Exploring Quantum Physics - Week 1 Question 7

Question:

Suppose a particle has wavefunction $\psi(x, t = 0) = Ae^{-\frac{x^2}{2l^2}}$. What is the average value (expectation value) of $\hat{p}$, $\langle \hat{p} \rangle$, for this state at $t = 0$ ?

Solution:

Recall the definition of expectation of an operator is given by $\langle \hat{p} \rangle = \int\limits_{-\infty}^{\infty}{\psi^{*}(x)\hat{p}\psi(x)dx}$, so again, the problem becomes evaluating this integral, so let's do it

$\begin{eqnarray*} \hat{p} &=& -i\hbar \frac{\partial}{\partial x} \\ \hat{p}\psi &=& -i\hbar \frac{\partial}{\partial x} A e^{-\frac{x^2}{2l^2}} \\ &=& -i\hbar A e^{-\frac{x^2}{2l^2}} \frac{-2x}{2l^2} \\ &=& \frac{i\hbar x}{l^2} A e^{-\frac{x^2}{2l^2}} \end{eqnarray*}$

With that, we can move on to evaluate the integral

$\begin{eqnarray*} \langle \hat{p} \rangle &=& \int\limits_{-\infty}^{\infty}{\psi^{*}(x)\hat{p}\psi(x)dx} \\ &=& \int\limits_{-\infty}^{\infty}{(Ae^{-\frac{x^2}{2l^2}})(\frac{i\hbar x}{l^2} A e^{-\frac{x^2}{2l^2}}) dx} \\ &=& \frac{A^2 \hbar i}{l^2}\int\limits_{-\infty}^{\infty}{(e^{-\frac{x^2}{2l^2}})(x e^{-\frac{x^2}{2l^2}}) dx} \\ &=& \frac{A^2 \hbar i}{l^2}\int\limits_{-\infty}^{\infty}{x e^{-\frac{x^2}{l^2}} dx} \\ &=& 0 \end{eqnarray*}$

Surprise! The last line comes from the fact that the integrand is an odd function. One could have think of that as the scaled mean of some Gaussian, too!