Exercício 3 Por que não podemos fazer a integração por partes diretamente em
puxando a derivada em ordem ao tempo para 
Porque integração por partes só pode ser usada quando a diferenciação e integração são feitas com a mesma variável.
Exercício 4 Calcular
Primeiro, vamos lembrar a equação de Schroedinger:
E seu complexo conjugado
para a evolução do tempo do valor médio de momento é
![{\begin{aligned} \dfrac{d<p>}{dt} &= \dfrac{d}{dt}\int\Psi ^* \dfrac{\hbar}{i}\dfrac{\partial \Psi}{\partial x}\, dx\\ &= \dfrac{\hbar}{i}\int \dfrac{\partial}{\partial t}\left( \Psi ^* \dfrac{\partial \Psi}{\partial x}\right)\, dx\\ &= \dfrac{\hbar}{i}\int\left( \dfrac{\partial \Psi^*}{\partial t}\dfrac{\partial \Psi}{\partial x}+\Psi^* \dfrac{\partial}{\partial x}\dfrac{\partial \Psi}{\partial t} \right) \, dx\\ &= \dfrac{\hbar}{i}\int \left[ \left( -\dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi^*}{\partial x^2}+\dfrac{i}{\hbar}V\Psi^* \right)\dfrac{\partial \Psi}{\partial x} + \Psi^*\dfrac{\partial}{\partial x}\left( \dfrac{i\hbar}{2m}\dfrac{\partial^2\Psi}{\partial x^2}-\dfrac{i}{\hbar}V\Psi \right)\right]\, dx\\\ &= \dfrac{\hbar}{i}\int \left[ -\dfrac{i\hbar}{2m}\left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3} \right)+\dfrac{i}{\hbar}\left( V\Psi ^*\dfrac{\partial\Psi}{\partial x}-\Psi ^*\dfrac{\partial (V\Psi)}{\partial x}\right)\right]\, dx \end{aligned}}](https://s0.wp.com/latex.php?latex=%7B%5Cbegin%7Baligned%7D+%5Cdfrac%7Bd%3Cp%3E%7D%7Bdt%7D+%26%3D+%5Cdfrac%7Bd%7D%7Bdt%7D%5Cint%5CPsi+%5E%2A+%5Cdfrac%7B%5Chbar%7D%7Bi%7D%5Cdfrac%7B%5Cpartial+%5CPsi%7D%7B%5Cpartial+x%7D%5C%2C+dx%5C%5C+%26%3D+%5Cdfrac%7B%5Chbar%7D%7Bi%7D%5Cint+%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial+t%7D%5Cleft%28+%5CPsi+%5E%2A+%5Cdfrac%7B%5Cpartial+%5CPsi%7D%7B%5Cpartial+x%7D%5Cright%29%5C%2C+dx%5C%5C+%26%3D+%5Cdfrac%7B%5Chbar%7D%7Bi%7D%5Cint%5Cleft%28+%5Cdfrac%7B%5Cpartial+%5CPsi%5E%2A%7D%7B%5Cpartial+t%7D%5Cdfrac%7B%5Cpartial+%5CPsi%7D%7B%5Cpartial+x%7D%2B%5CPsi%5E%2A+%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial+x%7D%5Cdfrac%7B%5Cpartial+%5CPsi%7D%7B%5Cpartial+t%7D+%5Cright%29+%5C%2C+dx%5C%5C+%26%3D+%5Cdfrac%7B%5Chbar%7D%7Bi%7D%5Cint+%5Cleft%5B+%5Cleft%28+-%5Cdfrac%7Bi%5Chbar%7D%7B2m%7D%5Cdfrac%7B%5Cpartial%5E2%5CPsi%5E%2A%7D%7B%5Cpartial+x%5E2%7D%2B%5Cdfrac%7Bi%7D%7B%5Chbar%7DV%5CPsi%5E%2A+%5Cright%29%5Cdfrac%7B%5Cpartial+%5CPsi%7D%7B%5Cpartial+x%7D+%2B+%5CPsi%5E%2A%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial+x%7D%5Cleft%28+%5Cdfrac%7Bi%5Chbar%7D%7B2m%7D%5Cdfrac%7B%5Cpartial%5E2%5CPsi%7D%7B%5Cpartial+x%5E2%7D-%5Cdfrac%7Bi%7D%7B%5Chbar%7DV%5CPsi+%5Cright%29%5Cright%5D%5C%2C+dx%5C%5C%5C+%26%3D+%5Cdfrac%7B%5Chbar%7D%7Bi%7D%5Cint+%5Cleft%5B+-%5Cdfrac%7Bi%5Chbar%7D%7B2m%7D%5Cleft%28%5Cdfrac%7B%5Cpartial%5E2%5CPsi%5E%2A%7D%7B%5Cpartial+x%5E2%7D%5Cdfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial+x%7D-%5CPsi%5E%2A%5Cdfrac%7B%5Cpartial+%5E3+%5CPsi%7D%7B%5Cpartial+x%5E3%7D+%5Cright%29%2B%5Cdfrac%7Bi%7D%7B%5Chbar%7D%5Cleft%28+V%5CPsi+%5E%2A%5Cdfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial+x%7D-%5CPsi+%5E%2A%5Cdfrac%7B%5Cpartial+%28V%5CPsi%29%7D%7B%5Cpartial+x%7D%5Cright%29%5Cright%5D%5C%2C+dx+%5Cend%7Baligned%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
Primeiro vamos calcular o primeiro termo da integral (ignorando os fatores constantes) fazendo a integração por partes (lembre-se que os pontos fronteira são nulos) duas vezes
![{\begin{aligned} \int \left(\dfrac{\partial^2\Psi^*}{\partial x^2}\dfrac{\partial\Psi}{\partial x}-\Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\right)\, dx &= \left[ \dfrac{\partial \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x}\right]-\int\dfrac{\partial \Psi^*}{\partial x}\dfrac{\partial ^2 \Psi}{\partial x^2}\, dx- \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &=-\left[ \Psi ^*\dfrac{\partial ^2 \Psi}{\partial x^2} \right]+\int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx - \int \Psi^*\dfrac{\partial ^3 \Psi}{\partial x^3}\, dx \\ &= 0 \end{aligned}}](https://s0.wp.com/latex.php?latex=%7B%5Cbegin%7Baligned%7D+%5Cint+%5Cleft%28%5Cdfrac%7B%5Cpartial%5E2%5CPsi%5E%2A%7D%7B%5Cpartial+x%5E2%7D%5Cdfrac%7B%5Cpartial%5CPsi%7D%7B%5Cpartial+x%7D-%5CPsi%5E%2A%5Cdfrac%7B%5Cpartial+%5E3+%5CPsi%7D%7B%5Cpartial+x%5E3%7D%5Cright%29%5C%2C+dx+%26%3D+%5Cleft%5B+%5Cdfrac%7B%5Cpartial+%5CPsi%5E%2A%7D%7B%5Cpartial+x%5E2%7D+%5Cdfrac%7B%5Cpartial+%5CPsi%7D%7B%5Cpartial+x%7D%5Cright%5D-%5Cint%5Cdfrac%7B%5Cpartial+%5CPsi%5E%2A%7D%7B%5Cpartial+x%7D%5Cdfrac%7B%5Cpartial+%5E2+%5CPsi%7D%7B%5Cpartial+x%5E2%7D%5C%2C+dx-+%5Cint+%5CPsi%5E%2A%5Cdfrac%7B%5Cpartial+%5E3+%5CPsi%7D%7B%5Cpartial+x%5E3%7D%5C%2C+dx+%5C%5C+%26%3D-%5Cleft%5B+%5CPsi+%5E%2A%5Cdfrac%7B%5Cpartial+%5E2+%5CPsi%7D%7B%5Cpartial+x%5E2%7D+%5Cright%5D%2B%5Cint+%5CPsi%5E%2A%5Cdfrac%7B%5Cpartial+%5E3+%5CPsi%7D%7B%5Cpartial+x%5E3%7D%5C%2C+dx+-+%5Cint+%5CPsi%5E%2A%5Cdfrac%7B%5Cpartial+%5E3+%5CPsi%7D%7B%5Cpartial+x%5E3%7D%5C%2C+dx+%5C%5C+%26%3D+0+%5Cend%7Baligned%7D%7D&bg=ffffff&fg=000000&s=0&c=20201002)
Depois vamos calcular o segundo termo do integral
Em conclusão, é
Portanto, o valor médio do operador momento obedece ao Segundo Axioma de Newton .
O resultado anterior pode ser generalizado e sua generalização é conhecida na literatura da Mecânica Quântica como o Teorema de Ehrenfest
Luso Academia 