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.

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

1 Answer 1


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.

  • $\begingroup$ 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

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