For a homework problem we are given the wave function
$$ \Psi(x) = \frac{N}{x^2 + a^2},\ a > 0 $$
and asked to normalize it. Then we are to find the expectation value of $x$. To do so, I normalized the wave function
$$ \Psi(x) = \frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)} $$
Then, following my notes computed
$$ \int_{-\infty}^\infty \Psi(x)^* \langle x \rangle \Psi(x)\,\mathrm dx $$
Given that the wave equation is currently in the $x$ basis I then computed
$$ \int_{-\infty}^\infty
\left(\frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)}\right)^*
x
\left(\frac{2 \sqrt{a^3}}{\sqrt{\pi}(x^2 + a^2)}\right)\,\mathrm dx $$
by plugging it into Wolfram Alpha. Though this results in a solution of $0$.
This does not seem correct. I feel like the solution should depend on $a$ in the very least. Is there some reason this is actually independent of $a$?
Best Answer
You've got the right answer! The wave-function is symmetric about $0$, while the operator $\langle x \rangle = x$ is odd. So you have the product of two even functions and an odd function in your integrand, which results in a composite odd function. Integrating an odd function over a symmetric interval will always give you $0$! Checking the symmetry of your functions is always a good way to figure out if your answer makes sense with these kinds of problems.
Edit: I meant to write $\hat{x} = x$, not $\langle x \rangle = x$. The correct expression for the expectation value is $\langle x \rangle = \int\psi^*\hat{x}\psi dx$.