Problem set 4
These are my solutions to the problem sets assigned to me during the graduate-level quantum mechanics course.
Problem 1 ¶ K ( x ′ ′ , t ; x ′ , t 0 ) = m 2 π i ℏ ( t − t 0 ) exp [ i m ( x ′ ′ − x ′ ) 2 2 ℏ ( t − t 0 ) ] K\left(x'',t;x',t_{0}\right)=\sqrt{\frac{m}{2\pi i\hbar\left(t-t_{0}\right)}}\exp\left[\frac{im\left(x''-x'\right)^{2}}{2\hbar\left(t-t_{0}\right)}\right] K ( x ′′ , t ; x ′ , t 0 ) = 2 πi ℏ ( t − t 0 ) m exp [ 2ℏ ( t − t 0 ) im ( x ′′ − x ′ ) 2 ] Show that the propagator of a one-dimensional free particle is given by
Queremos mostrar que o propagador de uma partícula livre unidimensional é dado por:
K ( x ′ ′ , t ; x ′ , t 0 ) = m 2 π i ℏ ( t − t 0 ) exp [ i m ( x ′ ′ − x ′ ) 2 2 ℏ ( t − t 0 ) ] K\left(x'',t;x',t_{0}\right)=\sqrt{\frac{m}{2\pi i\hbar\left(t-t_{0}\right)}}\exp\left[\frac{im\left(x''-x'\right)^{2}}{2\hbar\left(t-t_{0}\right)}\right] K ( x ′′ , t ; x ′ , t 0 ) = 2 πi ℏ ( t − t 0 ) m exp [ 2ℏ ( t − t 0 ) im ( x ′′ − x ′ ) 2 ] O propagador pode ser escrito elegantemente como:
K ( x ′ ′ , t ; x ′ , t 0 ) = ⟨ x ′ ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ x ′ ⟩ K\left(x'',t;x',t_{0}\right)=\left\langle x''\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|x'\right\rangle K ( x ′′ , t ; x ′ , t 0 ) = ⟨ x ′′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ x ′ ⟩ Sendo então o hamiltoniano de uma partícula livre dado por H = p 2 2 m H=\frac{p^{2}}{2m} H = 2 m p 2 ∣ p ′ ⟩ \left|p'\right\rangle ∣ p ′ ⟩ p p p H H H p ∣ p ′ ⟩ = p ′ ∣ p ′ ⟩ p\left|p'\right\rangle =p'\left|p'\right\rangle p ∣ p ′ ⟩ = p ′ ∣ p ′ ⟩ H ∣ p ′ ⟩ = p ′ 2 2 m ∣ p ′ ⟩ H\left|p'\right\rangle =\frac{p'^{2}}{2m}\left|p'\right\rangle H ∣ p ′ ⟩ = 2 m p ′ 2 ∣ p ′ ⟩
⟨ x ′ ∣ p ′ ⟩ = 1 2 π ℏ exp ( i p ′ x ′ ℏ ) \left\langle x'|p'\right\rangle =\sqrt{\frac{1}{2\pi\hbar}}\exp\left(\frac{ip'x'}{\hbar}\right) ⟨ x ′ ∣ p ′ ⟩ = 2 π ℏ 1 exp ( ℏ i p ′ x ′ ) E como ( H ∣ p ′ ⟩ ) † = ⟨ p ∣ H = ⟨ p ∣ p ′ 2 2 m \left(H\left|p'\right\rangle \right)^{\dagger}=\left\langle p\right|H=\left\langle p\right|\frac{p'^{2}}{2m} ( H ∣ p ′ ⟩ ) † = ⟨ p ∣ H = ⟨ p ∣ 2 m p ′ 2
K ( x ′ ′ , t ; x ′ , t 0 ) = ⟨ x ′ ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ x ′ ⟩ = ∫ d p ′ ⟨ x ′ ′ ∣ p ′ ⟩ ⟨ p ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ x ′ ⟩ = ∫ d p ′ 1 2 π ℏ exp ( i p ′ x ′ ′ ℏ ) ⟨ p ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ x ′ ⟩ = 1 2 π ℏ ∫ d p ′ exp ( i p ′ x ′ ′ ℏ ) ⟨ p ′ ∣ x ′ ⟩ exp ( − i p ′ 2 ( t − t 0 ) 2 ℏ m ) = 1 2 π ℏ ∫ d p ′ exp ( i p ′ ( x ′ ′ − x ′ ) ℏ ) exp ( − i p ′ 2 ( t − t 0 ) 2 ℏ m ) = 1 2 π ℏ ∫ d p ′ exp ( i p ′ ( x ′ ′ − x ′ ) ℏ − i p ′ 2 ( t − t 0 ) 2 ℏ m ) \begin{align*}
K\left(x'',t;x',t_{0}\right) & =\left\langle x''\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|x'\right\rangle \\
& =\int dp'\left\langle x''|p'\right\rangle \left\langle p'\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|x'\right\rangle \\
& =\int dp'\sqrt{\frac{1}{2\pi\hbar}}\exp\left(\frac{ip'x''}{\hbar}\right)\left\langle p'\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|x'\right\rangle \\
& =\sqrt{\frac{1}{2\pi\hbar}}\int dp'\exp\left(\frac{ip'x''}{\hbar}\right)\left\langle p'|x'\right\rangle \exp\left(\frac{-ip'^{2}\left(t-t_{0}\right)}{2\hbar m}\right)\\
& =\frac{1}{2\pi\hbar}\int dp'\exp\left(\frac{ip'\left(x''-x'\right)}{\hbar}\right)\exp\left(\frac{-ip'^{2}\left(t-t_{0}\right)}{2\hbar m}\right)\\
& =\frac{1}{2\pi\hbar}\int dp'\exp\left(\frac{ip'\left(x''-x'\right)}{\hbar}-\frac{ip'^{2}\left(t-t_{0}\right)}{2\hbar m}\right)
\end{align*} K ( x ′′ , t ; x ′ , t 0 ) = ⟨ x ′′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ x ′ ⟩ = ∫ d p ′ ⟨ x ′′ ∣ p ′ ⟩ ⟨ p ′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ x ′ ⟩ = ∫ d p ′ 2 π ℏ 1 exp ( ℏ i p ′ x ′′ ) ⟨ p ′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ x ′ ⟩ = 2 π ℏ 1 ∫ d p ′ exp ( ℏ i p ′ x ′′ ) ⟨ p ′ ∣ x ′ ⟩ exp ( 2ℏ m − i p ′ 2 ( t − t 0 ) ) = 2 π ℏ 1 ∫ d p ′ exp ( ℏ i p ′ ( x ′′ − x ′ ) ) exp ( 2ℏ m − i p ′ 2 ( t − t 0 ) ) = 2 π ℏ 1 ∫ d p ′ exp ( ℏ i p ′ ( x ′′ − x ′ ) − 2ℏ m i p ′ 2 ( t − t 0 ) ) Sendo − 2 α β = i ( x ′ ′ − x ′ ) ℏ -2\alpha\beta=\frac{i\left(x''-x'\right)}{\hbar} − 2 α β = ℏ i ( x ′′ − x ′ ) β 2 = i ( t ′ − t 0 ) 2 m ℏ \beta^{2}=\frac{i\left(t'-t_{0}\right)}{2m\hbar} β 2 = 2 m ℏ i ( t ′ − t 0 )
K ( x ′ ′ , t ; x ′ , t 0 ) = 1 2 π ℏ ∫ d p ′ exp ( − [ 2 α β p ′ + ( β p ′ ) 2 ] ) = 1 2 π ℏ ∫ d p ′ exp [ − ( 2 α β + β p ′ ) 2 + α 2 ] = 1 2 π ℏ e − α 2 ∫ d p ′ exp [ − ( α + β p ′ ) 2 ] = 1 2 π ℏ e α 2 β ∫ e − u 2 d u \begin{align*}
K\left(x'',t;x',t_{0}\right) & =\frac{1}{2\pi\hbar}\int dp'\exp\left(-\left[2\alpha\beta p'+\left(\beta p'\right)^{2}\right]\right)\\
& =\frac{1}{2\pi\hbar}\int dp'\exp\left[-\left(2\alpha\beta+\beta p'\right)^{2}+\alpha^{2}\right]\\
& =\frac{1}{2\pi\hbar}e^{-\alpha^{2}}\int dp'\exp\left[-\left(\alpha+\beta p'\right)^{2}\right]\\
& =\frac{1}{2\pi\hbar}\frac{e^{\alpha^{2}}}{\beta}\int e^{-u^{2}}du
\end{align*} K ( x ′′ , t ; x ′ , t 0 ) = 2 π ℏ 1 ∫ d p ′ exp ( − [ 2 α β p ′ + ( β p ′ ) 2 ] ) = 2 π ℏ 1 ∫ d p ′ exp [ − ( 2 α β + β p ′ ) 2 + α 2 ] = 2 π ℏ 1 e − α 2 ∫ d p ′ exp [ − ( α + β p ′ ) 2 ] = 2 π ℏ 1 β e α 2 ∫ e − u 2 d u Onde foi feito u = ( α + β p ′ ) u=\left(\alpha+\beta p'\right) u = ( α + β p ′ ) d u d p ′ = − β \frac{du}{dp'}=-\beta d p ′ d u = − β ∫ − ∞ + ∞ 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
K ( x ′ ′ , t ; x ′ , t 0 ) = 1 2 π ℏ e α 2 β \begin{align*}
K\left(x'',t;x',t_{0}\right) & =\frac{1}{2\sqrt{\pi}\hbar}\frac{e^{\alpha^{2}}}{\beta}
\end{align*} K ( x ′′ , t ; x ′ , t 0 ) = 2 π ℏ 1 β e α 2 Agora como:
− 2 α β = i ( x ′ ′ − x ′ ) ℏ α = − i ( x ′ ′ − x ′ ) 2 ℏ β α 2 = ( x ′ ′ − x ′ ) 2 2 2 ℏ 2 1 β 2 = ( x ′ ′ − x ′ ) 2 2 2 ℏ 2 ( 2 i m ℏ ( t ′ − t 0 ) ) = i m ( x ′ ′ − x ′ ) 2 2 ℏ ( t ′ − t 0 ) \begin{align*}
-2\alpha\beta & =\frac{i\left(x''-x'\right)}{\hbar}\\
\alpha & =-\frac{i\left(x''-x'\right)}{2\hbar\beta}\\
\alpha^{2} & =\frac{\left(x''-x'\right)^{2}}{2^{2}\hbar^{2}}\frac{1}{\beta^{2}}\\
& =\frac{\left(x''-x'\right)^{2}}{2^{2}\hbar^{2}}\left(\frac{2im\hbar}{\left(t'-t_{0}\right)}\right)\\
& =\frac{im\left(x''-x'\right)^{2}}{2\hbar\left(t'-t_{0}\right)}
\end{align*} − 2 α β α α 2 = ℏ i ( x ′′ − x ′ ) = − 2ℏ β i ( x ′′ − x ′ ) = 2 2 ℏ 2 ( x ′′ − x ′ ) 2 β 2 1 = 2 2 ℏ 2 ( x ′′ − x ′ ) 2 ( ( t ′ − t 0 ) 2 im ℏ ) = 2ℏ ( t ′ − t 0 ) im ( x ′′ − x ′ ) 2 Então:
K ( x ′ ′ , t ; x ′ , t 0 ) = 1 2 π ℏ e α 2 β = 1 2 π ℏ 2 m ℏ i ( t ′ − t 0 ) exp [ i m ( x ′ ′ − x ′ ) 2 2 ℏ ( t − t 0 ) ] \begin{align*}
K\left(x'',t;x',t_{0}\right) & =\frac{1}{2\sqrt{\pi}\hbar}\frac{e^{\alpha^{2}}}{\beta}\\
& =\frac{1}{2\sqrt{\pi}\hbar}\sqrt{\frac{2m\hbar}{i\left(t'-t_{0}\right)}}\exp\left[\frac{im\left(x''-x'\right)^{2}}{2\hbar\left(t-t_{0}\right)}\right]
\end{align*} K ( x ′′ , t ; x ′ , t 0 ) = 2 π ℏ 1 β e α 2 = 2 π ℏ 1 i ( t ′ − t 0 ) 2 m ℏ exp [ 2ℏ ( t − t 0 ) im ( x ′′ − x ′ ) 2 ] Então obtemos:
K ( x ′ ′ , t ; x ′ , t 0 ) = m 2 π i ℏ ( t − t 0 ) exp [ i m ( x ′ ′ − x ′ ) 2 2 ℏ ( t − t 0 ) ] K\left(x'',t;x',t_{0}\right)=\sqrt{\frac{m}{2\pi i\hbar\left(t-t_{0}\right)}}\exp\left[\frac{im\left(x''-x'\right)^{2}}{2\hbar\left(t-t_{0}\right)}\right] K ( x ′′ , t ; x ′ , t 0 ) = 2 πi ℏ ( t − t 0 ) m exp [ 2ℏ ( t − t 0 ) im ( x ′′ − x ′ ) 2 ] Que é o que queríamos demonstrar desde o começo.
Problem 2 ¶ Show, using the a , a † a,a^{\dagger} a , a †
K ( x ′ ′ , t ; x ′ , t 0 ) = m ω 2 π i ℏ sin [ ω t ( t − t 0 ) ] exp [ i m ω 2 ℏ sin [ ω ( t − t 0 ) ] ( ( x ′ ′ 2 + x ′ 2 ) cos [ ω ( t − t 0 ) ] − 2 x ′ ′ x ′ ) ] K\left(x'',t;x',t_{0}\right)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\left[\omega t\left(t-t_{0}\right)\right]}}\exp\left[\frac{im\omega}{2\hbar\sin\left[\omega\left(t-t_{0}\right)\right]}\left(\left(x''^{2}+x'^{2}\right)\cos\left[\omega\left(t-t_{0}\right)\right]-2x''x'\right)\right] K ( x ′′ , t ; x ′ , t 0 ) = 2 πi ℏ sin [ ω t ( t − t 0 ) ] mω exp [ 2ℏ sin [ ω ( t − t 0 ) ] imω ( ( x ′′ 2 + x ′ 2 ) cos [ ω ( t − t 0 ) ] − 2 x ′′ x ′ ) ] Usando o método do operador a , a † a,a^{\dagger} a , a †
K ( x ′ ′ , t ; x ′ , t 0 ) = m ω 2 π i ℏ sin [ ω t ( t − t 0 ) ] exp [ i m ω 2 ℏ sin [ ω ( t − t 0 ) ] ( ( x ′ ′ 2 + x ′ 2 ) cos [ ω ( t − t 0 ) ] − 2 x ′ ′ x ′ ) ] K\left(x'',t;x',t_{0}\right)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\left[\omega t\left(t-t_{0}\right)\right]}}\exp\left[\frac{im\omega}{2\hbar\sin\left[\omega\left(t-t_{0}\right)\right]}\left(\left(x''^{2}+x'^{2}\right)\cos\left[\omega\left(t-t_{0}\right)\right]-2x''x'\right)\right] K ( x ′′ , t ; x ′ , t 0 ) = 2 πi ℏ sin [ ω t ( t − t 0 ) ] mω exp [ 2ℏ sin [ ω ( t − t 0 ) ] imω ( ( x ′′ 2 + x ′ 2 ) cos [ ω ( t − t 0 ) ] − 2 x ′′ x ′ ) ] Partindodo propagador escrito como:
K ( x ′ ′ , t ; x ′ , t 0 ) = ⟨ x ′ ′ ∣ exp [ − i H ^ ( t − t 0 ) ℏ ] ∣ x ′ ⟩ K\left(x'',t;x',t_{0}\right)=\left\langle x''\left|\exp\left[\frac{-i\widehat{H}\left(t-t_{0}\right)}{\hbar}\right]\right|x'\right\rangle K ( x ′′ , t ; x ′ , t 0 ) = ⟨ x ′′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ x ′ ⟩ E sendo U ^ = U ^ ( t − t 0 ) = exp [ − i H ( t − t 0 ) ℏ ] \widehat{U}=\widehat{U}\left(t-t_{0}\right)=\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right] U = U ( t − t 0 ) = exp [ ℏ − i H ( t − t 0 ) ]
K ( x ′ ′ , t ; x ′ , t 0 ) = ⟨ x ′ ′ ∣ U ^ ∣ x ′ ⟩ K\left(x'',t;x',t_{0}\right)=\left\langle x''\left|\widehat{U}\right|x'\right\rangle K ( x ′′ , t ; x ′ , t 0 ) = ⟨ x ′′ ∣ ∣ U ∣ ∣ x ′ ⟩ Sendo então ∣ α ⟩ \left|\alpha\right\rangle ∣ α ⟩ a ^ \widehat{a} a a ^ ∣ α ⟩ = α ∣ α ⟩ \widehat{a}\left|\alpha\right\rangle =\alpha\left|\alpha\right\rangle a ∣ α ⟩ = α ∣ α ⟩ a ^ \widehat{a} a
a ^ = m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ \widehat{a}=\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p} a = 2ℏ mω x + 2ℏ mω i p Então, se α \alpha α ⟨ α ∣ a ^ ∣ α ⟩ = α \left\langle \alpha\left|\widehat{a}\right|\alpha\right\rangle =\alpha ⟨ α ∣ a ∣ α ⟩ = α x 0 = ⟨ α ∣ x ^ ∣ α ⟩ x_{0}=\left\langle \alpha\left|\widehat{x}\right|\alpha\right\rangle x 0 = ⟨ α ∣ x ∣ α ⟩ p 0 = ⟨ α ∣ p ^ ∣ α ⟩ p_{0}=\left\langle \alpha\left|\widehat{p}\right|\alpha\right\rangle p 0 = ⟨ α ∣ p ∣ α ⟩
α = ⟨ α ∣ a ^ ∣ α ⟩ = ⟨ α ∣ m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ ∣ α ⟩ = m ω 2 ℏ ⟨ α ∣ x ^ ∣ α ⟩ + i 2 ℏ m ω ⟨ α ∣ p ^ ∣ α ⟩ = m ω 2 ℏ x 0 + i 2 ℏ m ω p 0 \begin{align}
\alpha & =\left\langle \alpha\left|\widehat{a}\right|\alpha\right\rangle \\
& =\left\langle \alpha\left|\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right|\alpha\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}\left\langle \alpha\left|\widehat{x}\right|\alpha\right\rangle +\frac{i}{\sqrt{2\hbar m\omega}}\left\langle \alpha\left|\widehat{p}\right|\alpha\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}x_{0}+\frac{i}{\sqrt{2\hbar m\omega}}p_{0}
\end{align} α = ⟨ α ∣ a ∣ α ⟩ = ⟨ α ∣ ∣ 2ℏ mω x + 2ℏ mω i p ∣ ∣ α ⟩ = 2ℏ mω ⟨ α ∣ x ∣ α ⟩ + 2ℏ mω i ⟨ α ∣ p ∣ α ⟩ = 2ℏ mω x 0 + 2ℏ mω i p 0 Vamos utilizar o seguinte operador identidadeShao (2016) John Wiley & Sons, Ltd (n.d.)
1 2 π ℏ ∫ − ∞ + ∞ ∫ − ∞ + ∞ d x 0 d p 0 ∣ α ⟩ ⟨ α ∣ = I \frac{1}{2\pi\hbar}\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}dx_{0}dp_{0}\left|\alpha\right\rangle \left\langle \alpha\right|=\mathbb{I} 2 π ℏ 1 ∫ − ∞ + ∞ ∫ − ∞ + ∞ d x 0 d p 0 ∣ α ⟩ ⟨ α ∣ = I Inserindo esta identidade no propagador, temos que:
K ( x ′ ′ , t ; x ′ , t 0 ) = 1 2 π ℏ ∫ − ∞ + ∞ ∫ − ∞ + ∞ d x 0 d p 0 ⟨ x ′ ′ ∣ U ^ ∣ α ⟩ ⟨ α ∣ x ′ ⟩ K\left(x'',t;x',t_{0}\right)=\frac{1}{2\pi\hbar}\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}dx_{0}dp_{0}\left\langle x''\right|\widehat{U}\left|\alpha\right\rangle \left\langle \alpha|x'\right\rangle K ( x ′′ , t ; x ′ , t 0 ) = 2 π ℏ 1 ∫ − ∞ + ∞ ∫ − ∞ + ∞ d x 0 d p 0 ⟨ x ′′ ∣ U ∣ α ⟩ ⟨ α ∣ x ′ ⟩ Agora queremos reescrever U ^ ∣ α ⟩ \widehat{U}\left|\alpha\right\rangle U ∣ α ⟩ U ^ † a ^ U ^ \widehat{U}^{\dagger}\widehat{a}\widehat{U} U † a U
U ^ † a ^ U ^ = exp [ i ( t − t 0 ) ℏ H ^ ] a ^ exp [ − i ( t − t 0 ) ℏ H ^ ] = e λ A ^ B ^ e − λ A ^ \begin{align*}
\widehat{U}^{\dagger}\widehat{a}\widehat{U} & =\exp\left[\frac{i\left(t-t_{0}\right)}{\hbar}\widehat{H}\right]\widehat{a}\exp\left[-\frac{i\left(t-t_{0}\right)}{\hbar}\widehat{H}\right]\\
& =e^{\lambda\widehat{A}}\widehat{B}e^{-\lambda\widehat{A}}
\end{align*} U † a U = exp [ ℏ i ( t − t 0 ) H ] a exp [ − ℏ i ( t − t 0 ) H ] = e λ A B e − λ A Onde A ^ = H ^ \widehat{A}=\widehat{H} A = H B ^ = a ^ \widehat{B}=\widehat{a} B = a λ = i ( t − t 0 ) ℏ \lambda=\frac{i\left(t-t_{0}\right)}{\hbar} λ = ℏ i ( t − t 0 ) Shao (2016) Maziero (2021) Sakurai & Napolitano (2011)
e λ A ^ B ^ e − λ A ^ = B ^ + λ [ A ^ , B ^ ] + λ 2 2 ! [ A ^ , [ A ^ , B ^ ] ] + λ 3 3 ! [ A ^ , [ A ^ , [ A ^ , B ^ ] ] ] + … e^{\lambda\widehat{A}}\widehat{B}e^{-\lambda\widehat{A}}=\widehat{B}+\lambda\left[\widehat{A},\widehat{B}\right]+\frac{\lambda^{2}}{2!}\left[\widehat{A},\left[\widehat{A},\widehat{B}\right]\right]+\frac{\lambda^{3}}{3!}\left[\widehat{A},\left[\widehat{A},\left[\widehat{A},\widehat{B}\right]\right]\right]+\dots e λ A B e − λ A = B + λ [ A , B ] + 2 ! λ 2 [ A , [ A , B ] ] + 3 ! λ 3 [ A , [ A , [ A , B ] ] ] + … E como:
a ^ † a ^ = ( m ω 2 ℏ x ^ − i 2 ℏ m ω p ^ ) ( m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ ) = ( m ω 2 ℏ x ^ 2 + i 2 ℏ x ^ p ^ − i 2 ℏ p ^ x ^ + 1 2 ℏ m ω p ^ 2 ) = m ω 2 ℏ x ^ 2 + i 2 ℏ ( x ^ p ^ − p ^ x ^ ) + 1 2 ℏ m ω p ^ 2 = m ω 2 ℏ x ^ 2 + i 2 ℏ [ x ^ , p ^ ] + 1 2 ℏ m ω p ^ 2 \begin{align*}
\widehat{a}^{\dagger}\widehat{a} & =\left(\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}-\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right)\left(\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right)\\
& =\left(\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{i}{2\hbar}\widehat{x}\widehat{p}-\frac{i}{2\hbar}\widehat{p}\widehat{x}+\frac{1}{2\hbar m\omega}\widehat{p}^{2}\right)\\
& =\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{i}{2\hbar}\left(\widehat{x}\widehat{p}-\widehat{p}\widehat{x}\right)+\frac{1}{2\hbar m\omega}\widehat{p}^{2}\\
& =\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{i}{2\hbar}\left[\widehat{x},\widehat{p}\right]+\frac{1}{2\hbar m\omega}\widehat{p}^{2}
\end{align*} a † a = ( 2ℏ mω x − 2ℏ mω i p ) ( 2ℏ mω x + 2ℏ mω i p ) = ( 2ℏ mω x 2 + 2ℏ i x p − 2ℏ i p x + 2ℏ mω 1 p 2 ) = 2ℏ mω x 2 + 2ℏ i ( x p − p x ) + 2ℏ mω 1 p 2 = 2ℏ mω x 2 + 2ℏ i [ x , p ] + 2ℏ mω 1 p 2 Lembrando que [ x ^ , p ^ ] = i ℏ \left[\widehat{x},\widehat{p}\right]=i\hbar [ x , p ] = i ℏ
a ^ † a ^ = m ω 2 ℏ x ^ 2 + 1 2 ℏ m ω p ^ 2 − 1 2 = 1 ℏ ω ( 1 2 m p ^ 2 + m ω 2 2 x ^ 2 − ℏ ω 2 ) \begin{align*}
\widehat{a}^{\dagger}\widehat{a} & =\frac{m\omega}{2\hbar}\widehat{x}^{2}+\frac{1}{2\hbar m\omega}\widehat{p}^{2}-\frac{1}{2}\\
& =\frac{1}{\hbar\omega}\left(\frac{1}{2m}\widehat{p}^{2}+\frac{m\omega^{2}}{2}\widehat{x}^{2}-\frac{\hbar\omega}{2}\right)
\end{align*} a † a = 2ℏ mω x 2 + 2ℏ mω 1 p 2 − 2 1 = ℏ ω 1 ( 2 m 1 p 2 + 2 m ω 2 x 2 − 2 ℏ ω ) Então o Hamiltoniano para um oscilador harmônico pode ser escrito como:
H ^ = 1 2 m p ^ 2 + 1 2 m ω 2 x ^ 2 = ℏ ω ( a ^ † a ^ + 1 2 ) \begin{align*}
\widehat{H} & =\frac{1}{2m}\widehat{p}^{2}+\frac{1}{2}m\omega^{2}\widehat{x}^{2}\\
& =\hbar\omega\left(\widehat{a}^{\dagger}\widehat{a}+\frac{1}{2}\right)
\end{align*} H = 2 m 1 p 2 + 2 1 m ω 2 x 2 = ℏ ω ( a † a + 2 1 ) Definindo o operador número N ^ = a ^ † a ^ \widehat{N}=\widehat{a}^{\dagger}\widehat{a} N = a † a
H ^ = ℏ ω ( N ^ + 1 2 ) \widehat{H}=\hbar\omega\left(\widehat{N}+\frac{1}{2}\right) H = ℏ ω ( N + 2 1 ) E como e [ N ^ , a ^ ] = − a ^ \left[\widehat{N},\widehat{a}\right]=-\widehat{a} [ N , a ] = − a Wheeler (2017) ω ℏ 2 \frac{\omega\hbar}{2} 2 ω ℏ
e λ A ^ B ^ e − λ A ^ = a ^ + λ ℏ ω [ N ^ , a ^ ] + ( λ ℏ ω ) 2 2 ! [ N ^ , [ N ^ , a ^ ] ] + ( λ ℏ ω ) 3 3 ! [ N ^ , [ N ^ , [ N ^ , a ^ ] ] ] + … = a ^ + ( − λ ℏ ω ) a ^ − ( λ ℏ ω ) 2 2 ! [ N ^ , a ^ ] + ( − λ ℏ ω ) 3 3 ! [ N ^ , [ N ^ , a ^ ] ] + … = a ^ + ( − λ ℏ ω ) a ^ + ( λ ℏ ω ) 2 2 ! a ^ − ( − λ ℏ ω ) 3 3 ! [ N ^ , a ^ ] + … = a ^ + ( − λ ℏ ω ) a ^ + ( λ ℏ ω ) 2 2 ! a ^ + ( − λ ℏ ω ) 3 3 ! a ^ + … = a ^ [ 1 + ( − λ ℏ ω ) + ( λ ℏ ω ) 2 2 ! + ( − λ ℏ ω ) 3 3 ! + … ] = a ^ e − λ ℏ ω \begin{align*}
e^{\lambda\widehat{A}}\widehat{B}e^{-\lambda\widehat{A}} & =\widehat{a}+\lambda\hbar\omega\left[\widehat{N},\widehat{a}\right]+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\left[\widehat{N},\left[\widehat{N},\widehat{a}\right]\right]+\frac{\left(\lambda\hbar\omega\right)^{3}}{3!}\left[\widehat{N},\left[\widehat{N},\left[\widehat{N},\widehat{a}\right]\right]\right]+\dots\\
& =\widehat{a}+\left(-\lambda\hbar\omega\right)\widehat{a}-\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\left[\widehat{N},\widehat{a}\right]+\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}\left[\widehat{N},\left[\widehat{N},\widehat{a}\right]\right]+\dots\\
& =\widehat{a}+\left(-\lambda\hbar\omega\right)\widehat{a}+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\widehat{a}-\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}\left[\widehat{N},\widehat{a}\right]+\dots\\
& =\widehat{a}+\left(-\lambda\hbar\omega\right)\widehat{a}+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}\widehat{a}+\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}\widehat{a}+\dots\\
& =\widehat{a}\left[1+\left(-\lambda\hbar\omega\right)+\frac{\left(\lambda\hbar\omega\right)^{2}}{2!}+\frac{\left(-\lambda\hbar\omega\right)^{3}}{3!}+\dots\right]\\
& =\widehat{a}e^{-\lambda\hbar\omega}
\end{align*} e λ A B e − λ A = a + λ ℏ ω [ N , a ] + 2 ! ( λ ℏ ω ) 2 [ N , [ N , a ] ] + 3 ! ( λ ℏ ω ) 3 [ N , [ N , [ N , a ] ] ] + … = a + ( − λ ℏ ω ) a − 2 ! ( λ ℏ ω ) 2 [ N , a ] + 3 ! ( − λ ℏ ω ) 3 [ N , [ N , a ] ] + … = a + ( − λ ℏ ω ) a + 2 ! ( λ ℏ ω ) 2 a − 3 ! ( − λ ℏ ω ) 3 [ N , a ] + … = a + ( − λ ℏ ω ) a + 2 ! ( λ ℏ ω ) 2 a + 3 ! ( − λ ℏ ω ) 3 a + … = a [ 1 + ( − λ ℏ ω ) + 2 ! ( λ ℏ ω ) 2 + 3 ! ( − λ ℏ ω ) 3 + … ] = a e − λ ℏ ω E como λ = i ( t − t 0 ) ℏ \lambda=\frac{i\left(t-t_{0}\right)}{\hbar} λ = ℏ i ( t − t 0 )
U ^ † a ^ U ^ = a ^ e − i ω ( t − t 0 ) \widehat{U}^{\dagger}\widehat{a}\widehat{U}=\widehat{a}e^{-i\omega\left(t-t_{0}\right)} U † a U = a e − iω ( t − t 0 ) Ou então, aplicando o operador U ^ \widehat{U} U
a ^ = U ^ † a ^ U ^ e i ω ( t − t 0 ) U ^ a ^ = U ^ U ^ † a ^ U ^ e i ω ( t − t 0 ) = a ^ U ^ e i ω ( t − t 0 ) \begin{align*}
\widehat{a} & =\widehat{U}^{\dagger}\widehat{a}\widehat{U}e^{i\omega\left(t-t_{0}\right)}\\
\widehat{U}\widehat{a} & =\widehat{U}\widehat{U}^{\dagger}\widehat{a}\widehat{U}e^{i\omega\left(t-t_{0}\right)}\\
& =\widehat{a}\widehat{U}e^{i\omega\left(t-t_{0}\right)}
\end{align*} a U a = U † a U e iω ( t − t 0 ) = U U † a U e iω ( t − t 0 ) = a U e iω ( t − t 0 ) Então manipulando a ^ ∣ α ⟩ = α ∣ α ⟩ \widehat{a}\left|\alpha\right\rangle =\alpha\left|\alpha\right\rangle a ∣ α ⟩ = α ∣ α ⟩
a ^ ∣ α ⟩ = α ∣ α ⟩ U ^ a ^ ∣ α ⟩ = α U ^ ∣ α ⟩ a ^ e i ω ( t − t 0 ) U ^ ∣ α ⟩ = α U ^ ∣ α ⟩ a ^ ( U ^ ∣ α ⟩ ) = ( α e i ω ( t − t 0 ) ) ( U ^ ∣ α ⟩ ) a ^ ∣ α ( t ) ⟩ = α ( t ) ∣ α ( t ) ⟩ \begin{align*}
\widehat{a}\left|\alpha\right\rangle & =\alpha\left|\alpha\right\rangle \\
\widehat{U}\widehat{a}\left|\alpha\right\rangle & =\alpha\widehat{U}\left|\alpha\right\rangle \\
\widehat{a}e^{i\omega\left(t-t_{0}\right)}\widehat{U}\left|\alpha\right\rangle & =\alpha\widehat{U}\left|\alpha\right\rangle \\
\widehat{a}\left(\widehat{U}\left|\alpha\right\rangle \right) & =\left(\alpha e^{i\omega\left(t-t_{0}\right)}\right)\left(\widehat{U}\left|\alpha\right\rangle \right)\\
\widehat{a}\left|\alpha\left(t\right)\right\rangle & =\alpha\left(t\right)\left|\alpha\left(t\right)\right\rangle
\end{align*} a ∣ α ⟩ U a ∣ α ⟩ a e iω ( t − t 0 ) U ∣ α ⟩ a ( U ∣ α ⟩ ) a ∣ α ( t ) ⟩ = α ∣ α ⟩ = α U ∣ α ⟩ = α U ∣ α ⟩ = ( α e iω ( t − t 0 ) ) ( U ∣ α ⟩ ) = α ( t ) ∣ α ( t ) ⟩ Isto significa que o estado coerente permanece um estado coerente durante a evolução do tempo com um autovalor dependente do tempo. Voltando então ao propagador, podemos reescrever como:
K ( x ′ ′ , t ; x ′ , t 0 ) = 1 2 π ℏ ∫ − ∞ + ∞ ∫ − ∞ + ∞ d x 0 d p 0 ⟨ x ′ ′ ∣ α ( t ) ⟩ ⟨ α ∣ x ′ ⟩ K\left(x'',t;x',t_{0}\right)=\frac{1}{2\pi\hbar}\int^{+\infty}_{-\infty}\int^{+\infty}_{-\infty}dx_{0}dp_{0}\left\langle x''|\alpha\left(t\right)\right\rangle \left\langle \alpha|x'\right\rangle K ( x ′′ , t ; x ′ , t 0 ) = 2 π ℏ 1 ∫ − ∞ + ∞ ∫ − ∞ + ∞ d x 0 d p 0 ⟨ x ′′ ∣ α ( t ) ⟩ ⟨ α ∣ x ′ ⟩ Agora queremos manipular o termo ⟨ α ∣ x ′ ⟩ \left\langle \alpha|x'\right\rangle ⟨ α ∣ x ′ ⟩
⟨ x ∣ a ^ ∣ α ⟩ = ⟨ x ∣ m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ ∣ α ⟩ α ⟨ x ∣ α ⟩ = m ω 2 ℏ ⟨ x ∣ x ^ ∣ α ⟩ + i 2 ℏ m ω ⟨ x ∣ p ^ ∣ α ⟩ = m ω 2 ℏ x ⟨ x ∣ α ⟩ + ℏ 2 m ω ∂ ∂ x ⟨ x ∣ α ⟩ = m ω 2 ℏ x ⟨ x ∣ α ⟩ + ℏ 2 m ω ∂ ∂ x ⟨ x ∣ α ⟩ \begin{align*}
\left\langle x\left|\widehat{a}\right|\alpha\right\rangle & =\left\langle x\left|\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right|\alpha\right\rangle \\
\alpha\left\langle x|\alpha\right\rangle & =\sqrt{\frac{m\omega}{2\hbar}}\left\langle x\left|\widehat{x}\right|\alpha\right\rangle +\frac{i}{\sqrt{2\hbar m\omega}}\left\langle x\left|\widehat{p}\right|\alpha\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}x\left\langle x|\alpha\right\rangle +\sqrt{\frac{\hbar}{2m\omega}}\frac{\partial}{\partial x}\left\langle x|\alpha\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}x\left\langle x|\alpha\right\rangle +\sqrt{\frac{\hbar}{2m\omega}}\frac{\partial}{\partial x}\left\langle x|\alpha\right\rangle
\end{align*} ⟨ x ∣ a ∣ α ⟩ α ⟨ x ∣ α ⟩ = ⟨ x ∣ ∣ 2ℏ mω x + 2ℏ mω i p ∣ ∣ α ⟩ = 2ℏ mω ⟨ x ∣ x ∣ α ⟩ + 2ℏ mω i ⟨ x ∣ p ∣ α ⟩ = 2ℏ mω x ⟨ x ∣ α ⟩ + 2 mω ℏ ∂ x ∂ ⟨ x ∣ α ⟩ = 2ℏ mω x ⟨ x ∣ α ⟩ + 2 mω ℏ ∂ x ∂ ⟨ x ∣ α ⟩ Onde usamos o resultado obtido na aula 8 ⟨ x ∣ p ∣ a ⟩ = − i ℏ ∂ ∂ x ⟨ x ∣ a ⟩ \left\langle x\left|p\right|a\right\rangle =-i\hbar\frac{\partial}{\partial x}\left\langle x|a\right\rangle ⟨ x ∣ p ∣ a ⟩ = − i ℏ ∂ x ∂ ⟨ x ∣ a ⟩
ℏ 2 m ω ∂ ∂ x ⟨ x ∣ α ⟩ = − ( m ω 2 ℏ x − α ) ⟨ x ∣ α ⟩ \begin{align*}
\sqrt{\frac{\hbar}{2m\omega}}\frac{\partial}{\partial x}\left\langle x|\alpha\right\rangle & =-\left(\sqrt{\frac{m\omega}{2\hbar}}x-\alpha\right)\left\langle x|\alpha\right\rangle
\end{align*} 2 mω ℏ ∂ x ∂ ⟨ x ∣ α ⟩ = − ( 2ℏ mω x − α ) ⟨ x ∣ α ⟩ Em que a solução é:
⟨ x ∣ α ⟩ = C exp [ − ( m ω 2 ℏ x − α ) 2 ] \begin{align*}
\left\langle x|\alpha\right\rangle & =C\exp\left[-\left(\sqrt{\frac{m\omega}{2\hbar}}x-\alpha\right)^{2}\right]
\end{align*} ⟨ x ∣ α ⟩ = C exp [ − ( 2ℏ mω x − α ) 2 ] Confore pode ser verificado facilmente substituindo na equação diferencial. Abrindo mais temos:
⟨ x ∣ α ⟩ = C exp [ − ( m ω 2 ℏ x 2 + α 2 − 2 m ω ℏ x α ) ] \left\langle x|\alpha\right\rangle =C\exp\left[-\left(\frac{m\omega}{2\hbar}x^{2}+\alpha^{2}-\sqrt{\frac{2m\omega}{\hbar}}x\alpha\right)\right] ⟨ x ∣ α ⟩ = C exp [ − ( 2ℏ mω x 2 + α 2 − ℏ 2 mω xα ) ] Usando o obtido na equação 16
α 2 = ( m ω 2 ℏ x 0 + i 2 ℏ m ω p 0 ) 2 = m ω 2 ℏ x 0 2 − 1 2 ℏ m ω p 0 2 + i x 0 p 0 2 ℏ \begin{align*}
\alpha^{2} & =\left(\sqrt{\frac{m\omega}{2\hbar}}x_{0}+\frac{i}{\sqrt{2\hbar m\omega}}p_{0}\right)^{2}\\
& =\frac{m\omega}{2\hbar}x^{2}_{0}-\frac{1}{2\hbar m\omega}p^{2}_{0}+i\frac{x_{0}p_{0}}{2\hbar}
\end{align*} α 2 = ( 2ℏ mω x 0 + 2ℏ mω i p 0 ) 2 = 2ℏ mω x 0 2 − 2ℏ mω 1 p 0 2 + i 2ℏ x 0 p 0 Então:
⟨ x ∣ α ⟩ = C exp [ − m ω 2 ℏ x 2 + 2 m ω ℏ x α − ( m ω 2 ℏ x 0 2 − 1 2 ℏ m ω p 0 2 − i x 0 p 0 2 ℏ ) ] = C exp [ − m ω 2 ℏ ( x 2 + x 0 2 ) + 2 m ω ℏ x α + 1 2 ℏ m ω p 0 2 − i x 0 p 0 2 ℏ ] = C exp [ − m ω 2 ℏ ( x 2 + x 0 2 ) + 2 m ω ℏ x ( m ω 2 ℏ x 0 + i 2 ℏ m ω p 0 ) + 1 2 ℏ m ω p 0 2 − i x 0 p 0 2 ℏ ] = C exp [ − m ω 2 ℏ ( x 2 + x 0 2 ) + 2 m ω 2 ℏ x x 0 + i ℏ x p 0 + 1 2 ℏ m ω p 0 2 − i x 0 p 0 2 ℏ ] = = C exp [ − m ω 2 ℏ ( x 2 − 2 x x 0 + x 0 2 ) + i ℏ x p 0 + 1 2 ℏ m ω p 0 2 − i x 0 p 0 2 ℏ ] = C exp [ − m ω 2 ℏ ( x − x 0 ) 2 + p 0 2 2 ℏ m ω + i ( x p 0 ℏ − x 0 p 0 2 ℏ ) ] \begin{align*}
\left\langle x|\alpha\right\rangle & =C\exp\left[-\frac{m\omega}{2\hbar}x^{2}+\sqrt{\frac{2m\omega}{\hbar}}x\alpha-\left(\frac{m\omega}{2\hbar}x^{2}_{0}-\frac{1}{2\hbar m\omega}p^{2}_{0}-i\frac{x_{0}p_{0}}{2\hbar}\right)\right]\\
& =C\exp\left[-\frac{m\omega}{2\hbar}\left(x^{2}+x^{2}_{0}\right)+\sqrt{\frac{2m\omega}{\hbar}}x\alpha+\frac{1}{2\hbar m\omega}p^{2}_{0}-i\frac{x_{0}p_{0}}{2\hbar}\right]\\
& =C\exp\left[-\frac{m\omega}{2\hbar}\left(x^{2}+x^{2}_{0}\right)+\sqrt{\frac{2m\omega}{\hbar}}x\left(\sqrt{\frac{m\omega}{2\hbar}}x_{0}+\frac{i}{\sqrt{2\hbar m\omega}}p_{0}\right)+\frac{1}{2\hbar m\omega}p^{2}_{0}-i\frac{x_{0}p_{0}}{2\hbar}\right]\\
& =C\exp\left[-\frac{m\omega}{2\hbar}\left(x^{2}+x^{2}_{0}\right)+\frac{2m\omega}{2\hbar}xx_{0}+\frac{i}{\hbar}xp_{0}+\frac{1}{2\hbar m\omega}p^{2}_{0}-i\frac{x_{0}p_{0}}{2\hbar}\right]=\\
& =C\exp\left[-\frac{m\omega}{2\hbar}\left(x^{2}-2xx_{0}+x^{2}_{0}\right)+\frac{i}{\hbar}xp_{0}+\frac{1}{2\hbar m\omega}p^{2}_{0}-i\frac{x_{0}p_{0}}{2\hbar}\right]\\
& =C\exp\left[-\frac{m\omega}{2\hbar}\left(x-x_{0}\right)^{2}+\frac{p^{2}_{0}}{2\hbar m\omega}+i\left(\frac{xp_{0}}{\hbar}-\frac{x_{0}p_{0}}{2\hbar}\right)\right]
\end{align*} ⟨ x ∣ α ⟩ = C exp [ − 2ℏ mω x 2 + ℏ 2 mω xα − ( 2ℏ mω x 0 2 − 2ℏ mω 1 p 0 2 − i 2ℏ x 0 p 0 ) ] = C exp [ − 2ℏ mω ( x 2 + x 0 2 ) + ℏ 2 mω xα + 2ℏ mω 1 p 0 2 − i 2ℏ x 0 p 0 ] = C exp [ − 2ℏ mω ( x 2 + x 0 2 ) + ℏ 2 mω x ( 2ℏ mω x 0 + 2ℏ mω i p 0 ) + 2ℏ mω 1 p 0 2 − i 2ℏ x 0 p 0 ] = C exp [ − 2ℏ mω ( x 2 + x 0 2 ) + 2ℏ 2 mω x x 0 + ℏ i x p 0 + 2ℏ mω 1 p 0 2 − i 2ℏ x 0 p 0 ] = = C exp [ − 2ℏ mω ( x 2 − 2 x x 0 + x 0 2 ) + ℏ i x p 0 + 2ℏ mω 1 p 0 2 − i 2ℏ x 0 p 0 ] = C exp [ − 2ℏ mω ( x − x 0 ) 2 + 2ℏ mω p 0 2 + i ( ℏ x p 0 − 2ℏ x 0 p 0 ) ] Para a condição de normalização da função de onda, então temos que:
1 = ∫ − ∞ + ∞ ∣ ⟨ x ∣ α ⟩ ∣ 2 d x 1 = ∣ C ∣ 2 ∫ − ∞ + ∞ exp [ − m ω 2 ℏ ( x − x 0 ) 2 + p 0 2 ℏ m ω ] d x exp [ − p 0 2 ℏ m ω ] = ∣ C ∣ 2 ∫ − ∞ + ∞ exp [ − m ω 2 ℏ ( x − x 0 ) 2 ] d x \begin{align*}
1 & =\int^{+\infty}_{-\infty}\left|\left\langle x|\alpha\right\rangle \right|^{2}dx\\
1 & =\left|C\right|^{2}\int^{+\infty}_{-\infty}\exp\left[-\frac{m\omega}{2\hbar}\left(x-x_{0}\right)^{2}+\frac{p^{2}_{0}}{\hbar m\omega}\right]dx\\
\exp\left[-\frac{p^{2}_{0}}{\hbar m\omega}\right] & =\left|C\right|^{2}\int^{+\infty}_{-\infty}\exp\left[-\frac{m\omega}{2\hbar}\left(x-x_{0}\right)^{2}\right]dx
\end{align*} 1 1 exp [ − ℏ mω p 0 2 ] = ∫ − ∞ + ∞ ∣ ⟨ x ∣ α ⟩ ∣ 2 d x = ∣ C ∣ 2 ∫ − ∞ + ∞ exp [ − 2ℏ mω ( x − x 0 ) 2 + ℏ mω p 0 2 ] d x = ∣ C ∣ 2 ∫ − ∞ + ∞ exp [ − 2ℏ mω ( x − x 0 ) 2 ] d x Fazendo u = m ω 2 ℏ x − α u=\sqrt{\frac{m\omega}{2\hbar}}x-\alpha u = 2ℏ mω x − α d u d x = m ω 2 ℏ \frac{du}{dx}=\sqrt{\frac{m\omega}{2\hbar}} d x d u = 2ℏ mω
m ω 2 ℏ exp [ − p 0 2 ℏ m ω ] = ∣ C ∣ 2 ∫ − ∞ + ∞ e − 2 u 2 d u m ω 2 ℏ exp [ − p 0 2 ℏ m ω ] = ∣ C ∣ 2 π 2 ( m ω π ℏ ) 1 / 4 exp [ − p 0 2 2 ℏ m ω ] = C \begin{align*}
\sqrt{\frac{m\omega}{2\hbar}}\exp\left[-\frac{p^{2}_{0}}{\hbar m\omega}\right] & =\left|C\right|^{2}\int^{+\infty}_{-\infty}e^{-2u^{2}}du\\
\sqrt{\frac{m\omega}{2\hbar}}\exp\left[-\frac{p^{2}_{0}}{\hbar m\omega}\right] & =\left|C\right|^{2}\sqrt{\frac{\pi}{2}}\\
\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left[-\frac{p^{2}_{0}}{2\hbar m\omega}\right] & =C
\end{align*} 2ℏ mω exp [ − ℏ mω p 0 2 ] 2ℏ mω exp [ − ℏ mω p 0 2 ] ( π ℏ mω ) 1/4 exp [ − 2ℏ mω p 0 2 ] = ∣ C ∣ 2 ∫ − ∞ + ∞ e − 2 u 2 d u = ∣ C ∣ 2 2 π = C Onde usamos a integral de gauss ∫ − ∞ + ∞ 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 ∣ α ⟩ = ( m ω π ℏ ) 1 / 4 exp [ − p 0 2 2 ℏ m ω ] exp [ − m ω 2 ℏ ( x − x 0 ) 2 + p 0 2 2 ℏ m ω + i ( x p 0 ℏ − x 0 p 0 2 ℏ ) ] ⟨ x ∣ α ⟩ = ( m ω π ℏ ) 1 / 4 exp [ − m ω 2 ℏ ( x − x 0 ) 2 + i ( x p 0 ℏ − x 0 p 0 2 ℏ ) ] \begin{align*}
\left\langle x|\alpha\right\rangle & =\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left[-\frac{p^{2}_{0}}{2\hbar m\omega}\right]\exp\left[-\frac{m\omega}{2\hbar}\left(x-x_{0}\right)^{2}+\frac{p^{2}_{0}}{2\hbar m\omega}+i\left(\frac{xp_{0}}{\hbar}-\frac{x_{0}p_{0}}{2\hbar}\right)\right]\\
\left\langle x|\alpha\right\rangle & =\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left[-\frac{m\omega}{2\hbar}\left(x-x_{0}\right)^{2}+i\left(\frac{xp_{0}}{\hbar}-\frac{x_{0}p_{0}}{2\hbar}\right)\right]
\end{align*} ⟨ x ∣ α ⟩ ⟨ x ∣ α ⟩ = ( π ℏ mω ) 1/4 exp [ − 2ℏ mω p 0 2 ] exp [ − 2ℏ mω ( x − x 0 ) 2 + 2ℏ mω p 0 2 + i ( ℏ x p 0 − 2ℏ x 0 p 0 ) ] = ( π ℏ mω ) 1/4 exp [ − 2ℏ mω ( x − x 0 ) 2 + i ( ℏ x p 0 − 2ℏ x 0 p 0 ) ] Para ⟨ x ∣ α ( t ) ⟩ \left\langle x|\alpha\left(t\right)\right\rangle ⟨ x ∣ α ( t ) ⟩ p 0 p_{0} p 0 q 0 q_{0} q 0 p ( t ) = ⟨ α ( t ) ∣ p ^ ∣ α ( t ) ⟩ p\left(t\right)=\left\langle \alpha\left(t\right)\left|\widehat{p}\right|\alpha\left(t\right)\right\rangle p ( t ) = ⟨ α ( t ) ∣ p ∣ α ( t ) ⟩ q ( t ) = ⟨ α ( t ) ∣ q ^ ∣ α ( t ) ⟩ q\left(t\right)=\left\langle \alpha\left(t\right)\left|\widehat{q}\right|\alpha\left(t\right)\right\rangle q ( t ) = ⟨ α ( t ) ∣ q ∣ α ( t ) ⟩
⟨ α ( t ) ∣ a ^ ∣ α ( t ) ⟩ = ⟨ α ∣ U ^ † a ^ U ^ ∣ α ⟩ = ⟨ α ∣ a ^ e − i ω ( t − t 0 ) ∣ α ⟩ = e − i ω ( t − t 0 ) ⟨ α ∣ a ^ ∣ α ⟩ = e − i ω ( t − t 0 ) α \begin{align*}
\left\langle \alpha\left(t\right)\left|\widehat{a}\right|\alpha\left(t\right)\right\rangle & =\left\langle \alpha\left|\widehat{U}^{\dagger}\widehat{a}\widehat{U}\right|\alpha\right\rangle \\
& =\left\langle \alpha\left|\widehat{a}e^{-i\omega\left(t-t_{0}\right)}\right|\alpha\right\rangle \\
& =e^{-i\omega\left(t-t_{0}\right)}\left\langle \alpha\left|\widehat{a}\right|\alpha\right\rangle \\
& =e^{-i\omega\left(t-t_{0}\right)}\alpha
\end{align*} ⟨ α ( t ) ∣ a ∣ α ( t ) ⟩ = ⟨ α ∣ ∣ U † a U ∣ ∣ α ⟩ = ⟨ α ∣ ∣ a e − iω ( t − t 0 ) ∣ ∣ α ⟩ = e − iω ( t − t 0 ) ⟨ α ∣ a ∣ α ⟩ = e − iω ( t − t 0 ) α E usando a definição de a ^ \widehat{a} a
⟨ α ( t ) ∣ a ^ ∣ α ( t ) ⟩ = ⟨ α ( t ) ∣ m ω 2 ℏ x ^ + i 2 ℏ m ω p ^ ∣ α ( t ) ⟩ = m ω 2 ℏ ⟨ α ( t ) ∣ x ^ ∣ α ( t ) ⟩ + i 2 ℏ m ω ⟨ α ( t ) ∣ p ^ ∣ α ( t ) ⟩ = m ω 2 ℏ x ( t ) + i 2 ℏ m ω p ( t ) \begin{align*}
\left\langle \alpha\left(t\right)\left|\widehat{a}\right|\alpha\left(t\right)\right\rangle & =\left\langle \alpha\left(t\right)\left|\sqrt{\frac{m\omega}{2\hbar}}\widehat{x}+\frac{i}{\sqrt{2\hbar m\omega}}\widehat{p}\right|\alpha\left(t\right)\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}\left\langle \alpha\left(t\right)\left|\widehat{x}\right|\alpha\left(t\right)\right\rangle +\frac{i}{\sqrt{2\hbar m\omega}}\left\langle \alpha\left(t\right)\left|\widehat{p}\right|\alpha\left(t\right)\right\rangle \\
& =\sqrt{\frac{m\omega}{2\hbar}}x\left(t\right)+\frac{i}{\sqrt{2\hbar m\omega}}p\left(t\right)
\end{align*} ⟨ α ( t ) ∣ a ∣ α ( t ) ⟩ = ⟨ α ( t ) ∣ ∣ 2ℏ mω x + 2ℏ mω i p ∣ ∣ α ( t ) ⟩ = 2ℏ mω ⟨ α ( t ) ∣ x ∣ α ( t ) ⟩ + 2ℏ mω i ⟨ α ( t ) ∣ p ∣ α ( t ) ⟩ = 2ℏ mω x ( t ) + 2ℏ mω i p ( t ) Então:
α = e i ω ( t − t 0 ) ( m ω 2 ℏ x ( t ) + i 2 ℏ m ω p ( t ) ) \alpha=e^{i\omega\left(t-t_{0}\right)}\left(\sqrt{\frac{m\omega}{2\hbar}}x\left(t\right)+\frac{i}{\sqrt{2\hbar m\omega}}p\left(t\right)\right) α = e iω ( t − t 0 ) ( 2ℏ mω x ( t ) + 2ℏ mω i p ( t ) ) Para facilitar, vamos denotar τ = t − t 0 \tau=t-t_{0} τ = t − t 0
x ( τ ) ≡ x 0 cos ( ω τ ) + p 0 m ω sin ( ω τ ) p ( τ ) ≡ − m ω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) \begin{align*}
x\left(\tau\right) & \equiv x_{0}\cos\left(\omega\tau\right)+\frac{p_{0}}{m\omega}\sin\left(\omega\tau\right)\\
p\left(\tau\right) & \equiv-m\omega x_{0}\sin\left(\omega\tau\right)+p_{0}\cos\left(\omega\tau\right)
\end{align*} x ( τ ) p ( τ ) ≡ x 0 cos ( ω τ ) + mω p 0 sin ( ω τ ) ≡ − mω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) Pois:
α = e i ω τ [ m ω 2 ℏ ( x 0 cos ( ω τ ) + p 0 m ω sin ( ω τ ) ) + i 2 ℏ m ω ( − m ω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) ) ] = e i ω τ [ m ω 2 ℏ x 0 cos ( ω τ ) + p 0 2 m ω sin ( ω τ ) − i m ω 2 ℏ x 0 sin ( ω τ ) + p 0 2 ℏ m ω cos ( ω τ ) ] = e i ω τ [ m ω 2 ℏ x 0 ( e i ω τ + e − i ω τ 2 ) + p 0 2 m ω ( e i ω τ − e − i ω τ 2 i ) − i m ω 2 ℏ x 0 ( e i ω τ − e − i ω τ 2 i ) + p 0 2 ℏ m ω ( e i ω τ + e − i ω τ 2 ) ] = [ m ω 2 ℏ x 0 ( e 2 i ω τ + 1 2 ) − i p 0 2 m ω ( e 2 i ω τ − 1 2 ) − m ω 2 ℏ x 0 ( e 2 i ω τ − 1 2 ) + p 0 2 ℏ m ω ( e 2 i ω τ + 1 2 ) ] = m ω 2 ℏ x 0 + i p 0 2 m ω \begin{align}
\alpha & =e^{i\omega\tau}\left[\sqrt{\frac{m\omega}{2\hbar}}\left(x_{0}\cos\left(\omega\tau\right)+\frac{p_{0}}{m\omega}\sin\left(\omega\tau\right)\right)+\frac{i}{\sqrt{2\hbar m\omega}}\left(-m\omega x_{0}\sin\left(\omega\tau\right)+p_{0}\cos\left(\omega\tau\right)\right)\right]\\
& =e^{i\omega\tau}\left[\sqrt{\frac{m\omega}{2\hbar}}x_{0}\cos\left(\omega\tau\right)+\frac{p_{0}}{\sqrt{2m\omega}}\sin\left(\omega\tau\right)-i\sqrt{\frac{m\omega}{2\hbar}}x_{0}\sin\left(\omega\tau\right)+\frac{p_{0}}{\sqrt{2\hbar m\omega}}\cos\left(\omega\tau\right)\right]\\
& =e^{i\omega\tau}\left[\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{i\omega\tau}+e^{-i\omega\tau}}{2}\right)+\frac{p_{0}}{\sqrt{2m\omega}}\left(\frac{e^{i\omega\tau}-e^{-i\omega\tau}}{2i}\right)-i\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{i\omega\tau}-e^{-i\omega\tau}}{2i}\right)+\frac{p_{0}}{\sqrt{2\hbar m\omega}}\left(\frac{e^{i\omega\tau}+e^{-i\omega\tau}}{2}\right)\right]\\
& =\left[\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{2i\omega\tau}+1}{2}\right)-\frac{ip_{0}}{\sqrt{2m\omega}}\left(\frac{e^{2i\omega\tau}-1}{2}\right)-\sqrt{\frac{m\omega}{2\hbar}}x_{0}\left(\frac{e^{2i\omega\tau}-1}{2}\right)+\frac{p_{0}}{\sqrt{2\hbar m\omega}}\left(\frac{e^{2i\omega\tau}+1}{2}\right)\right]\\
& =\sqrt{\frac{m\omega}{2\hbar}}x_{0}+\frac{ip_{0}}{\sqrt{2m\omega}}
\end{align} α = e iω τ [ 2ℏ mω ( x 0 cos ( ω τ ) + mω p 0 sin ( ω τ ) ) + 2ℏ mω i ( − mω x 0 sin ( ω τ ) + p 0 cos ( ω τ ) ) ] = e iω τ [ 2ℏ mω x 0 cos ( ω τ ) + 2 mω p 0 sin ( ω τ ) − i 2ℏ mω x 0 sin ( ω τ ) + 2ℏ mω p 0 cos ( ω τ ) ] = e iω τ [ 2ℏ mω x 0 ( 2 e iω τ + e − iω τ ) + 2 mω p 0 ( 2 i e iω τ − e − iω τ ) − i 2ℏ mω x 0 ( 2 i e iω τ − e − iω τ ) + 2ℏ mω p 0 ( 2 e iω τ + e − iω τ ) ] = [ 2ℏ mω x 0 ( 2 e 2 iω τ + 1 ) − 2 mω i p 0 ( 2 e 2 iω τ − 1 ) − 2ℏ mω x 0 ( 2 e 2 iω τ − 1 ) + 2ℏ mω p 0 ( 2 e 2 iω τ + 1 ) ] = 2ℏ mω x 0 + 2 mω i p 0 Que é o resultado obtido na equação 16 x ( τ ) x\left(\tau\right) x ( τ ) p ( τ ) p\left(\tau\right) p ( τ ) ⟨ x ′ ′ ∣ α ( t ) ⟩ ⟨ α ∣ x ′ ⟩ = ⟨ x ′ ′ ∣ α ( t ) ⟩ ⟨ x ′ ∣ α ⟩ † \left\langle x''|\alpha\left(t\right)\right\rangle \left\langle \alpha|x'\right\rangle =\left\langle x''|\alpha\left(t\right)\right\rangle \left\langle x'|\alpha\right\rangle ^{\dagger} ⟨ x ′′ ∣ α ( t ) ⟩ ⟨ α ∣ x ′ ⟩ = ⟨ x ′′ ∣ α ( t ) ⟩ ⟨ x ′ ∣ α ⟩ † x ′ = x 1 x'=x_{1} x ′ = x 1 x ′ ′ = x 2 x''=x_{2} x ′′ = x 2
Subscript[Q,11]:=((m*w)/(Pi*h))^(1/4)*Exp[-(m*w)/(2*h)*(Subscript[x,1]-Subscript[x,0])^2
+I*Subscript[x,1]*Subscript[p,0]/h-I*Subscript[x,0]*Subscript[p,0]/(2*h)]
Subscript[Q,12]:=((m*w)/(Pi*h))^(1/4)*Exp[-(m*w)/(2*h)*(Subscript[x,1]-Subscript[x,0])^2
-I*Subscript[x,1]*Subscript[p,0]/h+I*Subscript[x,0]*Subscript[p,0]/(2*h)]
Subscript[Q,21]:=((m*w)/(Pi*h))^(1/4)*Exp[-(m*w)/(2*h)*(Subscript[x,2]-Subscript[x,t])^2
+I*Subscript[x,2]*Subscript[p,t]/h-I*Subscript[x,t]*Subscript[p,t]/(2*h)]
Subscript[Q,22]:=Subscript[Q,21]/.{Subscript[x,t]->Subscript[x,0]*Cos[w*t]+
Subscript[p,0]*Sin[w*t]/(m*w),
Subscript[p,t]->-m*w*Subscript[x,0]*Sin[w*t]+Subscript[p,0]*Cos[w*t]}
Subscript[Q,12]*Subscript[Q,22]/(2*Pi*h)Obtemos:
K ( x ′ ′ , t ; x ′ , t 0 ) = m ω ℏ 2 π 3 / 2 ℏ exp ( i x ′ ′ ( p 0 cos ( ω τ ) − m ω x 0 sin ( ω τ ) ) ℏ − m ω ( − p 0 sin ( ω τ ) m ω − x 0 cos ( ω τ ) + x ′ ′ ) 2 2 h ) × exp ( − i ( p 0 sin ( ω τ ) m ω + x 0 cos ( τ ω ) ) ( p 0 cos ( ω τ ) − m ω x 0 sin ( ω τ ) ) 2 ℏ − m ω ( x ′ − x 0 ) 2 2 ℏ + i p 0 x 0 2 ℏ − i p 0 x ′ ℏ ) \begin{align*}
K\left(x'',t;x',t_{0}\right) & =\frac{\sqrt{\frac{m\omega}{\hbar}}}{2\pi^{3/2}\hbar}\exp\left(\frac{ix''\left(p_{0}\cos(\omega\tau)-m\omega x_{0}\sin(\omega\tau)\right)}{\hbar}-\frac{m\omega\left(-\frac{p_{0}\sin(\omega\tau)}{m\omega}-x_{0}\cos(\omega\tau)+x''\right){}^{2}}{2h}\right)\\
& \times\exp\left(-\frac{i\left(\frac{p_{0}\sin(\omega\tau)}{m\omega}+x_{0}\cos(\tau\omega)\right)\left(p_{0}\cos(\omega\tau)-m\omega x_{0}\sin(\omega\tau)\right)}{2\hbar}-\frac{m\omega\left(x'-x_{0}\right){}^{2}}{2\hbar}+\frac{ip_{0}x_{0}}{2\hbar}-\frac{ip_{0}x'}{\hbar}\right)
\end{align*} K ( x ′′ , t ; x ′ , t 0 ) = 2 π 3/2 ℏ ℏ mω exp ⎝ ⎛ ℏ i x ′′ ( p 0 cos ( ω τ ) − mω x 0 sin ( ω τ ) ) − 2 h mω ( − mω p 0 s i n ( ω τ ) − x 0 cos ( ω τ ) + x ′′ ) 2 ⎠ ⎞ × exp ⎝ ⎛ − 2ℏ i ( mω p 0 s i n ( ω τ ) + x 0 cos ( τ ω ) ) ( p 0 cos ( ω τ ) − mω x 0 sin ( ω τ ) ) − 2ℏ mω ( x ′ − x 0 ) 2 + 2ℏ i p 0 x 0 − ℏ i p 0 x ′ ⎠ ⎞ Utilizando novamente o mathematica para simplificar o expoente com a função ’TrigToExp’, obtemos que:
β = p 0 2 e − 2 i ω τ 4 ℏ m ω − p 0 2 4 ℏ m ω + m ω x 0 x 2 e − i ω τ ℏ − m ω x 0 2 e − 2 i ω τ 4 ℏ + m ω x 0 x ′ ℏ − m ω x ′ 2 2 ℏ − m ω x ′ ′ 2 2 ℏ − 3 m ω x 0 2 4 ℏ + i p 0 x ′ ′ e − i ω τ ℏ − i p 0 x 0 e − 2 i ω τ 2 ℏ + i p 0 x 0 2 ℏ − i p 0 x ′ ℏ \begin{align*}
\beta & =\frac{p^{2}_{0}e^{-2i\omega\tau}}{4\hbar m\omega}-\frac{p^{2}_{0}}{4\hbar m\omega}+\frac{m\omega x_{0}x_{2}e^{-i\omega\tau}}{\hbar}-\frac{m\omega x^{2}_{0}e^{-2i\omega\tau}}{4\hbar}+\frac{m\omega x_{0}x'}{\hbar}-\frac{m\omega x'^{2}}{2\hbar}\\
& -\frac{m\omega x''^{2}}{2\hbar}-\frac{3m\omega x^{2}_{0}}{4\hbar}+\frac{ip_{0}x''e^{-i\omega\tau}}{\hbar}-\frac{ip_{0}x_{0}e^{-2i\omega\tau}}{2\hbar}+\frac{ip_{0}x_{0}}{2\hbar}-\frac{ip_{0}x'}{\hbar}
\end{align*} β = 4ℏ mω p 0 2 e − 2 iω τ − 4ℏ mω p 0 2 + ℏ mω x 0 x 2 e − iω τ − 4ℏ mω x 0 2 e − 2 iω τ + ℏ mω x 0 x ′ − 2ℏ mω x ′ 2 − 2ℏ mω x ′′ 2 − 4ℏ 3 mω x 0 2 + ℏ i p 0 x ′′ e − iω τ − 2ℏ i p 0 x 0 e − 2 iω τ + 2ℏ i p 0 x 0 − ℏ i p 0 x ′ Abrindo os termos e multiplicando o termo que aompanha p 0 2 p^{2}_{0} p 0 2 i i \frac{i}{i} i i
β = − m ω 2 ℏ ( x ′ 2 + x ′ ′ 2 ) − m ω 2 ℏ ( e − 2 i ω τ 2 + 1 2 + 1 ) x 0 2 − i 2 ℏ m ω ( 1 − e − 2 i ω τ 2 i ) p 0 2 + m ω ℏ ( x ′ ′ e − i ω τ + x ′ ) x 0 + i ℏ ( x ′ ′ e − i ω τ − x ′ ) p 0 − i ℏ ( e − 2 i ω τ 2 − 1 2 ) p 0 x 0 \begin{align*}
\beta & =-\frac{m\omega}{2\hbar}\left(x'^{2}+x''^{2}\right)-\frac{m\omega}{2\hbar}\left(\frac{e^{-2i\omega\tau}}{2}+\frac{1}{2}+1\right)x^{2}_{0}-\frac{i}{2\hbar m\omega}\left(\frac{1-e^{-2i\omega\tau}}{2i}\right)p^{2}_{0}\\
& +\frac{m\omega}{\hbar}\left(x''e^{-i\omega\tau}+x'\right)x_{0}+\frac{i}{\hbar}\left(x''e^{-i\omega\tau}-x'\right)p_{0}-\frac{i}{\hbar}\left(\frac{e^{-2i\omega\tau}}{2}-\frac{1}{2}\right)p_{0}x_{0}
\end{align*} β = − 2ℏ mω ( x ′ 2 + x ′′ 2 ) − 2ℏ mω ( 2 e − 2 iω τ + 2 1 + 1 ) x 0 2 − 2ℏ mω i ( 2 i 1 − e − 2 iω τ ) p 0 2 + ℏ mω ( x ′′ e − iω τ + x ′ ) x 0 + ℏ i ( x ′′ e − iω τ − x ′ ) p 0 − ℏ i ( 2 e − 2 iω τ − 2 1 ) p 0 x 0 Agora como
c o s ( ω t ) e − i ω t = ( e i ω t + e − i ω t 2 ) e − i ω t = ( 1 2 + e − 2 i ω t ) = ( 1 2 + e − 2 i ω t ) cos\left(\omega t\right)e^{-i\omega t}=\left(\frac{e^{i\omega t}+e^{-i\omega t}}{2}\right)e^{-i\omega t}=\left(\frac{1}{2}+e^{-2i\omega t}\right)=\left(\frac{1}{2}+e^{-2i\omega t}\right) cos ( ω t ) e − iω t = ( 2 e iω t + e − iω t ) e − iω t = ( 2 1 + e − 2 iω t ) = ( 2 1 + e − 2 iω t ) sin ( ω t ) e − ω i t = ( e i ω t − e − i ω t 2 i ) e − ω i t = ( 1 − e − 2 ω i t 2 i ) e − ω i t = i ( e − 2 ω i t 2 − 1 2 ) e − i ω t \sin\left(\omega t\right)e^{-\omega it}=\left(\frac{e^{i\omega t}-e^{-i\omega t}}{2i}\right)e^{-\omega it}=\left(\frac{1-e^{-2\omega it}}{2i}\right)e^{-\omega it}=i\left(\frac{e^{-2\omega it}}{2}-\frac{1}{2}\right)e^{-i\omega t} sin ( ω t ) e − ωi t = ( 2 i e iω t − e − iω t ) e − ωi t = ( 2 i 1 − e − 2 ωi t ) e − ωi t = i ( 2 e − 2 ωi t − 2 1 ) e − iω t Ajustando os sinais e manipulando:
β = − m ω 2 ℏ ( x ′ 2 + x ′ ′ 2 ) − [ m ω 2 ℏ ( 1 + c o s ( ω τ ) e − i ω τ ) ] x 0 2 − [ i 2 ℏ m ω ( sin ( ω τ ) e − i ω τ ) ] p 0 2 − [ 1 ℏ sin ( ω τ ) e − i ω τ ] p 0 x 0 − [ − m ω ℏ ( x ′ ′ e − i ω τ + x ′ ) ] x 0 − [ − i ℏ ( x ′ ′ e − i ω τ − x ′ ) ] p 0 \begin{align*}
\beta & =-\frac{m\omega}{2\hbar}\left(x'^{2}+x''^{2}\right)-\left[\frac{m\omega}{2\hbar}\left(1+cos\left(\omega\tau\right)e^{-i\omega\tau}\right)\right]x^{2}_{0}-\left[\frac{i}{2\hbar m\omega}\left(\sin\left(\omega\tau\right)e^{-i\omega\tau}\right)\right]p^{2}_{0}\\
& -\left[\frac{1}{\hbar}\sin\left(\omega\tau\right)e^{-i\omega\tau}\right]p_{0}x_{0}-\left[-\frac{m\omega}{\hbar}\left(x''e^{-i\omega\tau}+x'\right)\right]x_{0}-\left[-\frac{i}{\hbar}\left(x''e^{-i\omega\tau}-x'\right)\right]p_{0}
\end{align*} β = − 2ℏ mω ( x ′ 2 + x ′′ 2 ) − [ 2ℏ mω ( 1 + cos ( ω τ ) e − iω τ ) ] x 0 2 − [ 2ℏ mω i ( sin ( ω τ ) e − iω τ ) ] p 0 2 − [ ℏ 1 sin ( ω τ ) e − iω τ ] p 0 x 0 − [ − ℏ mω ( x ′′ e − iω τ + x ′ ) ] x 0 − [ − ℏ i ( x ′′ e − iω τ − x ′ ) ] p 0 Aqui precisamos notar que o term o c 22 c_{22} c 22 1 2 \frac{1}{2} 2 1 (37) do artigo, possívelmente um erro de digitação. Além disso na equação (34 ) foi negligenciado o termo que acompanha a exponencial, talvez intencionlmente. E rearranjando:
β = − m ω π ℏ ( x 1 2 + x 2 2 ) − c 11 x 0 2 − c 22 p 0 2 − c 1 x 0 − c 12 p 0 x 0 − c 1 p 0 \begin{align*}
\beta & =-\frac{m\omega}{\pi\hbar}\left(x^{2}_{1}+x^{2}_{2}\right)-c_{11}x^{2}_{0}-c_{22}p^{2}_{0}-c_{1}x_{0}-c_{12}p_{0}x_{0}-c_{1}p_{0}
\end{align*} β = − π ℏ mω ( x 1 2 + x 2 2 ) − c 11 x 0 2 − c 22 p 0 2 − c 1 x 0 − c 12 p 0 x 0 − c 1 p 0 Substituindo no propagador e integrando primeiro em p 0 p_{0} p 0
K ( x ′ ′ , t ; x ′ , t 0 ) = m ω π ℏ exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] ] 2 π ℏ ∫ − ∞ + ∞ exp [ − c 11 x 0 2 − c 1 x 0 ] ∫ − ∞ + ∞ exp [ − c 22 p 0 2 + ( − c 12 x 0 − c 2 ) p 0 ] d p 0 d x 0 K\left(x'',t;x',t_{0}\right)=\frac{\sqrt{\frac{m\omega}{\pi\hbar}}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]\right]}{2\pi\hbar}\int^{+\infty}_{-\infty}\exp\left[-c_{11}x^{2}_{0}-c_{1}x_{0}\right]\int^{+\infty}_{-\infty}\exp\left[-c_{22}p^{2}_{0}+\left(-c_{12}x_{0}-c_{2}\right)p_{0}\right]dp_{0}dx_{0} K ( x ′′ , t ; x ′ , t 0 ) = 2 π ℏ π ℏ mω exp [ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] ] ∫ − ∞ + ∞ exp [ − c 11 x 0 2 − c 1 x 0 ] ∫ − ∞ + ∞ exp [ − c 22 p 0 2 + ( − c 12 x 0 − c 2 ) p 0 ] d p 0 d x 0 Evidentemente poderíamos completar quadrado para realizar uma integração de gauss, mas usando uma forma alternativa generalizada como ∫ − ∞ + ∞ e − a x 2 + b x + c = π a e b 2 4 a + c \int^{+\infty}_{-\infty}e^{-ax^{2}+bx+c}=\sqrt{\frac{\pi}{a}}e^{\frac{b^{2}}{4a}+c} ∫ − ∞ + ∞ e − a x 2 + b x + c = a π e 4 a b 2 + c a = c 22 a=c_{22} a = c 22 b = ( c 12 x 0 + c 2 ) b=\left(c_{12}x_{0}+c_{2}\right) b = ( c 12 x 0 + c 2 ) c = 0 c=0 c = 0
K ( x ′ ′ , t ; x ′ , t 0 ) = m ω π ℏ exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] ] 2 π ℏ π c 22 ∫ − ∞ + ∞ exp [ − c 11 x 0 2 − c 1 x 0 + ( c 12 x 0 + c 2 ) 2 4 c 22 ] d x 0 = m ω ℏ c 22 1 2 π ℏ exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] ] ∫ − ∞ + ∞ exp [ − c 11 x 0 2 − c 1 x 0 + c 12 2 x 0 2 + c 2 2 + 2 c 12 c 2 x 0 4 c 22 ] d x 0 = m ω ℏ c 22 1 2 π ℏ exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] ] ∫ − ∞ + ∞ exp [ − ( c 11 − c 12 2 4 c 22 ) x 0 2 + ( c 12 c 2 2 c 22 − c 1 ) x 0 + ( c 2 2 4 c 22 ) ] d x 0 = m ω ℏ c 22 1 2 π ℏ π ( c 11 − 1 4 c 12 2 c 22 ) exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] + 1 4 ( c 12 c 2 2 c 22 − c 1 ) 2 ( c 11 − c 12 2 4 c 22 ) + 1 4 c 2 2 c 22 ] = m ω ℏ c 22 1 2 π ℏ 1 1 4 c 22 π ( 4 c 22 c 11 − c 12 2 ) exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] + 1 4 ( c 12 c 2 − 2 c 22 c 1 ) 2 1 4 c 22 2 ( 4 c 22 c 11 − c 12 2 ) 1 4 c 22 + 1 4 c 2 2 c 22 ] = m ω ℏ 1 π ℏ π ( 4 c 22 c 11 − c 12 2 ) exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] + 1 4 ( c 12 c 2 − 2 c 22 c 1 ) 2 ( 4 c 22 c 11 − c 12 2 ) c 22 + 1 4 c 2 2 c 22 ] = m ω π ℏ 1 ℏ 1 4 c 22 c 11 − c 12 2 exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] + 1 4 c 12 2 c 2 2 + 4 c 22 2 c 1 2 − 4 c 12 c 2 c 22 c 1 + 4 c 2 2 c 22 c 11 − c 2 2 c 12 2 ( 4 c 22 c 11 − c 12 2 ) c 22 ] = m ω π ℏ 1 ℏ 1 4 c 22 c 11 − c 12 2 exp [ − m ω 2 ℏ [ ( x 2 + x ′ 2 ) ] + c 2 2 c 11 − c 12 c 2 c 1 + c 22 2 c 1 2 4 c 22 c 11 − c 12 2 ] \begin{align*}
K\left(x'',t;x',t_{0}\right) & =\sqrt{\frac{m\omega}{\pi\hbar}}\frac{\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]\right]}{2\pi\hbar}\sqrt{\frac{\pi}{c_{22}}}\int^{+\infty}_{-\infty}\exp\left[-c_{11}x^{2}_{0}-c_{1}x_{0}+\frac{\left(c_{12}x_{0}+c_{2}\right)^{2}}{4c_{22}}\right]dx_{0}\\
& =\sqrt{\frac{m\omega}{\hbar c_{22}}}\frac{1}{2\pi\hbar}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]\right]\int^{+\infty}_{-\infty}\exp\left[-c_{11}x^{2}_{0}-c_{1}x_{0}+\frac{c^{2}_{12}x^{2}_{0}+c^{2}_{2}+2c_{12}c_{2}x_{0}}{4c_{22}}\right]dx_{0}\\
& =\sqrt{\frac{m\omega}{\hbar c_{22}}}\frac{1}{2\pi\hbar}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]\right]\int^{+\infty}_{-\infty}\exp\left[-\left(c_{11}-\frac{c^{2}_{12}}{4c_{22}}\right)x^{2}_{0}+\left(\frac{c_{12}c_{2}}{2c_{22}}-c_{1}\right)x_{0}+\left(\frac{c^{2}_{2}}{4c_{22}}\right)\right]dx_{0}\\
& =\sqrt{\frac{m\omega}{\hbar c_{22}}}\frac{1}{2\pi\hbar}\sqrt{\frac{\pi}{\left(c_{11}-\frac{1}{4}\frac{c^{2}_{12}}{c_{22}}\right)}}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]+\frac{1}{4}\frac{\left(\frac{c_{12}c_{2}}{2c_{22}}-c_{1}\right)^{2}}{\left(c_{11}-\frac{c^{2}_{12}}{4c_{22}}\right)}+\frac{1}{4}\frac{c^{2}_{2}}{c_{22}}\right]\\
& =\sqrt{\frac{m\omega}{\hbar c_{22}}}\frac{1}{2\pi\hbar}\frac{1}{\sqrt{\frac{1}{4c_{22}}}}\sqrt{\frac{\pi}{\left(4c_{22}c_{11}-c^{2}_{12}\right)}}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]+\frac{1}{4}\frac{\left(c_{12}c_{2}-2c_{22}c_{1}\right)^{2}\frac{1}{4c^{2}_{22}}}{\left(4c_{22}c_{11}-c^{2}_{12}\right)\frac{1}{4c_{22}}}+\frac{1}{4}\frac{c^{2}_{2}}{c_{22}}\right]\\
& =\sqrt{\frac{m\omega}{\hbar}}\frac{1}{\pi\hbar}\sqrt{\frac{\pi}{\left(4c_{22}c_{11}-c^{2}_{12}\right)}}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]+\frac{1}{4}\frac{\left(c_{12}c_{2}-2c_{22}c_{1}\right)^{2}}{\left(4c_{22}c_{11}-c^{2}_{12}\right)c_{22}}+\frac{1}{4}\frac{c^{2}_{2}}{c_{22}}\right]\\
& =\sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\hbar}\frac{1}{\sqrt{4c_{22}c_{11}-c^{2}_{12}}}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]+\frac{1}{4}\frac{c^{2}_{12}c^{2}_{2}+4c^{2}_{22}c^{2}_{1}-4c_{12}c_{2}c_{22}c_{1}+4c^{2}_{2}c_{22}c_{11}-c^{2}_{2}c^{2}_{12}}{\left(4c_{22}c_{11}-c^{2}_{12}\right)c_{22}}\right]\\
& =\sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\hbar}\frac{1}{\sqrt{4c_{22}c_{11}-c^{2}_{12}}}\exp\left[-\frac{m\omega}{2\hbar}\left[\left(x^{2}+x'^{2}\right)\right]+\frac{c^{2}_{2}c_{11}-c_{12}c_{2}c_{1}+c^{2}_{22}c^{2}_{1}}{4c_{22}c_{11}-c^{2}_{12}}\right]
\end{align*} K ( x ′′ , t ; x ′ , t 0 ) = π ℏ mω 2 π ℏ exp [ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] ] c 22 π ∫ − ∞ + ∞ exp [ − c 11 x 0 2 − c 1 x 0 + 4 c 22 ( c 12 x 0 + c 2 ) 2 ] d x 0 = ℏ c 22 mω 2 π ℏ 1 exp [ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] ] ∫ − ∞ + ∞ exp [ − c 11 x 0 2 − c 1 x 0 + 4 c 22 c 12 2 x 0 2 + c 2 2 + 2 c 12 c 2 x 0 ] d x 0 = ℏ c 22 mω 2 π ℏ 1 exp [ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] ] ∫ − ∞ + ∞ exp [ − ( c 11 − 4 c 22 c 12 2 ) x 0 2 + ( 2 c 22 c 12 c 2 − c 1 ) x 0 + ( 4 c 22 c 2 2 ) ] d x 0 = ℏ c 22 mω 2 π ℏ 1 ( c 11 − 4 1 c 22 c 12 2 ) π exp ⎣ ⎡ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] + 4 1 ( c 11 − 4 c 22 c 12 2 ) ( 2 c 22 c 12 c 2 − c 1 ) 2 + 4 1 c 22 c 2 2 ⎦ ⎤ = ℏ c 22 mω 2 π ℏ 1 4 c 22 1 1 ( 4 c 22 c 11 − c 12 2 ) π exp ⎣ ⎡ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] + 4 1 ( 4 c 22 c 11 − c 12 2 ) 4 c 22 1 ( c 12 c 2 − 2 c 22 c 1 ) 2 4 c 22 2 1 + 4 1 c 22 c 2 2 ⎦ ⎤ = ℏ mω π ℏ 1 ( 4 c 22 c 11 − c 12 2 ) π exp [ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] + 4 1 ( 4 c 22 c 11 − c 12 2 ) c 22 ( c 12 c 2 − 2 c 22 c 1 ) 2 + 4 1 c 22 c 2 2 ] = π ℏ mω ℏ 1 4 c 22 c 11 − c 12 2 1 exp [ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] + 4 1 ( 4 c 22 c 11 − c 12 2 ) c 22 c 12 2 c 2 2 + 4 c 22 2 c 1 2 − 4 c 12 c 2 c 22 c 1 + 4 c 2 2 c 22 c 11 − c 2 2 c 12 2 ] = π ℏ mω ℏ 1 4 c 22 c 11 − c 12 2 1 exp [ − 2ℏ mω [ ( x 2 + x ′ 2 ) ] + 4 c 22 c 11 − c 12 2 c 2 2 c 11 − c 12 c 2 c 1 + c 22 2 c 1 2 ] Manipulando então:
4 c 22 c 11 − c 12 2 = 4 m ω 2 ℏ ( 1 + cos ( ω τ ) e − i ω τ ) i sin ( ω τ ) e − i ω τ 2 ℏ m ω − ( 1 ℏ sin ( ω τ ) e − i ω τ ) 2 = 1 ℏ 2 [ i sin ( ω τ ) e − i ω τ i cos ( ω τ ) sin ( ω τ ) e − i 2 ω τ − sin 2 ( ω τ ) e − i 2 ω τ ] = 1 ℏ 2 [ i sin ( ω τ ) e − i ω τ + ( i cos ( ω τ ) − sin ( ω τ ) ) sin ( ω τ ) e − i 2 ω τ ] = 1 ℏ 2 [ sin ( ω τ ) e − i ω τ + ( cos ( ω τ ) + i sin ( ω τ ) ) sin ( ω τ ) e − i 2 ω τ ] = 1 ℏ 2 [ sin ( ω τ ) e − i ω τ + e − i ω t sin ( ω τ ) e − i 2 ω τ ] = − i sin ( ω τ ) h 2 [ e − i ω τ + e − i ω τ ] = 2 i sin ( ω t ) ℏ 2 e − i ω τ \begin{align}
\sqrt{4c_{22}c_{11}-c^{2}_{12}} & =4\frac{m\omega}{2\hbar}\left(1+\cos\left(\omega\tau\right)e^{-i\omega\tau}\right)i\frac{\sin\left(\omega\tau\right)e^{-i\omega\tau}}{2\hbar m\omega}-\left(\frac{1}{\hbar}\sin\left(\omega\tau\right)e^{-i\omega\tau}\right)^{2}\\
& =\frac{1}{\hbar^{2}}\left[i\sin\left(\omega\tau\right)e^{-i\omega\tau}i\cos\left(\omega\tau\right)\sin\left(\omega\tau\right)e^{-i2\omega\tau}-\sin^{2}\left(\omega\tau\right)e^{-i2\omega\tau}\right]\\
& =\frac{1}{\hbar^{2}}\left[i\sin\left(\omega\tau\right)e^{-i\omega\tau}+\left(i\cos\left(\omega\tau\right)-\sin\left(\omega\tau\right)\right)\sin\left(\omega\tau\right)e^{-i2\omega\tau}\right]\\
& =\frac{1}{\hbar^{2}}\left[\sin\left(\omega\tau\right)e^{-i\omega\tau}+\left(\cos\left(\omega\tau\right)+i\sin\left(\omega\tau\right)\right)\sin\left(\omega\tau\right)e^{-i2\omega\tau}\right]\\
& =\frac{1}{\hbar^{2}}\left[\sin\left(\omega\tau\right)e^{-i\omega\tau}+e^{-i\omega t}\sin\left(\omega\tau\right)e^{-i2\omega\tau}\right]\\
& =-\frac{i\sin\left(\omega\tau\right)}{h^{2}}\left[e^{-i\omega\tau}+e^{-i\omega\tau}\right]\\
& =2\frac{i\sin\left(\omega t\right)}{\hbar^{2}}e^{-i\omega\tau}
\end{align} 4 c 22 c 11 − c 12 2 = 4 2ℏ mω ( 1 + cos ( ω τ ) e − iω τ ) i 2ℏ mω sin ( ω τ ) e − iω τ − ( ℏ 1 sin ( ω τ ) e − iω τ ) 2 = ℏ 2 1 [ i sin ( ω τ ) e − iω τ i cos ( ω τ ) sin ( ω τ ) e − i 2 ω τ − sin 2 ( ω τ ) e − i 2 ω τ ] = ℏ 2 1 [ i sin ( ω τ ) e − iω τ + ( i cos ( ω τ ) − sin ( ω τ ) ) sin ( ω τ ) e − i 2 ω τ ] = ℏ 2 1 [ sin ( ω τ ) e − iω τ + ( cos ( ω τ ) + i sin ( ω τ ) ) sin ( ω τ ) e − i 2 ω τ ] = ℏ 2 1 [ sin ( ω τ ) e − iω τ + e − iω t sin ( ω τ ) e − i 2 ω τ ] = − h 2 i sin ( ω τ ) [ e − iω τ + e − iω τ ] = 2 ℏ 2 i sin ( ω t ) e − iω τ Substituindo então
m ω π ℏ 1 ℏ 1 4 c 22 c 11 − c 12 2 = m ω π ℏ 1 ℏ 1 2 i sin ( ω τ ) h 2 e − i ω τ = m ω 2 π ℏ i sin ( ω τ ) e − ω τ \begin{align*}
\sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\hbar}\frac{1}{\sqrt{4c_{22}c_{11}-c^{2}_{12}}} & =\sqrt{\frac{m\omega}{\pi\hbar}}\frac{1}{\hbar}\frac{1}{\sqrt{2\frac{i\sin\left(\omega\tau\right)}{h^{2}}e^{-i\omega\tau}}}\\
& =\sqrt{\frac{m\omega}{2\pi\hbar i\sin\left(\omega\tau\right)e^{-\omega\tau}}}
\end{align*} π ℏ mω ℏ 1 4 c 22 c 11 − c 12 2 1 = π ℏ mω ℏ 1 2 h 2 i s i n ( ω τ ) e − iω τ 1 = 2 π ℏ i sin ( ω τ ) e − ω τ mω Que é o resultado que estávamos buscando, o termo e ω τ / 2 e^{\omega\tau/2} e ω τ /2 Shao (2016)
(\!\(\*SubsuperscriptBox[\(c\),\(2\),\(2\)]\ \*SubscriptBox[\(c\),\(11\)]\)-
Subscript[c,1]Subscript[c,2]Subscript[c,12]+\!\( \*SubsuperscriptBox[\(c\),\(1\),
\(2\)]\ \*SubscriptBox[\(c\),\(22\)]\))/(\!\( \*SubsuperscriptBox[\(c\),\(12\),
\(2\)]-\(4\ \*SubscriptBox[\(c\),\(11\)]\ \*SubscriptBox[\(c\),\(22\)]\)\))/.{Subscript[c,11]
-> m*w*(1+Cos[w*t]*Exp[-I*w*t])/(2*h), Subscript[c,22]->I*Sin[w*t]*Exp[-I*w*t]/(2*h*m*w),
Subscript[c,12]->Sin[w*t]*Exp[-I*w*t]/(h), Subscript[c,1]->-m*w*(Subscript[x,2]*Exp[-I*w*t]
+Subscript[x,1])/(h), Subscript[c,2]->-I*(Subscript[x,2]*Exp[-I*w*t]-Subscript[x,1])/(h)}E então usar o comando ’Simplify’ , onde obtemos:
β = − m ω 2 ℏ [ ( x 2 2 + x 1 2 ) ] + m ω ( − 2 x ′ ′ x ′ e i ω τ + x ′ 2 + x ′ ′ 2 ) ℏ ( − 1 + e 2 i t w ) = − m ω 2 ℏ [ ( x 2 2 + x 1 2 ) ] + m ω ( x ′ 2 + x ′ ′ 2 − e i ω τ 2 x ′ ′ x ′ ) e i ω τ ℏ ( e i ω τ − e − i t w ) = − m ω 2 ℏ [ ( x 2 2 + x 1 2 ) ] + i m ω ( x ′ 2 + x ′ ′ 2 − e i ω τ 2 x ′ ′ x ′ ) ℏ e i ω τ 2 sin ( ω τ ) = i m ω 2 ℏ sin ( ω τ ) [ i sin ( ω τ ) [ ( x ′ ′ 2 + x ′ 2 ) ] + e − i ω τ ( x ′ 2 + x ′ ′ 2 − e i ω τ 2 x ′ ′ x ′ ) ] = i m ω 2 ℏ sin ( ω τ ) [ i sin ( ω τ ) [ ( x ′ ′ 2 + x ′ 2 ) ] + e − i ω τ ( x ′ ′ 2 + x ′ 2 ) − 2 x ′ ′ x ′ ] = i m ω 2 ℏ sin ( ω τ ) [ ( e i ω τ − e − i ω τ ) 2 [ ( x ′ ′ 2 + x ′ 2 ) ] + e − i ω τ ( x ′ ′ 2 + x ′ 2 ) − 2 x ′ ′ x ′ ] = i m ω 2 ℏ sin ( ω τ ) [ ( e i ω τ + e − i ω τ ) 2 [ ( x ′ ′ 2 + x ′ 2 ) ] − 2 x ′ ′ x ′ ] = i m ω 2 ℏ sin ( ω τ ) [ cos ( ω τ ) [ ( x ′ ′ 2 + x ′ 2 ) ] − 2 x ′ ′ x ′ ] \begin{align*}
\beta & =-\frac{m\omega}{2\hbar}\left[\left(x^{2}_{2}+x^{2}_{1}\right)\right]+\frac{m\omega\left(-2x''x'e^{i\omega\tau}+x'^{2}+x''^{2}\right)}{\hbar\left(-1+e^{2itw}\right)}\\
& =-\frac{m\omega}{2\hbar}\left[\left(x^{2}_{2}+x^{2}_{1}\right)\right]+\frac{m\omega\left(x'^{2}+x''^{2}-e^{i\omega\tau}2x''x'\right)}{e^{i\omega\tau}\hbar\left(e^{i\omega\tau}-e^{-itw}\right)}\\
& =-\frac{m\omega}{2\hbar}\left[\left(x^{2}_{2}+x^{2}_{1}\right)\right]+\frac{im\omega\left(x'^{2}+x''^{2}-e^{i\omega\tau}2x''x'\right)}{\hbar e^{i\omega\tau}2\sin\left(\omega\tau\right)}\\
& =\frac{im\omega}{2\hbar\sin\left(\omega\tau\right)}\left[i\sin\left(\omega\tau\right)\left[\left(x''^{2}+x'^{2}\right)\right]+e^{-i\omega\tau}\left(x'^{2}+x''^{2}-e^{i\omega\tau}2x''x'\right)\right]\\
& =\frac{im\omega}{2\hbar\sin\left(\omega\tau\right)}\left[i\sin\left(\omega\tau\right)\left[\left(x''^{2}+x'^{2}\right)\right]+e^{-i\omega\tau}\left(x''^{2}+x'^{2}\right)-2x''x'\right]\\
& =\frac{im\omega}{2\hbar\sin\left(\omega\tau\right)}\left[\frac{\left(e^{i\omega\tau}-e^{-i\omega\tau}\right)}{2}\left[\left(x''^{2}+x'^{2}\right)\right]+e^{-i\omega\tau}\left(x''^{2}+x'^{2}\right)-2x''x'\right]\\
& =\frac{im\omega}{2\hbar\sin\left(\omega\tau\right)}\left[\frac{\left(e^{i\omega\tau}+e^{-i\omega\tau}\right)}{2}\left[\left(x''^{2}+x'^{2}\right)\right]-2x''x'\right]\\
& =\frac{im\omega}{2\hbar\sin\left(\omega\tau\right)}\left[\cos\left(\omega\tau\right)\left[\left(x''^{2}+x'^{2}\right)\right]-2x''x'\right]
\end{align*} β = − 2ℏ mω [ ( x 2 2 + x 1 2 ) ] + ℏ ( − 1 + e 2 i tw ) mω ( − 2 x ′′ x ′ e iω τ + x ′ 2 + x ′′ 2 ) = − 2ℏ mω [ ( x 2 2 + x 1 2 ) ] + e iω τ ℏ ( e iω τ − e − i tw ) mω ( x ′ 2 + x ′′ 2 − e iω τ 2 x ′′ x ′ ) = − 2ℏ mω [ ( x 2 2 + x 1 2 ) ] + ℏ e iω τ 2 sin ( ω τ ) imω ( x ′ 2 + x ′′ 2 − e iω τ 2 x ′′ x ′ ) = 2ℏ sin ( ω τ ) imω [ i sin ( ω τ ) [ ( x ′′ 2 + x ′ 2 ) ] + e − iω τ ( x ′ 2 + x ′′ 2 − e iω τ 2 x ′′ x ′ ) ] = 2ℏ sin ( ω τ ) imω [ i sin ( ω τ ) [ ( x ′′ 2 + x ′ 2 ) ] + e − iω τ ( x ′′ 2 + x ′ 2 ) − 2 x ′′ x ′ ] = 2ℏ sin ( ω τ ) imω [ 2 ( e iω τ − e − iω τ ) [ ( x ′′ 2 + x ′ 2 ) ] + e − iω τ ( x ′′ 2 + x ′ 2 ) − 2 x ′′ x ′ ] = 2ℏ sin ( ω τ ) imω [ 2 ( e iω τ + e − iω τ ) [ ( x ′′ 2 + x ′ 2 ) ] − 2 x ′′ x ′ ] = 2ℏ sin ( ω τ ) imω [ cos ( ω τ ) [ ( x ′′ 2 + x ′ 2 ) ] − 2 x ′′ x ′ ] Lembrando que τ = t − t 0 \tau=t-t_{0} τ = t − t 0
K ( x ′ ′ , t ; x ′ , t 0 ) = m ω 2 π i ℏ sin [ ω t ( t − t 0 ) ] exp [ i m ω 2 ℏ sin [ ω ( t − t 0 ) ] ( ( x ′ ′ 2 + x ′ 2 ) cos [ ω ( t − t 0 ) ] − 2 x ′ ′ x ′ ) ] K\left(x'',t;x',t_{0}\right)=\sqrt{\frac{m\omega}{2\pi i\hbar\sin\left[\omega t\left(t-t_{0}\right)\right]}}\exp\left[\frac{im\omega}{2\hbar\sin\left[\omega\left(t-t_{0}\right)\right]}\left(\left(x''^{2}+x'^{2}\right)\cos\left[\omega\left(t-t_{0}\right)\right]-2x''x'\right)\right] K ( x ′′ , t ; x ′ , t 0 ) = 2 πi ℏ sin [ ω t ( t − t 0 ) ] mω exp [ 2ℏ sin [ ω ( t − t 0 ) ] imω ( ( x ′′ 2 + x ′ 2 ) cos [ ω ( t − t 0 ) ] − 2 x ′′ x ′ ) ] Problem 3 ¶ The propagator can also be written as
K ( x ′ ′ , t ; x ′ , t 0 ) = ⟨ x ′ ′ , t ∣ x ′ , t 9 ⟩ K\left(x'',t;x',t_{0}\right)=\left\langle x'',t|x',t_{9}\right\rangle K ( x ′′ , t ; x ′ , t 0 ) = ⟨ x ′′ , t ∣ x ′ , t 9 ⟩ where ∣ x ′ , t 0 ⟩ \left|x',t_{0}\right\rangle ∣ x ′ , t 0 ⟩ ⟨ x ′ ′ , t ∣ \left\langle x'',t\right| ⟨ x ′′ , t ∣ ( 1 ) (1) ( 1 ) ⟨ p ′ ′ , t ∣ p ′ , t 0 ⟩ \left\langle p'',t|p',t_{0}\right\rangle ⟨ p ′′ , t ∣ p ′ , t 0 ⟩ ⟨ p ′ ′ , t ∣ p ′ , t 0 ⟩ \left\langle p'',t|p',t_{0}\right\rangle ⟨ p ′′ , t ∣ p ′ , t 0 ⟩
Análogo ao que foi feito em aula, podemos fazer então:
⟨ p ′ ′ ∣ α , t 0 ; t ⟩ = ⟨ p ′ ′ ∣ e − H ( t − t 0 ) / ℏ ∣ α , t 0 ⟩ = ∑ a ′ ⟨ p ′ ′ ∣ a ′ ⟩ ⟨ a ′ ∣ α , t 0 ⟩ e − E a ′ ( t − t 0 ) / ℏ = ∑ a ′ ⟨ p ′ ′ ∣ a ′ ⟩ ⟨ a ′ ∣ α , t 0 ⟩ e − E a ′ ( t − t 0 ) / ℏ = ∫ d 3 p ′ ∑ a ′ ⟨ p ′ ′ ∣ a ′ ⟩ ⟨ a ′ ∣ p ′ ⟩ ⟨ p ′ ∣ α , t 0 ⟩ e − E a ′ ( t − t 0 ) / ℏ = ∫ d 3 p ′ [ ∑ a ′ ⟨ p ′ ′ ∣ a ′ ⟩ ⟨ a ′ ∣ p ′ ⟩ e − E a ′ ( t − t 0 ) / ℏ ] ⟨ p ′ ∣ α , t 0 ⟩ = ∫ d 3 p ′ K ( p ′ ′ , t ; p ′ , t 0 ) ⟨ p ′ ∣ α , t 0 ⟩ \begin{align*}
\left\langle p''|\alpha,t_{0};t\right\rangle & =\left\langle p''\left|e^{-H\left(t-t_{0}\right)/\hbar}\right|\alpha,t_{0}\right\rangle \\
& =\sum_{a'}\left\langle p''|a'\right\rangle \left\langle a'|\alpha,t_{0}\right\rangle e^{-E_{a'}\left(t-t_{0}\right)/\hbar}\\
& =\sum_{a'}\left\langle p''|a'\right\rangle \left\langle a'|\alpha,t_{0}\right\rangle e^{-E_{a'}\left(t-t_{0}\right)/\hbar}\\
& =\int d^{3}p'\sum_{a'}\left\langle p''|a'\right\rangle \left\langle a'|p'\right\rangle \left\langle p'|\alpha,t_{0}\right\rangle e^{-E_{a'}\left(t-t_{0}\right)/\hbar}\\
& =\int d^{3}p'\left[\sum_{a'}\left\langle p''|a'\right\rangle \left\langle a'|p'\right\rangle e^{-E_{a'}\left(t-t_{0}\right)/\hbar}\right]\left\langle p'|\alpha,t_{0}\right\rangle \\
& =\int d^{3}p'K\left(p'',t;p',t_{0}\right)\left\langle p'|\alpha,t_{0}\right\rangle
\end{align*} ⟨ p ′′ ∣ α , t 0 ; t ⟩ = ⟨ p ′′ ∣ ∣ e − H ( t − t 0 ) /ℏ ∣ ∣ α , t 0 ⟩ = a ′ ∑ ⟨ p ′′ ∣ a ′ ⟩ ⟨ a ′ ∣ α , t 0 ⟩ e − E a ′ ( t − t 0 ) /ℏ = a ′ ∑ ⟨ p ′′ ∣ a ′ ⟩ ⟨ a ′ ∣ α , t 0 ⟩ e − E a ′ ( t − t 0 ) /ℏ = ∫ d 3 p ′ a ′ ∑ ⟨ p ′′ ∣ a ′ ⟩ ⟨ a ′ ∣ p ′ ⟩ ⟨ p ′ ∣ α , t 0 ⟩ e − E a ′ ( t − t 0 ) /ℏ = ∫ d 3 p ′ [ a ′ ∑ ⟨ p ′′ ∣ a ′ ⟩ ⟨ a ′ ∣ p ′ ⟩ e − E a ′ ( t − t 0 ) /ℏ ] ⟨ p ′ ∣ α , t 0 ⟩ = ∫ d 3 p ′ K ( p ′′ , t ; p ′ , t 0 ) ⟨ p ′ ∣ α , t 0 ⟩ Temos então o propagador:
K ( p ′ ′ , t ; p ′ , t 0 ) = ∑ a ′ ⟨ p ′ ′ ∣ a ′ ⟩ ⟨ a ′ ∣ p ′ ⟩ e − E a ′ ( t − t 0 ) / ℏ K\left(p'',t;p',t_{0}\right)=\sum_{a'}\left\langle p''|a'\right\rangle \left\langle a'|p'\right\rangle e^{-E_{a'}\left(t-t_{0}\right)/\hbar} K ( p ′′ , t ; p ′ , t 0 ) = a ′ ∑ ⟨ p ′′ ∣ a ′ ⟩ ⟨ a ′ ∣ p ′ ⟩ e − E a ′ ( t − t 0 ) /ℏ Para derivar uma expressão explícita para o caso da partícula livre, podemos escrever então mais elegantemente como:
K ( p ′ ′ , t ; p ′ , t 0 ) = ⟨ p ′ ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ p ′ ⟩ K\left(p'',t;p',t_{0}\right)=\left\langle p''\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|p'\right\rangle K ( p ′′ , t ; p ′ , t 0 ) = ⟨ p ′′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ p ′ ⟩ Fazendo análogo à primeira questão, onde o hamiltoniano de uma partícula livre é dado por H = p 2 2 m H=\frac{p^{2}}{2m} H = 2 m p 2 ∣ p ′ ⟩ \left|p'\right\rangle ∣ p ′ ⟩ p p p H H H p ∣ p ′ ⟩ = p ′ ∣ p ′ ⟩ p\left|p'\right\rangle =p'\left|p'\right\rangle p ∣ p ′ ⟩ = p ′ ∣ p ′ ⟩ H ∣ p ′ ⟩ = ( p ′ 2 2 m ) ∣ p ′ ⟩ H\left|p'\right\rangle =\left(\frac{p'^{2}}{2m}\right)\left|p'\right\rangle H ∣ p ′ ⟩ = ( 2 m p ′ 2 ) ∣ p ′ ⟩
K ( p ′ ′ , t ; p ′ , t 0 ) = ⟨ p ′ ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ p ′ ⟩ = ∫ d p ′ ′ ′ ⟨ p ′ ′ ∣ p ′ ′ ′ ⟩ ⟨ p ′ ′ ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ p ′ ⟩ = ∫ d p ′ ′ ′ δ ( p ′ ′ − p ′ ′ ′ ) ⟨ p ′ ′ ′ ′ ∣ exp [ − i H ( t − t 0 ) ℏ ] ∣ p ′ ⟩ = ∫ ⟨ p ′ ′ ′ ∣ p ′ ⟩ δ ( p ′ ′ − p ′ ′ ′ ) d p ′ ′ ′ exp ( − i p ′ 2 ( t − t 0 ) ℏ 2 m ) = ∫ δ ( p ′ ′ ′ − p ′ ′ ) δ ( p ′ ′ ′ − p ′ ) d p ′ ′ ′ exp ( − i p ′ 2 ( t − t 0 ) ℏ 2 m ) \begin{align*}
K\left(p'',t;p',t_{0}\right) & =\left\langle p''\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|p'\right\rangle \\
& =\int dp'''\left\langle p''|p'''\right\rangle \left\langle p'''\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|p'\right\rangle \\
& =\int dp'''\delta\left(p''-p'''\right)\left\langle p''''\left|\exp\left[\frac{-iH\left(t-t_{0}\right)}{\hbar}\right]\right|p'\right\rangle \\
& =\int\left\langle p'''|p'\right\rangle \delta\left(p''-p'''\right)dp'''\exp\left(\frac{-ip'^{2}\left(t-t_{0}\right)}{\hbar2m}\right)\\
& =\int\delta\left(p'''-p''\right)\delta\left(p'''-p'\right)dp'''\exp\left(\frac{-ip'^{2}\left(t-t_{0}\right)}{\hbar2m}\right)
\end{align*} K ( p ′′ , t ; p ′ , t 0 ) = ⟨ p ′′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ p ′ ⟩ = ∫ d p ′′′ ⟨ p ′′ ∣ p ′′′ ⟩ ⟨ p ′′′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ p ′ ⟩ = ∫ d p ′′′ δ ( p ′′ − p ′′′ ) ⟨ p ′′′′ ∣ ∣ exp [ ℏ − i H ( t − t 0 ) ] ∣ ∣ p ′ ⟩ = ∫ ⟨ p ′′′ ∣ p ′ ⟩ δ ( p ′′ − p ′′′ ) d p ′′′ exp ( ℏ2 m − i p ′ 2 ( t − t 0 ) ) = ∫ δ ( p ′′′ − p ′′ ) δ ( p ′′′ − p ′ ) d p ′′′ exp ( ℏ2 m − i p ′ 2 ( t − t 0 ) ) Agora sendo f ( p ′ ′ ′ ) = δ ( p ′ ′ ′ − p ′ ′ ) f\left(p'''\right)=\delta\left(p'''-p''\right) f ( p ′′′ ) = δ ( p ′′′ − p ′′ ) ∫ f ( p ′ ′ ′ ) δ ( p ′ ′ ′ − p ′ ) d p ′ ′ ′ = f ( p ′ ) = δ ( p ′ − p ′ ′ ) \int f\left(p'''\right)\delta\left(p'''-p'\right)dp'''=f\left(p'\right)=\delta\left(p'-p''\right) ∫ f ( p ′′′ ) δ ( p ′′′ − p ′ ) d p ′′′ = f ( p ′ ) = δ ( p ′ − p ′′ )
K ( p ′ ′ , t ; p ′ , t 0 ) = δ ( p ′ − p ′ ′ ) exp ( − i p ′ 2 ( t − t 0 ) 2 m ℏ ) K\left(p'',t;p',t_{0}\right)=\delta\left(p'-p''\right)\exp\left(\frac{-ip'^{2}\left(t-t_{0}\right)}{2m\hbar}\right) K ( p ′′ , t ; p ′ , t 0 ) = δ ( p ′ − p ′′ ) exp ( 2 m ℏ − i p ′ 2 ( t − t 0 ) )
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