}\left(x\right)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}H_{n}\left(\varepsilon\right)e^{-\varepsilon^{2}/2} ψ n ( x ) = ( π ℏ mω ) 1/4 2 n n ! 1 H n ( ε ) e − ε 2 /2 Oh seja:
ψ n ( x ) = a n H n ( ε ) e − ε 2 / 2 \psi_{n}\left(x\right)=a_{n}H_{n}\left(\varepsilon\right)e^{-\varepsilon^{2}/2} ψ n ( x ) = a n H n ( ε ) e − ε 2 /2 sendo a n = ( m ω π ℏ ) 1 / 4 1 2 n n ! a_{n}=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}} a n = ( π ℏ mω ) 1/4 2 n n ! 1 n n n n n n