Poço quadrado infinito
Nosso potencial agora é:
V ( x ) = { 0 , 0 ≤ x ≤ a ∞ outro caso V\left(x\right)=\begin{cases}
0, & 0\leq x\leq a\\
\infty & \text{outro caso}
\end{cases} V ( x ) = { 0 , ∞ 0 ≤ x ≤ a outro caso A partícula é livre entre dois extremos onde uma força infinita previne que ela escape. Fora do poço temos ψ ( x ) = 0 \psi\left(x\right)=0 ψ ( x ) = 0 V = 0 V=0 V = 0
− ℏ 2 2 m ∂ 2 ψ ∂ x 2 + V ψ = E ψ ∂ 2 ψ ∂ x 2 = − 2 m E ℏ 2 ψ ∂ 2 ψ ∂ x 2 = − k 2 ψ \begin{align}
-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+V\psi & =E\psi\\
\frac{\partial^{2}\psi}{\partial x^{2}} & =-\frac{2mE}{\hbar^{2}}\psi\\
\frac{\partial^{2}\psi}{\partial x^{2}} & =-k^{2}\psi
\end{align} − 2 m ℏ 2 ∂ x 2 ∂ 2 ψ + V ψ ∂ x 2 ∂ 2 ψ ∂ x 2 ∂ 2 ψ = E ψ = − ℏ 2 2 m E ψ = − k 2 ψ Essa é uma equação de um oscilador harmônico simples com uma solução geral do tipo:
ψ ( x ) = A sin ( k x ) + B cos k ( x ) \psi\left(x\right)=A\sin\left(kx\right)+B\cos k\left(x\right) ψ ( x ) = A sin ( k x ) + B cos k ( x ) Pois:
d ψ ( x ) d x = k A cos ( k x ) − k B sin k ( x ) d 2 ψ ( x ) d x = − k 2 A sin ( k x ) − k 2 B cos k ( x ) \begin{align}
\frac{d\psi\left(x\right)}{dx} & =kA\cos\left(kx\right)-kB\sin k\left(x\right)\\
\frac{d^{2}\psi\left(x\right)}{dx} & =-k^{2}A\sin\left(kx\right)-k^{2}B\cos k\left(x\right)
\end{align} d x d ψ ( x ) d x d 2 ψ ( x ) = k A cos ( k x ) − k B sin k ( x ) = − k 2 A sin ( k x ) − k 2 B cos k ( x ) Ou então:
d 2 ψ ( x ) d x = − k 2 ( A sin ( k x ) + B cos k ( x ) ) = − k 2 ψ ( x ) \frac{d^{2}\psi\left(x\right)}{dx}=-k^{2}\left(A\sin\left(kx\right)+B\cos k\left(x\right)\right)=-k^{2}\psi\left(x\right) d x d 2 ψ ( x ) = − k 2 ( A sin ( k x ) + B cos k ( x ) ) = − k 2 ψ ( x ) Queremos agora achar A A A B B B
ψ ( 0 ) = ψ ( a ) = 0 \psi\left(0\right)=\psi\left(a\right)=0 ψ ( 0 ) = ψ ( a ) = 0 Tipicamente ψ \psi ψ ( x ) \left(x\right) ( x ) Ψ ( x , t ) \Psi\left(x,t\right) Ψ ( x , t )
ψ ( 0 ) = A sin ( 0 ) + B cos k ( 0 ) 0 = B \begin{align}
\psi\left(0\right) & =A\sin\left(0\right)+B\cos k\left(0\right)\\
0 & =B
\end{align} ψ ( 0 ) 0 = A sin ( 0 ) + B cos k ( 0 ) = B Então temos:
ψ ( x ) = A sin ( k x ) \psi\left(x\right)=A\sin\left(kx\right) ψ ( x ) = A sin ( k x ) A outra condição de contorno é:
ψ ( a ) = A sin ( k a ) = 0 \begin{align}
\psi\left(a\right) & =A\sin\left(ka\right)=0
\end{align} ψ ( a ) = A sin ( ka ) = 0 Como A = 0 A=0 A = 0 k = 2 m E ℏ = 0 k=\frac{\sqrt{2mE}}{\hbar}=0 k = ℏ 2 m E = 0 k a ka ka π \pi π
k a = n π → k n = n π a ka=n\pi\rightarrow k_{n}=\frac{n\pi}{a} ka = nπ → k n = a nπ Onde usamos k n k_{n} k n k k k n = 1 , 2 , 3 , … n=1,2,3,\dots n = 1 , 2 , 3 , … E n E_{n} E n
k n = 2 m E n ℏ k n 2 = 2 m E n ℏ 2 ℏ 2 k n 2 2 m = E n E n = ℏ 2 n 2 π 2 2 m \begin{alignat}{1}
k_{n} & =\frac{\sqrt{2mE_{n}}}{\hbar}\\
k^{2}_{n} & =\frac{2mE_{n}}{\hbar^{2}}\\
\frac{\hbar^{2}k^{2}_{n}}{2m} & =E_{n}\\
E_{n} & =\frac{\hbar^{2}n^{2}\pi^{2}}{2m}
\end{alignat} k n k n 2 2 m ℏ 2 k n 2 E n = ℏ 2 m E n = ℏ 2 2 m E n = E n = 2 m ℏ 2 n 2 π 2 Mas ainda não obtemos A A A A A A
∫ − ∞ + ∞ ψ ∗ ψ d x = ∫ 0 a ( A sin ( k x ) ) ∗ ( A sin ( k x ) ) d x = ∣ A ∣ 2 ∫ 0 a sin 2 ( k x ) d x \begin{align}
\int^{+\infty}_{-\infty}\psi^{*}\psi dx & =\int^{a}_{0}\left(A\sin\left(kx\right)\right)^{*}\left(A\sin\left(kx\right)\right)dx\\
& =\left|A\right|^{2}\int^{a}_{0}\sin^{2}\left(kx\right)dx
\end{align} ∫ − ∞ + ∞ ψ ∗ ψ d x = ∫ 0 a ( A sin ( k x ) ) ∗ ( A sin ( k x ) ) d x = ∣ A ∣ 2 ∫ 0 a sin 2 ( k x ) d x Vamos usar a identidade trigonométrica 2 sin 2 ( x ) = 1 − cos ( 2 x ) 2\sin^{2}\left(x\right)=1-\cos\left(2x\right) 2 sin 2 ( x ) = 1 − cos ( 2 x )
∫ − ∞ + ∞ ψ ∗ ψ d x = ∣ A ∣ 2 2 ∫ 0 a ( 1 − cos ( 2 k x ) ) d x = ∣ A ∣ 2 2 [ ∫ 0 a d x − ∫ 0 a cos ( 2 k x ) d x ] \begin{align}
\int^{+\infty}_{-\infty}\psi^{*}\psi dx & =\frac{\left|A\right|^{2}}{2}\int^{a}_{0}\left(1-\cos\left(2kx\right)\right)dx\\
& =\frac{\left|A\right|^{2}}{2}\left[\int^{a}_{0}dx-\int^{a}_{0}\cos\left(2kx\right)dx\right]
\end{align} ∫ − ∞ + ∞ ψ ∗ ψ d x = 2 ∣ A ∣ 2 ∫ 0 a ( 1 − cos ( 2 k x ) ) d x = 2 ∣ A ∣ 2 [ ∫ 0 a d x − ∫ 0 a cos ( 2 k x ) d x ] e substitui u = 2 k x u=2kx u = 2 k x 2 k a 2ka 2 ka d u d x = 2 k \frac{du}{dx}=2k d x d u = 2 k
∫ − ∞ + ∞ ψ ∗ ψ d x = ∣ A ∣ 2 2 [ a − 1 2 k ∫ 0 2 k a cos ( u ) d u ] = ∣ A ∣ 2 2 [ a − 1 2 k [ sin ( u ) ] 0 2 k a ] = ∣ A ∣ 2 2 [ a − 1 2 k [ sin ( 2 k a ) − sin ( 0 ) ] ] = ∣ A ∣ 2 2 [ a − sin ( 2 k a ) 2 k ] \begin{align}
\int^{+\infty}_{-\infty}\psi^{*}\psi dx & =\frac{\left|A\right|^{2}}{2}\left[a-\frac{1}{2k}\int^{2ka}_{0}\cos\left(u\right)du\right]\\
& =\frac{\left|A\right|^{2}}{2}\left[a-\frac{1}{2k}\left[\sin\left(u\right)\right]^{2ka}_{0}\right]\\
& =\frac{\left|A\right|^{2}}{2}\left[a-\frac{1}{2k}\left[\sin\left(2ka\right)-\sin\left(0\right)\right]\right]\\
& =\frac{\left|A\right|^{2}}{2}\left[a-\frac{\sin\left(2ka\right)}{2k}\right]
\end{align} ∫ − ∞ + ∞ ψ ∗ ψ d x = 2 ∣ A ∣ 2 [ a − 2 k 1 ∫ 0 2 ka cos ( u ) d u ] = 2 ∣ A ∣ 2 [ a − 2 k 1 [ sin ( u ) ] 0 2 ka ] = 2 ∣ A ∣ 2 [ a − 2 k 1 [ sin ( 2 ka ) − sin ( 0 ) ] ] = 2 ∣ A ∣ 2 [ a − 2 k sin ( 2 ka ) ] Substituindo k = n π a k=\frac{n\pi}{a} k = a nπ
2 k a = n 2 π a a = n 2 π 2ka=n\frac{2\pi a}{a}=n2\pi 2 ka = n a 2 πa = n 2 π Como independente do n n n n 2 π → 2 π n2\pi\rightarrow2\pi n 2 π → 2 π sin ( 2 π ) = 0 \sin\left(2\pi\right)=0 sin ( 2 π ) = 0
∫ − ∞ + ∞ ψ ∗ ψ d x = a ∣ A ∣ 2 2 = 1 \int^{+\infty}_{-\infty}\psi^{*}\psi dx=\frac{a\left|A\right|^{2}}{2}=1 ∫ − ∞ + ∞ ψ ∗ ψ d x = 2 a ∣ A ∣ 2 = 1 Então ∣ A ∣ 2 = 2 / a \left|A\right|^{2}=2/a ∣ A ∣ 2 = 2/ a A A A A A A A = 2 a A=\sqrt{\frac{2}{a}} A = a 2
ψ n ( x ) = A sin ( k n x ) = 2 a sin ( n π a x ) \begin{align}
\psi_{n}\left(x\right) & =A\sin\left(k_{n}x\right)\\
& =\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)
\end{align} ψ n ( x ) = A sin ( k n x ) = a 2 sin ( a nπ x ) O conjunto de soluções tem algumas propriedades interessantes. Em primeiro lugar, elas são alternativamente par e ímpar em respeito ao centro. Se o centro é em a 2 \frac{a}{2} 2 a
ψ n ( x − a 2 ) = 2 a sin ( n π a [ x − a 2 ] ) = 2 a sin ( n π a x − n π 2 ) \begin{align}
\psi_{n}\left(x-\frac{a}{2}\right) & =\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}\left[x-\frac{a}{2}\right]\right)\\
& =\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x-\frac{n\pi}{2}\right)
\end{align} ψ n ( x − 2 a ) = a 2 sin ( a nπ [ x − 2 a ] ) = a 2 sin ( a nπ x − 2 nπ ) Usando a propriedade trigonométrica sin ( a − b ) = sin a cos b − cos a sin b \sin\left(a-b\right)=\sin a\cos b-\cos a\sin b sin ( a − b ) = sin a cos b − cos a sin b
ψ n ( x − a 2 ) = 2 a [ sin ( n π a x ) cos ( n π 2 ) − cos ( n π a x ) sin ( n π 2 ) ] = \begin{align}
\psi_{n}\left(x-\frac{a}{2}\right) & =\sqrt{\frac{2}{a}}\left[\sin\left(\frac{n\pi}{a}x\right)\cos\left(\frac{n\pi}{2}\right)-\cos\left(\frac{n\pi}{a}x\right)\sin\left(\frac{n\pi}{2}\right)\right]\\
& =
\end{align} ψ n ( x − 2 a ) = a 2 [ sin ( a nπ x ) cos ( 2 nπ ) − cos ( a nπ x ) sin ( 2 nπ ) ] = Sempre que n n n sin ( n π / 2 ) = 0 \sin\left(n\pi/2\right)=0 sin ( nπ /2 ) = 0 n n n cos ( n π / 2 ) = 0 \cos\left(n\pi/2\right)=0 cos ( nπ /2 ) = 0 n n n n n n
Em segundo lugar, também temos que pra cada vez que aumentamos o estado de energia, temos mais um nó, ou seja um ponto onde a função de onda atravessa o eixo. Para n = 1 n=1 n = 1 n = 2 n=2 n = 2
∫ ψ m ( x ) ∗ ψ n ( x ) d x = 0 \int\psi_{m}\left(x\right)^{*}\psi_{n}\left(x\right)dx=0 ∫ ψ m ( x ) ∗ ψ n ( x ) d x = 0 para m ≠ n m\neq n m = n m = n m=n m = n
∫ ψ ( x ) ∗ ψ ( x ) d x = ∫ ∣ ψ ( x ) ∣ 2 d x = 1 \int\psi\left(x\right)^{*}\psi\left(x\right)dx=\int\left|\psi\left(x\right)\right|^{2}dx=1 ∫ ψ ( x ) ∗ ψ ( x ) d x = ∫ ∣ ψ ( x ) ∣ 2 d x = 1 Ou seja:
∫ ψ m ( x ) ∗ ψ n ( x ) d x = δ m n \int\psi_{m}\left(x\right)^{*}\psi_{n}\left(x\right)dx=\delta_{mn} ∫ ψ m ( x ) ∗ ψ n ( x ) d x = δ mn de forma geral, ou seja, são ortonormais. Onde δ m n \delta_{mn} δ mn
δ m n = { 0 , m ≠ n 1 , m = n \delta_{mn}=\begin{cases}
0, & m\neq n\\
1, & m=n
\end{cases} δ mn = { 0 , 1 , m = n m = n Exemplo: Prove a ortogonalidade da função de onda ψ n ( x ) \psi_{n}\left(x\right) ψ n ( x )
Podemos provar a ortogonalidade:
∫ ψ m ( x ) ∗ ψ n ( x ) d x = ∫ ( 2 a sin ( m π a x ) ) ∗ ( 2 a sin ( n π a x ) ) d x = 2 a ∫ sin ( m π a x ) sin ( n π a x ) d x \begin{align}
\int\psi_{m}\left(x\right)^{*}\psi_{n}\left(x\right)dx & =\int\left(\sqrt{\frac{2}{a}}\sin\left(\frac{m\pi}{a}x\right)\right)^{*}\left(\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\right)dx\\
& =\frac{2}{a}\int\sin\left(\frac{m\pi}{a}x\right)\sin\left(\frac{n\pi}{a}x\right)dx
\end{align} ∫ ψ m ( x ) ∗ ψ n ( x ) d x = ∫ ( a 2 sin ( a mπ x ) ) ∗ ( a 2 sin ( a nπ x ) ) d x = a 2 ∫ sin ( a mπ x ) sin ( a nπ x ) d x Vamos usar outra propriedade trigonométrica agora:
cos ( a ± b ) = cos a cos b ∓ sin a sin b \begin{align}
\cos\left(a\pm b\right) & =\cos a\cos b\mp\sin a\sin b
\end{align} cos ( a ± b ) = cos a cos b ∓ sin a sin b Então:
cos ( a − b ) − cos ( a + b ) = ( cos a cos b + sin a sin b ) − ( cos a cos b − sin a sin b ) = 2 sin a sin b \begin{align}
\cos\left(a-b\right)-\cos\left(a+b\right) & =\left(\cos a\cos b+\sin a\sin b\right)-\left(\cos a\cos b-\sin a\sin b\right)\\
& =2\sin a\sin b
\end{align} cos ( a − b ) − cos ( a + b ) = ( cos a cos b + sin a sin b ) − ( cos a cos b − sin a sin b ) = 2 sin a sin b Então:
∫ ψ m ( x ) ∗ ψ n ( x ) d x = 1 a ∫ [ cos ( m π a x − n π a x ) − cos ( m π a x + n π a x ) ] d x = 1 a ∫ [ cos ( ( m − n ) π x a ) − cos ( ( m + n ) π x a ) ] d x \begin{align}
\int\psi_{m}\left(x\right)^{*}\psi_{n}\left(x\right)dx & =\frac{1}{a}\int\left[\cos\left(\frac{m\pi}{a}x-\frac{n\pi}{a}x\right)-\cos\left(\frac{m\pi}{a}x+\frac{n\pi}{a}x\right)\right]dx\\
& =\frac{1}{a}\int\left[\cos\left(\left(m-n\right)\frac{\pi x}{a}\right)-\cos\left(\left(m+n\right)\frac{\pi x}{a}\right)\right]dx
\end{align} ∫ ψ m ( x ) ∗ ψ n ( x ) d x = a 1 ∫ [ cos ( a mπ x − a nπ x ) − cos ( a mπ x + a nπ x ) ] d x = a 1 ∫ [ cos ( ( m − n ) a π x ) − cos ( ( m + n ) a π x ) ] d x Substituindo u = α ± π x a u=\frac{\alpha_{\pm}\pi x}{a} u = a α ± π x α − = m − n \alpha_{-}=m-n α − = m − n α + = m + n \alpha_{+}=m+n α + = m + n a a a α ± π a a = α ± π \alpha_{\pm}\pi\frac{a}{a}=\alpha_{\pm}\pi α ± π a a = α ± π d u d x = α ± π a \frac{du}{dx}=\frac{\alpha_{\pm}\pi}{a} d x d u = a α ± π a α ± π d u = d x \frac{a}{\alpha_{\pm}\pi}du=dx α ± π a d u = d x
∫ 0 a cos ( α ± π x a ) d x = a α ± π ∫ 0 α ± π cos ( u ) d u = a α ± π [ sin ( u ) ] 0 α ± π = a α ± π [ sin ( α ± π ) − sin ( 0 ) ] = a sin ( α ± π ) α ± π \begin{align}
\int^{a}_{0}\cos\left(\frac{\alpha_{\pm}\pi x}{a}\right)dx & =\frac{a}{\alpha_{\pm}\pi}\int^{\alpha_{\pm}\pi}_{0}\cos\left(u\right)du\\
& =\frac{a}{\alpha_{\pm}\pi}\left[\sin\left(u\right)\right]^{\alpha_{\pm}\pi}_{0}\\
& =\frac{a}{\alpha_{\pm}\pi}\left[\sin\left(\alpha_{\pm}\pi\right)-\sin\left(0\right)\right]\\
& =\frac{a\sin\left(\alpha_{\pm}\pi\right)}{\alpha_{\pm}\pi}
\end{align} ∫ 0 a cos ( a α ± π x ) d x = α ± π a ∫ 0 α ± π cos ( u ) d u = α ± π a [ sin ( u ) ] 0 α ± π = α ± π a [ sin ( α ± π ) − sin ( 0 ) ] = α ± π a sin ( α ± π ) Então:
∫ ψ m ( x ) ∗ ψ n ( x ) d x = 1 a ∫ [ cos ( ( m − n ) π x a ) − cos ( ( m + n ) π x a ) ] d x = 1 a ( a sin ( α − π ) α − π − a sin ( α + π ) α + π ) = 1 π ( sin ( ( m − n ) π ) ( m − n ) − sin ( ( m + n ) π ) ( m + n ) ) \begin{align}
\int\psi_{m}\left(x\right)^{*}\psi_{n}\left(x\right)dx & =\frac{1}{a}\int\left[\cos\left(\left(m-n\right)\frac{\pi x}{a}\right)-\cos\left(\left(m+n\right)\frac{\pi x}{a}\right)\right]dx\\
& =\frac{1}{a}\left(\frac{a\sin\left(\alpha_{-}\pi\right)}{\alpha_{-}\pi}-\frac{a\sin\left(\alpha_{+}\pi\right)}{\alpha_{+}\pi}\right)\\
& =\frac{1}{\pi}\left(\frac{\sin\left(\left(m-n\right)\pi\right)}{\left(m-n\right)}-\frac{\sin\left(\left(m+n\right)\pi\right)}{\left(m+n\right)}\right)
\end{align} ∫ ψ m ( x ) ∗ ψ n ( x ) d x = a 1 ∫ [ cos ( ( m − n ) a π x ) − cos ( ( m + n ) a π x ) ] d x = a 1 ( α − π a sin ( α − π ) − α + π a sin ( α + π ) ) = π 1 ( ( m − n ) sin ( ( m − n ) π ) − ( m + n ) sin ( ( m + n ) π ) ) Como m m m n n n m ± n = l m\pm n=l m ± n = l sin ( l π ) \sin\left(l\pi\right) sin ( l π ) l l l sin ( l π ) = 0 \sin\left(l\pi\right)=0 sin ( l π ) = 0
∫ ψ m ( x ) ∗ ψ n ( x ) d x = 0 \int\psi_{m}\left(x\right)^{*}\psi_{n}\left(x\right)dx=0 ∫ ψ m ( x ) ∗ ψ n ( x ) d x = 0 E a quarta e última característica que queremos destacar é que o conjunto de soluções é dito também completo no sentido de que qualquer outra função f ( x ) f\left(x\right) f ( x )
f ( x ) = ∑ n = 1 ∞ c n ψ n ( x ) = ∑ n = 1 ∞ c n 2 a sin ( n π a x ) = ∑ n = 1 ∞ c n sin ( n π a x ) f\left(x\right)=\sum^{\infty}_{n=1}c_{n}\psi_{n}\left(x\right)=\sum^{\infty}_{n=1}c_{n}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)=\sum^{\infty}_{n=1}c_{n}\sin\left(\frac{n\pi}{a}x\right) f ( x ) = n = 1 ∑ ∞ c n ψ n ( x ) = n = 1 ∑ ∞ c n a 2 sin ( a nπ x ) = n = 1 ∑ ∞ c n sin ( a nπ x ) Não vamos provar aqui, mas esta é basicamente a série de Fourier para f ( x ) f\left(x\right) f ( x ) c n c_{n} c n f ( x ) f\left(x\right) f ( x ) ψ \psi ψ ∗ ^{*} ∗
f ( x ) = ∑ n = 1 ∞ c n ψ n ∫ ψ m ∗ f ( x ) d x = ∫ ψ m ∗ ∑ n = 1 ∞ c n ψ n d x ∫ ψ m ∗ f ( x ) d x = ∑ n = 1 ∞ c n ( ∫ ψ m ∗ ψ n d x ) ∫ ψ m ∗ f ( x ) d x = ∑ n = 1 ∞ c n δ m n ∫ ψ m ∗ f ( x ) d x = c m \begin{align}
f\left(x\right) & =\sum^{\infty}_{n=1}c_{n}\psi_{n}\\
\int\psi^{*}_{m}f\left(x\right)dx & =\int\psi^{*}_{m}\sum^{\infty}_{n=1}c_{n}\psi_{n}dx\\
\int\psi^{*}_{m}f\left(x\right)dx & =\sum^{\infty}_{n=1}c_{n}\left(\int\psi^{*}_{m}\psi_{n}dx\right)\\
\int\psi^{*}_{m}f\left(x\right)dx & =\sum^{\infty}_{n=1}c_{n}\delta_{mn}\\
\int\psi^{*}_{m}f\left(x\right)dx & =c_{m}
\end{align} f ( x ) ∫ ψ m ∗ f ( x ) d x ∫ ψ m ∗ f ( x ) d x ∫ ψ m ∗ f ( x ) d x ∫ ψ m ∗ f ( x ) d x = n = 1 ∑ ∞ c n ψ n = ∫ ψ m ∗ n = 1 ∑ ∞ c n ψ n d x = n = 1 ∑ ∞ c n ( ∫ ψ m ∗ ψ n d x ) = n = 1 ∑ ∞ c n δ mn = c m Ou como são apenas rótulos sem significados , podemos mudar os termos m → n m\rightarrow n m → n
∫ ψ n ∗ f ( x ) d x = c n \int\psi^{*}_{n}f\left(x\right)dx=c_{n} ∫ ψ n ∗ f ( x ) d x = c n Essas propriedades não se restringem ao problema do poço quadrado infinito.
A primeira é verdade sempre que o potencial for uma função simétrica.
A segunda é verdade independente do formato do potencial.
Ortogonalidade é uma propriedade geral para qualquer potencial que estamos trabalhando onde o hamiltoniano é hermitiano.
Completeza , segundo Griffiths vale “para todos potenciais que provavelmente vamos encontrar”, apesar da sua prova ser chata e trabalhosa, por isso ele supõe que a maioria dos físicos assume isso e torce pelo melhor. Uma vez que o Griffhits não se propôs a demonstrar isso, e é uma questão mais matemática que física, vamos seguir o mesmo procedimento por hora e confiar nos matemáticos.
Uma solução então da equação de Schrödinger pode ser dada por:
Ψ n ( t ) = ψ n ( x ) φ n ( t ) = 2 a sin ( n π a x ) e − i E n t / ℏ = 2 a sin ( n π a x ) exp [ − i n 2 π 2 h 2 t 2 m a 2 ℏ ] = 2 a sin ( n π a x ) exp [ − i n 2 π 2 h t 2 m a 2 ] \begin{align}
\Psi_{n}\left(t\right) & =\psi_{n}\left(x\right)\varphi_{n}\left(t\right)\\
& =\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)e^{-iE_{n}t/\hbar}\\
& =\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-\frac{in^{2}\pi^{2}h^{2}t}{2ma^{2}\hbar}\right]\\
& =\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-\frac{in^{2}\pi^{2}ht}{2ma^{2}}\right]
\end{align} Ψ n ( t ) = ψ n ( x ) φ n ( t ) = a 2 sin ( a nπ x ) e − i E n t /ℏ = a 2 sin ( a nπ x ) exp [ − 2 m a 2 ℏ i n 2 π 2 h 2 t ] = a 2 sin ( a nπ x ) exp [ − 2 m a 2 i n 2 π 2 h t ] E a solução mais geral é então:
Ψ ( x . t ) = ∑ n = 1 ∞ c n Ψ n ( x , t ) = ∑ n = 1 ∞ c n ψ n ( x ) φ n ( t ) = ∑ n = 1 ∞ c n 2 a sin ( n π a x ) exp [ − i n 2 π 2 h t 2 m a 2 ] \Psi\left(x.t\right)=\sum^{\infty}_{n=1}c_{n}\Psi_{n}\left(x,t\right)=\sum^{\infty}_{n=1}c_{n}\psi_{n}\left(x\right)\varphi_{n}\left(t\right)=\sum^{\infty}_{n=1}c_{n}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-\frac{in^{2}\pi^{2}ht}{2ma^{2}}\right] Ψ ( x . t ) = n = 1 ∑ ∞ c n Ψ n ( x , t ) = n = 1 ∑ ∞ c n ψ n ( x ) φ n ( t ) = n = 1 ∑ ∞ c n a 2 sin ( a nπ x ) exp [ − 2 m a 2 i n 2 π 2 h t ] Resta então apenas demonstrar que sempre podemos expressar a condição inicial Ψ ( x , 0 ) \Psi\left(x,0\right) Ψ ( x , 0 ) c n c_{n} c n ψ \psi ψ
c n = ∫ ψ n ∗ f ( x ) d x = ∫ 0 a 2 a sin ( n π a x ) Ψ ( x , 0 ) d x c_{n}=\int\psi^{*}_{n}f\left(x\right)dx=\int^{a}_{0}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\Psi\left(x,0\right)dx c n = ∫ ψ n ∗ f ( x ) d x = ∫ 0 a a 2 sin ( a nπ x ) Ψ ( x , 0 ) d x Poderíamos facilmente reescrever tudo de forma que c n c_{n} c n 2 / a \sqrt{2/a} 2/ a a n = 2 / a c n a_{n}=\sqrt{2/a}c_{n} a n = 2/ a c n ψ ′ = a / 2 ψ \psi'=\sqrt{a/2}\psi ψ ′ = a /2 ψ
Exemplo: Uma partícula no potencial quadrado infinito possui a seguinte função de onda inicial
Ψ ( x , 0 ) = A x ( a − x ) \Psi\left(x,0\right)=Ax\left(a-x\right) Ψ ( x , 0 ) = A x ( a − x ) Queremos achar a Ψ ( x , t ) \Psi\left(x,t\right) Ψ ( x , t )
Precisamos começar achando a constante de normalização
∫ 0 a ∣ Ψ ( x , 0 ) ∣ 2 d x = ∫ 0 a A 2 x 2 ( a − x ) 2 d x 1 = A 2 ∫ x 2 ( a 2 − 2 a x + x 2 ) d x = A 2 ( a 2 ∫ 0 a x 2 d x − 2 a ∫ 0 a x 3 d x + ∫ 0 a x 4 d x ) = A 2 ( a 2 x 3 3 − 2 a x 4 4 + x 5 5 ) 0 a = A 2 ( a 2 a 3 3 − 2 a 5 a 4 4 ⋅ 5 + 4 a 5 4 ⋅ 5 ) = A 2 a 5 ( 1 3 + 4 − 10 20 ) = A 2 a 5 ( 20 3 ⋅ 20 − 3 ⋅ 6 20 ⋅ 3 ) 1 = A 2 a 5 2 60 \begin{align}
\int^{a}_{0}\left|\Psi\left(x,0\right)\right|^{2}dx & =\int^{a}_{0}A^{2}x^{2}\left(a-x\right)^{2}dx\\
1 & =A^{2}\int x^{2}\left(a^{2}-2ax+x^{2}\right)dx\\
& =A^{2}\left(a^{2}\int^{a}_{0}x^{2}dx-2a\int^{a}_{0}x^{3}dx+\int^{a}_{0}x^{4}dx\right)\\
& =A^{2}\left(a^{2}\frac{x^{3}}{3}-2a\frac{x^{4}}{4}+\frac{x^{5}}{5}\right)^{a}_{0}\\
& =A^{2}\left(a^{2}\frac{a^{3}}{3}-2a\frac{5a^{4}}{4\cdot5}+\frac{4a^{5}}{4\cdot5}\right)\\
& =A^{2}a^{5}\left(\frac{1}{3}+\frac{4-10}{20}\right)\\
& =A^{2}a^{5}\left(\frac{20}{3\cdot20}-\frac{3\cdot6}{20\cdot3}\right)\\
1 & =A^{2}a^{5}\frac{2}{60}
\end{align} ∫ 0 a ∣ Ψ ( x , 0 ) ∣ 2 d x 1 1 = ∫ 0 a A 2 x 2 ( a − x ) 2 d x = A 2 ∫ x 2 ( a 2 − 2 a x + x 2 ) d x = A 2 ( a 2 ∫ 0 a x 2 d x − 2 a ∫ 0 a x 3 d x + ∫ 0 a x 4 d x ) = A 2 ( a 2 3 x 3 − 2 a 4 x 4 + 5 x 5 ) 0 a = A 2 ( a 2 3 a 3 − 2 a 4 ⋅ 5 5 a 4 + 4 ⋅ 5 4 a 5 ) = A 2 a 5 ( 3 1 + 20 4 − 10 ) = A 2 a 5 ( 3 ⋅ 20 20 − 20 ⋅ 3 3 ⋅ 6 ) = A 2 a 5 60 2 Então:
A 2 = 60 2 a 5 → A = 30 a 5 A^{2}=\frac{60}{2a^{5}}\rightarrow A=\sqrt{\frac{30}{a^{5}}} A 2 = 2 a 5 60 → A = a 5 30 E então o coeficiente da equação é dado por:
c n = ∫ ψ n ∗ f ( x ) d x = ∫ 0 a 2 a sin ( n π a x ) Ψ ( x , 0 ) d x = 30 a 5 ∫ 0 a 2 a sin ( n π a x ) x ( a − x ) d x = 60 a 6 ∫ 0 a sin ( n π a x ) ( x a − x 2 ) d x = 60 a 6 ( a ∫ 0 a x sin ( n π a x ) d x − ∫ 0 a x 2 sin ( n π a x ) d x ) = 60 a 6 ( a ∫ 0 n π u ( a n π ) sin ( u ) ( a n π ) d u − ∫ 0 n π u 2 ( a n π ) 2 sin ( u ) ( a n π ) d u ) = 60 a 3 ( a 3 n 2 π 2 ∫ 0 n π u sin ( u ) d u − a 3 n 3 π 3 ∫ 0 n π u 2 sin ( u ) d u ) = 60 n 2 π 2 ( ∫ 0 n π u sin ( u ) d u − 1 n π ∫ 0 n π u 2 sin ( u ) d u ) \begin{align}
c_{n} & =\int\psi^{*}_{n}f\left(x\right)dx\\
& =\int^{a}_{0}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\Psi\left(x,0\right)dx\\
& =\sqrt{\frac{30}{a^{5}}}\int^{a}_{0}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)x\left(a-x\right)dx\\
& =\sqrt{\frac{60}{a^{6}}}\int^{a}_{0}\sin\left(\frac{n\pi}{a}x\right)\left(xa-x^{2}\right)dx\\
& =\sqrt{\frac{60}{a^{6}}}\left(a\int^{a}_{0}x\sin\left(\frac{n\pi}{a}x\right)dx-\int^{a}_{0}x^{2}\sin\left(\frac{n\pi}{a}x\right)dx\right)\\
& =\sqrt{\frac{60}{a^{6}}}\left(a\int^{n\pi}_{0}u\left(\frac{a}{n\pi}\right)\sin\left(u\right)\left(\frac{a}{n\pi}\right)du-\int^{n\pi}_{0}u^{2}\left(\frac{a}{n\pi}\right)^{2}\sin\left(u\right)\left(\frac{a}{n\pi}\right)du\right)\\
& =\frac{\sqrt{60}}{a^{3}}\left(\frac{a^{3}}{n^{2}\pi^{2}}\int^{n\pi}_{0}u\sin\left(u\right)du-\frac{a^{3}}{n^{3}\pi^{3}}\int^{n\pi}_{0}u^{2}\sin\left(u\right)du\right)\\
& =\frac{\sqrt{60}}{n^{2}\pi^{2}}\left(\int^{n\pi}_{0}u\sin\left(u\right)du-\frac{1}{n\pi}\int^{n\pi}_{0}u^{2}\sin\left(u\right)du\right)
\end{align} c n = ∫ ψ n ∗ f ( x ) d x = ∫ 0 a a 2 sin ( a nπ x ) Ψ ( x , 0 ) d x = a 5 30 ∫ 0 a a 2 sin ( a nπ x ) x ( a − x ) d x = a 6 60 ∫ 0 a sin ( a nπ x ) ( x a − x 2 ) d x = a 6 60 ( a ∫ 0 a x sin ( a nπ x ) d x − ∫ 0 a x 2 sin ( a nπ x ) d x ) = a 6 60 ( a ∫ 0 nπ u ( nπ a ) sin ( u ) ( nπ a ) d u − ∫ 0 nπ u 2 ( nπ a ) 2 sin ( u ) ( nπ a ) d u ) = a 3 60 ( n 2 π 2 a 3 ∫ 0 nπ u sin ( u ) d u − n 3 π 3 a 3 ∫ 0 nπ u 2 sin ( u ) d u ) = n 2 π 2 60 ( ∫ 0 nπ u sin ( u ) d u − nπ 1 ∫ 0 nπ u 2 sin ( u ) d u ) Primeiro fizemos a substituição n π a x = u \frac{n\pi}{a}x=u a nπ x = u d x = a n π d u dx=\frac{a}{n\pi}du d x = nπ a d u x 2 = u 2 ( a / n π ) 2 x^{2}=u^{2}(a/n\pi)^{2} x 2 = u 2 ( a / nπ ) 2 u = a ( n π / a ) = n π u=a(n\pi/a)=n\pi u = a ( nπ / a ) = nπ f ( u ) = u 2 f\left(u\right)=u^{2} f ( u ) = u 2 g ′ ( x ) = sin ( u ) g'\left(x\right)=\sin\left(u\right) g ′ ( x ) = sin ( u )
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ 0 n π u 2 sin ( u ) d u = − u 2 cos ( u ) ∣ 0 n π + 2 ∫ 0 n π cos ( u ) u d u = − n 2 π 2 cos ( n π ) + 2 ∫ 0 n π u cos ( u ) d u \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int^{n\pi}_{0}u^{2}\sin\left(u\right)du & =\left.-u^{2}\cos\left(u\right)\right|^{n\pi}_{0}+2\int^{n\pi}_{0}\cos\left(u\right)udu\\
& =-n^{2}\pi^{2}\cos\left(n\pi\right)+2\int^{n\pi}_{0}u\cos\left(u\right)du
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ 0 nπ u 2 sin ( u ) d u = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = − u 2 cos ( u ) ∣ ∣ 0 nπ + 2 ∫ 0 nπ cos ( u ) u d u = − n 2 π 2 cos ( nπ ) + 2 ∫ 0 nπ u cos ( u ) d u Precisamos aplicar novamente na segunda integral agora f ( u ) = u f\left(u\right)=u f ( u ) = u
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ 0 n π cos ( u ) u d u = u sin ( u ) ∣ 0 n π − ∫ 0 n π sin ( u ) d u = [ cos ( u ) ] 0 n π = [ cos ( n π ) − cos ( 0 ) ] = cos ( n π ) − 1 \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int^{n\pi}_{0}\cos\left(u\right)udu & =\left.u\sin\left(u\right)\right|^{n\pi}_{0}-\int^{n\pi}_{0}\sin\left(u\right)du\\
& =\left[\cos\left(u\right)\right]^{n\pi}_{0}\\
& =\left[\cos\left(n\pi\right)-\cos\left(0\right)\right]\\
& =\cos\left(n\pi\right)-1
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ 0 nπ cos ( u ) u d u = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = u sin ( u ) ∣ 0 nπ − ∫ 0 nπ sin ( u ) d u = [ cos ( u ) ] 0 nπ = [ cos ( nπ ) − cos ( 0 ) ] = cos ( nπ ) − 1 Então a segunda integral é:
∫ 0 n π u 2 sin ( u ) d u = − n 2 π 2 cos ( n π ) + 2 ( cos ( n π ) − 1 ) = − n 2 π 2 cos ( n π ) + 2 cos ( n π ) − 2 = ( − n 2 π 2 + 2 ) cos ( n π ) − 2 \begin{align}
\int^{n\pi}_{0}u^{2}\sin\left(u\right)du= & -n^{2}\pi^{2}\cos\left(n\pi\right)+2\left(\cos\left(n\pi\right)-1\right)\\
& =-n^{2}\pi^{2}\cos\left(n\pi\right)+2\cos\left(n\pi\right)-2\\
& =\left(-n^{2}\pi^{2}+2\right)\cos\left(n\pi\right)-2
\end{align} ∫ 0 nπ u 2 sin ( u ) d u = − n 2 π 2 cos ( nπ ) + 2 ( cos ( nπ ) − 1 ) = − n 2 π 2 cos ( nπ ) + 2 cos ( nπ ) − 2 = ( − n 2 π 2 + 2 ) cos ( nπ ) − 2 E a primeira integral é dada por, onde f ( x ) = u f\left(x\right)=u f ( x ) = u
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ 0 n π sin ( u ) u d u = − u cos ( u ) ∣ 0 n π + ∫ 0 n π cos ( u ) d u = − n π cos ( n π ) + [ sin ( u ) ] 0 n π = − n π cos ( n π ) \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int^{n\pi}_{0}\sin\left(u\right)udu & =\left.-u\cos\left(u\right)\right|^{n\pi}_{0}+\int^{n\pi}_{0}\cos\left(u\right)du\\
& =-n\pi\cos\left(n\pi\right)+\left[\sin\left(u\right)\right]^{n\pi}_{0}\\
& =-n\pi\cos\left(n\pi\right)
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ 0 nπ sin ( u ) u d u = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = − u cos ( u ) ∣ 0 nπ + ∫ 0 nπ cos ( u ) d u = − nπ cos ( nπ ) + [ sin ( u ) ] 0 nπ = − nπ cos ( nπ ) Então o coeficiente pode ser dado por:
c n = 60 n 2 π 2 ( ∫ 0 n π u sin ( u ) d u − 1 n π ∫ 0 n π u 2 sin ( u ) d u ) = 60 n 2 π 2 ( − n π cos ( n π ) − 1 n π [ ( − n 2 π 2 + 2 ) cos ( n π ) − 2 ] ) = 60 n 2 π 2 ( − n π cos ( n π ) + ( n 2 π 2 − 2 ) n π cos ( n π ) + 2 n π ) = 60 n 2 π 2 ( ( − n 2 π 2 + n 2 π 2 − 2 ) n π cos ( n π ) + 2 n π ) = 60 n 2 π 2 ( − 2 n π cos ( n π ) + 2 n π ) = 2 60 n 3 π 3 ( 1 − cos ( n π ) ) = 2 60 n 3 π 3 ( 1 − ( − 1 ) n ) \begin{align}
c_{n} & =\frac{\sqrt{60}}{n^{2}\pi^{2}}\left(\int^{n\pi}_{0}u\sin\left(u\right)du-\frac{1}{n\pi}\int^{n\pi}_{0}u^{2}\sin\left(u\right)du\right)\\
& =\frac{\sqrt{60}}{n^{2}\pi^{2}}\left(-n\pi\cos\left(n\pi\right)-\frac{1}{n\pi}\left[\left(-n^{2}\pi^{2}+2\right)\cos\left(n\pi\right)-2\right]\right)\\
& =\frac{\sqrt{60}}{n^{2}\pi^{2}}\left(-n\pi\cos\left(n\pi\right)+\frac{\left(n^{2}\pi^{2}-2\right)}{n\pi}\cos\left(n\pi\right)+\frac{2}{n\pi}\right)\\
& =\frac{\sqrt{60}}{n^{2}\pi^{2}}\left(\frac{\left(-n^{2}\pi^{2}+n^{2}\pi^{2}-2\right)}{n\pi}\cos\left(n\pi\right)+\frac{2}{n\pi}\right)\\
& =\frac{\sqrt{60}}{n^{2}\pi^{2}}\left(\frac{-2}{n\pi}\cos\left(n\pi\right)+\frac{2}{n\pi}\right)\\
& =\frac{2\sqrt{60}}{n^{3}\pi^{3}}\left(1-\cos\left(n\pi\right)\right)\\
& =\frac{2\sqrt{60}}{n^{3}\pi^{3}}\left(1-\left(-1\right)^{n}\right)
\end{align} c n = n 2 π 2 60 ( ∫ 0 nπ u sin ( u ) d u − nπ 1 ∫ 0 nπ u 2 sin ( u ) d u ) = n 2 π 2 60 ( − nπ cos ( nπ ) − nπ 1 [ ( − n 2 π 2 + 2 ) cos ( nπ ) − 2 ] ) = n 2 π 2 60 ( − nπ cos ( nπ ) + nπ ( n 2 π 2 − 2 ) cos ( nπ ) + nπ 2 ) = n 2 π 2 60 ( nπ ( − n 2 π 2 + n 2 π 2 − 2 ) cos ( nπ ) + nπ 2 ) = n 2 π 2 60 ( nπ − 2 cos ( nπ ) + nπ 2 ) = n 3 π 3 2 60 ( 1 − cos ( nπ ) ) = n 3 π 3 2 60 ( 1 − ( − 1 ) n ) Então se n n n cos ( n π ) = − 1 \cos\left(n\pi\right)=-1 cos ( nπ ) = − 1 n n n cos ( n π ) = 1 \cos\left(n\pi\right)=1 cos ( nπ ) = 1 cos ( n π ) = ( − 1 ) n \cos\left(n\pi\right)=\left(-1\right)^{n} cos ( nπ ) = ( − 1 ) n
c n = { 0 ,se n par 4 60 n 3 π 3 ,se n ı ˊ mpar c_{n}=\begin{cases}
0 & \text{,se n par}\\
\frac{4\sqrt{60}}{n^{3}\pi^{3}} & \text{,se n ímpar}
\end{cases} c n = { 0 n 3 π 3 4 60 ,se n par ,se n ı ˊ mpar Então a solução é:
Ψ ( x . t ) = ∑ n = 1 ∞ c n Ψ n ( x , t ) = ∑ n = 1 ∞ c n 2 a sin ( n π a x ) exp [ − i n 2 π 2 h t 2 m a 2 ] = ∑ n = 1 , 3 , 5 ∞ 4 60 n 3 π 3 2 a sin ( n π a x ) exp [ − i n 2 π 2 h t 2 m a 2 ] = 4 π 3 120 a ∑ n = 1 , 3 , 5 ∞ 1 n 3 sin ( n π a x ) exp [ − i n 2 π 2 h t 2 m a 2 ] \begin{align}
\Psi\left(x.t\right) & =\sum^{\infty}_{n=1}c_{n}\Psi_{n}\left(x,t\right)\\
& =\sum^{\infty}_{n=1}c_{n}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-\frac{in^{2}\pi^{2}ht}{2ma^{2}}\right]\\
& =\sum^{\infty}_{n=1,3,5}\frac{4\sqrt{60}}{n^{3}\pi^{3}}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-\frac{in^{2}\pi^{2}ht}{2ma^{2}}\right]\\
& =\frac{4}{\pi^{3}}\sqrt{\frac{120}{a}}\sum^{\infty}_{n=1,3,5}\frac{1}{n^{3}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-\frac{in^{2}\pi^{2}ht}{2ma^{2}}\right]
\end{align} Ψ ( x . t ) = n = 1 ∑ ∞ c n Ψ n ( x , t ) = n = 1 ∑ ∞ c n a 2 sin ( a nπ x ) exp [ − 2 m a 2 i n 2 π 2 h t ] = n = 1 , 3 , 5 ∑ ∞ n 3 π 3 4 60 a 2 sin ( a nπ x ) exp [ − 2 m a 2 i n 2 π 2 h t ] = π 3 4 a 120 n = 1 , 3 , 5 ∑ ∞ n 3 1 sin ( a nπ x ) exp [ − 2 m a 2 i n 2 π 2 h t ] Ou se 120 = 4 ⋅ 30 = 2 2 ⋅ 30 120=4\cdot30=2^{2}\cdot30 120 = 4 ⋅ 30 = 2 2 ⋅ 30 120 = 2 2 ⋅ 30 = 2 2 \sqrt{120}=\sqrt{2^{2}\cdot30}=\sqrt{2^{2}} 120 = 2 2 ⋅ 30 = 2 2 30 = 2 30 \sqrt{30}=2\sqrt{30} 30 = 2 30 4 120 = 8 30 4\sqrt{120}=8\sqrt{30} 4 120 = 8 30 8 = 2 3 8=2^{3} 8 = 2 3
Ψ ( x , t ) = 30 a ( 2 π ) 3 ∑ n = 1 , 3 , 5 ∞ 1 n 3 sin ( n π a x ) exp [ − i n 2 π 2 h t 2 m a 2 ] \Psi\left(x,t\right)=\sqrt{\frac{30}{a}}\left(\frac{2}{\pi}\right)^{3}\sum^{\infty}_{n=1,3,5}\frac{1}{n^{3}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-\frac{in^{2}\pi^{2}ht}{2ma^{2}}\right] Ψ ( x , t ) = a 30 ( π 2 ) 3 n = 1 , 3 , 5 ∑ ∞ n 3 1 sin ( a nπ x ) exp [ − 2 m a 2 i n 2 π 2 h t ] Exemplo: Mostre que ∑ n = 1 ∞ ∣ c n ∣ 2 = 1 \sum^{\infty}_{n=1}\left|c_{n}\right|^{2}=1 ∑ n = 1 ∞ ∣ c n ∣ 2 = 1
Recebemos a sugestão de utilizar as seguintes séries:
1 1 6 + 1 3 6 + 1 5 6 + ⋯ = π 6 960 1 1 4 + 1 3 4 + 1 5 4 + ⋯ = π 4 96 \frac{1}{1^{6}}+\frac{1}{3^{6}}+\frac{1}{5^{6}}+\dots=\frac{\pi^{6}}{960}\qquad\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\dots=\frac{\pi^{4}}{96} 1 6 1 + 3 6 1 + 5 6 1 + ⋯ = 960 π 6 1 4 1 + 3 4 1 + 5 4 1 + ⋯ = 96 π 4 Queremos demonstrar que ∑ n = 1 ∞ ∣ c n ∣ 2 = 1 \sum^{\infty}_{n=1}\left|c_{n}\right|^{2}=1 ∑ n = 1 ∞ ∣ c n ∣ 2 = 1
∑ n = 1 ∞ ∣ c n ∣ 2 = ∑ n = 1 , 3 , 5 … ∞ ( 4 60 n 3 π 3 ) 2 = ( 4 60 π 3 ) 2 ∑ n = 1 , 3 , 5 … ∞ 1 n 6 = 16 ⋅ 60 π 6 ∑ n = 1 , 3 , 5 … ∞ 1 n 6 = 960 π 6 ∑ n = 1 , 3 , 5 … ∞ 1 n 6 \begin{align}
\sum^{\infty}_{n=1}\left|c_{n}\right|^{2} & =\sum^{\infty}_{n=1,3,5\dots}\left(\frac{4\sqrt{60}}{n^{3}\pi^{3}}\right)^{2}=\left(\frac{4\sqrt{60}}{\pi^{3}}\right)^{2}\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{6}}=\frac{16\cdot60}{\pi^{6}}\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{6}}=\frac{960}{\pi^{6}}\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{6}}
\end{align} n = 1 ∑ ∞ ∣ c n ∣ 2 = n = 1 , 3 , 5 … ∑ ∞ ( n 3 π 3 4 60 ) 2 = ( π 3 4 60 ) 2 n = 1 , 3 , 5 … ∑ ∞ n 6 1 = π 6 16 ⋅ 60 n = 1 , 3 , 5 … ∑ ∞ n 6 1 = π 6 960 n = 1 , 3 , 5 … ∑ ∞ n 6 1 Então usando a série que nos foi fornecida:
∑ n = 1 ∞ ∣ c n ∣ 2 = 960 π 6 ( π 6 960 ) = 1 \begin{align}
\sum^{\infty}_{n=1}\left|c_{n}\right|^{2} & =\frac{960}{\pi^{6}}\left(\frac{\pi^{6}}{960}\right)=1
\end{align} n = 1 ∑ ∞ ∣ c n ∣ 2 = π 6 960 ( 960 π 6 ) = 1 E o valor esperado da anergia , sendo E n = ℏ 2 n 2 π 2 2 m E_{n}=\frac{\hbar^{2}n^{2}\pi^{2}}{2m} E n = 2 m ℏ 2 n 2 π 2
⟨ H ⟩ = ∑ n = 1 ∞ E n ∣ c n ∣ 2 = ∑ n = 1 , 3 , 5 … ∞ ℏ 2 n 2 π 2 2 m ( 4 60 n 3 π 3 ) 2 = ℏ 2 π 2 2 m 960 π 6 ∑ n = 1 , 3 , 5 … ∞ n 2 n 6 \left\langle H\right\rangle =\sum^{\infty}_{n=1}E_{n}\left|c_{n}\right|^{2}=\sum^{\infty}_{n=1,3,5\dots}\frac{\hbar^{2}n^{2}\pi^{2}}{2m}\left(\frac{4\sqrt{60}}{n^{3}\pi^{3}}\right)^{2}=\frac{\hbar^{2}\pi^{2}}{2m}\frac{960}{\pi^{6}}\sum^{\infty}_{n=1,3,5\dots}\frac{n^{2}}{n^{6}} ⟨ H ⟩ = n = 1 ∑ ∞ E n ∣ c n ∣ 2 = n = 1 , 3 , 5 … ∑ ∞ 2 m ℏ 2 n 2 π 2 ( n 3 π 3 4 60 ) 2 = 2 m ℏ 2 π 2 π 6 960 n = 1 , 3 , 5 … ∑ ∞ n 6 n 2 Usando a série que nos foi fornecida:
⟨ H ⟩ = ℏ 2 π 2 2 m ( 10 π 2 ) ( 96 π 4 ) ( π 4 96 ) = ℏ 2 π 2 2 m ( 10 π 2 ) \left\langle H\right\rangle =\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{10}{\pi^{2}}\right)\left(\frac{96}{\pi^{4}}\right)\left(\frac{\pi^{4}}{96}\right)=\frac{\hbar^{2}\pi^{2}}{2m}\left(\frac{10}{\pi^{2}}\right) ⟨ H ⟩ = 2 m ℏ 2 π 2 ( π 2 10 ) ( π 4 96 ) ( 96 π 4 ) = 2 m ℏ 2 π 2 ( π 2 10 ) Isso é muito próximo da energia para n = 1 n=1 n = 1 E 1 = ℏ 2 π 2 2 m E_{1}=\frac{\hbar^{2}\pi^{2}}{2m} E 1 = 2 m ℏ 2 π 2
⟨ H ⟩ = 10 π 2 E 1 = 1 , 01 E 1 \left\langle H\right\rangle =\frac{10}{\pi^{2}}E_{1}=1,01E_{1} ⟨ H ⟩ = π 2 10 E 1 = 1 , 01 E 1 Isso também pode ser visto fazendo:
∣ c 1 ∣ 2 = ( 4 60 π 3 ) 2 = 960 π 6 = 0.998 \left|c_{1}\right|^{2}=\left(\frac{4\sqrt{60}}{\pi^{3}}\right)^{2}=\frac{960}{\pi^{6}}=0.998 ∣ c 1 ∣ 2 = ( π 3 4 60 ) 2 = π 6 960 = 0.998 Ou seja, isso indica que o primeiro termo ’domina’ nosso somatório de soluções, já que todos os outros termos farão a diferença para garantir a normalização. Digamos que c n c_{n} c n ψ n \psi_{n} ψ n Ψ \Psi Ψ 99.8 % 99.8\% 99.8% E 1 E_{1} E 1
Podemos verificar a normalização do somatório de ∣ c n ∣ 2 \left|c_{n}\right|^{2} ∣ c n ∣ 2 t = 0 t=0 t = 0 t t t
1 = ∫ ∣ Ψ ( x , 0 ) ∣ 2 d x = ∫ Ψ ∗ ( x , 0 ) Ψ ( x , 0 ) d x = ∫ ( ∑ n = 1 ∞ c n ψ n ) ∗ ( ∑ m = 1 ∞ c m ψ m ) d x = ∑ n = 1 ∞ ∑ m = 1 ∞ ∫ c n ∗ ψ n ∗ c m ψ m d x = ∑ n = 1 ∞ ∑ m = 1 ∞ c n ∗ c m ∫ ψ n ∗ ψ m d x = ∑ n , m = 1 ∞ c n ∗ c m δ n m = ∑ n = 1 ∞ c n ∗ c n \begin{align}
1 & =\int\left|\Psi\left(x,0\right)\right|^{2}dx\\
& =\int\Psi^{*}\left(x,0\right)\Psi\left(x,0\right)dx\\
& =\int\left(\sum^{\infty}_{n=1}c_{n}\psi_{n}\right)^{*}\left(\sum^{\infty}_{m=1}c_{m}\psi_{m}\right)dx\\
& =\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}\int c^{*}_{n}\psi^{*}_{n}c_{m}\psi_{m}dx\\
& =\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}c^{*}_{n}c_{m}\int\psi^{*}_{n}\psi_{m}dx\\
& =\sum^{\infty}_{n,m=1}c^{*}_{n}c_{m}\delta_{nm}\\
& =\sum^{\infty}_{n=1}c^{*}_{n}c_{n}
\end{align} 1 = ∫ ∣ Ψ ( x , 0 ) ∣ 2 d x = ∫ Ψ ∗ ( x , 0 ) Ψ ( x , 0 ) d x = ∫ ( n = 1 ∑ ∞ c n ψ n ) ∗ ( m = 1 ∑ ∞ c m ψ m ) d x = n = 1 ∑ ∞ m = 1 ∑ ∞ ∫ c n ∗ ψ n ∗ c m ψ m d x = n = 1 ∑ ∞ m = 1 ∑ ∞ c n ∗ c m ∫ ψ n ∗ ψ m d x = n , m = 1 ∑ ∞ c n ∗ c m δ nm = n = 1 ∑ ∞ c n ∗ c n Então:
∑ n = 1 ∞ ∣ c n ∣ 2 = 1 \sum^{\infty}_{n=1}\left|c_{n}\right|^{2}=1 n = 1 ∑ ∞ ∣ c n ∣ 2 = 1 De modo análogo podemos checar o valor esperado de energia, utilizando equação de Schrödinger independente do tempo H ^ ψ n = E n ψ n \widehat{H}\psi_{n}=E_{n}\psi_{n} H ψ n = E n ψ n
⟨ H ⟩ = ∫ Ψ H ^ Ψ d x = ∫ ( ∑ n = 1 ∞ c n ψ n ) ∗ H ^ ( ∑ m = 1 ∞ c m ψ m ) d x = ∑ n , m = 1 ∞ c n ∗ c m ∫ ψ n ∗ ( H ^ ψ m ) d x = ∑ n , m = 1 ∞ c n ∗ c m ∫ ψ n ∗ ( E m ψ m ) d x = ∑ n , m = 1 ∞ c n ∗ c m E m ∫ ψ n ∗ ψ m d x = ∑ n , m = 1 ∞ c n ∗ c m E m δ n m \begin{align}
\left\langle H\right\rangle & =\int\Psi\widehat{H}\Psi dx\\
& =\int\left(\sum^{\infty}_{n=1}c_{n}\psi_{n}\right)^{*}\widehat{H}\left(\sum^{\infty}_{m=1}c_{m}\psi_{m}\right)dx\\
& =\sum^{\infty}_{n,m=1}c^{*}_{n}c_{m}\int\psi^{*}_{n}\left(\widehat{H}\psi_{m}\right)dx\\
& =\sum^{\infty}_{n,m=1}c^{*}_{n}c_{m}\int\psi^{*}_{n}\left(E_{m}\psi_{m}\right)dx\\
& =\sum^{\infty}_{n,m=1}c^{*}_{n}c_{m}E_{m}\int\psi^{*}_{n}\psi_{m}dx\\
& =\sum^{\infty}_{n,m=1}c^{*}_{n}c_{m}E_{m}\delta_{nm}
\end{align} ⟨ H ⟩ = ∫ Ψ H Ψ d x = ∫ ( n = 1 ∑ ∞ c n ψ n ) ∗ H ( m = 1 ∑ ∞ c m ψ m ) d x = n , m = 1 ∑ ∞ c n ∗ c m ∫ ψ n ∗ ( H ψ m ) d x = n , m = 1 ∑ ∞ c n ∗ c m ∫ ψ n ∗ ( E m ψ m ) d x = n , m = 1 ∑ ∞ c n ∗ c m E m ∫ ψ n ∗ ψ m d x = n , m = 1 ∑ ∞ c n ∗ c m E m δ nm Então:
⟨ H ⟩ = ∑ n ∞ ∣ c n ∣ 2 E n \left\langle H\right\rangle =\sum^{\infty}_{n}\left|c_{n}\right|^{2}E_{n} ⟨ H ⟩ = n ∑ ∞ ∣ c n ∣ 2 E n Exemplo: Calcule ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩ ⟨ x 2 ⟩ \left\langle x^{2}\right\rangle ⟨ x 2 ⟩ ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩ ⟨ p ⟩ 2 \left\langle p\right\rangle ^{2} ⟨ p ⟩ 2 σ x \sigma_{x} σ x σ p \sigma_{p} σ p
Primeiro vamos recuperar a solução já conhecida no interior do poço que é 17
ψ n ( x ) = 2 a sin ( n π a x ) \psi_{n}\left(x\right)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right) ψ n ( x ) = a 2 sin ( a nπ x ) Então:
⟨ x ⟩ = ∫ 0 a x ∣ ψ n ∣ 2 d x = 2 a ∫ 0 a x sin 2 ( n π a x ) d x = 2 a ∫ 0 a ( a n π ) u sin 2 ( u ) ( a n π ) d u = 2 a n 2 π 2 ∫ 0 n π u sin 2 u d u \begin{align}
\left\langle x\right\rangle & =\int^{a}_{0}x\left|\psi_{n}\right|^{2}dx\\
& =\frac{2}{a}\int^{a}_{0}x\sin^{2}\left(\frac{n\pi}{a}x\right)dx\\
& =\frac{2}{a}\int^{a}_{0}\left(\frac{a}{n\pi}\right)u\sin^{2}\left(u\right)\left(\frac{a}{n\pi}\right)du\\
& =\frac{2a}{n^{2}\pi^{2}}\int^{n\pi}_{0}u\sin^{2}udu
\end{align} ⟨ x ⟩ = ∫ 0 a x ∣ ψ n ∣ 2 d x = a 2 ∫ 0 a x sin 2 ( a nπ x ) d x = a 2 ∫ 0 a ( nπ a ) u sin 2 ( u ) ( nπ a ) d u = n 2 π 2 2 a ∫ 0 nπ u sin 2 u d u Onde substituímos u = n π a x u=\frac{n\pi}{a}x u = a nπ x x = a n π u x=\frac{a}{n\pi}u x = nπ a u d x = a n π d u dx=\frac{a}{n\pi}du d x = nπ a d u x = 0 → u = 0 x=0\rightarrow u=0 x = 0 → u = 0 x = a → u = n π x=a\rightarrow u=n\pi x = a → u = nπ f ( u ) = u f\left(u\right)=u f ( u ) = u sin 2 ( x ) \sin^{2}\left(x\right) sin 2 ( x ) 2 sin 2 ( x ) = 1 − cos ( 2 x ) 2\sin^{2}\left(x\right)=1-\cos\left(2x\right) 2 sin 2 ( x ) = 1 − cos ( 2 x ) 12 ∫ sin 2 x d x = x / 2 − sin ( 2 x ) / 4 + C \int\sin^{2}xdx=x/2-\sin\left(2x\right)/4+C ∫ sin 2 x d x = x /2 − sin ( 2 x ) /4 + C
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ 0 n π u sin 2 u d u = u ( u 2 − sin ( 2 u ) 4 ) ∣ 0 n π − ∫ 0 n π ( u 2 − sin ( 2 u ) 4 ) d u = ( n 2 π 2 2 − n π sin ( 2 n π ) 4 ) − 1 2 ∫ 0 n π u d u + 1 4 ∫ 0 n π sin ( 2 u ) d u = n 2 π 2 2 − 1 4 [ u 2 ] 0 n π + 1 4 ∫ 0 2 n π sin ( v ) d v = 2 n 2 π 2 4 − n 2 π 2 4 − 1 4 [ cos v ] 0 2 n π = n 2 π 2 4 − 1 4 [ cos 2 π − cos 0 ] = n 2 π 2 4 − 1 4 ( 1 − 1 ) = n 2 π 2 4 \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int^{n\pi}_{0}u\sin^{2}udu & =\left.u\left(\frac{u}{2}-\frac{\sin\left(2u\right)}{4}\right)\right|^{n\pi}_{0}-\int^{n\pi}_{0}\left(\frac{u}{2}-\frac{\sin\left(2u\right)}{4}\right)du\\
& =\left(\frac{n^{2}\pi^{2}}{2}-\frac{n\pi\sin\left(2n\pi\right)}{4}\right)-\frac{1}{2}\int^{n\pi}_{0}udu+\frac{1}{4}\int^{n\pi}_{0}\sin\left(2u\right)du\\
& =\frac{n^{2}\pi^{2}}{2}-\frac{1}{4}\left[u^{2}\right]^{n\pi}_{0}+\frac{1}{4}\int^{2n\pi}_{0}\sin\left(v\right)dv\\
& =\frac{2n^{2}\pi^{2}}{4}-\frac{n^{2}\pi^{2}}{4}-\frac{1}{4}\left[\cos v\right]^{2n\pi}_{0}\\
& =\frac{n^{2}\pi^{2}}{4}-\frac{1}{4}\left[\cos2\pi-\cos0\right]\\
& =\frac{n^{2}\pi^{2}}{4}-\frac{1}{4}\left(1-1\right)\\
& =\frac{n^{2}\pi^{2}}{4}
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ 0 nπ u sin 2 u d u = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = u ( 2 u − 4 sin ( 2 u ) ) ∣ ∣ 0 nπ − ∫ 0 nπ ( 2 u − 4 sin ( 2 u ) ) d u = ( 2 n 2 π 2 − 4 nπ sin ( 2 nπ ) ) − 2 1 ∫ 0 nπ u d u + 4 1 ∫ 0 nπ sin ( 2 u ) d u = 2 n 2 π 2 − 4 1 [ u 2 ] 0 nπ + 4 1 ∫ 0 2 nπ sin ( v ) d v = 4 2 n 2 π 2 − 4 n 2 π 2 − 4 1 [ cos v ] 0 2 nπ = 4 n 2 π 2 − 4 1 [ cos 2 π − cos 0 ] = 4 n 2 π 2 − 4 1 ( 1 − 1 ) = 4 n 2 π 2 Onde na integral foi substituído v = 2 u v=2u v = 2 u d v 2 = d u \frac{dv}{2}=du 2 d v = d u 2 n π 2n\pi 2 nπ
⟨ x ⟩ = 2 a n 2 π 2 ∫ 0 n π u sin 2 u d u = 2 a n 2 π 2 ( n 2 π 2 4 ) = a 2 \begin{align}
\left\langle x\right\rangle & =\frac{2a}{n^{2}\pi^{2}}\int^{n\pi}_{0}u\sin^{2}udu=\frac{2a}{n^{2}\pi^{2}}\left(\frac{n^{2}\pi^{2}}{4}\right)=\frac{a}{2}
\end{align} ⟨ x ⟩ = n 2 π 2 2 a ∫ 0 nπ u sin 2 u d u = n 2 π 2 2 a ( 4 n 2 π 2 ) = 2 a Agora para ⟨ x 2 ⟩ \left\langle x^{2}\right\rangle ⟨ x 2 ⟩
⟨ x 2 ⟩ = ∫ 0 a x 2 sin 2 ( n π a x ) d x = 2 a ∫ 0 a ( a n π ) 2 u sin 2 ( u ) ( a n π ) d u = 2 a 2 n 3 π 3 ∫ 0 n π u 2 sin 2 u d u \begin{align}
\left\langle x^{2}\right\rangle & =\int^{a}_{0}x^{2}\sin^{2}\left(\frac{n\pi}{a}x\right)dx=\frac{2}{a}\int^{a}_{0}\left(\frac{a}{n\pi}\right)^{2}u\sin^{2}\left(u\right)\left(\frac{a}{n\pi}\right)du=\frac{2a^{2}}{n^{3}\pi^{3}}\int^{n\pi}_{0}u^{2}\sin^{2}udu
\end{align} ⟨ x 2 ⟩ = ∫ 0 a x 2 sin 2 ( a nπ x ) d x = a 2 ∫ 0 a ( nπ a ) 2 u sin 2 ( u ) ( nπ a ) d u = n 3 π 3 2 a 2 ∫ 0 nπ u 2 sin 2 u d u Onde fizemos a mesma substituição u = n π a x u=\frac{n\pi}{a}x u = a nπ x f ( u ) = u 2 : f\left(u\right)=u^{2}: f ( u ) = u 2 :
∫ a b f ( x ) g ′ ( x ) d x = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x ∫ 0 n π u 2 sin 2 u d u = u 2 ( u 2 − sin ( 2 u ) 4 ) ∣ 0 n π − ∫ 0 n π 2 u ( u 2 − sin ( 2 u ) 4 ) d u = n 3 π 3 2 − n 2 π 2 sin ( 2 n π ) 4 − ∫ 0 n π u 2 d u + 1 2 ∫ 0 n π u sin ( 2 u ) d u = n 3 π 3 2 − [ u 3 3 ] 0 n π + 1 2 ∫ 0 2 n π v 2 sin ( v ) d v 2 = 3 n 3 π 3 6 − 2 n 3 π 3 6 + 1 8 ∫ 0 2 n π v sin ( v ) d v = n 3 π 3 6 + 1 8 ∫ 0 2 n π v sin ( v ) d v \begin{align}
\int^{b}_{a}f\left(x\right)g'\left(x\right)dx & =\left.f\left(x\right)g\left(x\right)\right|^{b}_{a}-\int^{b}_{a}g\left(x\right)f'\left(x\right)dx\\
\int^{n\pi}_{0}u^{2}\sin^{2}udu & =\left.u^{2}\left(\frac{u}{2}-\frac{\sin\left(2u\right)}{4}\right)\right|^{n\pi}_{0}-\int^{n\pi}_{0}2u\left(\frac{u}{2}-\frac{\sin\left(2u\right)}{4}\right)du\\
& =\frac{n^{3}\pi^{3}}{2}-\frac{n^{2}\pi^{2}\sin\left(2n\pi\right)}{4}-\int^{n\pi}_{0}u^{2}du+\frac{1}{2}\int^{n\pi}_{0}u\sin\left(2u\right)du\\
& =\frac{n^{3}\pi^{3}}{2}-\left[\frac{u^{3}}{3}\right]^{n\pi}_{0}+\frac{1}{2}\int^{2n\pi}_{0}\frac{v}{2}\sin\left(v\right)\frac{dv}{2}\\
& =\frac{3n^{3}\pi^{3}}{6}-\frac{2n^{3}\pi^{3}}{6}+\frac{1}{8}\int^{2n\pi}_{0}v\sin\left(v\right)dv\\
& =\frac{n^{3}\pi^{3}}{6}+\frac{1}{8}\int^{2n\pi}_{0}v\sin\left(v\right)dv
\end{align} ∫ a b f ( x ) g ′ ( x ) d x ∫ 0 nπ u 2 sin 2 u d u = f ( x ) g ( x ) ∣ a b − ∫ a b g ( x ) f ′ ( x ) d x = u 2 ( 2 u − 4 sin ( 2 u ) ) ∣ ∣ 0 nπ − ∫ 0 nπ 2 u ( 2 u − 4 sin ( 2 u ) ) d u = 2 n 3 π 3 − 4 n 2 π 2 sin ( 2 nπ ) − ∫ 0 nπ u 2 d u + 2 1 ∫ 0 nπ u sin ( 2 u ) d u = 2 n 3 π 3 − [ 3 u 3 ] 0 nπ + 2 1 ∫ 0 2 nπ 2 v sin ( v ) 2 d v = 6 3 n 3 π 3 − 6 2 n 3 π 3 + 8 1 ∫ 0 2 nπ v sin ( v ) d v = 6 n 3 π 3 + 8 1 ∫ 0 2 nπ v sin ( v ) d v Onde fizemos v = 2 u v=2u v = 2 u d u = d v / 2 du=dv/2 d u = d v /2 2 n π 2n\pi 2 nπ ∫ 0 a u sin ( u ) d u = − a cos ( a ) \int^{a}_{0}u\sin\left(u\right)du=-a\cos\left(a\right) ∫ 0 a u sin ( u ) d u = − a cos ( a ) 44
∫ 0 n π u 2 sin 2 u d u = n 3 π 3 6 + 1 8 ( − 2 n π cos ( 2 n π ) ) = n 3 π 3 6 − n π 4 \begin{align}
\int^{n\pi}_{0}u^{2}\sin^{2}udu & =\frac{n^{3}\pi^{3}}{6}+\frac{1}{8}\left(-2n\pi\cos\left(2n\pi\right)\right)\\
& =\frac{n^{3}\pi^{3}}{6}-\frac{n\pi}{4}
\end{align} ∫ 0 nπ u 2 sin 2 u d u = 6 n 3 π 3 + 8 1 ( − 2 nπ cos ( 2 nπ ) ) = 6 n 3 π 3 − 4 nπ Então:
⟨ x 2 ⟩ = 2 a 2 n 3 π 3 ∫ 0 n π u 2 sin 2 u d u = 2 a 2 n 3 π 3 ( n 3 π 3 6 − n π 4 ) = a 2 ( 1 3 − 1 2 n 2 π 2 ) \begin{align}
\left\langle x^{2}\right\rangle & =\frac{2a^{2}}{n^{3}\pi^{3}}\int^{n\pi}_{0}u^{2}\sin^{2}udu=\frac{2a^{2}}{n^{3}\pi^{3}}\left(\frac{n^{3}\pi^{3}}{6}-\frac{n\pi}{4}\right)=a^{2}\left(\frac{1}{3}-\frac{1}{2n^{2}\pi^{2}}\right)
\end{align} ⟨ x 2 ⟩ = n 3 π 3 2 a 2 ∫ 0 nπ u 2 sin 2 u d u = n 3 π 3 2 a 2 ( 6 n 3 π 3 − 4 nπ ) = a 2 ( 3 1 − 2 n 2 π 2 1 ) A forma mais fácil de calcularmos ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩
⟨ p ⟩ = m d ⟨ x ⟩ d t = m d d t ( a 2 ) = 0 \left\langle p\right\rangle =m\frac{d\left\langle x\right\rangle }{dt}=m\frac{d}{dt}\left(\frac{a}{2}\right)=0 ⟨ p ⟩ = m d t d ⟨ x ⟩ = m d t d ( 2 a ) = 0 E por fim então:
⟨ p 2 ⟩ = ∫ ψ n ∗ ( ℏ i d d x ) 2 ψ n d x = − ℏ 2 ∫ ψ n ∗ d 2 ψ n d x 2 d x \begin{align}
\left\langle p^{2}\right\rangle & =\int\psi^{*}_{n}\left(\frac{\hbar}{i}\frac{d}{dx}\right)^{2}\psi_{n}dx=-\hbar^{2}\int\psi^{*}_{n}\frac{d^{2}\psi_{n}}{dx^{2}}dx
\end{align} ⟨ p 2 ⟩ = ∫ ψ n ∗ ( i ℏ d x d ) 2 ψ n d x = − ℏ 2 ∫ ψ n ∗ d x 2 d 2 ψ n d x E sendo a equação de Schrödinger neste caso dado pela combinação das equações 2 10
∂ 2 ψ n ∂ x 2 = − k n 2 ψ n = − ( n π a ) 2 ψ n \frac{\partial^{2}\psi_{n}}{\partial x^{2}}=-k^{2}_{n}\psi_{n}=-\left(\frac{n\pi}{a}\right)^{2}\psi_{n} ∂ x 2 ∂ 2 ψ n = − k n 2 ψ n = − ( a nπ ) 2 ψ n Então:
⟨ p 2 ⟩ = − ℏ 2 ∫ ψ n ∗ d 2 ψ n d x 2 d x = − ℏ 2 ∫ ψ n ∗ ( − ( n π a ) 2 ) ψ n d x = ( ℏ n π a ) 2 ∫ ψ n ∗ ψ n d x = ( ℏ n π a ) 2 \begin{align}
\left\langle p^{2}\right\rangle & =-\hbar^{2}\int\psi^{*}_{n}\frac{d^{2}\psi_{n}}{dx^{2}}dx=-\hbar^{2}\int\psi^{*}_{n}\left(-\left(\frac{n\pi}{a}\right)^{2}\right)\psi_{n}dx=\left(\frac{\hbar n\pi}{a}\right)^{2}\int\psi^{*}_{n}\psi_{n}dx=\left(\frac{\hbar n\pi}{a}\right)^{2}
\end{align} ⟨ p 2 ⟩ = − ℏ 2 ∫ ψ n ∗ d x 2 d 2 ψ n d x = − ℏ 2 ∫ ψ n ∗ ( − ( a nπ ) 2 ) ψ n d x = ( a ℏ nπ ) 2 ∫ ψ n ∗ ψ n d x = ( a ℏ nπ ) 2 Então o desvio padrão da posição é dado por:
σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = a 2 ( 1 3 − 1 2 n 2 π 2 ) − ( a 2 ) 2 = a 4 12 − 2 4 n 2 π 2 − 3 12 = a 1 4 ⋅ 3 − 2 4 n 2 π 2 \begin{align}
\sigma_{x} & =\sqrt{\left\langle x^{2}\right\rangle -\left\langle x\right\rangle ^{2}}\\
& =\sqrt{a^{2}\left(\frac{1}{3}-\frac{1}{2n^{2}\pi^{2}}\right)-\left(\frac{a}{2}\right)^{2}}\\
& =a\sqrt{\frac{4}{12}-\frac{2}{4n^{2}\pi^{2}}-\frac{3}{12}}\\
& =a\sqrt{\frac{1}{4\cdot3}-\frac{2}{4n^{2}\pi^{2}}}
\end{align} σ x = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 = a 2 ( 3 1 − 2 n 2 π 2 1 ) − ( 2 a ) 2 = a 12 4 − 4 n 2 π 2 2 − 12 3 = a 4 ⋅ 3 1 − 4 n 2 π 2 2 Ou seja:
σ x = a 2 1 3 − 2 n 2 π 2 \sigma_{x}=\frac{a}{2}\sqrt{\frac{1}{3}-\frac{2}{n^{2}\pi^{2}}} σ x = 2 a 3 1 − n 2 π 2 2 De modo análogo temos para σ p \sigma_{p} σ p
σ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ( ℏ n π a ) 2 − ( 0 ) 2 = ( ℏ n π a ) 2 = ℏ n π a \begin{align}
\sigma_{p} & =\sqrt{\left\langle p^{2}\right\rangle -\left\langle p\right\rangle ^{2}}=\sqrt{\left(\frac{\hbar n\pi}{a}\right)^{2}-\left(0\right)^{2}}=\sqrt{\left(\frac{\hbar n\pi}{a}\right)^{2}}=\frac{\hbar n\pi}{a}
\end{align} σ p = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ( a ℏ nπ ) 2 − ( 0 ) 2 = ( a ℏ nπ ) 2 = a ℏ nπ A relação de incerteza é então:
σ x σ p = a 2 1 3 − 2 n 2 π 2 ℏ n π a = ℏ n π 2 1 3 − 2 n 2 π 2 = ℏ 2 ( n π ) 2 ( 1 3 − 2 n 2 π 2 ) \sigma_{x}\sigma_{p}=\frac{a}{2}\sqrt{\frac{1}{3}-\frac{2}{n^{2}\pi^{2}}}\frac{\hbar n\pi}{a}=\frac{\hbar n\pi}{2}\sqrt{\frac{1}{3}-\frac{2}{n^{2}\pi^{2}}}=\frac{\hbar}{2}\sqrt{\left(n\pi\right)^{2}\left(\frac{1}{3}-\frac{2}{n^{2}\pi^{2}}\right)} σ x σ p = 2 a 3 1 − n 2 π 2 2 a ℏ nπ = 2 ℏ nπ 3 1 − n 2 π 2 2 = 2 ℏ ( nπ ) 2 ( 3 1 − n 2 π 2 2 ) Então:
σ x σ p = ℏ 2 n 2 π 2 3 − 2 \sigma_{x}\sigma_{p}=\frac{\hbar}{2}\sqrt{\frac{n^{2}\pi^{2}}{3}-2} σ x σ p = 2 ℏ 3 n 2 π 2 − 2 E podemos ver que quanto maior n n n n = 1 n=1 n = 1
σ x σ p = ℏ 2 π 2 3 − 2 \sigma_{x}\sigma_{p}=\frac{\hbar}{2}\sqrt{\frac{\pi^{2}}{3}-2} σ x σ p = 2 ℏ 3 π 2 − 2 E como temos π 2 3 − 2 = 1.13 \sqrt{\frac{\pi^{2}}{3}-2}=1.13 3 π 2 − 2 = 1.13 σ x σ p = 1.14 ℏ 2 \sigma_{x}\sigma_{p}=1.14\frac{\hbar}{2} σ x σ p = 1.14 2 ℏ σ x σ p ≥ ℏ 2 \sigma_{x}\sigma_{p}\geq\frac{\hbar}{2} σ x σ p ≥ 2 ℏ
Exemplo: Uma partícula em um poço quadrado infinito tem a seguinte função de onda inicial:
ψ ( x , 0 ) = A [ ψ 1 ( x ) + ψ 2 ( x ) ] \psi\left(x,0\right)=A\left[\psi_{1}\left(x\right)+\psi_{2}\left(x\right)\right] ψ ( x , 0 ) = A [ ψ 1 ( x ) + ψ 2 ( x ) ] a) Normalize ψ ( x , 0 ) \psi\left(x,0\right) ψ ( x , 0 )
b) Encontre ψ \psi ψ ( x , t ) \left(x,t\right) ( x , t ) ∣ ψ ( x , t ) ∣ 2 \left|\psi\left(x,t\right)\right|^{2} ∣ ψ ( x , t ) ∣ 2 ω ≡ π 2 h / 2 m a 2 \omega\equiv\pi^{2}h/2ma^{2} ω ≡ π 2 h /2 m a 2
c) Calcule ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩
d) Calcular ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩
e) Se você medir a energia dessa partícula, quais valores você espera obter e com qual probabilidade? Encontre o valor esperado de H H H E 1 E_{1} E 1 E 2 E_{2} E 2
Pela notação podemos assumir que ψ n ( x ) \psi_{n}\left(x\right) ψ n ( x ) Ψ ( x , t ) \Psi\left(x,t\right) Ψ ( x , t ) Ψ ( x , 0 ) \Psi\left(x,0\right) Ψ ( x , 0 )
∣ Ψ ∣ 2 = ( A [ ψ 1 ( x ) + ψ 2 ( x ) ] ) ∗ ( A [ ψ 1 ( x ) + ψ 2 ( x ) ] ) = A ∗ [ ψ 1 ∗ + ψ 2 ∗ ] A [ ψ 1 + ψ 2 ] = A ∗ A ( ψ 1 ∗ ψ 1 + ψ 1 ∗ ψ 2 + ψ 2 ∗ ψ 1 + ψ 2 ∗ ψ 2 ) = ∣ A ∣ 2 ( ∣ ψ 1 ∣ 2 + ∣ ψ ∣ 2 2 + ψ 1 ∗ ψ 2 + ψ 2 ∗ ψ 1 ) \begin{align}
\left|\Psi\right|^{2} & =\left(A\left[\psi_{1}\left(x\right)+\psi_{2}\left(x\right)\right]\right)^{*}\left(A\left[\psi_{1}\left(x\right)+\psi_{2}\left(x\right)\right]\right)\\
& =A^{*}\left[\psi^{*}_{1}+\psi^{*}_{2}\right]A\left[\psi_{1}+\psi_{2}\right]\\
& =A^{*}A\left(\psi^{*}_{1}\psi_{1}+\psi^{*}_{1}\psi_{2}+\psi^{*}_{2}\psi_{1}+\psi^{*}_{2}\psi_{2}\right)\\
& =\left|A\right|^{2}\left(\left|\psi_{1}\right|^{2}+\left|\psi\right|^{2}_{2}+\psi^{*}_{1}\psi_{2}+\psi^{*}_{2}\psi_{1}\right)
\end{align} ∣ Ψ ∣ 2 = ( A [ ψ 1 ( x ) + ψ 2 ( x ) ] ) ∗ ( A [ ψ 1 ( x ) + ψ 2 ( x ) ] ) = A ∗ [ ψ 1 ∗ + ψ 2 ∗ ] A [ ψ 1 + ψ 2 ] = A ∗ A ( ψ 1 ∗ ψ 1 + ψ 1 ∗ ψ 2 + ψ 2 ∗ ψ 1 + ψ 2 ∗ ψ 2 ) = ∣ A ∣ 2 ( ∣ ψ 1 ∣ 2 + ∣ ψ ∣ 2 2 + ψ 1 ∗ ψ 2 + ψ 2 ∗ ψ 1 ) Integrando então, pela condição de normalização:
1 = ∫ ∣ Ψ ∣ 2 d x = ∣ A ∣ 2 ( ∫ ∣ ψ 1 ∣ 2 d x + ∫ ∣ ψ ∣ 2 2 d x + ∫ ψ 1 ∗ ψ 2 d x + ∫ ψ 2 ∗ ψ 1 d x ) = ∣ A ∣ 2 ( 1 + 1 ) = 2 ∣ A ∣ 2 1=\int\left|\Psi\right|^{2}dx=\left|A\right|^{2}\left(\int\left|\psi_{1}\right|^{2}dx+\int\left|\psi\right|^{2}_{2}dx+\int\psi^{*}_{1}\psi_{2}dx+\int\psi^{*}_{2}\psi_{1}dx\right)=\left|A\right|^{2}\left(1+1\right)=2\left|A\right|^{2} 1 = ∫ ∣ Ψ ∣ 2 d x = ∣ A ∣ 2 ( ∫ ∣ ψ 1 ∣ 2 d x + ∫ ∣ ψ ∣ 2 2 d x + ∫ ψ 1 ∗ ψ 2 d x + ∫ ψ 2 ∗ ψ 1 d x ) = ∣ A ∣ 2 ( 1 + 1 ) = 2 ∣ A ∣ 2 Então:
A = 1 2 A=\frac{1}{\sqrt{2}} A = 2 1 Queremos agora calcular Ψ ( x , t ) \Psi\left(x,t\right) Ψ ( x , t )
Ψ ( x , 0 ) = ∑ n = 1 ∞ c n ψ n ( x ) \Psi\left(x,0\right)=\sum^{\infty}_{n=1}c_{n}\psi_{n}\left(x\right) Ψ ( x , 0 ) = n = 1 ∑ ∞ c n ψ n ( x ) temos claramente c 1 = c 2 = 1 / 2 c_{1}=c_{2}=1/\sqrt{2} c 1 = c 2 = 1/ 2 c n = 0 c_{n}=0 c n = 0 n > 2 n>2 n > 2
Ψ ( x , t ) = ∑ n = 1 ∞ c n Ψ n ( x , t ) = ∑ n = 1 ∞ c n ψ n ( x ) e − i E n t / ℏ \Psi\left(x,t\right)=\sum^{\infty}_{n=1}c_{n}\Psi_{n}\left(x,t\right)=\sum^{\infty}_{n=1}c_{n}\psi_{n}\left(x\right)e^{-iE_{n}t/\hbar} Ψ ( x , t ) = n = 1 ∑ ∞ c n Ψ n ( x , t ) = n = 1 ∑ ∞ c n ψ n ( x ) e − i E n t /ℏ Uma vez que c n c_{n} c n ψ n ( x ) \psi_{n}\left(x\right) ψ n ( x ) ψ n \psi_{n} ψ n n > 2 n>2 n > 2
Ψ ( x , t ) = c 1 ψ 1 ( x ) e − i E 1 t / ℏ + c 2 ψ 2 ( x ) e − i E 2 t / ℏ \Psi\left(x,t\right)=c_{1}\psi_{1}\left(x\right)e^{-iE_{1}t/\hbar}+c_{2}\psi_{2}\left(x\right)e^{-iE_{2}t/\hbar} Ψ ( x , t ) = c 1 ψ 1 ( x ) e − i E 1 t /ℏ + c 2 ψ 2 ( x ) e − i E 2 t /ℏ Usando os valores de c 1 c_{1} c 1 ω ≡ π 2 ℏ / 2 m a 2 \omega\equiv\pi^{2}\hbar/2ma^{2} ω ≡ π 2 ℏ/2 m a 2
E n = n 2 π 2 ℏ 2 2 m a 2 = ω ℏ n 2 E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}=\omega\hbar n^{2} E n = 2 m a 2 n 2 π 2 ℏ 2 = ω ℏ n 2 Então:
Ψ ( x , t ) = 1 2 ψ 1 ( x ) e − i ω ℏ 1 2 t / ℏ + 1 2 ψ 2 ( x ) e − i ω ℏ 2 2 t / ℏ = 1 2 ( ψ 1 ( x ) e − i ω t + ψ 2 ( x ) e − i 4 ω t ) = e − i ω t 2 ( ψ 1 ( x ) + ψ 2 ( x ) e − i 3 ω t ) \begin{align}
\Psi\left(x,t\right) & =\frac{1}{\sqrt{2}}\psi_{1}\left(x\right)e^{-i\omega\hbar1^{2}t/\hbar}+\frac{1}{\sqrt{2}}\psi_{2}\left(x\right)e^{-i\omega\hbar2^{2}t/\hbar}\\
& =\frac{1}{\sqrt{2}}\left(\psi_{1}\left(x\right)e^{-i\omega t}+\psi_{2}\left(x\right)e^{-i4\omega t}\right)\\
& =\frac{e^{-i\omega t}}{\sqrt{2}}\left(\psi_{1}\left(x\right)+\psi_{2}\left(x\right)e^{-i3\omega t}\right)
\end{align} Ψ ( x , t ) = 2 1 ψ 1 ( x ) e − iω ℏ 1 2 t /ℏ + 2 1 ψ 2 ( x ) e − iω ℏ 2 2 t /ℏ = 2 1 ( ψ 1 ( x ) e − iω t + ψ 2 ( x ) e − i 4 ω t ) = 2 e − iω t ( ψ 1 ( x ) + ψ 2 ( x ) e − i 3 ω t ) Também já calculamos que ψ n ( x ) = 2 a sin ( n π x a ) \psi_{n}\left(x\right)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right) ψ n ( x ) = a 2 sin ( a nπ x )
Ψ ( x , t ) = e − i ω t 2 ( 2 a sin ( π x a ) + 2 a sin ( 2 π x a ) e − i 3 ω t ) = e − i ω t a ( sin ( π x a ) + sin ( 2 π x a ) e − i 3 ω t ) \begin{align}
\Psi\left(x,t\right) & =\frac{e^{-i\omega t}}{\sqrt{2}}\left(\sqrt{\frac{2}{a}}\sin\left(\frac{\pi x}{a}\right)+\sqrt{\frac{2}{a}}\sin\left(\frac{2\pi x}{a}\right)e^{-i3\omega t}\right)\\
& =\frac{e^{-i\omega t}}{\sqrt{a}}\left(\sin\left(\frac{\pi x}{a}\right)+\sin\left(\frac{2\pi x}{a}\right)e^{-i3\omega t}\right)
\end{align} Ψ ( x , t ) = 2 e − iω t ( a 2 sin ( a π x ) + a 2 sin ( a 2 π x ) e − i 3 ω t ) = a e − iω t ( sin ( a π x ) + sin ( a 2 π x ) e − i 3 ω t ) Então:
∣ Ψ ∣ 2 = [ e − i ω t a ( sin ( π x a ) + sin ( 2 π x a ) e − i 3 ω t ) ] ∗ [ e − i ω t a ( sin ( π x a ) + sin ( 2 π x a ) e − i 3 ω t ) ] = e + i ω t a ( sin ( π x a ) + sin ( 2 π x a ) e + i 3 ω t ) e − i ω t a ( sin ( π x a ) + sin ( 2 π x a ) e − i 3 ω t ) = 1 a ( sin 2 ( π x a ) + sin ( π x a ) sin ( 2 π x a ) ( e − i 3 ω t + e + i 3 ω t ) + sin 2 ( 2 π x a ) ) \begin{align}
\left|\Psi\right|^{2} & =\left[\frac{e^{-i\omega t}}{\sqrt{a}}\left(\sin\left(\frac{\pi x}{a}\right)+\sin\left(\frac{2\pi x}{a}\right)e^{-i3\omega t}\right)\right]^{*}\left[\frac{e^{-i\omega t}}{\sqrt{a}}\left(\sin\left(\frac{\pi x}{a}\right)+\sin\left(\frac{2\pi x}{a}\right)e^{-i3\omega t}\right)\right]\\
& =\frac{e^{+i\omega t}}{\sqrt{a}}\left(\sin\left(\frac{\pi x}{a}\right)+\sin\left(\frac{2\pi x}{a}\right)e^{+i3\omega t}\right)\frac{e^{-i\omega t}}{\sqrt{a}}\left(\sin\left(\frac{\pi x}{a}\right)+\sin\left(\frac{2\pi x}{a}\right)e^{-i3\omega t}\right)\\
& =\frac{1}{a}\left(\sin^{2}\left(\frac{\pi x}{a}\right)+\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)\left(e^{-i3\omega t}+e^{+i3\omega t}\right)+\sin^{2}\left(\frac{2\pi x}{a}\right)\right)
\end{align} ∣ Ψ ∣ 2 = [ a e − iω t ( sin ( a π x ) + sin ( a 2 π x ) e − i 3 ω t ) ] ∗ [ a e − iω t ( sin ( a π x ) + sin ( a 2 π x ) e − i 3 ω t ) ] = a e + iω t ( sin ( a π x ) + sin ( a 2 π x ) e + i 3 ω t ) a e − iω t ( sin ( a π x ) + sin ( a 2 π x ) e − i 3 ω t ) = a 1 ( sin 2 ( a π x ) + sin ( a π x ) sin ( a 2 π x ) ( e − i 3 ω t + e + i 3 ω t ) + sin 2 ( a 2 π x ) ) E sendo cos ( x ) = e i x + e i x 2 \cos\left(x\right)=\frac{e^{ix}+e^{ix}}{2} cos ( x ) = 2 e i x + e i x
∣ Ψ ∣ 2 = 1 a ( sin 2 ( π x a ) + 2 sin ( π x a ) sin ( 2 π x a ) cos ( 3 ω t ) + sin 2 ( 2 π x a ) ) \begin{align}
\left|\Psi\right|^{2} & =\frac{1}{a}\left(\sin^{2}\left(\frac{\pi x}{a}\right)+2\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)\cos\left(3\omega t\right)+\sin^{2}\left(\frac{2\pi x}{a}\right)\right)
\end{align} ∣ Ψ ∣ 2 = a 1 ( sin 2 ( a π x ) + 2 sin ( a π x ) sin ( a 2 π x ) cos ( 3 ω t ) + sin 2 ( a 2 π x ) ) Queremos então computar ⟨ x ⟩ \left\langle x\right\rangle ⟨ x ⟩
⟨ x ⟩ = ∫ x ∣ Ψ ∣ 2 d x = 1 a ( ∫ x sin 2 ( π x a ) d x + ∫ 2 sin ( π x a ) sin ( 2 π x a ) cos ( 3 ω t ) d x + ∫ x sin 2 ( 2 π x a ) d x ) = 1 a ( ∫ x sin 2 ( π x a ) d x + 2 cos ( 3 ω t ) ∫ x sin ( π x a ) sin ( 2 π x a ) d x + ∫ x sin 2 ( 2 π x a ) d x ) \begin{align}
\left\langle x\right\rangle & =\int x\left|\Psi\right|^{2}dx\\
& =\frac{1}{a}\left(\int x\sin^{2}\left(\frac{\pi x}{a}\right)dx+\int2\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)\cos\left(3\omega t\right)dx+\int x\sin^{2}\left(\frac{2\pi x}{a}\right)dx\right)\\
& =\frac{1}{a}\left(\int x\sin^{2}\left(\frac{\pi x}{a}\right)dx+2\cos\left(3\omega t\right)\int x\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)dx+\int x\sin^{2}\left(\frac{2\pi x}{a}\right)dx\right)
\end{align} ⟨ x ⟩ = ∫ x ∣ Ψ ∣ 2 d x = a 1 ( ∫ x sin 2 ( a π x ) d x + ∫ 2 sin ( a π x ) sin ( a 2 π x ) cos ( 3 ω t ) d x + ∫ x sin 2 ( a 2 π x ) d x ) = a 1 ( ∫ x sin 2 ( a π x ) d x + 2 cos ( 3 ω t ) ∫ x sin ( a π x ) sin ( a 2 π x ) d x + ∫ x sin 2 ( a 2 π x ) d x ) Onde lembramos que a integral é na posição, logo cos ( 3 ω t ) \cos\left(3\omega t\right) cos ( 3 ω t ) u = π x a u=\frac{\pi x}{a} u = a π x x = 0 → u = 0 x=0\rightarrow u=0 x = 0 → u = 0 x = a → u = π x=a\rightarrow u=\pi x = a → u = π
∫ 0 π u ( a π ) sin 2 ( u ) ( a π ) d x = a 2 π 2 ∫ 0 π u sin 2 ( u ) d x = a 2 π 2 π 2 4 = ( a 2 ) 2 \int^{\pi}_{0}u\left(\frac{a}{\pi}\right)\sin^{2}\left(u\right)\left(\frac{a}{\pi}\right)dx=\frac{a^{2}}{\pi^{2}}\int^{\pi}_{0}u\sin^{2}\left(u\right)dx=\frac{a^{2}}{\pi^{2}}\frac{\pi^{2}}{4}=\left(\frac{a}{2}\right)^{2} ∫ 0 π u ( π a ) sin 2 ( u ) ( π a ) d x = π 2 a 2 ∫ 0 π u sin 2 ( u ) d x = π 2 a 2 4 π 2 = ( 2 a ) 2 Essa é uma integral que já tínhamos calculado 62 u = 2 π x a u=\frac{2\pi x}{a} u = a 2 π x 2 π 2\pi 2 π
∫ 0 2 π u ( a 2 π ) sin 2 ( u ) ( a 2 π ) d u = a 2 4 π 2 ∫ 0 2 π u sin 2 ( u ) d u = a 2 4 π 2 ( 2 π ) 2 4 = ( a 2 ) 2 \int^{2\pi}_{0}u\left(\frac{a}{2\pi}\right)\sin^{2}\left(u\right)\left(\frac{a}{2\pi}\right)du=\frac{a^{2}}{4\pi^{2}}\int^{2\pi}_{0}u\sin^{2}\left(u\right)du=\frac{a^{2}}{4\pi^{2}}\frac{\left(2\pi\right)^{2}}{4}=\left(\frac{a}{2}\right)^{2} ∫ 0 2 π u ( 2 π a ) sin 2 ( u ) ( 2 π a ) d u = 4 π 2 a 2 ∫ 0 2 π u sin 2 ( u ) d u = 4 π 2 a 2 4 ( 2 π ) 2 = ( 2 a ) 2 Onde obtemos a mesma integral e o mesmo resultado. Agora para a integral do meio, vamos utilizar a seguinte propriedade trigonométrica:
cos ( A ± B ) = cos A cos B ∓ sin A sin B \begin{align}
\cos\left(A\pm B\right) & =\cos A\cos B\mp\sin A\sin B
\end{align} cos ( A ± B ) = cos A cos B ∓ sin A sin B Então:
cos ( A + B ) − cos ( A − B ) = [ ( cos A cos B − sin A sin B ) + … ⋯ − ( cos A cos B + sin A sin B ) ] cos ( A + B ) − cos ( A − B ) = − 2 sin A sin B ( cos ( A − B ) − cos ( A + B ) ) 1 2 = sin A sin B ( cos ( π x a − 2 π x a ) − cos ( π x a + 2 π x a ) ) 1 2 = sin ( π x a ) sin ( 2 π x a ) ( cos ( − π x a ) − cos ( 4 π x a ) ) 1 2 = sin ( π x a ) sin ( 2 π x a ) ( cos ( π x a ) − cos ( 3 π x a ) ) 1 2 = sin ( π x a ) sin ( 2 π x a ) \begin{align}
\cos\left(A+B\right)-\cos\left(A-B\right) & =\left[\left(\cos A\cos B-\sin A\sin B\right)+\dots\right.\\
& \left.\dots-\left(\cos A\cos B+\sin A\sin B\right)\right]\\
\cos\left(A+B\right)-\cos\left(A-B\right) & =-2\sin A\sin B\\
\left(\cos\left(A-B\right)-\cos\left(A+B\right)\right)\frac{1}{2} & =\sin A\sin B\\
\left(\cos\left(\frac{\pi x}{a}-\frac{2\pi x}{a}\right)-\cos\left(\frac{\pi x}{a}+\frac{2\pi x}{a}\right)\right)\frac{1}{2} & =\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)\\
\left(\cos\left(-\frac{\pi x}{a}\right)-\cos\left(\frac{4\pi x}{a}\right)\right)\frac{1}{2} & =\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)\\
\left(\cos\left(\frac{\pi x}{a}\right)-\cos\left(\frac{3\pi x}{a}\right)\right)\frac{1}{2} & =\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)
\end{align} cos ( A + B ) − cos ( A − B ) cos ( A + B ) − cos ( A − B ) ( cos ( A − B ) − cos ( A + B ) ) 2 1 ( cos ( a π x − a 2 π x ) − cos ( a π x + a 2 π x ) ) 2 1 ( cos ( − a π x ) − cos ( a 4 π x ) ) 2 1 ( cos ( a π x ) − cos ( a 3 π x ) ) 2 1 = [ ( cos A cos B − sin A sin B ) + … ⋯ − ( cos A cos B + sin A sin B ) ] = − 2 sin A sin B = sin A sin B = sin ( a π x ) sin ( a 2 π x ) = sin ( a π x ) sin ( a 2 π x ) = sin ( a π x ) sin ( a 2 π x ) Então a segunda integral se torna:
∫ x sin ( π x a ) sin ( 2 π x a ) d x = 1 2 ( ∫ x cos ( π x a ) d x − ∫ x cos ( 3 π x a ) d x ) 2 ∫ x sin ( π x a ) sin ( 2 π x a ) d = ∫ 0 π u ( a π ) cos ( u ) ( a π ) d x − ∫ 0 3 π u ( a 3 π ) cos ( u ) ( a 3 π ) d u = a 2 π 2 ∫ 0 π u cos ( u ) d x − a 2 3 2 π 2 ∫ 0 3 π u cos ( u ) d u \begin{align}
\int x\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)dx & =\frac{1}{2}\left(\int x\cos\left(\frac{\pi x}{a}\right)dx-\int x\cos\left(\frac{3\pi x}{a}\right)dx\right)\\
2\int x\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)d & =\int^{\pi}_{0}u\left(\frac{a}{\pi}\right)\cos\left(u\right)\left(\frac{a}{\pi}\right)dx-\int^{3\pi}_{0}u\left(\frac{a}{3\pi}\right)\cos\left(u\right)\left(\frac{a}{3\pi}\right)du\\
& =\frac{a^{2}}{\pi^{2}}\int^{\pi}_{0}u\cos\left(u\right)dx-\frac{a^{2}}{3^{2}\pi^{2}}\int^{3\pi}_{0}u\cos\left(u\right)du
\end{align} ∫ x sin ( a π x ) sin ( a 2 π x ) d x 2 ∫ x sin ( a π x ) sin ( a 2 π x ) d = 2 1 ( ∫ x cos ( a π x ) d x − ∫ x cos ( a 3 π x ) d x ) = ∫ 0 π u ( π a ) cos ( u ) ( π a ) d x − ∫ 0 3 π u ( 3 π a ) cos ( u ) ( 3 π a ) d u = π 2 a 2 ∫ 0 π u cos ( u ) d x − 3 2 π 2 a 2 ∫ 0 3 π u cos ( u ) d u Onde fizemos a substituição u = π x a u=\frac{\pi x}{a} u = a π x π \pi π u = 4 π x a u=\frac{4\pi x}{a} u = a 4 π x 4 π 4\pi 4 π 42
2 ∫ x sin ( π x a ) sin ( 2 π x a ) d x = a 2 π 2 ∫ 0 π u cos ( u ) d x − a 2 3 2 π 2 ∫ 0 3 π u cos ( u ) d u = a 2 π 2 ( cos ( π ) − 1 ) − a 2 3 2 π 2 ( cos ( 3 π ) − 1 ) = a 2 π 2 ( − 1 − 1 ) − a 2 3 2 π 2 ( − 1 − 1 ) = − 2 [ 9 a 2 9 π 2 − a 2 9 π 2 ] = − 2 8 a 2 9 π 2 = − 16 a 2 9 π 2 \begin{align}
2\int x\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)dx & =\frac{a^{2}}{\pi^{2}}\int^{\pi}_{0}u\cos\left(u\right)dx-\frac{a^{2}}{3^{2}\pi^{2}}\int^{3\pi}_{0}u\cos\left(u\right)du\\
& =\frac{a^{2}}{\pi^{2}}\left(\cos\left(\pi\right)-1\right)-\frac{a^{2}}{3^{2}\pi^{2}}\left(\cos\left(3\pi\right)-1\right)\\
& =\frac{a^{2}}{\pi^{2}}\left(-1-1\right)-\frac{a^{2}}{3^{2}\pi^{2}}\left(-1-1\right)\\
& =-2\left[\frac{9a^{2}}{9\pi^{2}}-\frac{a^{2}}{9\pi^{2}}\right]\\
& =-2\frac{8a^{2}}{9\pi^{2}}\\
& =-\frac{16a^{2}}{9\pi^{2}}
\end{align} 2 ∫ x sin ( a π x ) sin ( a 2 π x ) d x = π 2 a 2 ∫ 0 π u cos ( u ) d x − 3 2 π 2 a 2 ∫ 0 3 π u cos ( u ) d u = π 2 a 2 ( cos ( π ) − 1 ) − 3 2 π 2 a 2 ( cos ( 3 π ) − 1 ) = π 2 a 2 ( − 1 − 1 ) − 3 2 π 2 a 2 ( − 1 − 1 ) = − 2 [ 9 π 2 9 a 2 − 9 π 2 a 2 ] = − 2 9 π 2 8 a 2 = − 9 π 2 16 a 2 Então:
⟨ x ⟩ = 1 a ( ∫ x sin 2 ( π x a ) d x + cos ( 3 ω t ) [ 2 ∫ x sin ( π x a ) sin ( 2 π x a ) d x ] + ∫ x sin 2 ( 2 π x a ) d x ) = 1 a ( ( a 2 ) 2 + cos ( 3 ω t ) ( − 16 a 2 9 π 2 ) + ( a 2 ) 2 ) = a 2 a ( 1 4 − cos ( 3 ω t ) 16 9 π 2 + 1 4 ) = a ( 2 4 − cos ( 3 ω t ) 16 9 π 2 ) = a ( 1 2 − 2 2 cos ( 3 ω t ) 16 9 π 2 ) = a 2 ( 1 − 32 9 π 2 cos ( 3 ω t ) ) \begin{align}
\left\langle x\right\rangle & =\frac{1}{a}\left(\int x\sin^{2}\left(\frac{\pi x}{a}\right)dx+\cos\left(3\omega t\right)\left[2\int x\sin\left(\frac{\pi x}{a}\right)\sin\left(\frac{2\pi x}{a}\right)dx\right]+\int x\sin^{2}\left(\frac{2\pi x}{a}\right)dx\right)\\
& =\frac{1}{a}\left(\left(\frac{a}{2}\right)^{2}+\cos\left(3\omega t\right)\left(-\frac{16a^{2}}{9\pi^{2}}\right)+\left(\frac{a}{2}\right)^{2}\right)\\
& =\frac{a^{2}}{a}\left(\frac{1}{4}-\cos\left(3\omega t\right)\frac{16}{9\pi^{2}}+\frac{1}{4}\right)\\
& =a\left(\frac{2}{4}-\cos\left(3\omega t\right)\frac{16}{9\pi^{2}}\right)\\
& =a\left(\frac{1}{2}-\frac{2}{2}\cos\left(3\omega t\right)\frac{16}{9\pi^{2}}\right)\\
& =\frac{a}{2}\left(1-\frac{32}{9\pi^{2}}\cos\left(3\omega t\right)\right)
\end{align} ⟨ x ⟩ = a 1 ( ∫ x sin 2 ( a π x ) d x + cos ( 3 ω t ) [ 2 ∫ x sin ( a π x ) sin ( a 2 π x ) d x ] + ∫ x sin 2 ( a 2 π x ) d x ) = a 1 ( ( 2 a ) 2 + cos ( 3 ω t ) ( − 9 π 2 16 a 2 ) + ( 2 a ) 2 ) = a a 2 ( 4 1 − cos ( 3 ω t ) 9 π 2 16 + 4 1 ) = a ( 4 2 − cos ( 3 ω t ) 9 π 2 16 ) = a ( 2 1 − 2 2 cos ( 3 ω t ) 9 π 2 16 ) = 2 a ( 1 − 9 π 2 32 cos ( 3 ω t ) ) Sobre a amplitude podemos notar que como calculamos anteriormente, a posição da partícula oscila em torno de a 2 \frac{a}{2} 2 a a 2 \frac{a}{2} 2 a 32 9 π 2 a 2 = 0.36 a 2 < a 2 \frac{32}{9\pi^{2}}\frac{a}{2}=0.36\frac{a}{2}<\frac{a}{2} 9 π 2 32 2 a = 0.36 2 a < 2 a 3 ω = 3 π 2 ℏ 2 m a 2 3\omega=\frac{3\pi^{2}\hbar}{2ma^{2}} 3 ω = 2 m a 2 3 π 2 ℏ ⟨ p ⟩ \left\langle p\right\rangle ⟨ p ⟩
⟨ p ⟩ = m d ⟨ x ⟩ d t = m d d t ( a 2 [ 1 − 32 9 π 2 cos ( 3 ω t ) ] ) = m a 3 ⋅ 16 9 π 2 ω sin ( 3 ω t ) = m a 16 3 π 2 ω sin ( 3 ω t ) \left\langle p\right\rangle =m\frac{d\left\langle x\right\rangle }{dt}=m\frac{d}{dt}\left(\frac{a}{2}\left[1-\frac{32}{9\pi^{2}}\cos\left(3\omega t\right)\right]\right)=ma\frac{3\cdot16}{9\pi^{2}}\omega\sin\left(3\omega t\right)=ma\frac{16}{3\pi^{2}}\omega\sin\left(3\omega t\right) ⟨ p ⟩ = m d t d ⟨ x ⟩ = m d t d ( 2 a [ 1 − 9 π 2 32 cos ( 3 ω t ) ] ) = ma 9 π 2 3 ⋅ 16 ω sin ( 3 ω t ) = ma 3 π 2 16 ω sin ( 3 ω t ) E comoω = π 2 ℏ 2 m a 2 \omega=\frac{\pi^{2}\hbar}{2ma^{2}} ω = 2 m a 2 π 2 ℏ
⟨ p ⟩ = m a 16 3 π 2 π 2 ℏ 2 m a 2 sin ( 3 ω t ) = 8 ℏ 3 a sin ( 3 ω t ) \left\langle p\right\rangle =ma\frac{16}{3\pi^{2}}\frac{\pi^{2}\hbar}{2ma^{2}}\sin\left(3\omega t\right)=\frac{8\hbar}{3a}\sin\left(3\omega t\right) ⟨ p ⟩ = ma 3 π 2 16 2 m a 2 π 2 ℏ sin ( 3 ω t ) = 3 a 8ℏ sin ( 3 ω t ) Relembrando nossa função de onda:
Ψ ( x , t ) = 1 2 ψ 1 ( x ) e − i ω ℏ 1 2 t / ℏ + 1 2 ψ 2 ( x ) e − i ω ℏ 2 2 t / ℏ = 1 2 Ψ 1 ( x , t ) + 1 2 Ψ 2 ( x , t ) \begin{align}
\Psi\left(x,t\right) & =\frac{1}{\sqrt{2}}\psi_{1}\left(x\right)e^{-i\omega\hbar1^{2}t/\hbar}+\frac{1}{\sqrt{2}}\psi_{2}\left(x\right)e^{-i\omega\hbar2^{2}t/\hbar}=\frac{1}{\sqrt{2}}\Psi_{1}\left(x,t\right)+\frac{1}{\sqrt{2}}\Psi_{2}\left(x,t\right)
\end{align} Ψ ( x , t ) = 2 1 ψ 1 ( x ) e − iω ℏ 1 2 t /ℏ + 2 1 ψ 2 ( x ) e − iω ℏ 2 2 t /ℏ = 2 1 Ψ 1 ( x , t ) + 2 1 Ψ 2 ( x , t ) Evidentemente podemos obter então E 1 E_{1} E 1 E 2 E_{2} E 2 ∣ c 1 ∣ 2 = ∣ c 2 ∣ 2 = ( 1 2 ) 2 = 1 2 \left|c_{1}\right|^{2}=\left|c_{2}\right|^{2}=\left(\frac{1}{\sqrt{2}}\right)^{2}=\frac{1}{2} ∣ c 1 ∣ 2 = ∣ c 2 ∣ 2 = ( 2 1 ) 2 = 2 1 ⟨ H ⟩ \left\langle H\right\rangle ⟨ H ⟩
⟨ H ⟩ = ∑ ∣ c n ∣ 2 E n = E 1 2 + E 2 2 = 1 2 ( 1 π 2 ℏ 2 2 m a 2 + 4 π 2 ℏ 2 2 m a 2 ) = 5 π 2 ℏ 2 4 m a 2 \left\langle H\right\rangle =\sum\left|c_{n}\right|^{2}E_{n}=\frac{E_{1}}{2}+\frac{E_{2}}{2}=\frac{1}{2}\left(\frac{1\pi^{2}\hbar^{2}}{2ma^{2}}+\frac{4\pi^{2}\hbar^{2}}{2ma^{2}}\right)=\frac{5\pi^{2}\hbar^{2}}{4ma^{2}} ⟨ H ⟩ = ∑ ∣ c n ∣ 2 E n = 2 E 1 + 2 E 2 = 2 1 ( 2 m a 2 1 π 2 ℏ 2 + 2 m a 2 4 π 2 ℏ 2 ) = 4 m a 2 5 π 2 ℏ 2 Onde usamos fato de que E n = n 2 π 2 ℏ 2 2 m a 2 E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} E n = 2 m a 2 n 2 π 2 ℏ 2 E 1 E_{1} E 1 E 2 E_{2} E 2
Exemplo: Uma partícula em um poço quadrado infinito tem a seguinte função de onda inicial:
Ψ ( x , 0 ) = { A x 0 ≤ x ≤ a / 2 A ( a − x ) a / 2 ≤ x ≤ a \Psi\left(x,0\right)=\begin{cases}
Ax & 0\leq x\leq a/2\\
A\left(a-x\right) & a/2\leq x\leq a
\end{cases} Ψ ( x , 0 ) = { A x A ( a − x ) 0 ≤ x ≤ a /2 a /2 ≤ x ≤ a a) Determine a constante A A A
b) Encontre Ψ ( x , t ) \Psi\left(x,t\right) Ψ ( x , t )
c) Encontre a probabilidade de que a medição de energia retorne o valor E 1 E_{1} E 1
d) Qual o valor esperado de energia usando ⟨ H ⟩ = ∑ n = 1 ∞ ∣ c n ∣ 2 E n \left\langle H\right\rangle =\sum^{\infty}_{n=1}\left|c_{n}\right|^{2}E_{n} ⟨ H ⟩ = ∑ n = 1 ∞ ∣ c n ∣ 2 E n
Vamos começar novamente determinando a constante A A A
1 = ∫ − ∞ + ∞ ∣ Ψ ∣ 2 d x = ∫ 0 a / 2 A 2 x 2 d x + ∫ a / 2 a A 2 ( a − x ) 2 d x = A 2 [ x 3 3 ] 0 a / 2 − A 2 ∫ a / 2 0 u 2 d u = A 2 a 3 2 3 3 − A 2 [ u 3 3 ] a / 2 0 = A 2 a 3 2 3 3 + A 2 a 3 2 3 3 a / 2 0 = A 2 a 3 12 \begin{align}
1= & \int^{+\infty}_{-\infty}\left|\Psi\right|^{2}dx\\
= & \int^{a/2}_{0}A^{2}x^{2}dx+\int^{a}_{a/2}A^{2}\left(a-x\right)^{2}dx\\
= & A^{2}\left[\frac{x^{3}}{3}\right]^{a/2}_{0}-A^{2}\int^{0}_{a/2}u^{2}du\\
= & A^{2}\frac{a^{3}}{2^{3}3}-A^{2}\left[\frac{u^{3}}{3}\right]^{0}_{a/2}\\
= & A^{2}\frac{a^{3}}{2^{3}3}+A^{2}\frac{a^{3}}{2^{3}3}^{0}_{a/2}\\
= & A^{2}\frac{a^{3}}{12}
\end{align} 1 = = = = = = ∫ − ∞ + ∞ ∣ Ψ ∣ 2 d x ∫ 0 a /2 A 2 x 2 d x + ∫ a /2 a A 2 ( a − x ) 2 d x A 2 [ 3 x 3 ] 0 a /2 − A 2 ∫ a /2 0 u 2 d u A 2 2 3 3 a 3 − A 2 [ 3 u 3 ] a /2 0 A 2 2 3 3 a 3 + A 2 2 3 3 a 3 a /2 0 A 2 12 a 3 Onde fizemos a substituição u = a − x u=a-x u = a − x x = a / 2 → u = a / 2 x=a/2\rightarrow u=a/2 x = a /2 → u = a /2 x = a → u = 0 x=a\rightarrow u=0 x = a → u = 0 d u = − d x du=-dx d u = − d x 12 = 2 2 ⋅ 3 12=2^{2}\cdot3 12 = 2 2 ⋅ 3
A = 2 2 3 a 3 = 2 3 a 3 A=\sqrt{\frac{2^{2}3}{a^{3}}}=2\sqrt{\frac{3}{a^{3}}} A = a 3 2 2 3 = 2 a 3 3 Queremos agora encontrar Ψ ( x , t ) = ∑ c n Ψ n ( x , t ) = ∑ n = 1 ∞ c n ψ n ( x ) e − i E n t / ℏ \Psi\left(x,t\right)=\sum c_{n}\Psi_{n}\left(x,t\right)=\sum^{\infty}_{n=1}c_{n}\psi_{n}\left(x\right)e^{-iE_{n}t/\hbar} Ψ ( x , t ) = ∑ c n Ψ n ( x , t ) = ∑ n = 1 ∞ c n ψ n ( x ) e − i E n t /ℏ 17 ψ n ( x ) = 2 a sin ( n π a x ) \psi_{n}\left(x\right)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right) ψ n ( x ) = a 2 sin ( a nπ x ) c n c_{n} c n E n = n 2 π 2 ℏ 2 2 m a 2 E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} E n = 2 m a 2 n 2 π 2 ℏ 2 36
c n = 2 a ∫ 0 a sin ( n π a x ) Ψ ( x , 0 ) d x = 2 a ( ∫ 0 a / 2 sin ( n π a x ) A x d x + ∫ a / 2 a sin ( n π a x ) A ( a − x ) d x ) = 2 a ( A ∫ 0 a / 2 x sin ( n π a x ) d x + A a ∫ a / 2 a sin ( n π a x ) d x − A ∫ a / 2 a x sin ( n π a x ) d x ) \begin{align}
c_{n} & =\sqrt{\frac{2}{a}}\int^{a}_{0}\sin\left(\frac{n\pi}{a}x\right)\Psi\left(x,0\right)dx\\
& =\sqrt{\frac{2}{a}}\left(\int^{a/2}_{0}\sin\left(\frac{n\pi}{a}x\right)Axdx+\int^{a}_{a/2}\sin\left(\frac{n\pi}{a}x\right)A\left(a-x\right)dx\right)\\
& =\sqrt{\frac{2}{a}}\left(A\int^{a/2}_{0}x\sin\left(\frac{n\pi}{a}x\right)dx+Aa\int^{a}_{a/2}\sin\left(\frac{n\pi}{a}x\right)dx-A\int^{a}_{a/2}x\sin\left(\frac{n\pi}{a}x\right)dx\right)
\end{align} c n = a 2 ∫ 0 a sin ( a nπ x ) Ψ ( x , 0 ) d x = a 2 ( ∫ 0 a /2 sin ( a nπ x ) A x d x + ∫ a /2 a sin ( a nπ x ) A ( a − x ) d x ) = a 2 ( A ∫ 0 a /2 x sin ( a nπ x ) d x + A a ∫ a /2 a sin ( a nπ x ) d x − A ∫ a /2 a x sin ( a nπ x ) d x ) Temos três integrais pra resolver. Primeiro vamos substituir u = n π a x u=\frac{n\pi}{a}x u = a nπ x
I 1 = ∫ 0 n π / 2 u ( a n π ) sin ( u ) ( a n π ) d u = ( a n π ) 2 ∫ 0 n π / 2 u sin ( u ) d u I_{1}=\int^{n\pi/2}_{0}u\left(\frac{a}{n\pi}\right)\sin\left(u\right)\left(\frac{a}{n\pi}\right)du=\left(\frac{a}{n\pi}\right)^{2}\int^{n\pi/2}_{0}u\sin\left(u\right)du I 1 = ∫ 0 nπ /2 u ( nπ a ) sin ( u ) ( nπ a ) d u = ( nπ a ) 2 ∫ 0 nπ /2 u sin ( u ) d u Isso é parecido com o que já calculamos em 44 n π 2 \frac{n\pi}{2} 2 nπ n π n\pi nπ
∫ a b sin ( u ) u d u = sin ( u ) − u cos ( u ) ∣ a b \int^{b}_{a}\sin\left(u\right)udu=\left.\sin\left(u\right)-u\cos\left(u\right)\right|^{b}_{a} ∫ a b sin ( u ) u d u = sin ( u ) − u cos ( u ) ∣ a b Com esse resultado temos:
I 1 = ( a n π ) 2 [ sin ( u ) − u cos ( u ) ] ∣ 0 n π / 2 = a 2 n 2 π 2 [ sin ( n π 2 ) − n π 2 cos ( n π 2 ) ] I_{1}=\left(\frac{a}{n\pi}\right)^{2}\left.\left[\sin\left(u\right)-u\cos\left(u\right)\right]\right|^{n\pi/2}_{0}=\frac{a^{2}}{n^{2}\pi^{2}}\left[\sin\left(\frac{n\pi}{2}\right)-\frac{n\pi}{2}\cos\left(\frac{n\pi}{2}\right)\right] I 1 = ( nπ a ) 2 [ sin ( u ) − u cos ( u ) ] ∣ 0 nπ /2 = n 2 π 2 a 2 [ sin ( 2 nπ ) − 2 nπ cos ( 2 nπ ) ] Repetindo a substituição, com a diferença que agora os limites serão entre x = a 2 → u = n π 2 x=\frac{a}{2}\rightarrow u=\frac{n\pi}{2} x = 2 a → u = 2 nπ x = a → u = n π x=a\rightarrow u=n\pi x = a → u = nπ
I 3 = ( a n π ) 2 ∫ n π / 2 n π u sin ( u ) d u = ( a n π ) 2 [ − n π cos ( n π ) + n π 2 cos ( n π 2 ) + sin ( n π ) − sin ( n π 2 ) ] = ( a n π ) 2 [ − n π cos ( n π ) + n π 2 cos ( n π 2 ) − sin ( n π 2 ) ] = a 2 n π [ cos ( n π / 2 ) 2 − cos ( n π ) − sin ( n π / 2 ) n π ] \begin{align}
I_{3} & =\left(\frac{a}{n\pi}\right)^{2}\int^{n\pi}_{n\pi/2}u\sin\left(u\right)du\\
& =\left(\frac{a}{n\pi}\right)^{2}\left[-n\pi\cos\left(n\pi\right)+\frac{n\pi}{2}\cos\left(\frac{n\pi}{2}\right)+\sin\left(n\pi\right)-\sin\left(\frac{n\pi}{2}\right)\right]\\
& =\left(\frac{a}{n\pi}\right)^{2}\left[-n\pi\cos\left(n\pi\right)+\frac{n\pi}{2}\cos\left(\frac{n\pi}{2}\right)-\sin\left(\frac{n\pi}{2}\right)\right]\\
& =\frac{a^{2}}{n\pi}\left[\frac{\cos\left(n\pi/2\right)}{2}-\cos\left(n\pi\right)-\frac{\sin\left(n\pi/2\right)}{n\pi}\right]
\end{align} I 3 = ( nπ a ) 2 ∫ nπ /2 nπ u sin ( u ) d u = ( nπ a ) 2 [ − nπ cos ( nπ ) + 2 nπ cos ( 2 nπ ) + sin ( nπ ) − sin ( 2 nπ ) ] = ( nπ a ) 2 [ − nπ cos ( nπ ) + 2 nπ cos ( 2 nπ ) − sin ( 2 nπ ) ] = nπ a 2 [ 2 cos ( nπ /2 ) − cos ( nπ ) − nπ sin ( nπ /2 ) ] A segunda integral é, com a mesma substituição u = n π a x u=\frac{n\pi}{a}x u = a nπ x
I 2 = ∫ n π / 2 n π sin ( u ) ( a n π ) d u = a n π ∫ n π / 2 n π sin ( u ) d x = − a n π [ cos u ] n π / 2 n π = a n π [ cos ( n π 2 ) − cos n π ] I_{2}=\int^{n\pi}_{n\pi/2}\sin\left(u\right)\left(\frac{a}{n\pi}\right)du=\frac{a}{n\pi}\int^{n\pi}_{n\pi/2}\sin\left(u\right)dx=-\frac{a}{n\pi}\left[\cos u\right]^{n\pi}_{n\pi/2}=\frac{a}{n\pi}\left[\cos\left(\frac{n\pi}{2}\right)-\cos n\pi\right] I 2 = ∫ nπ /2 nπ sin ( u ) ( nπ a ) d u = nπ a ∫ nπ /2 nπ sin ( u ) d x = − nπ a [ cos u ] nπ /2 nπ = nπ a [ cos ( 2 nπ ) − cos nπ ] Então:
c n = A 2 a ( ∫ 0 a / 2 x sin ( n π a x ) d x + a ∫ a / 2 a sin ( n π a x ) d x − ∫ a / 2 a x sin ( n π a x ) d x ) = A 2 a ( a 2 n 2 π 2 [ sin ( n π 2 ) − n π 2 cos ( n π 2 ) ] + a a n π [ cos ( n π 2 ) − cos n π ] … ⋯ − a 2 n π [ cos ( n π / 2 ) 2 − cos ( n π ) − sin ( n π / 2 ) n π ] ) = A 2 a a 2 n 2 π 2 ( sin ( n π 2 ) − n π 2 cos ( n π 2 ) + n π cos ( n π 2 ) − n π cos n π … ⋯ − n π 2 cos ( n π 2 ) + n π cos ( n π ) + sin ( n π 2 ) ) = A 2 a 3 n 2 π 2 2 sin ( n π 2 ) \begin{align}
c_{n} & =A\sqrt{\frac{2}{a}}\left(\int^{a/2}_{0}x\sin\left(\frac{n\pi}{a}x\right)dx+a\int^{a}_{a/2}\sin\left(\frac{n\pi}{a}x\right)dx-\int^{a}_{a/2}x\sin\left(\frac{n\pi}{a}x\right)dx\right)\\
& =A\sqrt{\frac{2}{a}}\left(\frac{a^{2}}{n^{2}\pi^{2}}\left[\sin\left(\frac{n\pi}{2}\right)-\frac{n\pi}{2}\cos\left(\frac{n\pi}{2}\right)\right]+a\frac{a}{n\pi}\left[\cos\left(\frac{n\pi}{2}\right)-\cos n\pi\right]\dots\right.\\
& \left.\dots-\frac{a^{2}}{n\pi}\left[\frac{\cos\left(n\pi/2\right)}{2}-\cos\left(n\pi\right)-\frac{\sin\left(n\pi/2\right)}{n\pi}\right]\right)\\
& =A\sqrt{\frac{2}{a}}\frac{a^{2}}{n^{2}\pi^{2}}\left(\sin\left(\frac{n\pi}{2}\right)-\frac{n\pi}{2}\cos\left(\frac{n\pi}{2}\right)+n\pi\cos\left(\frac{n\pi}{2}\right)-n\pi\cos n\pi\dots\right.\\
& \left.\dots-\frac{n\pi}{2}\cos\left(\frac{n\pi}{2}\right)+n\pi\cos\left(n\pi\right)+\sin\left(\frac{n\pi}{2}\right)\right)\\
& =A\frac{\sqrt{2a^{3}}}{n^{2}\pi^{2}}2\sin\left(\frac{n\pi}{2}\right)
\end{align} c n = A a 2 ( ∫ 0 a /2 x sin ( a nπ x ) d x + a ∫ a /2 a sin ( a nπ x ) d x − ∫ a /2 a x sin ( a nπ x ) d x ) = A a 2 ( n 2 π 2 a 2 [ sin ( 2 nπ ) − 2 nπ cos ( 2 nπ ) ] + a nπ a [ cos ( 2 nπ ) − cos nπ ] … ⋯ − nπ a 2 [ 2 cos ( nπ /2 ) − cos ( nπ ) − nπ sin ( nπ /2 ) ] ) = A a 2 n 2 π 2 a 2 ( sin ( 2 nπ ) − 2 nπ cos ( 2 nπ ) + nπ cos ( 2 nπ ) − nπ cos nπ … ⋯ − 2 nπ cos ( 2 nπ ) + nπ cos ( nπ ) + sin ( 2 nπ ) ) = A n 2 π 2 2 a 3 2 sin ( 2 nπ ) Substituindo o A = 2 3 a 3 A=2\sqrt{\frac{3}{a^{3}}} A = 2 a 3 3
A 2 2 a 3 n 2 π 2 = 2 3 a 3 2 2 a 3 n 2 π 2 = 4 6 n 2 π 2 \begin{align}
A\frac{2\sqrt{2a^{3}}}{n^{2}\pi^{2}} & =2\sqrt{\frac{3}{a^{3}}}\frac{2\sqrt{2a^{3}}}{n^{2}\pi^{2}}=\frac{4\sqrt{6}}{n^{2}\pi^{2}}
\end{align} A n 2 π 2 2 2 a 3 = 2 a 3 3 n 2 π 2 2 2 a 3 = n 2 π 2 4 6 Agora quando n n n n = 0 , 2 , 4 , … n=0,2,4,\dots n = 0 , 2 , 4 , …
sin ( 0 ) = sin ( n π ) = sin ( 2 n π ) = 0 \sin\left(0\right)=\sin\left(n\pi\right)=\sin\left(2n\pi\right)=0 sin ( 0 ) = sin ( nπ ) = sin ( 2 nπ ) = 0 Então só resta n n n n = 1 , 3 , 5 , n=1,3,5, n = 1 , 3 , 5 ,
sin ( π 2 ) = 1 , sin ( 3 π 2 ) = − 1 , sin ( 5 π 2 ) = 1 \sin\left(\frac{\pi}{2}\right)=1,\quad\sin\left(\frac{3\pi}{2}\right)=-1,\quad\sin\left(\frac{5\pi}{2}\right)=1 sin ( 2 π ) = 1 , sin ( 2 3 π ) = − 1 , sin ( 2 5 π ) = 1 Então queremos uma transformação do tipo 1 → 0 1\rightarrow0 1 → 0 3 → 1 3\rightarrow1 3 → 1 5 → 2 … n → p 5\rightarrow2\dots n\rightarrow p 5 → 2 … n → p ( − 1 ) p \left(-1\right)^{p} ( − 1 ) p n = 2 p + 1 n=2p+1 n = 2 p + 1 2 p 2p 2 p p = 0 , 1 , 2 , 3 , … p=0,1,2,3,\dots p = 0 , 1 , 2 , 3 , … n n n p p p n n n
n = 2 p + 1 → p = n − 1 2 n=2p+1\rightarrow p=\frac{n-1}{2} n = 2 p + 1 → p = 2 n − 1 De forma que temos
sin ( n π 2 ) = ( − 1 ) n − 1 2 \sin\left(\frac{n\pi}{2}\right)=\left(-1\right)^{\frac{n-1}{2}} sin ( 2 nπ ) = ( − 1 ) 2 n − 1 Então:
c n = { 0 , n par 4 6 n 2 π 2 ( − 1 ) n − 1 2 n , ı ˊ mpar c_{n}=\begin{cases}
0 & ,n\text{ par}\\
\frac{4\sqrt{6}}{n^{2}\pi^{2}}\left(-1\right)^{\frac{n-1}{2}} & n,\text{ ímpar}
\end{cases} c n = { 0 n 2 π 2 4 6 ( − 1 ) 2 n − 1 , n par n , ı ˊ mpar Como já calculamos que a solução no interior do posso é 17 ψ n ( x ) = 2 a sin ( n π a x ) \psi_{n}\left(x\right)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right) ψ n ( x ) = a 2 sin ( a nπ x ) E n = n 2 π 2 ℏ 2 2 m a 2 E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} E n = 2 m a 2 n 2 π 2 ℏ 2
Ψ ( x , t ) = ∑ n = 1 ∞ c n ψ n ( x ) e − i E n t / ℏ = ∑ n = 1 , 3 , 5 … ∞ 4 6 n 2 π 2 ( − 1 ) n − 1 2 2 a sin ( n π a x ) exp [ − i E n t / ℏ ] = 4 6 π 2 2 a ∑ n = 1 , 3 , 5 … ∞ ( − 1 ) n − 1 2 1 n 2 sin ( n π a x ) exp [ − i E n t / ℏ ] = 4 π 2 12 a ∑ n = 1 , 3 , 5 … ∞ ( − 1 ) n − 1 2 1 n 2 sin ( n π a x ) exp [ − i E n t / ℏ ] \begin{align}
\Psi\left(x,t\right) & =\sum^{\infty}_{n=1}c_{n}\psi_{n}\left(x\right)e^{-iE_{n}t/\hbar}\\
& =\sum^{\infty}_{n=1,3,5\dots}\frac{4\sqrt{6}}{n^{2}\pi^{2}}\left(-1\right)^{\frac{n-1}{2}}\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-iE_{n}t/\hbar\right]\\
& =\frac{4\sqrt{6}}{\pi^{2}}\sqrt{\frac{2}{a}}\sum^{\infty}_{n=1,3,5\dots}\left(-1\right)^{\frac{n-1}{2}}\frac{1}{n^{2}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-iE_{n}t/\hbar\right]\\
& =\frac{4}{\pi^{2}}\sqrt{\frac{12}{a}}\sum^{\infty}_{n=1,3,5\dots}\left(-1\right)^{\frac{n-1}{2}}\frac{1}{n^{2}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-iE_{n}t/\hbar\right]
\end{align} Ψ ( x , t ) = n = 1 ∑ ∞ c n ψ n ( x ) e − i E n t /ℏ = n = 1 , 3 , 5 … ∑ ∞ n 2 π 2 4 6 ( − 1 ) 2 n − 1 a 2 sin ( a nπ x ) exp [ − i E n t /ℏ ] = π 2 4 6 a 2 n = 1 , 3 , 5 … ∑ ∞ ( − 1 ) 2 n − 1 n 2 1 sin ( a nπ x ) exp [ − i E n t /ℏ ] = π 2 4 a 12 n = 1 , 3 , 5 … ∑ ∞ ( − 1 ) 2 n − 1 n 2 1 sin ( a nπ x ) exp [ − i E n t /ℏ ] Ou como 12 = 2 2 3 = 2 3 \sqrt{12}=\sqrt{2^{2}3}=2\sqrt{3} 12 = 2 2 3 = 2 3
Ψ ( x , t ) = 8 π 2 3 a ∑ n = 1 , 3 , 5 … ∞ ( − 1 ) ( n − 1 ) / 2 1 n 2 sin ( n π a x ) exp [ − i E n t / ℏ ] \Psi\left(x,t\right)=\frac{8}{\pi^{2}}\sqrt{\frac{3}{a}}\sum^{\infty}_{n=1,3,5\dots}\left(-1\right)^{\left(n-1\right)/2}\frac{1}{n^{2}}\sin\left(\frac{n\pi}{a}x\right)\exp\left[-iE_{n}t/\hbar\right] Ψ ( x , t ) = π 2 8 a 3 n = 1 , 3 , 5 … ∑ ∞ ( − 1 ) ( n − 1 ) /2 n 2 1 sin ( a nπ x ) exp [ − i E n t /ℏ ] Para obtermos a probabilidade de que a medida da energia seja E 1 E_{1} E 1 P 1 = ∣ c 1 ∣ 2 P_{1}=\left|c_{1}\right|^{2} P 1 = ∣ c 1 ∣ 2
P 1 = ∣ c 1 ∣ 2 = ∣ 4 6 1 2 π 2 ( − 1 ) 1 − 1 2 ∣ 2 = ∣ 4 6 π 2 ∣ 2 = 0.9855 \begin{align}
P_{1} & =\left|c_{1}\right|^{2}\\
& =\left|\frac{4\sqrt{6}}{1^{2}\pi^{2}}\left(-1\right)^{\frac{1-1}{2}}\right|^{2}\\
& =\left|\frac{4\sqrt{6}}{\pi^{2}}\right|^{2}\\
& =0.9855
\end{align} P 1 = ∣ c 1 ∣ 2 = ∣ ∣ 1 2 π 2 4 6 ( − 1 ) 2 1 − 1 ∣ ∣ 2 = ∣ ∣ π 2 4 6 ∣ ∣ 2 = 0.9855 Ou seja P 1 = 98 , 55 % P_{1}=98,55\% P 1 = 98 , 55%
⟨ H ⟩ = ∑ n = 1 ∞ ∣ c n ∣ 2 E n = ∑ n = 1 , 2 , 3 … ∞ ∣ 4 6 n 2 π 2 ( − 1 ) n − 1 2 ∣ 2 n 2 π 2 ℏ 2 2 m a 2 = 4 6 π 2 π 2 ℏ 2 2 m a 2 ∑ n = 1 , 3 , 5 … ∞ ∣ ( − 1 ) n − 1 2 n 2 ∣ 2 n 2 = 2 6 ℏ 2 m a 2 ∑ n = 1 , 3 , 5 … ∞ ( − 1 ) n − 1 n 2 = 2 6 ℏ 2 m a 2 ( ( − 1 ) 0 1 2 + ( − 1 ) 2 3 2 + ( − 1 ) 4 5 2 + … ) = 2 6 ℏ 2 m a 2 ( 1 1 2 + 1 3 2 + 1 5 2 + … ) \begin{align}
\left\langle H\right\rangle & =\sum^{\infty}_{n=1}\left|c_{n}\right|^{2}E_{n}\\
& =\sum^{\infty}_{n=1,2,3\dots}\left|\frac{4\sqrt{6}}{n^{2}\pi^{2}}\left(-1\right)^{\frac{n-1}{2}}\right|^{2}\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}\\
& =\frac{4\sqrt{6}}{\pi^{2}}\frac{\pi^{2}\hbar^{2}}{2ma^{2}}\sum^{\infty}_{n=1,3,5\dots}\left|\frac{\left(-1\right)^{\frac{n-1}{2}}}{n^{2}}\right|^{2}n^{2}\\
& =\frac{2\sqrt{6}\hbar^{2}}{ma^{2}}\sum^{\infty}_{n=1,3,5\dots}\frac{\left(-1\right)^{n-1}}{n^{2}}\\
& =\frac{2\sqrt{6}\hbar^{2}}{ma^{2}}\left(\frac{\left(-1\right)^{0}}{1^{2}}+\frac{\left(-1\right)^{2}}{3^{2}}+\frac{\left(-1\right)^{4}}{5^{2}}+\dots\right)\\
& =\frac{2\sqrt{6}\hbar^{2}}{ma^{2}}\left(\frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\dots\right)
\end{align} ⟨ H ⟩ = n = 1 ∑ ∞ ∣ c n ∣ 2 E n = n = 1 , 2 , 3 … ∑ ∞ ∣ ∣ n 2 π 2 4 6 ( − 1 ) 2 n − 1 ∣ ∣ 2 2 m a 2 n 2 π 2 ℏ 2 = π 2 4 6 2 m a 2 π 2 ℏ 2 n = 1 , 3 , 5 … ∑ ∞ ∣ ∣ n 2 ( − 1 ) 2 n − 1 ∣ ∣ 2 n 2 = m a 2 2 6 ℏ 2 n = 1 , 3 , 5 … ∑ ∞ n 2 ( − 1 ) n − 1 = m a 2 2 6 ℏ 2 ( 1 2 ( − 1 ) 0 + 3 2 ( − 1 ) 2 + 5 2 ( − 1 ) 4 + … ) = m a 2 2 6 ℏ 2 ( 1 2 1 + 3 2 1 + 5 2 1 + … ) Onde usamos o valor c n c_{n} c n E n E_{n} E n
1 1 2 + 1 3 2 + 1 5 2 + ⋯ = ∑ n = 1 , 3 , 5 … ∞ 1 n 2 = π 2 8 \frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\dots=\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{2}}=\frac{\pi^{2}}{8} 1 2 1 + 3 2 1 + 5 2 1 + ⋯ = n = 1 , 3 , 5 … ∑ ∞ n 2 1 = 8 π 2 Então:
⟨ H ⟩ = 2 6 ℏ 2 m a 2 π 2 8 = π 2 6 ℏ 2 8 m a 2 \begin{align}
\left\langle H\right\rangle & =\frac{2\sqrt{6}\hbar^{2}}{ma^{2}}\frac{\pi^{2}}{8}\\
& =\frac{\pi^{2}\sqrt{6}\hbar^{2}}{8ma^{2}}
\end{align} ⟨ H ⟩ = m a 2 2 6 ℏ 2 8 π 2 = 8 m a 2 π 2 6 ℏ 2 Um comentário do livro que acho que vale a pena trazer é que não há restrição sobre o formato da função de onda, ela só precisa ser normalizável. Ψ ( x , 0 ) \Psi\left(x,0\right) Ψ ( x , 0 ) ⟨ H ⟩ \left\langle H\right\rangle ⟨ H ⟩ Ψ ( x , 0 ) \Psi\left(x,0\right) Ψ ( x , 0 )
S = ∑ n = 1 , 3 , 5 … ∞ 1 n k S=\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{k}} S = n = 1 , 3 , 5 … ∑ ∞ n k 1 Começamos com a função Zeta de Riemman dada por:
ζ ( k ) = ∑ n = 1 ∞ 1 n k \zeta\left(k\right)=\sum^{\infty}_{n=1}\frac{1}{n^{k}} ζ ( k ) = n = 1 ∑ ∞ n k 1 Vamos dividir em pares e ímpares:
ζ ( k ) = ∑ n = 2 , 4 , 6 … ∞ 1 n k + ∑ n = 1 , 3 , 5 … ∞ 1 n k \zeta\left(k\right)=\sum^{\infty}_{n=2,4,6\dots}\frac{1}{n^{k}}+\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{k}} ζ ( k ) = n = 2 , 4 , 6 … ∑ ∞ n k 1 + n = 1 , 3 , 5 … ∑ ∞ n k 1 Vamos reescrever os termos pares como n = 2 m n=2m n = 2 m
∑ n = 2 , 4 , 6 … ∞ 1 n k = ∑ n = 1 ∞ 1 ( 2 m ) k \sum^{\infty}_{n=2,4,6\dots}\frac{1}{n^{k}}=\sum^{\infty}_{n=1}\frac{1}{\left(2m\right)^{k}} n = 2 , 4 , 6 … ∑ ∞ n k 1 = n = 1 ∑ ∞ ( 2 m ) k 1 Pois em ambos os casos:
∑ n = 1 ∞ 1 ( 2 m ) k = 1 ( 2 ⋅ 1 ) k + 1 ( 2 ⋅ 2 ) k + 1 ( 2 ⋅ 3 ) k + … = 1 2 k + 1 4 k + 1 6 k + … = ∑ n = 2 , 4 , 6 … ∞ 1 n k \begin{align}
\sum^{\infty}_{n=1}\frac{1}{\left(2m\right)^{k}} & =\frac{1}{\left(2\cdot1\right)^{k}}+\frac{1}{\left(2\cdot2\right)^{k}}+\frac{1}{\left(2\cdot3\right)^{k}}+\dots\\
& =\frac{1}{2^{k}}+\frac{1}{4^{k}}+\frac{1}{6^{k}}+\dots\\
& =\sum^{\infty}_{n=2,4,6\dots}\frac{1}{n^{k}}
\end{align} n = 1 ∑ ∞ ( 2 m ) k 1 = ( 2 ⋅ 1 ) k 1 + ( 2 ⋅ 2 ) k 1 + ( 2 ⋅ 3 ) k 1 + … = 2 k 1 + 4 k 1 + 6 k 1 + … = n = 2 , 4 , 6 … ∑ ∞ n k 1 E como:
∑ n = 1 ∞ 1 ( 2 m ) k = ∑ n = 1 ∞ 1 2 k m k = 1 2 k ∑ n = 1 ∞ 1 m k = ζ ( k ) 2 k \sum^{\infty}_{n=1}\frac{1}{\left(2m\right)^{k}}=\sum^{\infty}_{n=1}\frac{1}{2^{k}m^{k}}=\frac{1}{2^{k}}\sum^{\infty}_{n=1}\frac{1}{m^{k}}=\frac{\zeta\left(k\right)}{2^{k}} n = 1 ∑ ∞ ( 2 m ) k 1 = n = 1 ∑ ∞ 2 k m k 1 = 2 k 1 n = 1 ∑ ∞ m k 1 = 2 k ζ ( k ) Então:
ζ ( k ) = ∑ n = 2 , 4 , 6 … ∞ 1 n k + ∑ n = 1 , 3 , 5 … ∞ 1 n k ζ ( k ) = ζ ( k ) 2 k + ∑ n = 1 , 3 , 5 … ∞ 1 n k ∑ n = 1 , 3 , 5 … ∞ 1 n k = ζ ( k ) − ζ ( k ) 2 k = ζ ( k ) ( 1 − 2 − k ) \begin{align}
\zeta\left(k\right) & =\sum^{\infty}_{n=2,4,6\dots}\frac{1}{n^{k}}+\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{k}}\\
\zeta\left(k\right) & =\frac{\zeta\left(k\right)}{2^{k}}+\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{k}}\\
\sum^{\infty}_{n=1,3,5\dots}\frac{1}{n^{k}} & =\zeta\left(k\right)-\frac{\zeta\left(k\right)}{2^{k}}\\
& =\zeta\left(k\right)\left(1-2^{-k}\right)
\end{align} ζ ( k ) ζ ( k ) n = 1 , 3 , 5 … ∑ ∞ n k 1 = n = 2 , 4 , 6 … ∑ ∞ n k 1 + n = 1 , 3 , 5 … ∑ ∞ n k 1 = 2 k ζ ( k ) + n = 1 , 3 , 5 … ∑ ∞ n k 1 = ζ ( k ) − 2 k ζ ( k ) = ζ ( k ) ( 1 − 2 − k ) Agora para obtermos os diferentes valores dos somatórios para diferentes valores de k k k k = 2 , 4 k=2,4 k = 2 , 4 k = 6 k=6 k = 6 ζ ( k ) \zeta\left(k\right) ζ ( k ) ζ ( k ) \zeta\left(k\right) ζ ( k ) k k k k = 2 k=2 k = 2 ζ ( k ) \zeta\left(k\right) ζ ( k )
ζ ( 2 n ) = ( − 1 ) n − 1 ( 2 π ) 2 n 2 ⋅ ( 2 n ) ! B 2 n \zeta\left(2n\right)=\frac{\left(-1\right)^{n-1}\left(2\pi\right)^{2n}}{2\cdot\left(2n\right)!}B_{2n} ζ ( 2 n ) = 2 ⋅ ( 2 n ) ! ( − 1 ) n − 1 ( 2 π ) 2 n B 2 n Mas novamente para obter essa solução, agora precisaríamos ver que é número de Bernoulli. Como meu foco não é a matemática, eu vou aceitar nos resultados que os matemáticos forneceram, se não sinto que estamos entrando em uma toca de coelho infinita.