This is the problem:(adsbygoogle = window.adsbygoogle || []).push({});

Calculate:

[tex]

\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}

\frac {d \mean{p}}_{dt} [/tex]

Here's a few more points to keep in mind....

(A) The assumption is that <p> is defined as:

[tex]

\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}

\mean{p} = -i \hbar \int \left( \psi^* \frac {\partial {\psi}}_{\partial {x}} \right) dx

[/itex]

(B) The problem states that the solution is:

[tex]

\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}

\frac {d \mean{p}}_{dt}} = \mean {\frac {- \partial {V}}_{\partial {x}}}

[/tex]

(C) In working the problem, I am able to reach this point (based on previous examples), but am not sure how to approach moving forward:

[tex]

\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}

\frac {d \mean{p}}_{dt}} = \left( \frac {\hbar^2}_{2m} \right) \int_{-\infty}^{+\infty} \left( \frac {\partial}_{\partial x} \right) \left( \frac {\partial }_{\partial {x}} \right) \left( \psi^* \left( \frac {\partial {\psi}}_{\partial {x}} \right) - \left( \frac {\partial {\psi^*}}_{\partial {x}} \right) \psi \right) dx

[/tex]

Any ideas for what to do next? The book shows a similiar example for finding the time derivative of <x>, but the difference for that approach is that the partial derivative was canceled by an x on the inside of the integral. Now I have two of them, and no ideas for how to simplify.

Thanks.

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# Time deritivate of the expectation value of p

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