2
$\begingroup$

The expectation value of momentum is given by:

$$ \langle p\rangle = \int_{-\infty}^{\infty}\psi^{*}(x)\left(-i\hbar\frac{\partial}{\partial x}\right)\psi(x)dx $$

How can I show that the above expression is equivalent to this? $$ \langle p\rangle = \int_{-\infty}^{\infty}p|\tilde\psi(p)|^{2}dp $$

I have tried to use that

$$\psi(x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi(p)e^{ipx / \hbar}dp$$ and $$\psi^{*}(x)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi^{*}(p)e^{-ipx / \hbar}dp$$

Then $$ \langle p \rangle = \int_{-\infty}^{\infty} \left[\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi^{*}(p)e^{-ipx / \hbar}dp \right ]\left(-i\hbar\frac{\partial}{\partial x}\right )\left[\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\tilde\psi(p)e^{ipx / \hbar}dp \right]dx$$ $$=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\left [ \left (\int_{-\infty}^{\infty} \tilde\psi^{*}(p)e^{-ipx / \hbar}dp \right) (-i\hbar) \left (\int_{-\infty}^{\infty} \frac{\partial}{\partial x}\tilde\psi(p)e^{ipx / \hbar}dp \right)\right]dx$$ $$=\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty}\left [ \left (\int_{-\infty}^{\infty} \tilde\psi^{*}(p)e^{-ipx / \hbar}dp \right) (-i\hbar) \left (\int_{-\infty}^{\infty} \frac{ip}{\hbar}\tilde\psi(p)e^{ipx / \hbar}dp \right)\right]dx $$

But I don't know if this is the right approach or if I'm doing the right thing.

$\endgroup$
1
  • $\begingroup$ Yes please see my edit. $\endgroup$
    – Thiago
    Mar 10, 2014 at 22:19

1 Answer 1

6
$\begingroup$

If you represent the wave function $\psi(x)$ with it's fourier transform,

\begin{eqnarray*} \psi(x) &=& \frac{1}{\sqrt{2\pi \hbar}}\int \tilde{\psi}(p)e^{\frac{ipx}{\hbar}}dp\\ \psi(x)^\star &=& \frac{1}{\sqrt{2\pi \hbar}} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}dq \end{eqnarray*}

(where p and q are almost like "dummy" momenta), then you can rewrite the expectation value of momentum as follows:

\begin{eqnarray} \langle p \rangle &=& \int \psi^\star \left(-i\hbar \frac{\partial}{\partial x}\right)\psi dx\\ &=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(q)e^{\frac{-iqx}{\hbar}}\left(-i\hbar \frac{\partial}{\partial x}\right) \tilde{\psi}^\star(p)e^{\frac{ipx}{\hbar}} dpdqdx \end{eqnarray}

Now if you apply the derivative with respect to $x$, you'll spit out a $p$ in the integrand

\begin{eqnarray} &=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(p) \tilde{\psi}^\star(q)e^{\frac{i(q-p)x}{\hbar}} \left(p\right) dpdqdx \end{eqnarray}

and exchanging integration order to integrate over $x$ first -- since we know these functions to be $L^2$ integrable --yields the (scaled) dirac delta function:

\begin{eqnarray} &=& \frac{1}{2\pi \hbar} \int \tilde{\psi}^\star(p) \tilde{\psi}^\star(q)\hbar \delta(q-p) \left(p\right) dpdq \\ \langle p \rangle &=&\frac{1}{2\pi } \int \vert\tilde{\psi}(p)\vert^2 p dp \end{eqnarray}

There's a missing factor of $2\pi$ in there, but I trust you'll find it if you do it carefully by hand.

$\endgroup$
2
  • $\begingroup$ Sorry, but where does $q$ come from? $\endgroup$
    – Thiago
    Mar 10, 2014 at 22:40
  • 2
    $\begingroup$ p and q both mean momentum. When your first equation is rewritten, an expression with p is inserted in two places. You would not know which p goes with which integral. So one of the expressions was rewritten using q. $\endgroup$
    – mmesser314
    Mar 11, 2014 at 4:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.