Mostre que as primeiras 4 funções de onda do OHQ podem ser escrita como:
\begin{array}{ll} \psi_{0}(x)=\sqrt{\frac{\rho}{\sqrt{\pi}}} e^{-\frac{1}{2} \rho^{2} x^{2}} & \psi_{1}(x)=\sqrt{\frac{\rho}{2 \sqrt{\pi}}}(2 \rho x) e^{-\frac{1}{2} \rho^{2} x^{2}} \\\ \psi_{2}(x)=\sqrt{\frac{\rho}{8 \sqrt{\pi}}}\left(4 \rho^{2} x^{2}-2\right) e^{-\frac{1}{2} \rho^{2} x^{2}} & \psi_{3}(x)=\sqrt{\frac{\rho}{48 \sqrt{\pi}}}\left(8 \rho^{3} x^{3}-12 \rho x\right) e^{-\frac{1}{2} \rho^{2} x^{2}} \end{array}

Passo 1

O enunciado pede para mostrar que as primeiras quatro funções de onda do oscilador harmônico quântico em uma dimensão podem ser escritas como
\begin{array}{ll} \psi_{0}(x)=\sqrt{\frac{\rho}{\sqrt{\pi}}}\,e^{-\tfrac12\rho^{2}x^{2}},& \psi_{1}(x)=\sqrt{\frac{\rho}{2\sqrt{\pi}}}\,(2\rho x)\,e^{-\tfrac12\rho^{2}x^{2}},\\\ \psi_{2}(x)=\sqrt{\frac{\rho}{8\sqrt{\pi}}}\,\bigl(4\rho^{2}x^{2}-2\bigr)\,e^{-\tfrac12\rho^{2}x^{2}},& \psi_{3}(x)=\sqrt{\frac{\rho}{48\sqrt{\pi}}}\,\bigl(8\rho^{3}x^{3}-12\rho x\bigr)\,e^{-\tfrac12\rho^{2}x^{2}}. \end{array}
O conteúdo usado será a forma geral das funções de Hermite normalizadas e as expressões dos primeiros polinômios de Hermite.

Passo 2

A forma geral da função de onda do nível n é
\psi_{n}(x) =\Bigl(\frac{\rho}{\sqrt{\pi}\,2^{n}n!}\Bigr)^{\!1/2} \,H_{n}\bigl(\rho x\bigr)\,e^{-\tfrac12\rho^{2}x^{2}},
onde \rho=\sqrt{\tfrac{m\omega}{\hbar}} e H_{n}(z) são os polinômios de Hermite.

Passo 3

Recordamos os primeiros polinômios de Hermite na convenção física:
H_{0}(z)=1,\quad H_{1}(z)=2z,\quad H_{2}(z)=4z^{2}-2,\quad H_{3}(z)=8z^{3}-12z.

Passo 4

Para n=0 temos
\psi_{0}(x) =\Bigl(\tfrac{\rho}{\sqrt{\pi}\,2^{0}0!}\Bigr)^{\!1/2}\,H_{0}(\rho x)\,e^{-\tfrac12\rho^{2}x^{2}} =\sqrt{\frac{\rho}{\sqrt{\pi}}}\,e^{-\tfrac12\rho^{2}x^{2}}.

Passo 5

Para n=1 temos
\psi_{1}(x) =\Bigl(\tfrac{\rho}{\sqrt{\pi}\,2^{1}1!}\Bigr)^{\!1/2}\,H_{1}(\rho x)\,e^{-\tfrac12\rho^{2}x^{2}} =\sqrt{\frac{\rho}{2\sqrt{\pi}}}\,(2\rho x)\,e^{-\tfrac12\rho^{2}x^{2}}.

Passo 6

Para n=2 obtemos
\psi_{2}(x) =\Bigl(\tfrac{\rho}{\sqrt{\pi}\,2^{2}2!}\Bigr)^{\!1/2}\,H_{2}(\rho x)\,e^{-\tfrac12\rho^{2}x^{2}} =\sqrt{\frac{\rho}{8\sqrt{\pi}}}\,\bigl(4\rho^{2}x^{2}-2\bigr)\,e^{-\tfrac12\rho^{2}x^{2}}.

Passo 7

Para n=3 segue
\psi_{3}(x) =\Bigl(\tfrac{\rho}{\sqrt{\pi}\,2^{3}3!}\Bigr)^{\!1/2}\,H_{3}(\rho x)\,e^{-\tfrac12\rho^{2}x^{2}} =\sqrt{\frac{\rho}{48\sqrt{\pi}}}\,\bigl(8\rho^{3}x^{3}-12\rho x\bigr)\,e^{-\tfrac12\rho^{2}x^{2}}.

Resposta

\begin{array}{ll} \psi_{0}(x)=\sqrt{\frac{\rho}{\sqrt{\pi}}}\,e^{-\tfrac12\rho^{2}x^{2}},& \psi_{1}(x)=\sqrt{\frac{\rho}{2\sqrt{\pi}}}\,(2\rho x)\,e^{-\tfrac12\rho^{2}x^{2}},\\\ \psi_{2}(x)=\sqrt{\frac{\rho}{8\sqrt{\pi}}}\,\bigl(4\rho^{2}x^{2}-2\bigr)\,e^{-\tfrac12\rho^{2}x^{2}},& \psi_{3}(x)=\sqrt{\frac{\rho}{48\sqrt{\pi}}}\,\bigl(8\rho^{3}x^{3}-12\rho x\bigr)\,e^{-\tfrac12\rho^{2}x^{2}}. \end{array}

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