Problem set 3
These are my solutions to the problem sets assigned to me during the graduate-level quantum mechanics course.
Problem 1 ¶ 1. In a one-dimensional problem, consider a particle of potential energy V ( X ) = − f X V\left(X\right)=-fX V ( X ) = − f X f f f
a) Write Ehrenfest’s theorem for the mean values of the position X X X P P P
b) Show that the root-mean-square deviation Δ P \Delta P Δ P
c) Write the Schrodinger equation in the ∣ p ⟩ \left|p\right\rangle ∣ p ⟩ ∂ ∂ t ∣ ⟨ p ∣ Ψ ( t ) ⟩ ∣ 2 \frac{\partial}{\partial t}\left|\left\langle p|\Psi\left(t\right)\right\rangle \right|^{2} ∂ t ∂ ∣ ⟨ p ∣Ψ ( t ) ⟩ ∣ 2 ∂ ∂ p ∣ ⟨ p ∣ Ψ ( t ) ⟩ ∣ 2 \frac{\partial}{\partial p}\left|\left\langle p|\Psi\left(t\right)\right\rangle \right|^{2} ∂ p ∂ ∣ ⟨ p ∣Ψ ( t ) ⟩ ∣ 2
Dados do enunciado:
a)
A encontrar:
Equações:
H = P 2 2 m − f X H=\frac{P^{2}}{2m}-fX H = 2 m P 2 − f X
d ⟨ O ⟩ d t = − i ℏ ⟨ [ O , H ] ⟩ + ⟨ ∂ O ∂ t ⟩ \frac{d\left\langle O\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[O,H\right]\right\rangle +\left\langle \frac{\partial O}{\partial t}\right\rangle d t d ⟨ O ⟩ = − ℏ i ⟨ [ O , H ] ⟩ + ⟨ ∂ t ∂ O ⟩ O O O
[ X , O ( P ) ] = i ℏ ∂ O ∂ P \left[X,O\left(P\right)\right]=i\hbar\frac{\partial O}{\partial P} [ X , O ( P ) ] = i ℏ ∂ P ∂ O
[ P , O ( X ) ] = − i ℏ ∂ O ∂ X \left[P,O\left(X\right)\right]=-i\hbar\frac{\partial O}{\partial X} [ P , O ( X ) ] = − i ℏ ∂ X ∂ O
Então para X X X
d ⟨ X ⟩ d t = − i ℏ ⟨ [ X , H ] ⟩ + ⟨ ∂ X ∂ t ⟩ \frac{d\left\langle X\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[X,H\right]\right\rangle +\left\langle \frac{\partial X}{\partial t}\right\rangle d t d ⟨ X ⟩ = − ℏ i ⟨ [ X , H ] ⟩ + ⟨ ∂ t ∂ X ⟩ Como o operador X X X
d ⟨ X ⟩ d t = − i ℏ ⟨ [ X , H ] ⟩ = ⟨ ∂ H ∂ P ⟩ = ⟨ P ⟩ m \begin{align*}
\frac{d\left\langle X\right\rangle }{dt}= & -\frac{i}{\hbar}\left\langle \left[X,H\right]\right\rangle =\left\langle \frac{\partial H}{\partial P}\right\rangle =\frac{\left\langle P\right\rangle }{m}
\end{align*} d t d ⟨ X ⟩ = − ℏ i ⟨ [ X , H ] ⟩ = ⟨ ∂ P ∂ H ⟩ = m ⟨ P ⟩ E de modo análogo para P P P
d ⟨ P ⟩ d t = − i ℏ ⟨ [ P , H ] ⟩ + ⟨ ∂ P ∂ t ⟩ = − i ℏ ⟨ [ P , H ] ⟩ = − ⟨ ∂ H ∂ X ⟩ = − ⟨ ∂ V ∂ X ⟩ = f \frac{d\left\langle P\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[P,H\right]\right\rangle +\left\langle \frac{\partial P}{\partial t}\right\rangle =-\frac{i}{\hbar}\left\langle \left[P,H\right]\right\rangle =-\left\langle \frac{\partial H}{\partial X}\right\rangle =-\left\langle \frac{\partial V}{\partial X}\right\rangle =f d t d ⟨ P ⟩ = − ℏ i ⟨ [ P , H ] ⟩ + ⟨ ∂ t ∂ P ⟩ = − ℏ i ⟨ [ P , H ] ⟩ = − ⟨ ∂ X ∂ H ⟩ = − ⟨ ∂ X ∂ V ⟩ = f Logo obtemos então:
m d ⟨ X ⟩ d t = ⟨ P ⟩ d ⟨ P ⟩ d t = − ⟨ ∂ V ∂ X ⟩ m\frac{d\left\langle X\right\rangle }{dt}=\left\langle P\right\rangle \qquad\frac{d\left\langle P\right\rangle }{dt}=-\left\langle \frac{\partial V}{\partial X}\right\rangle m d t d ⟨ X ⟩ = ⟨ P ⟩ d t d ⟨ P ⟩ = − ⟨ ∂ X ∂ V ⟩ Uma vez que ∂ V ∂ X = − f I \frac{\partial V}{\partial X}=-f\mathbb{I} ∂ X ∂ V = − f I I \mathbb{I} I t = 0 t=0 t = 0 t t t ⟨ P ⟩ 0 = ⟨ P ⟩ t = 0 \left\langle P\right\rangle _{0}=\left\langle P\right\rangle _{t=0} ⟨ P ⟩ 0 = ⟨ P ⟩ t = 0
d ⟨ P ⟩ = f d t ⟨ P ⟩ − ⟨ P ⟩ 0 = f t ⟨ P ⟩ = ⟨ P ⟩ 0 + f t \begin{align*}
d\left\langle P\right\rangle & =fdt\\
\left\langle P\right\rangle -\left\langle P\right\rangle _{0} & =ft\\
\left\langle P\right\rangle & =\left\langle P\right\rangle _{0}+ft
\end{align*} d ⟨ P ⟩ ⟨ P ⟩ − ⟨ P ⟩ 0 ⟨ P ⟩ = fd t = f t = ⟨ P ⟩ 0 + f t E resolvendo então para a posição, temos:
d ⟨ X ⟩ = ⟨ P ⟩ m d t d ⟨ X ⟩ = ( ⟨ P ⟩ 0 + f t ) 1 m d t ⟨ X ⟩ − ⟨ X ⟩ 0 = ⟨ P ⟩ 0 t m + f t 2 2 m ⟨ X ⟩ = ⟨ X ⟩ 0 + ⟨ P ⟩ 0 t m + f t 2 2 m \begin{align*}
d\left\langle X\right\rangle & =\frac{\left\langle P\right\rangle }{m}dt\\
d\left\langle X\right\rangle & =\left(\left\langle P\right\rangle _{0}+ft\right)\frac{1}{m}dt\\
\left\langle X\right\rangle -\left\langle X\right\rangle _{0} & =\left\langle P\right\rangle _{0}\frac{t}{m}+\frac{ft^{2}}{2m}\\
\left\langle X\right\rangle & =\left\langle X\right\rangle _{0}+\left\langle P\right\rangle _{0}\frac{t}{m}+\frac{ft^{2}}{2m}
\end{align*} d ⟨ X ⟩ d ⟨ X ⟩ ⟨ X ⟩ − ⟨ X ⟩ 0 ⟨ X ⟩ = m ⟨ P ⟩ d t = ( ⟨ P ⟩ 0 + f t ) m 1 d t = ⟨ P ⟩ 0 m t + 2 m f t 2 = ⟨ X ⟩ 0 + ⟨ P ⟩ 0 m t + 2 m f t 2 Então os resultados são:
⟨ P ⟩ = ⟨ P ⟩ 0 + f t ⟨ X ⟩ = ⟨ X ⟩ 0 + ⟨ P ⟩ 0 t m + f t 2 2 m \left\langle P\right\rangle =\left\langle P\right\rangle _{0}+ft\qquad\left\langle X\right\rangle =\left\langle X\right\rangle _{0}+\left\langle P\right\rangle _{0}\frac{t}{m}+\frac{ft^{2}}{2m} ⟨ P ⟩ = ⟨ P ⟩ 0 + f t ⟨ X ⟩ = ⟨ X ⟩ 0 + ⟨ P ⟩ 0 m t + 2 m f t 2 Estes resultados tem a forma idêntica quando comparado com o movimento clássico:
b)
A encontrar:
Equações:
H = P 2 2 m − f X H=\frac{P^{2}}{2m}-fX H = 2 m P 2 − f X
( Δ P ) 2 = ⟨ P 2 ⟩ − ⟨ P ⟩ 2 \left(\Delta P\right)^{2}=\left\langle P^{2}\right\rangle -\left\langle P\right\rangle ^{2} ( Δ P ) 2 = ⟨ P 2 ⟩ − ⟨ P ⟩ 2
d ⟨ O ⟩ d t = − i ℏ ⟨ [ O , H ] ⟩ + ⟨ ∂ O ∂ t ⟩ \frac{d\left\langle O\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[O,H\right]\right\rangle +\left\langle \frac{\partial O}{\partial t}\right\rangle d t d ⟨ O ⟩ = − ℏ i ⟨ [ O , H ] ⟩ + ⟨ ∂ t ∂ O ⟩
Soluções obtidas anteriormente:
Então, a partir do Teorema generalizado de Ehrenfest, tendo O = H O=H O = H d ⟨ H ⟩ d t = ⟨ [ O , H ] ⟩ = ⟨ ∂ H ∂ t ⟩ = 0 \frac{d\left\langle H\right\rangle }{dt}=\left\langle \left[O,H\right]\right\rangle =\left\langle \frac{\partial H}{\partial t}\right\rangle =0 d t d ⟨ H ⟩ = ⟨ [ O , H ] ⟩ = ⟨ ∂ t ∂ H ⟩ = 0 ⟨ X ⟩ \left\langle X\right\rangle ⟨ X ⟩
d ⟨ H ⟩ d t = d d t ⟨ P 2 2 m − f X ⟩ 0 = 1 2 m d d t ⟨ P 2 ⟩ − f d d t ⟨ X ⟩ 0 = 1 2 m d d t ⟨ P 2 ⟩ − f ⟨ P ⟩ m d d t ⟨ P 2 ⟩ = 2 f ⟨ P ⟩ m \begin{align*}
\frac{d\left\langle H\right\rangle }{dt} & =\frac{d}{dt}\left\langle \frac{P^{2}}{2m}-fX\right\rangle \\
0 & =\frac{1}{2m}\frac{d}{dt}\left\langle P^{2}\right\rangle -f\frac{d}{dt}\left\langle X\right\rangle \\
0 & =\frac{1}{2m}\frac{d}{dt}\left\langle P^{2}\right\rangle -f\frac{\left\langle P\right\rangle }{m}\\
\frac{d}{dt}\left\langle P^{2}\right\rangle & =2f\frac{\left\langle P\right\rangle }{m}
\end{align*} d t d ⟨ H ⟩ 0 0 d t d ⟨ P 2 ⟩ = d t d ⟨ 2 m P 2 − f X ⟩ = 2 m 1 d t d ⟨ P 2 ⟩ − f d t d ⟨ X ⟩ = 2 m 1 d t d ⟨ P 2 ⟩ − f m ⟨ P ⟩ = 2 f m ⟨ P ⟩ E também do resultado anterior obtemos a derivada temporal de ⟨ P ⟩ \left\langle P\right\rangle ⟨ P ⟩
d d t ⟨ P ⟩ 2 = 2 ⟨ P ⟩ d d t ⟨ P ⟩ = 2 f ⟨ P ⟩ \begin{align*}
\frac{d}{dt}\left\langle P\right\rangle ^{2} & =2\left\langle P\right\rangle \frac{d}{dt}\left\langle P\right\rangle \\
& =2f\left\langle P\right\rangle
\end{align*} d t d ⟨ P ⟩ 2 = 2 ⟨ P ⟩ d t d ⟨ P ⟩ = 2 f ⟨ P ⟩ Temos que o quadrado do desvio padrão é:
( Δ P ) 2 = ⟨ P 2 ⟩ − ⟨ P ⟩ 2 \left(\Delta P\right)^{2}=\left\langle P^{2}\right\rangle -\left\langle P\right\rangle ^{2} ( Δ P ) 2 = ⟨ P 2 ⟩ − ⟨ P ⟩ 2 Derivando e substituindo então os resultados encontrados:
d d t ( Δ P ) 2 = d d t ( ⟨ P 2 ⟩ − ⟨ P ⟩ 2 ) ( Δ P ) d d t ( Δ P ) = 2 f ⟨ P ⟩ m − 2 f ⟨ P ⟩ d d t ( Δ P ) = 0 \begin{align*}
\frac{d}{dt}\left(\Delta P\right)^{2} & =\frac{d}{dt}\left(\left\langle P^{2}\right\rangle -\left\langle P\right\rangle ^{2}\right)\\
\left(\Delta P\right)\frac{d}{dt}\left(\Delta P\right) & =2f\frac{\left\langle P\right\rangle }{m}-2f\left\langle P\right\rangle \\
\frac{d}{dt}\left(\Delta P\right) & =0
\end{align*} d t d ( Δ P ) 2 ( Δ P ) d t d ( Δ P ) d t d ( Δ P ) = d t d ( ⟨ P 2 ⟩ − ⟨ P ⟩ 2 ) = 2 f m ⟨ P ⟩ − 2 f ⟨ P ⟩ = 0 Logo Δ P \Delta P Δ P
c)
A encontrar:
Escrever a equaçaõ de Schrodinger na representação { ∣ p ⟩ } \left\{ \left|p\right\rangle \right\} { ∣ p ⟩ }
Deduzir uma relação entre ∂ ∂ t ∣ ⟨ p ∣ Ψ ( t ) ⟩ ∣ 2 \frac{\partial}{\partial t}\left|\left\langle p|\Psi\left(t\right)\right\rangle \right|^{2} ∂ t ∂ ∣ ⟨ p ∣Ψ ( t ) ⟩ ∣ 2 ∂ ∂ p ∣ ⟨ p ∣ Ψ ( t ) ⟩ ∣ 2 \frac{\partial}{\partial p}\left|\left\langle p|\Psi\left(t\right)\right\rangle \right|^{2} ∂ p ∂ ∣ ⟨ p ∣Ψ ( t ) ⟩ ∣ 2
Integrar a equação obtida
Obter uma interpretação física
Equações:
H = P 2 2 m − f X H=\frac{P^{2}}{2m}-fX H = 2 m P 2 − f X
i ℏ d d t ∣ Ψ ( t ) ⟩ = H ∣ Ψ ( t ) ⟩ i\hbar\frac{d}{dt}\left|\Psi\left(t\right)\right\rangle =H\left|\Psi\left(t\right)\right\rangle i ℏ d t d ∣ Ψ ( t ) ⟩ = H ∣ Ψ ( t ) ⟩
A equação de Schrodinger então, na representação { ∣ p ⟩ } \left\{ \left|p\right\rangle \right\} { ∣ p ⟩ }
i ℏ ∂ ∂ t ⟨ p ∣ Ψ ( t ) ⟩ = ⟨ p ∣ H ∣ Ψ ( t ) ⟩ = ⟨ p ∣ ( P 2 2 m − f X ) ∣ Ψ ( t ) ⟩ = ⟨ p ∣ P 2 ∣ Ψ ( t ) ⟩ 2 m − f ⟨ p ∣ X ∣ Ψ ( t ) ⟩ \begin{align*}
i\hbar\frac{\partial}{\partial t}\left\langle p|\Psi\left(t\right)\right\rangle & =\left\langle p\right|H\left|\Psi\left(t\right)\right\rangle \\
& =\left\langle p\right|\left(\frac{P^{2}}{2m}-fX\right)\left|\Psi\left(t\right)\right\rangle \\
& =\frac{\left\langle p\left|P^{2}\right|\Psi\left(t\right)\right\rangle }{2m}-f\left\langle p\left|X\right|\Psi\left(t\right)\right\rangle
\end{align*} i ℏ ∂ t ∂ ⟨ p ∣Ψ ( t ) ⟩ = ⟨ p ∣ H ∣ Ψ ( t ) ⟩ = ⟨ p ∣ ( 2 m P 2 − f X ) ∣ Ψ ( t ) ⟩ = 2 m ⟨ p ∣ ∣ P 2 ∣ ∣ Ψ ( t ) ⟩ − f ⟨ p ∣ X ∣ Ψ ( t ) ⟩ Sendo então a função de onda na representação { ∣ p ⟩ } \left\{ \left|p\right\rangle \right\} { ∣ p ⟩ } ψ ( p , t ) ≡ ⟨ p ∣ Ψ ( t ) ⟩ \psi\left(p,t\right)\equiv\left\langle p|\Psi\left(t\right)\right\rangle ψ ( p , t ) ≡ ⟨ p ∣Ψ ( t ) ⟩ P P P P ∣ p ⟩ = p ∣ p ⟩ P\left|p\right\rangle =p\left|p\right\rangle P ∣ p ⟩ = p ∣ p ⟩ X = i ℏ ∂ ∂ p X=i\hbar\frac{\partial}{\partial p} X = i ℏ ∂ p ∂
i ℏ d ⟨ p ∣ Ψ ( t ) ⟩ d t = p 2 ⟨ p ∣ Ψ ( t ) ⟩ 2 m − f i ℏ ∂ ∂ p ⟨ p ∣ Ψ ( t ) ⟩ = ( p 2 2 m − f i ℏ ∂ ∂ p ) ⟨ p ∣ Ψ ( t ) ⟩ i ℏ d ψ ( p , t ) d t = ( p 2 2 m − f i ℏ ∂ ∂ p ) ψ ( p , t ) \begin{align*}
i\hbar\frac{d\left\langle p|\Psi\left(t\right)\right\rangle }{dt} & =p^{2}\frac{\left\langle p|\Psi\left(t\right)\right\rangle }{2m}-fi\hbar\frac{\partial}{\partial p}\left\langle p|\Psi\left(t\right)\right\rangle \\
& =\left(\frac{p^{2}}{2m}-fi\hbar\frac{\partial}{\partial p}\right)\left\langle p|\Psi\left(t\right)\right\rangle \\
i\hbar\frac{d\psi\left(p,t\right)}{dt} & =\left(\frac{p^{2}}{2m}-fi\hbar\frac{\partial}{\partial p}\right)\psi\left(p,t\right)
\end{align*} i ℏ d t d ⟨ p ∣Ψ ( t ) ⟩ i ℏ d t d ψ ( p , t ) = p 2 2 m ⟨ p ∣Ψ ( t ) ⟩ − f i ℏ ∂ p ∂ ⟨ p ∣Ψ ( t ) ⟩ = ( 2 m p 2 − f i ℏ ∂ p ∂ ) ⟨ p ∣Ψ ( t ) ⟩ = ( 2 m p 2 − f i ℏ ∂ p ∂ ) ψ ( p , t ) Então agora queremos deduzir uma relação entre ∂ ∂ t ∣ ψ ( p , t ) ∣ 2 \frac{\partial}{\partial t}\left|\psi\left(p,t\right)\right|^{2} ∂ t ∂ ∣ ψ ( p , t ) ∣ 2 ∂ ∂ p ∣ ψ ( p , t ) ∣ 2 \frac{\partial}{\partial p}\left|\psi\left(p,t\right)\right|^{2} ∂ p ∂ ∣ ψ ( p , t ) ∣ 2
∂ ∂ t ∣ ψ ( p , t ) ∣ 2 = ∂ ∂ t ( ψ ∗ ( p , t ) ψ ( p , t ) ) = ψ ∗ ( p , t ) ∂ ∂ t ψ ( p , t ) + ψ ( p , t ) ∂ ∂ t ψ ∗ ( p , t ) = ψ ∗ ( p , t ) ( − i p 2 2 m ℏ − f ∂ ∂ p ) ψ ( p , t ) + ψ ( p , t ) ( i p 2 2 m ℏ − f ∂ ∂ p ) ψ ∗ ( p , t ) = − f ψ ∗ ( p , t ) ∂ ∂ p ψ ( p , t ) − f ψ ( p , t ) ∂ ∂ p ψ ∗ ( p , t ) = − f [ ψ ∗ ( p , t ) ∂ ∂ p ψ ( p , t ) + ψ ( p , t ) ∂ ∂ p ψ ∗ ( p , t ) ] = − f [ ∂ ∂ p ( ψ ∗ ( p , t ) ψ ( p , t ) ) ] = − f ∂ ∂ p ∣ ψ ( p , t ) ∣ 2 \begin{align*}
\frac{\partial}{\partial t}\left|\psi\left(p,t\right)\right|^{2} & =\frac{\partial}{\partial t}\left(\psi^{*}\left(p,t\right)\psi\left(p,t\right)\right)\\
& =\psi^{*}\left(p,t\right)\frac{\partial}{\partial t}\psi\left(p,t\right)+\psi\left(p,t\right)\frac{\partial}{\partial t}\psi^{*}\left(p,t\right)\\
& =\psi^{*}\left(p,t\right)\left(-\frac{ip^{2}}{2m\hbar}-f\frac{\partial}{\partial p}\right)\psi\left(p,t\right)+\psi\left(p,t\right)\left(\frac{ip^{2}}{2m\hbar}-f\frac{\partial}{\partial p}\right)\psi^{*}\left(p,t\right)\\
& =-f\psi^{*}\left(p,t\right)\frac{\partial}{\partial p}\psi\left(p,t\right)-f\psi\left(p,t\right)\frac{\partial}{\partial p}\psi^{*}\left(p,t\right)\\
& =-f\left[\psi^{*}\left(p,t\right)\frac{\partial}{\partial p}\psi\left(p,t\right)+\psi\left(p,t\right)\frac{\partial}{\partial p}\psi^{*}\left(p,t\right)\right]\\
& =-f\left[\frac{\partial}{\partial p}\left(\psi^{*}\left(p,t\right)\psi\left(p,t\right)\right)\right]\\
& =-f\frac{\partial}{\partial p}\left|\psi\left(p,t\right)\right|^{2}
\end{align*} ∂ t ∂ ∣ ψ ( p , t ) ∣ 2 = ∂ t ∂ ( ψ ∗ ( p , t ) ψ ( p , t ) ) = ψ ∗ ( p , t ) ∂ t ∂ ψ ( p , t ) + ψ ( p , t ) ∂ t ∂ ψ ∗ ( p , t ) = ψ ∗ ( p , t ) ( − 2 m ℏ i p 2 − f ∂ p ∂ ) ψ ( p , t ) + ψ ( p , t ) ( 2 m ℏ i p 2 − f ∂ p ∂ ) ψ ∗ ( p , t ) = − f ψ ∗ ( p , t ) ∂ p ∂ ψ ( p , t ) − f ψ ( p , t ) ∂ p ∂ ψ ∗ ( p , t ) = − f [ ψ ∗ ( p , t ) ∂ p ∂ ψ ( p , t ) + ψ ( p , t ) ∂ p ∂ ψ ∗ ( p , t ) ] = − f [ ∂ p ∂ ( ψ ∗ ( p , t ) ψ ( p , t ) ) ] = − f ∂ p ∂ ∣ ψ ( p , t ) ∣ 2 Denotando então Φ ( p , t ) ≡ ∣ ψ ( p , t ) ∣ 2 \Phi\left(p,t\right)\equiv\left|\psi\left(p,t\right)\right|^{2} Φ ( p , t ) ≡ ∣ ψ ( p , t ) ∣ 2
∂ Φ ( p , t ) ∂ t = − f ∂ Φ ( p , t ) ∂ p \frac{\partial\Phi\left(p,t\right)}{\partial t}=-f\frac{\partial\Phi\left(p,t\right)}{\partial p} ∂ t ∂ Φ ( p , t ) = − f ∂ p ∂ Φ ( p , t ) Fazendo as seguintes mudanças de variáveis:
r ≡ p − f t s ≡ p + f t r\equiv p-ft\qquad s\equiv p+ft r ≡ p − f t s ≡ p + f t Então:
∂ Φ ( p , t ) ∂ t = ∂ Φ ( p , t ) ∂ r ∂ r ∂ t + ∂ Φ ( p , t ) ∂ s ∂ s ∂ t = f ( ∂ Φ ( p , t ) ∂ s − ∂ Φ ( p , t ) ∂ r ) \begin{align*}
\frac{\partial\Phi\left(p,t\right)}{\partial t} & =\frac{\partial\Phi\left(p,t\right)}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial\Phi\left(p,t\right)}{\partial s}\frac{\partial s}{\partial t}\\
& =f\left(\frac{\partial\Phi\left(p,t\right)}{\partial s}-\frac{\partial\Phi\left(p,t\right)}{\partial r}\right)
\end{align*} ∂ t ∂ Φ ( p , t ) = ∂ r ∂ Φ ( p , t ) ∂ t ∂ r + ∂ s ∂ Φ ( p , t ) ∂ t ∂ s = f ( ∂ s ∂ Φ ( p , t ) − ∂ r ∂ Φ ( p , t ) ) E de modo análogo:
∂ Φ ( p , t ) ∂ p = ∂ Φ ( p , t ) ∂ r ∂ r ∂ p + ∂ Φ ( p , t ) ∂ s ∂ s ∂ p = ∂ Φ ( p , t ) ∂ s + ∂ Φ ( p , t ) ∂ r \begin{align*}
\frac{\partial\Phi\left(p,t\right)}{\partial p} & =\frac{\partial\Phi\left(p,t\right)}{\partial r}\frac{\partial r}{\partial p}+\frac{\partial\Phi\left(p,t\right)}{\partial s}\frac{\partial s}{\partial p}\\
& =\frac{\partial\Phi\left(p,t\right)}{\partial s}+\frac{\partial\Phi\left(p,t\right)}{\partial r}
\end{align*} ∂ p ∂ Φ ( p , t ) = ∂ r ∂ Φ ( p , t ) ∂ p ∂ r + ∂ s ∂ Φ ( p , t ) ∂ p ∂ s = ∂ s ∂ Φ ( p , t ) + ∂ r ∂ Φ ( p , t ) Substituindo então:
∂ Φ ( p , t ) ∂ t = − f ∂ Φ ( p , t ) ∂ p f ( ∂ Φ ( p , t ) ∂ s − ∂ Φ ( p , t ) ∂ r ) = − f ( ∂ Φ ( p , t ) ∂ s + ∂ Φ ( p , t ) ∂ r ) f ∂ Φ ( p , t ) ∂ s = − f ∂ Φ ( p , t ) ∂ s ∂ Φ ( p , t ) ∂ s = 0 \begin{align*}
\frac{\partial\Phi\left(p,t\right)}{\partial t} & =-f\frac{\partial\Phi\left(p,t\right)}{\partial p}\\
f\left(\frac{\partial\Phi\left(p,t\right)}{\partial s}-\frac{\partial\Phi\left(p,t\right)}{\partial r}\right) & =-f\left(\frac{\partial\Phi\left(p,t\right)}{\partial s}+\frac{\partial\Phi\left(p,t\right)}{\partial r}\right)\\
f\frac{\partial\Phi\left(p,t\right)}{\partial s} & =-f\frac{\partial\Phi\left(p,t\right)}{\partial s}\\
\frac{\partial\Phi\left(p,t\right)}{\partial s} & =0
\end{align*} ∂ t ∂ Φ ( p , t ) f ( ∂ s ∂ Φ ( p , t ) − ∂ r ∂ Φ ( p , t ) ) f ∂ s ∂ Φ ( p , t ) ∂ s ∂ Φ ( p , t ) = − f ∂ p ∂ Φ ( p , t ) = − f ( ∂ s ∂ Φ ( p , t ) + ∂ r ∂ Φ ( p , t ) ) = − f ∂ s ∂ Φ ( p , t ) = 0 Logo Φ \Phi Φ r r r
Φ ( p , t ) = g ( r ) = g ( p − f t ) \Phi\left(p,t\right)=g\left(r\right)=g\left(p-ft\right) Φ ( p , t ) = g ( r ) = g ( p − f t ) Onde g ( r ) g\left(r\right) g ( r )
Φ ( p − f t , 0 ) = g ( r ) = g ( p − f t ) \Phi\left(p-ft,0\right)=g\left(r\right)=g\left(p-ft\right) Φ ( p − f t , 0 ) = g ( r ) = g ( p − f t ) Então:
Φ ( p , t ) = Φ ( p − f t , 0 ) \Phi\left(p,t\right)=\Phi\left(p-ft,0\right) Φ ( p , t ) = Φ ( p − f t , 0 ) E uma vez que Φ ( p , t ) ≡ ∣ ψ ( p , t ) ∣ 2 \Phi\left(p,t\right)\equiv\left|\psi\left(p,t\right)\right|^{2} Φ ( p , t ) ≡ ∣ ψ ( p , t ) ∣ 2 p p p p = p 0 + f t p=p_{0}+ft p = p 0 + f t 15 Lima (2013)
− ∂ Φ ( p , t ) ∂ t 1 f = ∂ Φ ( p , t ) ∂ p -\frac{\partial\Phi\left(p,t\right)}{\partial t}\frac{1}{f}=\frac{\partial\Phi\left(p,t\right)}{\partial p} − ∂ t ∂ Φ ( p , t ) f 1 = ∂ p ∂ Φ ( p , t ) Supondo uma solução do tipo Φ ( p , t ) = P ( p ) T ( t ) \Phi\left(p,t\right)=P\left(p\right)T\left(t\right) Φ ( p , t ) = P ( p ) T ( t )
− 1 f ∂ P ( p ) T ( t ) ∂ t = ∂ P ( p ) T ( t ) ∂ p − 1 f T ( t ) ∂ T ( t ) ∂ t = 1 P ( p ) ∂ P ( p ) ∂ p \begin{align*}
-\frac{1}{f}\frac{\partial P\left(p\right)T\left(t\right)}{\partial t} & =\frac{\partial P\left(p\right)T\left(t\right)}{\partial p}\\
-\frac{1}{fT\left(t\right)}\frac{\partial T\left(t\right)}{\partial t} & =\frac{1}{P\left(p\right)}\frac{\partial P\left(p\right)}{\partial p}
\end{align*} − f 1 ∂ t ∂ P ( p ) T ( t ) − f T ( t ) 1 ∂ t ∂ T ( t ) = ∂ p ∂ P ( p ) T ( t ) = P ( p ) 1 ∂ p ∂ P ( p ) Então:
− 1 T ( t ) ∂ T ( t ) ∂ t = f P ( p ) ∂ P ( p ) ∂ p = λ -\frac{1}{T\left(t\right)}\frac{\partial T\left(t\right)}{\partial t}=\frac{f}{P\left(p\right)}\frac{\partial P\left(p\right)}{\partial p}=\lambda − T ( t ) 1 ∂ t ∂ T ( t ) = P ( p ) f ∂ p ∂ P ( p ) = λ Onde λ \lambda λ
∂ T ( t ) ∂ t = − f λ T \frac{\partial T\left(t\right)}{\partial t}=-f\lambda T ∂ t ∂ T ( t ) = − f λ T Temos: T ( t ) = A e ( − λ f ) t T\left(t\right)=Ae^{\left(-\lambda f\right)t} T ( t ) = A e ( − λ f ) t
∂ P ( p ) ∂ p = λ P \frac{\partial P\left(p\right)}{\partial p}=\lambda P ∂ p ∂ P ( p ) = λ P Então P ( p ) = B e λ p P\left(p\right)=Be^{\lambda p} P ( p ) = B e λ p
Φ ( p , t ) = A e λ p e − λ f t = A exp [ λ ( p − f t ) ] \Phi\left(p,t\right)=Ae^{\lambda p}e^{-\lambda ft}=A\exp\left[\lambda\left(p-ft\right)\right] Φ ( p , t ) = A e λ p e − λ f t = A exp [ λ ( p − f t ) ] Como esperado:
Φ ( p − f t , 0 ) = A exp [ λ ( p − f t ) ] = Φ ( p , t ) \Phi\left(p-ft,0\right)=A\exp\left[\lambda\left(p-ft\right)\right]=\Phi\left(p,t\right) Φ ( p − f t , 0 ) = A exp [ λ ( p − f t ) ] = Φ ( p , t ) Para facilitar a compreensão podemos fixar um p = p ′ p=p' p = p ′ Φ ( p ′ − f t ′ , 0 ) \Phi\left(p'-ft',0\right) Φ ( p ′ − f t ′ , 0 ) p = p ′ − f t ′ p=p'-ft' p = p ′ − f t ′ t = 0 t=0 t = 0 p = p ′ p=p' p = p ′ t = t ′ t=t' t = t ′ f f f
Problem 2 ¶ Consider a free particle.
a) Show, applying Ehrenfest’s theorem, that ⟨ X ⟩ \left\langle X\right\rangle ⟨ X ⟩ ⟨ P ⟩ \left\langle P\right\rangle ⟨ P ⟩
b) Write the equations of motion for the mean values ⟨ X 2 ⟩ \left\langle X^{2}\right\rangle ⟨ X 2 ⟩ ⟨ X P + P X ⟩ \left\langle XP+PX\right\rangle ⟨ XP + PX ⟩
c) Show that, with a suitable choice of the time origin, the root-mean-square deviation Δ X \Delta X Δ X
( Δ X ) 2 = 1 m 2 ( Δ P ) 0 2 t 2 + ( Δ X ) 0 2 \left(\Delta X\right)^{2}=\frac{1}{m^{2}}\left(\Delta P\right)^{2}_{0}t^{2}+\left(\Delta X\right)^{2}_{0} ( Δ X ) 2 = m 2 1 ( Δ P ) 0 2 t 2 + ( Δ X ) 0 2 where ( Δ X ) 0 \left(\Delta X\right)_{0} ( Δ X ) 0 ( Δ P ) 0 \left(\Delta P\right)_{0} ( Δ P ) 0
a)
A encontrar:
Equações:
Temos então que:
d d t ⟨ X ⟩ = ⟨ P ⟩ m d d t ⟨ P ⟩ = 0 \frac{d}{dt}\left\langle X\right\rangle =\frac{\left\langle P\right\rangle }{m}\qquad\frac{d}{dt}\left\langle P\right\rangle =0 d t d ⟨ X ⟩ = m ⟨ P ⟩ d t d ⟨ P ⟩ = 0 Dessa forma já mostramos que ⟨ P ⟩ \left\langle P\right\rangle ⟨ P ⟩ ⟨ X ⟩ \left\langle X\right\rangle ⟨ X ⟩
d ⟨ X ⟩ = ⟨ P ⟩ m d t ⟨ X ⟩ − ⟨ X ⟩ 0 = ⟨ P ⟩ m t ⟨ X ⟩ = ⟨ X ⟩ 0 + ⟨ P ⟩ 0 m t \begin{align*}
d\left\langle X\right\rangle & =\frac{\left\langle P\right\rangle }{m}dt\\
\left\langle X\right\rangle -\left\langle X\right\rangle _{0} & =\frac{\left\langle P\right\rangle }{m}t\\
\left\langle X\right\rangle & =\left\langle X\right\rangle _{0}+\frac{\left\langle P\right\rangle _{0}}{m}t
\end{align*} d ⟨ X ⟩ ⟨ X ⟩ − ⟨ X ⟩ 0 ⟨ X ⟩ = m ⟨ P ⟩ d t = m ⟨ P ⟩ t = ⟨ X ⟩ 0 + m ⟨ P ⟩ 0 t Onde denotamos ⟨ P ⟩ 0 = ⟨ P ⟩ \left\langle P\right\rangle _{0}=\left\langle P\right\rangle ⟨ P ⟩ 0 = ⟨ P ⟩ ⟨ X ⟩ \left\langle X\right\rangle ⟨ X ⟩
b)
A encontrar:
Equações:
H = P 2 2 m H=\frac{P^{2}}{2m} H = 2 m P 2
d ⟨ O ⟩ d t = − i ℏ ⟨ [ O , H ] ⟩ + ⟨ ∂ O ∂ t ⟩ \frac{d\left\langle O\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[O,H\right]\right\rangle +\left\langle \frac{\partial O}{\partial t}\right\rangle d t d ⟨ O ⟩ = − ℏ i ⟨ [ O , H ] ⟩ + ⟨ ∂ t ∂ O ⟩
X ∣ ψ ⟩ = x ∣ ψ ⟩ X\left|\psi\right\rangle =x\left|\psi\right\rangle X ∣ ψ ⟩ = x ∣ ψ ⟩
⟨ x ∣ P ∣ ψ ⟩ = − i ∂ ∂ x ⟨ x ∣ ψ ⟩ , \left\langle x\left|P\right|\psi\right\rangle =-i\frac{\partial}{\partial x}\left\langle x|\psi\right\rangle , ⟨ x ∣ P ∣ ψ ⟩ = − i ∂ x ∂ ⟨ x ∣ ψ ⟩ ,
Do teorema generalizado então, para o operador O = X 2 O=X^{2} O = X 2
d ⟨ O ⟩ d t = − i ℏ ⟨ [ O , H ] ⟩ + ⟨ ∂ O ∂ t ⟩ d ⟨ X 2 ⟩ d t = − i ℏ ⟨ [ X 2 , P 2 2 m ] ⟩ + ⟨ ∂ X 2 ∂ t ⟩ = − i 2 m ℏ ⟨ [ X 2 , P 2 ] ⟩ + 0 \begin{align*}
\frac{d\left\langle O\right\rangle }{dt}= & -\frac{i}{\hbar}\left\langle \left[O,H\right]\right\rangle +\left\langle \frac{\partial O}{\partial t}\right\rangle \\
\frac{d\left\langle X^{2}\right\rangle }{dt}= & -\frac{i}{\hbar}\left\langle \left[X^{2},\frac{P^{2}}{2m}\right]\right\rangle +\left\langle \frac{\partial X^{2}}{\partial t}\right\rangle \\
= & -\frac{i}{2m\hbar}\left\langle \left[X^{2},P^{2}\right]\right\rangle +0
\end{align*} d t d ⟨ O ⟩ = d t d ⟨ X 2 ⟩ = = − ℏ i ⟨ [ O , H ] ⟩ + ⟨ ∂ t ∂ O ⟩ − ℏ i ⟨ [ X 2 , 2 m P 2 ] ⟩ + ⟨ ∂ t ∂ X 2 ⟩ − 2 m ℏ i ⟨ [ X 2 , P 2 ] ⟩ + 0 Sabendo que [ X , P ] = i ℏ \left[X,P\right]=i\hbar [ X , P ] = i ℏ
[ X 2 , P 2 ] = X X P 2 − P 2 X X = ( X X P 2 − X P 2 X ) + ( X P 2 X − P 2 X X ) = X ( X P 2 − P 2 X ) + ( X P 2 − P 2 X ) X = X ( X P 2 − P 2 X ) + ( X P 2 − P 2 X ) X = X ( [ X P P − P X P ] + [ P X P − P P X ] ) + ( [ X P P − P X P ] + [ P X P − P P X ] ) X = X ( [ X P − P X ] P + P [ X P − P X ] ) + ( [ X P − P X ] P + P [ X P − P X ] ) X = X ( [ X , P ] P + P [ X , P ] ) + ( [ X , P ] P + P [ X , P ] ) X = i ℏ ( X ( P + P ) + ( P + P ) X ) = 2 i ℏ ( X P + P X ) \begin{align*}
\left[X^{2},P^{2}\right] & =XXP^{2}-P^{2}XX\\
& =\left(XXP^{2}-XP^{2}X\right)+\left(XP^{2}X-P^{2}XX\right)\\
& =X\left(XP^{2}-P^{2}X\right)+\left(XP^{2}-P^{2}X\right)X\\
& =X\left(XP^{2}-P^{2}X\right)+\left(XP^{2}-P^{2}X\right)X\\
& =X\left(\left[XPP-PXP\right]+\left[PXP-PPX\right]\right)+\left(\left[XPP-PXP\right]+\left[PXP-PPX\right]\right)X\\
& =X\left(\left[XP-PX\right]P+P\left[XP-PX\right]\right)+\left(\left[XP-PX\right]P+P\left[XP-PX\right]\right)X\\
& =X\left(\left[X,P\right]P+P\left[X,P\right]\right)+\left(\left[X,P\right]P+P\left[X,P\right]\right)X\\
& =i\hbar\left(X\left(P+P\right)+\left(P+P\right)X\right)\\
& =2i\hbar\left(XP+PX\right)
\end{align*} [ X 2 , P 2 ] = XX P 2 − P 2 XX = ( XX P 2 − X P 2 X ) + ( X P 2 X − P 2 XX ) = X ( X P 2 − P 2 X ) + ( X P 2 − P 2 X ) X = X ( X P 2 − P 2 X ) + ( X P 2 − P 2 X ) X = X ( [ XPP − PXP ] + [ PXP − PPX ] ) + ( [ XPP − PXP ] + [ PXP − PPX ] ) X = X ( [ XP − PX ] P + P [ XP − PX ] ) + ( [ XP − PX ] P + P [ XP − PX ] ) X = X ( [ X , P ] P + P [ X , P ] ) + ( [ X , P ] P + P [ X , P ] ) X = i ℏ ( X ( P + P ) + ( P + P ) X ) = 2 i ℏ ( XP + PX ) Então:
d ⟨ X 2 ⟩ d t = − i 2 m ℏ ⟨ [ X 2 , P 2 ] ⟩ = 1 m ⟨ X P + P X ⟩ \begin{align*}
\frac{d\left\langle X^{2}\right\rangle }{dt}= & -\frac{i}{2m\hbar}\left\langle \left[X^{2},P^{2}\right]\right\rangle \\
= & \frac{1}{m}\left\langle XP+PX\right\rangle
\end{align*} d t d ⟨ X 2 ⟩ = = − 2 m ℏ i ⟨ [ X 2 , P 2 ] ⟩ m 1 ⟨ XP + PX ⟩ E para ⟨ X P + P X ⟩ \left\langle XP+PX\right\rangle ⟨ XP + PX ⟩
d ⟨ X P + P X ⟩ d t = − i ℏ ⟨ [ X P + P X , P 2 2 m ] ⟩ = − i 2 m ℏ ⟨ [ X P + P X , P 2 ] ⟩ \begin{align*}
\frac{d\left\langle XP+PX\right\rangle }{dt}= & -\frac{i}{\hbar}\left\langle \left[XP+PX,\frac{P^{2}}{2m}\right]\right\rangle \\
= & -\frac{i}{2m\hbar}\left\langle \left[XP+PX,P^{2}\right]\right\rangle
\end{align*} d t d ⟨ XP + PX ⟩ = = − ℏ i ⟨ [ XP + PX , 2 m P 2 ] ⟩ − 2 m ℏ i ⟨ [ XP + PX , P 2 ] ⟩ Então:
[ X P + P X , P 2 ] = ( X P + P X ) P 2 − P 2 ( X P + P X ) = ( [ X P − P X ] + P X + P X ) P 2 − P 2 ( [ X P − X P ] + X P + P X ) = ( [ X , ‘ P ] + P X + P X ) P 2 − P 2 ( [ X , ‘ P ] + X P + P X ) = ( i ℏ + 2 P X ) P 2 − P 2 ( i ℏ + 2 P X ) = 2 [ P X P 2 − P 2 P X ] = 2 P [ X P P − P P X ] = 2 P [ ( X P P − P X P ) + ( P X P − P P X ) ] = 2 P [ ( X P − P X ) P + P ( X P − P X ) ] = 2 P [ [ X , P ] P + P [ X , P ] ] = 2 P [ i ℏ P + i ℏ P ] = 4 i ℏ P 2 \begin{align*}
\left[XP+PX,P^{2}\right] & =\left(XP+PX\right)P^{2}-P^{2}\left(XP+PX\right)\\
& =\left(\left[XP-PX\right]+PX+PX\right)P^{2}-P^{2}\left(\left[XP-XP\right]+XP+PX\right)\\
& =\left(\left[X,`P\right]+PX+PX\right)P^{2}-P^{2}\left(\left[X,`P\right]+XP+PX\right)\\
& =\left(i\hbar+2PX\right)P^{2}-P^{2}\left(i\hbar+2PX\right)\\
& =2\left[PXP^{2}-P^{2}PX\right]\\
& =2P\left[XPP-PPX\right]\\
& =2P\left[\left(XPP-PXP\right)+\left(PXP-PPX\right)\right]\\
& =2P\left[\left(XP-PX\right)P+P\left(XP-PX\right)\right]\\
& =2P\left[\left[X,P\right]P+P\left[X,P\right]\right]\\
& =2P\left[i\hbar P+i\hbar P\right]\\
& =4i\hbar P^{2}
\end{align*} [ XP + PX , P 2 ] = ( XP + PX ) P 2 − P 2 ( XP + PX ) = ( [ XP − PX ] + PX + PX ) P 2 − P 2 ( [ XP − XP ] + XP + PX ) = ( [ X , ‘ P ] + PX + PX ) P 2 − P 2 ( [ X , ‘ P ] + XP + PX ) = ( i ℏ + 2 PX ) P 2 − P 2 ( i ℏ + 2 PX ) = 2 [ PX P 2 − P 2 PX ] = 2 P [ XPP − PPX ] = 2 P [ ( XPP − PXP ) + ( PXP − PPX ) ] = 2 P [ ( XP − PX ) P + P ( XP − PX ) ] = 2 P [ [ X , P ] P + P [ X , P ] ] = 2 P [ i ℏ P + i ℏ P ] = 4 i ℏ P 2 Então:
d ⟨ X P + P X ⟩ d t = − i ℏ ⟨ 4 i ℏ P 2 ⟩ = 2 m ⟨ P 2 ⟩ \begin{align*}
\frac{d\left\langle XP+PX\right\rangle }{dt}= & -\frac{i}{\hbar}\left\langle 4i\hbar P^{2}\right\rangle \\
= & \frac{2}{m}\left\langle P^{2}\right\rangle
\end{align*} d t d ⟨ XP + PX ⟩ = = − ℏ i ⟨ 4 i ℏ P 2 ⟩ m 2 ⟨ P 2 ⟩ Integrando temos então para ⟨ X P + P X ⟩ \left\langle XP+PX\right\rangle ⟨ XP + PX ⟩
⟨ X P + P X ⟩ = ⟨ X P + P X ⟩ 0 + 2 t m ⟨ P 2 ⟩ \left\langle XP+PX\right\rangle =\left\langle XP+PX\right\rangle _{0}+\frac{2t}{m}\left\langle P^{2}\right\rangle ⟨ XP + PX ⟩ = ⟨ XP + PX ⟩ 0 + m 2 t ⟨ P 2 ⟩ E subtituindo o resultado para ⟨ X 2 ⟩ \left\langle X^{2}\right\rangle ⟨ X 2 ⟩
d ⟨ X 2 ⟩ d t = 1 m ⟨ X P + P X ⟩ 0 + 2 t m 2 ⟨ P 2 ⟩ \frac{d\left\langle X^{2}\right\rangle }{dt}=\frac{1}{m}\left\langle XP+PX\right\rangle _{0}+\frac{2t}{m^{2}}\left\langle P^{2}\right\rangle d t d ⟨ X 2 ⟩ = m 1 ⟨ XP + PX ⟩ 0 + m 2 2 t ⟨ P 2 ⟩ Logo:
⟨ X 2 ⟩ = ⟨ X 2 ⟩ 0 + t m ⟨ X P + P X ⟩ 0 + ⟨ P 2 ⟩ t 2 m 2 \left\langle X^{2}\right\rangle =\left\langle X^{2}\right\rangle _{0}+\frac{t}{m}\left\langle XP+PX\right\rangle _{0}+\left\langle P^{2}\right\rangle \frac{t^{2}}{m^{2}} ⟨ X 2 ⟩ = ⟨ X 2 ⟩ 0 + m t ⟨ XP + PX ⟩ 0 + ⟨ P 2 ⟩ m 2 t 2 Então as equações do movimento são:
⟨ X P + P X ⟩ = ⟨ X P + P X ⟩ 0 + 2 t m ⟨ P 2 ⟩ ⟨ X 2 ⟩ = ⟨ X 2 ⟩ 0 + t m ⟨ X P + P X ⟩ 0 + ⟨ P 2 ⟩ t 2 m 2 \left\langle XP+PX\right\rangle =\left\langle XP+PX\right\rangle _{0}+\frac{2t}{m}\left\langle P^{2}\right\rangle \qquad\left\langle X^{2}\right\rangle =\left\langle X^{2}\right\rangle _{0}+\frac{t}{m}\left\langle XP+PX\right\rangle _{0}+\left\langle P^{2}\right\rangle \frac{t^{2}}{m^{2}} ⟨ XP + PX ⟩ = ⟨ XP + PX ⟩ 0 + m 2 t ⟨ P 2 ⟩ ⟨ X 2 ⟩ = ⟨ X 2 ⟩ 0 + m t ⟨ XP + PX ⟩ 0 + ⟨ P 2 ⟩ m 2 t 2 c)
A encontrar:
Equações:
H = P 2 2 m H=\frac{P^{2}}{2m} H = 2 m P 2
d ⟨ O ⟩ d t = − i ℏ ⟨ [ O , H ] ⟩ + ⟨ ∂ O ∂ t ⟩ \frac{d\left\langle O\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[O,H\right]\right\rangle +\left\langle \frac{\partial O}{\partial t}\right\rangle d t d ⟨ O ⟩ = − ℏ i ⟨ [ O , H ] ⟩ + ⟨ ∂ t ∂ O ⟩
( Δ X ) 2 = ⟨ X 2 ⟩ − ⟨ X ⟩ 2 \left(\Delta X\right)^{2}=\left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2} ( Δ X ) 2 = ⟨ X 2 ⟩ − ⟨ X ⟩ 2
Soluções obtidas anteriormente:
⟨ X ⟩ = ⟨ X ⟩ 0 + ⟨ P ⟩ 0 m t \left\langle X\right\rangle =\left\langle X\right\rangle _{0}+\frac{\left\langle P\right\rangle _{0}}{m}t ⟨ X ⟩ = ⟨ X ⟩ 0 + m ⟨ P ⟩ 0 t
⟨ X 2 ⟩ = ⟨ X 2 ⟩ 0 + t m ⟨ X P + P X ⟩ 0 + ⟨ P 2 ⟩ t 2 m 2 \left\langle X^{2}\right\rangle =\left\langle X^{2}\right\rangle _{0}+\frac{t}{m}\left\langle XP+PX\right\rangle _{0}+\left\langle P^{2}\right\rangle \frac{t^{2}}{m^{2}} ⟨ X 2 ⟩ = ⟨ X 2 ⟩ 0 + m t ⟨ XP + PX ⟩ 0 + ⟨ P 2 ⟩ m 2 t 2
Lembrando então que o devio padrão obedece:
( Δ X ) 2 = ⟨ X 2 ⟩ − ⟨ X ⟩ 2 = ⟨ X 2 ⟩ 0 + t m ⟨ X P + P X ⟩ 0 + ⟨ P 2 ⟩ t 2 m 2 − ( ⟨ X ⟩ 0 + ⟨ P ⟩ 0 m t ) ( ⟨ X ⟩ 0 + ⟨ P ⟩ 0 m t ) = ⟨ X 2 ⟩ 0 + t m ⟨ X P + P X ⟩ 0 + ⟨ P 2 ⟩ t 2 m 2 − ( ⟨ X ⟩ 0 2 + 2 ⟨ X ⟩ 0 ⟨ P ⟩ 0 m t + ⟨ P ⟩ 0 2 m 2 t 2 ) = ( ⟨ P 2 ⟩ 0 − ⟨ P ⟩ 0 2 ) t 2 m 2 + ( ⟨ X 2 ⟩ 0 − ⟨ X ⟩ 0 2 ) + t m ⟨ X P + P X ⟩ 0 − 2 ⟨ X ⟩ 0 ⟨ P ⟩ 0 m t = ( Δ P ) 0 2 t 2 m 2 + ( Δ X ) 0 2 + [ t m ⟨ X P + P X ⟩ 0 − 2 t m ⟨ X ⟩ 0 ⟨ P ⟩ 0 ] = ( Δ P ) 0 2 t 2 m 2 + ( Δ X ) 0 2 + t m [ ⟨ X P + P X ⟩ 0 − 2 ⟨ X ⟩ 0 ⟨ P ⟩ 0 ] \begin{align*}
\left(\Delta X\right)^{2} & =\left\langle X^{2}\right\rangle -\left\langle X\right\rangle ^{2}\\
& =\left\langle X^{2}\right\rangle _{0}+\frac{t}{m}\left\langle XP+PX\right\rangle _{0}+\left\langle P^{2}\right\rangle \frac{t^{2}}{m^{2}}-\left(\left\langle X\right\rangle _{0}+\frac{\left\langle P\right\rangle _{0}}{m}t\right)\left(\left\langle X\right\rangle _{0}+\frac{\left\langle P\right\rangle _{0}}{m}t\right)\\
& =\left\langle X^{2}\right\rangle _{0}+\frac{t}{m}\left\langle XP+PX\right\rangle _{0}+\left\langle P^{2}\right\rangle \frac{t^{2}}{m^{2}}-\left(\left\langle X\right\rangle ^{2}_{0}+2\frac{\left\langle X\right\rangle _{0}\left\langle P\right\rangle _{0}}{m}t+\frac{\left\langle P\right\rangle ^{2}_{0}}{m^{2}}t^{2}\right)\\
& =\left(\left\langle P^{2}\right\rangle _{0}-\left\langle P\right\rangle ^{2}_{0}\right)\frac{t^{2}}{m^{2}}+\left(\left\langle X^{2}\right\rangle _{0}-\left\langle X\right\rangle ^{2}_{0}\right)+\frac{t}{m}\left\langle XP+PX\right\rangle _{0}-2\frac{\left\langle X\right\rangle _{0}\left\langle P\right\rangle _{0}}{m}t\\
& =\left(\Delta P\right)^{2}_{0}\frac{t^{2}}{m^{2}}+\left(\Delta X\right)^{2}_{0}+\left[\frac{t}{m}\left\langle XP+PX\right\rangle _{0}-\frac{2t}{m}\left\langle X\right\rangle _{0}\left\langle P\right\rangle _{0}\right]\\
& =\left(\Delta P\right)^{2}_{0}\frac{t^{2}}{m^{2}}+\left(\Delta X\right)^{2}_{0}+\frac{t}{m}\left[\left\langle XP+PX\right\rangle _{0}-2\left\langle X\right\rangle _{0}\left\langle P\right\rangle _{0}\right]
\end{align*} ( Δ X ) 2 = ⟨ X 2 ⟩ − ⟨ X ⟩ 2 = ⟨ X 2 ⟩ 0 + m t ⟨ XP + PX ⟩ 0 + ⟨ P 2 ⟩ m 2 t 2 − ( ⟨ X ⟩ 0 + m ⟨ P ⟩ 0 t ) ( ⟨ X ⟩ 0 + m ⟨ P ⟩ 0 t ) = ⟨ X 2 ⟩ 0 + m t ⟨ XP + PX ⟩ 0 + ⟨ P 2 ⟩ m 2 t 2 − ( ⟨ X ⟩ 0 2 + 2 m ⟨ X ⟩ 0 ⟨ P ⟩ 0 t + m 2 ⟨ P ⟩ 0 2 t 2 ) = ( ⟨ P 2 ⟩ 0 − ⟨ P ⟩ 0 2 ) m 2 t 2 + ( ⟨ X 2 ⟩ 0 − ⟨ X ⟩ 0 2 ) + m t ⟨ XP + PX ⟩ 0 − 2 m ⟨ X ⟩ 0 ⟨ P ⟩ 0 t = ( Δ P ) 0 2 m 2 t 2 + ( Δ X ) 0 2 + [ m t ⟨ XP + PX ⟩ 0 − m 2 t ⟨ X ⟩ 0 ⟨ P ⟩ 0 ] = ( Δ P ) 0 2 m 2 t 2 + ( Δ X ) 0 2 + m t [ ⟨ XP + PX ⟩ 0 − 2 ⟨ X ⟩ 0 ⟨ P ⟩ 0 ] Temos que ⟨ P 2 ⟩ 0 = ⟨ P 2 ⟩ = constante \left\langle P^{2}\right\rangle _{0}=\left\langle P^{2}\right\rangle =\text{constante} ⟨ P 2 ⟩ 0 = ⟨ P 2 ⟩ = constante
d ⟨ P 2 ⟩ d t = − i ℏ ⟨ [ P 2 , H ] ⟩ = − i 2 m ℏ ⟨ [ P 2 , P 2 ] ⟩ = 0 \begin{align*}
\frac{d\left\langle P^{2}\right\rangle }{dt}= & -\frac{i}{\hbar}\left\langle \left[P^{2},H\right]\right\rangle \\
= & -\frac{i}{2m\hbar}\left\langle \left[P^{2},P^{2}\right]\right\rangle \\
= & 0
\end{align*} d t d ⟨ P 2 ⟩ = = = − ℏ i ⟨ [ P 2 , H ] ⟩ − 2 m ℏ i ⟨ [ P 2 , P 2 ] ⟩ 0 Para o caso de uma partícula livre como estsamos tratando, existe um t t t ⟨ X P + P X ⟩ 0 − 1 2 ⟨ X ⟩ 0 ⟨ P ⟩ 0 = 0 \left\langle XP+PX\right\rangle _{0}-\frac{1}{2}\left\langle X\right\rangle _{0}\left\langle P\right\rangle _{0}=0 ⟨ XP + PX ⟩ 0 − 2 1 ⟨ X ⟩ 0 ⟨ P ⟩ 0 = 0 Wheeler (1998)
Problem 3 ¶ In a one-dimensional problem, consider a particle with the Hamiltonian
H = P 2 2 m + V ( X ) H=\frac{P^{2}}{2m}+V\left(X\right) H = 2 m P 2 + V ( X ) where V ( X ) = λ X n V\left(X\right)=\lambda X^{n} V ( X ) = λ X n
a) Calculate the commutator [ H , X P ] \left[H,XP\right] [ H , XP ] ∣ φ ⟩ \left|\varphi\right\rangle ∣ φ ⟩ in the potential V V V ⟨ T ⟩ \left\langle T\right\rangle ⟨ T ⟩ ⟨ V ⟩ \left\langle V\right\rangle ⟨ V ⟩ 2 ⟨ T ⟩ = n ⟨ V ⟩ 2\left\langle T\right\rangle =n\left\langle V\right\rangle 2 ⟨ T ⟩ = n ⟨ V ⟩
b) In a three-dimensional problem, H is written:
H = P 2 2 m + V ( R ) H=\frac{P^{2}}{2m}+V\left(R\right) H = 2 m P 2 + V ( R ) Calculate the commutator [ H , R ⋅ P ] \left[H,R\cdot P\right] [ H , R ⋅ P ] V ( R ) V\left(R\right) V ( R ) n n n X X X Y Y Y Z Z Z
and satisfies Euler’s identity:
x ∂ V ∂ x + y ∂ V ∂ y + z ∂ V ∂ z = n V ( x , y , z ) x\frac{\partial V}{\partial x}+y\frac{\partial V}{\partial y}+z\frac{\partial V}{\partial z}=nV\left(x,y,z\right) x ∂ x ∂ V + y ∂ y ∂ V + z ∂ z ∂ V = nV ( x , y , z ) Dados do enunciado:
H = P 2 2 m + V ( X ) H=\frac{P^{2}}{2m}+V\left(X\right) H = 2 m P 2 + V ( X )
V ( X ) = λ X n V\left(X\right)=\lambda X^{n} V ( X ) = λ X n
a)
A encontrar:
O comutador [ H , X P ] \left[H,XP\right] [ H , XP ]
Mostrar que se existe estados estacionários ∣ φ ⟩ \left|\varphi\right\rangle ∣ φ ⟩ V V V 2 ⟨ T ⟩ = n ⟨ V ⟩ 2\left\langle T\right\rangle =n\left\langle V\right\rangle 2 ⟨ T ⟩ = n ⟨ V ⟩
Equações:
d ⟨ O ⟩ d t = − i ℏ ⟨ [ O , H ] ⟩ + ⟨ ∂ O ∂ t ⟩ \frac{d\left\langle O\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[O,H\right]\right\rangle +\left\langle \frac{\partial O}{\partial t}\right\rangle d t d ⟨ O ⟩ = − ℏ i ⟨ [ O , H ] ⟩ + ⟨ ∂ t ∂ O ⟩
[ A , B ] = − [ B , A ] \left[A,B\right]=-\left[B,A\right] [ A , B ] = − [ B , A ]
[ X , O ( P ) ] = i ℏ ∂ O ∂ P \left[X,O\left(P\right)\right]=i\hbar\frac{\partial O}{\partial P} [ X , O ( P ) ] = i ℏ ∂ P ∂ O
[ P , O ( X ) ] = − i ℏ ∂ O ∂ X \left[P,O\left(X\right)\right]=-i\hbar\frac{\partial O}{\partial X} [ P , O ( X ) ] = − i ℏ ∂ X ∂ O
T = P 2 2 m T=\frac{P^{2}}{2m} T = 2 m P 2
Calculando o comutador:
[ H , X P ] = H X P − X P H = ( P 2 2 m + λ X n ) X P − X P ( P 2 2 m + λ X n ) = P 2 X P 2 m + λ X n X P − X P P 2 2 m + λ X P X n = [ P 2 X P 2 m − X P 2 P 2 m ] + [ λ X n X P − λ X X n P ] + [ X P 2 P 2 m − X P P 2 2 m ] + [ λ X X n P − λ X P X n ] = [ P 2 X − X P 2 ] P 2 m + X 2 m [ P 2 P − P P 2 ] + λ [ X n X − X X n ] P + λ X [ X n P − P X n ] = [ P 2 , X ] P 2 m + 0 + 0 + λ X [ X n , P ] = − ( [ X , P 2 ] P 2 m + λ X [ P , X n ] ) = − ( i ℏ ∂ P 2 ∂ P P 2 m − λ X i ℏ ∂ X n ∂ X ) = − ( 2 i ℏ P 2 2 m − λ n i ℏ X n ) = − i ℏ ( P 2 m − n λ X n ) \begin{align*}
\left[H,XP\right] & =HXP-XPH\\
& =\left(\frac{P^{2}}{2m}+\lambda X^{n}\right)XP-XP\left(\frac{P^{2}}{2m}+\lambda X^{n}\right)\\
& =\frac{P^{2}XP}{2m}+\lambda X^{n}XP-XP\frac{P^{2}}{2m}+\lambda XPX^{n}\\
& =\left[\frac{P^{2}XP}{2m}-\frac{XP^{2}P}{2m}\right]+\left[\lambda X^{n}XP-\lambda XX^{n}P\right]+\left[\frac{XP^{2}P}{2m}-XP\frac{P^{2}}{2m}\right]+\left[\lambda XX^{n}P-\lambda XPX^{n}\right]\\
& =\left[P^{2}X-XP^{2}\right]\frac{P}{2m}+\frac{X}{2m}\left[P^{2}P-PP^{2}\right]+\lambda\left[X^{n}X-XX^{n}\right]P+\lambda X\left[X^{n}P-PX^{n}\right]\\
& =\left[P^{2},X\right]\frac{P}{2m}+0+0+\lambda X\left[X^{n},P\right]\\
& =-\left(\left[X,P^{2}\right]\frac{P}{2m}+\lambda X\left[P,X^{n}\right]\right)\\
& =-\left(i\hbar\frac{\partial P^{2}}{\partial P}\frac{P}{2m}-\lambda Xi\hbar\frac{\partial X^{n}}{\partial X}\right)\\
& =-\left(2i\hbar\frac{P^{2}}{2m}-\lambda ni\hbar X^{n}\right)\\
& =-i\hbar\left(\frac{P^{2}}{m}-n\lambda X^{n}\right)
\end{align*} [ H , XP ] = H XP − XP H = ( 2 m P 2 + λ X n ) XP − XP ( 2 m P 2 + λ X n ) = 2 m P 2 XP + λ X n XP − XP 2 m P 2 + λ XP X n = [ 2 m P 2 XP − 2 m X P 2 P ] + [ λ X n XP − λ X X n P ] + [ 2 m X P 2 P − XP 2 m P 2 ] + [ λ X X n P − λ XP X n ] = [ P 2 X − X P 2 ] 2 m P + 2 m X [ P 2 P − P P 2 ] + λ [ X n X − X X n ] P + λ X [ X n P − P X n ] = [ P 2 , X ] 2 m P + 0 + 0 + λ X [ X n , P ] = − ( [ X , P 2 ] 2 m P + λ X [ P , X n ] ) = − ( i ℏ ∂ P ∂ P 2 2 m P − λ X i ℏ ∂ X ∂ X n ) = − ( 2 i ℏ 2 m P 2 − λni ℏ X n ) = − i ℏ ( m P 2 − nλ X n ) Então o comutador é:
[ H , X P ] = − i ℏ ( P 2 m − n λ X n ) \left[H,XP\right]=-i\hbar\left(\frac{P^{2}}{m}-n\lambda X^{n}\right) [ H , XP ] = − i ℏ ( m P 2 − nλ X n ) E uma vez que os operadores não dependem explicitamente do tempo:
d ⟨ X P ⟩ d t = − i ℏ ⟨ [ X P , H ] ⟩ + ⟨ ∂ X P ∂ t ⟩ = i ℏ ⟨ [ H , X P ] ⟩ + 0 = i ℏ ( − i ℏ ) ⟨ ( P 2 m − n λ X n ) ⟩ = ⟨ P 2 m ⟩ − n ⟨ λ X n ⟩ = 2 ⟨ T ⟩ − n ⟨ V ⟩ \begin{align*}
\frac{d\left\langle XP\right\rangle }{dt} & =-\frac{i}{\hbar}\left\langle \left[XP,H\right]\right\rangle +\left\langle \frac{\partial XP}{\partial t}\right\rangle \\
& =\frac{i}{\hbar}\left\langle \left[H,XP\right]\right\rangle +0\\
& =\frac{i}{\hbar}\left(-i\hbar\right)\left\langle \left(\frac{P^{2}}{m}-n\lambda X^{n}\right)\right\rangle \\
& =\left\langle \frac{P^{2}}{m}\right\rangle -n\left\langle \lambda X^{n}\right\rangle \\
& =2\left\langle T\right\rangle -n\left\langle V\right\rangle
\end{align*} d t d ⟨ XP ⟩ = − ℏ i ⟨ [ XP , H ] ⟩ + ⟨ ∂ t ∂ XP ⟩ = ℏ i ⟨ [ H , XP ] ⟩ + 0 = ℏ i ( − i ℏ ) ⟨ ( m P 2 − nλ X n ) ⟩ = ⟨ m P 2 ⟩ − n ⟨ λ X n ⟩ = 2 ⟨ T ⟩ − n ⟨ V ⟩ E uma vez que estamos lidando com um estado estacionário, o valor esperado deve ser constante (também pode ser demonstrado explicitamente que d ⟨ X P ⟩ d t = 0 \frac{d\left\langle XP\right\rangle }{dt}=0 d t d ⟨ XP ⟩ = 0 Pace (2005)
0 = 2 ⟨ T ⟩ − n ⟨ V ⟩ 2 ⟨ T ⟩ = n ⟨ V ⟩ \begin{align*}
0 & =2\left\langle T\right\rangle -n\left\langle V\right\rangle \\
2\left\langle T\right\rangle & =n\left\langle V\right\rangle
\end{align*} 0 2 ⟨ T ⟩ = 2 ⟨ T ⟩ − n ⟨ V ⟩ = n ⟨ V ⟩ b)
Dados do enunciado:
H = P 2 2 m + V ( R ) H=\frac{\boldsymbol{P}^{2}}{2m}+V\left(\boldsymbol{R}\right) H = 2 m P 2 + V ( R )
V ( R ) V\left(\boldsymbol{R}\right) V ( R ) n n n X , Y , Z X,Y,Z X , Y , Z
V ( α x , α y , α z ) = α n V ( x , y , z ) V\left(\alpha x,\alpha y,\alpha z\right)=\alpha^{n}V\left(x,y,z\right) V ( αx , α y , α z ) = α n V ( x , y , z )
x ∂ V ∂ x + y ∂ V ∂ y + z ∂ V ∂ z = n V ( x , y , z ) x\frac{\partial V}{\partial x}+y\frac{\partial V}{\partial y}+z\frac{\partial V}{\partial z}=nV\left(x,y,z\right) x ∂ x ∂ V + y ∂ y ∂ V + z ∂ z ∂ V = nV ( x , y , z )
A encontrar:
O comutador [ H , R ⋅ P ] \left[H,\boldsymbol{R}\cdot\boldsymbol{P}\right] [ H , R ⋅ P ]
Relação entre a energia cinética média e o potencial médio da partícula no estado estacionário
Equações:
d ⟨ O ⟩ d t = − i ℏ ⟨ [ O , H ] ⟩ + ⟨ ∂ O ∂ t ⟩ \frac{d\left\langle O\right\rangle }{dt}=-\frac{i}{\hbar}\left\langle \left[O,H\right]\right\rangle +\left\langle \frac{\partial O}{\partial t}\right\rangle d t d ⟨ O ⟩ = − ℏ i ⟨ [ O , H ] ⟩ + ⟨ ∂ t ∂ O ⟩
[ A , B ] = − [ B , A ] \left[A,B\right]=-\left[B,A\right] [ A , B ] = − [ B , A ]
[ X , O ( P ) ] = i ℏ ∂ O ∂ P \left[X,O\left(P\right)\right]=i\hbar\frac{\partial O}{\partial P} [ X , O ( P ) ] = i ℏ ∂ P ∂ O
[ P , O ( X ) ] = − i ℏ ∂ O ∂ X \left[P,O\left(X\right)\right]=-i\hbar\frac{\partial O}{\partial X} [ P , O ( X ) ] = − i ℏ ∂ X ∂ O
T = P 2 2 m T=\frac{P^{2}}{2m} T = 2 m P 2
Fazendo um cálculo análogo ao caso anterior, temos:
[ H , R ⋅ P ] = H ( R ⋅ P ) − ( R ⋅ P ) H = ∑ i = 1 3 [ H X i P i − X i P i H ] = ∑ i = 1 3 [ H , X i P i ] \begin{align*}
\left[H,\boldsymbol{R}\cdot\boldsymbol{P}\right] & =H\left(\boldsymbol{R}\cdot\boldsymbol{P}\right)-\left(\boldsymbol{R}\cdot\boldsymbol{P}\right)H\\
& =\sum^{3}_{i=1}\left[HX_{i}P_{i}-X_{i}P_{i}H\right]\\
& =\sum^{3}_{i=1}\left[H,X_{i}P_{i}\right]
\end{align*} [ H , R ⋅ P ] = H ( R ⋅ P ) − ( R ⋅ P ) H = i = 1 ∑ 3 [ H X i P i − X i P i H ] = i = 1 ∑ 3 [ H , X i P i ] Então análogo ao caso anterior temos:
[ H , X i P i ] = H X i P i − X i P i H = ( P 2 2 m + V ( r ) ) X i P i − X i P i ( P 2 2 m + V ( r ) ) = P 2 X i P i 2 m + V X i P i − X i P i P 2 2 m + X i P i V = [ P 2 X i P i 2 m − X i P 2 P i 2 m ] + [ V X i P i − X i V P i ] + [ X i P 2 P i 2 m − X i P i P 2 2 m ] + [ X i V P i − X i P i V ] = [ P 2 X i − X i P 2 ] P i 2 m + X i 2 m [ P 2 P i − P i P 2 ] + [ V X i − X i V ] P i + X i [ V P i − P i V ] = [ P 2 , X i ] P i 2 m + X i 2 m [ P 2 , P i ] + [ V , X i ] P i + X i [ V , P i ] = − ( [ X i , P 2 ] P i 2 m + X i 2 m [ P i , P 2 ] + [ X i , V ] P i + X i [ P i , V ] ) = − ( i ℏ ∂ P 2 ∂ P i P i 2 m − X i 2 m i ℏ ∂ P 2 ∂ X i + i ℏ ∂ V 2 ∂ P i P i − X i i ℏ ∂ V ∂ X i ) = − ( 2 i ℏ P i 2 2 m − X i i ℏ ∂ V ∂ X i ) \begin{align*}
\left[H,X_{i}P_{i}\right] & =HX_{i}P_{i}-X_{i}P_{i}H\\
& =\left(\frac{\boldsymbol{P}^{2}}{2m}+V\left(\boldsymbol{r}\right)\right)X_{i}P_{i}-X_{i}P_{i}\left(\frac{\boldsymbol{P}^{2}}{2m}+V\left(\boldsymbol{r}\right)\right)\\
& =\frac{\boldsymbol{P}^{2}X_{i}P_{i}}{2m}+VX_{i}P_{i}-X_{i}P_{i}\frac{\boldsymbol{P}^{2}}{2m}+X_{i}P_{i}V\\
& =\left[\frac{\boldsymbol{P}^{2}X_{i}P_{i}}{2m}-\frac{X_{i}\boldsymbol{P}^{2}P_{i}}{2m}\right]+\left[VX_{i}P_{i}-X_{i}VP_{i}\right]+\left[\frac{X_{i}\boldsymbol{P}^{2}P_{i}}{2m}-X_{i}P_{i}\frac{\boldsymbol{P}^{2}}{2m}\right]+\left[X_{i}VP_{i}-X_{i}P_{i}V\right]\\
& =\left[\boldsymbol{P}^{2}X_{i}-X_{i}\boldsymbol{P}^{2}\right]\frac{P_{i}}{2m}+\frac{X_{i}}{2m}\left[\boldsymbol{P}^{2}P_{i}-P_{i}\boldsymbol{P}^{2}\right]+\left[VX_{i}-X_{i}V\right]P_{i}+X_{i}\left[VP_{i}-P_{i}V\right]\\
& =\left[\boldsymbol{P}^{2},X_{i}\right]\frac{P_{i}}{2m}+\frac{X_{i}}{2m}\left[\boldsymbol{P}^{2},P_{i}\right]+\left[V,X_{i}\right]P_{i}+X_{i}\left[V,P_{i}\right]\\
& =-\left(\left[X_{i},\boldsymbol{P}^{2}\right]\frac{P_{i}}{2m}+\frac{X_{i}}{2m}\left[P_{i},\boldsymbol{P}^{2}\right]+\left[X_{i},V\right]P_{i}+X_{i}\left[P_{i},V\right]\right)\\
& =-\left(i\hbar\frac{\partial\boldsymbol{P}^{2}}{\partial P_{i}}\frac{P_{i}}{2m}-\frac{X_{i}}{2m}i\hbar\frac{\partial\boldsymbol{P}^{2}}{\partial X_{i}}+i\hbar\frac{\partial V^{2}}{\partial P_{i}}P_{i}-X_{i}i\hbar\frac{\partial V}{\partial X_{i}}\right)\\
& =-\left(2i\hbar\frac{P^{2}_{i}}{2m}-X_{i}i\hbar\frac{\partial V}{\partial X_{i}}\right)
\end{align*} [ H , X i P i ] = H X i P i − X i P i H = ( 2 m P 2 + V ( r ) ) X i P i − X i P i ( 2 m P 2 + V ( r ) ) = 2 m P 2 X i P i + V X i P i − X i P i 2 m P 2 + X i P i V = [ 2 m P 2 X i P i − 2 m X i P 2 P i ] + [ V X i P i − X i V P i ] + [ 2 m X i P 2 P i − X i P i 2 m P 2 ] + [ X i V P i − X i P i V ] = [ P 2 X i − X i P 2 ] 2 m P i + 2 m X i [ P 2 P i − P i P 2 ] + [ V X i − X i V ] P i + X i [ V P i − P i V ] = [ P 2 , X i ] 2 m P i + 2 m X i [ P 2 , P i ] + [ V , X i ] P i + X i [ V , P i ] = − ( [ X i , P 2 ] 2 m P i + 2 m X i [ P i , P 2 ] + [ X i , V ] P i + X i [ P i , V ] ) = − ( i ℏ ∂ P i ∂ P 2 2 m P i − 2 m X i i ℏ ∂ X i ∂ P 2 + i ℏ ∂ P i ∂ V 2 P i − X i i ℏ ∂ X i ∂ V ) = − ( 2 i ℏ 2 m P i 2 − X i i ℏ ∂ X i ∂ V ) Então:
[ H , R ⋅ P ] = − ∑ i = 1 3 ( 2 i ℏ P i 2 2 m − X i i ℏ ∂ V ∂ X i ) = − i ℏ ( P 2 m − ∑ i = 1 3 X i ∂ V ∂ X i ) = − i ℏ ( P 2 m − n V ( r ) ) \begin{align*}
\left[H,\boldsymbol{R}\cdot\boldsymbol{P}\right] & =-\sum^{3}_{i=1}\left(2i\hbar\frac{P^{2}_{i}}{2m}-X_{i}i\hbar\frac{\partial V}{\partial X_{i}}\right)\\
& =-i\hbar\left(\frac{\boldsymbol{P}^{2}}{m}-\sum^{3}_{i=1}X_{i}\frac{\partial V}{\partial X_{i}}\right)\\
& =-i\hbar\left(\frac{\boldsymbol{P}^{2}}{m}-nV\left(\boldsymbol{r}\right)\right)
\end{align*} [ H , R ⋅ P ] = − i = 1 ∑ 3 ( 2 i ℏ 2 m P i 2 − X i i ℏ ∂ X i ∂ V ) = − i ℏ ( m P 2 − i = 1 ∑ 3 X i ∂ X i ∂ V ) = − i ℏ ( m P 2 − nV ( r ) ) Ficamos então com o seguinte resultado para o comutador:
[ H , R ⋅ P ] = − i ℏ ( P 2 m − n V ( r ) ) \left[H,\boldsymbol{R}\cdot\boldsymbol{P}\right]=-i\hbar\left(\frac{\boldsymbol{P}^{2}}{m}-nV\left(\boldsymbol{r}\right)\right) [ H , R ⋅ P ] = − i ℏ ( m P 2 − nV ( r ) ) Resultado análogo ao que obtemos no cenário anterior. Então para obter as relações entre as energias, ficamos com:
d ⟨ R ⋅ P ⟩ d t = − i ℏ ⟨ [ R ⋅ P , H ] ⟩ + ⟨ ∂ R ⋅ P ∂ t ⟩ = i ℏ ⟨ [ H , R ⋅ P ] ⟩ + 0 = i ℏ ( − i ℏ ) ⟨ ( P 2 m − n V ( r ) ) ⟩ = ⟨ P 2 m ⟩ − n ⟨ V ( r ) ⟩ = 2 ⟨ T ⟩ − n ⟨ V ( r ) ⟩ \begin{align*}
\frac{d\left\langle \boldsymbol{R}\cdot\boldsymbol{P}\right\rangle }{dt} & =-\frac{i}{\hbar}\left\langle \left[\boldsymbol{R}\cdot\boldsymbol{P},H\right]\right\rangle +\left\langle \frac{\partial\boldsymbol{R}\cdot\boldsymbol{P}}{\partial t}\right\rangle \\
& =\frac{i}{\hbar}\left\langle \left[H,\boldsymbol{R}\cdot\boldsymbol{P}\right]\right\rangle +0\\
& =\frac{i}{\hbar}\left(-i\hbar\right)\left\langle \left(\frac{\boldsymbol{P}^{2}}{m}-nV\left(\boldsymbol{r}\right)\right)\right\rangle \\
& =\left\langle \frac{\boldsymbol{P}^{2}}{m}\right\rangle -n\left\langle V\left(\boldsymbol{r}\right)\right\rangle \\
& =2\left\langle T\right\rangle -n\left\langle V\left(\boldsymbol{r}\right)\right\rangle
\end{align*} d t d ⟨ R ⋅ P ⟩ = − ℏ i ⟨ [ R ⋅ P , H ] ⟩ + ⟨ ∂ t ∂ R ⋅ P ⟩ = ℏ i ⟨ [ H , R ⋅ P ] ⟩ + 0 = ℏ i ( − i ℏ ) ⟨ ( m P 2 − nV ( r ) ) ⟩ = ⟨ m P 2 ⟩ − n ⟨ V ( r ) ⟩ = 2 ⟨ T ⟩ − n ⟨ V ( r ) ⟩ De modo análogo então ao caso anterior, obtemos que:
2 ⟨ T ⟩ = n ⟨ V ( r ) ⟩ 2\left\langle T\right\rangle =n\left\langle V\left(\boldsymbol{r}\right)\right\rangle 2 ⟨ T ⟩ = n ⟨ V ( r ) ⟩ Podemos ainda observar que usando a propriedade de função homogênea, obtemos o teorema virial:
2 ⟨ T ⟩ = ⟨ x ∂ V ∂ x + y ∂ V ∂ y + z ∂ V ∂ z ⟩ 2\left\langle T\right\rangle =\left\langle x\frac{\partial V}{\partial x}+y\frac{\partial V}{\partial y}+z\frac{\partial V}{\partial z}\right\rangle 2 ⟨ T ⟩ = ⟨ x ∂ x ∂ V + y ∂ y ∂ V + z ∂ z ∂ V ⟩ Ou:
2 ⟨ T ⟩ = ⟨ r ⋅ ∇ V ⟩ 2\left\langle T\right\rangle =\left\langle \boldsymbol{r}\cdot\nabla V\right\rangle 2 ⟨ T ⟩ = ⟨ r ⋅ ∇ V ⟩
Lima, P. C. de. (2013). Equações Diferenciais Parciais I . http://150.164.25.15/ lima/apostilas/EDP1-aulas-22-8-13.pdf Wheeler, N. (1998). Ehrenfest’s Theorem . https://www.reed.edu/physics/faculty/wheeler/documents/Quantum%20Mechanics/Miscellaneous%20Essays/Ehrenfest%27s%20Theorem.pdf Pace, D. (2005). Problem Set 5 Solutions . http://lief.if.ufrgs.br/ ambusher/griffiths/sol5-Quantum_Mechanics_-_Griffiths_.pdf