Seja a função de estado de uma partícula de massa m dada por:
\psi_{n}(x)=A_{n} \operatorname{sen}\left(\frac{n \pi}{L} x\right), \quad \operatorname{com} n=1,2,3, \ldots
Considere \psi_{n}(x)=0 nos intervalos abertos de -\infty a 0 e de L a +\infty.
(a) Mostre que a constante de normalização A_{n}=\sqrt{\frac{2}{L}} independentemente de n.
(b) Calcule \langle x\rangle.
(c) Calcule \left\langle x^{2}\right\rangle.
(d) Calcule \sigma_{x}.
(e) Calcule \langle p\rangle.
(f) Calcule \left\langle p^{2}\right\rangle.
(g) Calcule \sigma_{p}.
(h) Calcule a incerteza dada por \sigma_{x} \sigma_{p}.

Passo 1

O enunciado trata da partícula num poço infinito de largura L, com autoestados
\psi_{n}(x)=A_{n}\sin\!\Bigl(\frac{n\pi}{L}x\Bigr)
para 0<x<L e zero fora desse intervalo.
Conteúdo: normalização de funções de onda, operadores posição \hat x e momento \hat p=-\,i\hbar\frac{d}{dx}, e cálculo de valores esperados e dispersões.

Passo 2

Para achar A_{n} fazemos
1=\int_{-\infty}^{\infty}|\psi_n|^2\,dx =\int_{0}^{L}A_n^2\sin^2\!\Bigl(\frac{n\pi x}{L}\Bigr)dx.
Como \int_{0}^{L}\sin^2(\tfrac{n\pi x}{L})dx=L/2, vem
A_n^2\frac{L}{2}=1 \quad\Longrightarrow\quad A_n=\sqrt{\frac{2}{L}}.

Passo 3

Cálculo de \langle x\rangle:
\langle x\rangle =\frac{2}{L}\int_{0}^{L}x\sin^2\!\Bigl(\frac{n\pi x}{L}\Bigr)dx =\frac{2}{L}\int_{0}^{L}x\frac{1-\cos(\tfrac{2n\pi x}{L})}{2}dx.
Como \int_{0}^{L}x\,dx=L^2/2 e \int_{0}^{L}x\cos(\tfrac{2n\pi x}{L})dx=0, resulta
\langle x\rangle=\frac{2}{L}\cdot\frac{L^2}{4}=\frac{L}{2}.

Passo 4

Cálculo de \langle x^2\rangle:
\langle x^2\rangle =\frac{2}{L}\int_{0}^{L}x^2\sin^2\!\Bigl(\frac{n\pi x}{L}\Bigr)dx =\frac{2}{L}\int_{0}^{L}x^2\frac{1-\cos(\tfrac{2n\pi x}{L})}{2}dx.
Usando \int_0^L x^2dx=L^3/3 e \int_0^L x^2\cos(\tfrac{2n\pi x}{L})dx =\tfrac{L^3}{2n^2\pi^2}, obtém-se
\langle x^2\rangle =\frac{2}{L}\Bigl(\frac{L^3}{6}-\frac{L^3}{4n^2\pi^2}\Bigr) =\frac{L^2}{3}-\frac{L^2}{2n^2\pi^2}.

Passo 5

Dispersão em x:
\sigma_x =\sqrt{\langle x^2\rangle-\langle x\rangle^2} =\sqrt{\Bigl(\frac{L^2}{3}-\frac{L^2}{2n^2\pi^2}\Bigr)-\frac{L^2}{4}} =L\sqrt{\frac{1}{12}-\frac{1}{2n^2\pi^2}}.

Passo 6

Cálculo de \langle p\rangle:
\langle p\rangle =-i\hbar\int_{0}^{L}\psi_n^*\frac{d\psi_n}{dx}dx \propto\int_0^L\sin\Bigl(\tfrac{n\pi x}{L}\Bigr)\cos\Bigl(\tfrac{n\pi x}{L}\Bigr)dx =\frac{1}{2}\int_0^L\sin\Bigl(\tfrac{2n\pi x}{L}\Bigr)dx=0.

Passo 7

Cálculo de \langle p^2\rangle:
\langle p^2\rangle =\int_0^L\psi_n^*\bigl(-\hbar^2\tfrac{d^2}{dx^2}\bigr)\psi_n\,dx =\hbar^2\Bigl(\frac{n\pi}{L}\Bigr)^2.

Passo 8

Dispersão em p:
\sigma_p =\sqrt{\langle p^2\rangle-\langle p\rangle^2} =\hbar\frac{n\pi}{L}.

Passo 9

Incerteza \sigma_x\sigma_p:
\sigma_x\sigma_p =\Bigl(L\sqrt{\tfrac{1}{12}-\tfrac{1}{2n^2\pi^2}}\Bigr)\Bigl(\hbar\frac{n\pi}{L}\Bigr) =\hbar\,n\pi\sqrt{\frac{1}{12}-\frac{1}{2n^2\pi^2}} =\hbar\sqrt{\frac{n^2\pi^2-6}{12}}.

Resposta

(a) A_n=\sqrt{\tfrac{2}{L}}
(b) \langle x\rangle=\tfrac{L}{2}
(c) \langle x^2\rangle=\tfrac{L^2}{3}-\tfrac{L^2}{2n^2\pi^2}
(d) \sigma_x=L\sqrt{\tfrac{1}{12}-\tfrac{1}{2n^2\pi^2}}
(e) \langle p\rangle=0
(f) \langle p^2\rangle=\hbar^2\bigl(\tfrac{n\pi}{L}\bigr)^2
(g) \sigma_p=\hbar\tfrac{n\pi}{L}
(h) \sigma_x\sigma_p=\hbar\sqrt{\tfrac{n^2\pi^2-6}{12}}

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