How to find the position expectation value from the Fourier transform of a wave function?

Question

Here's my reasoning:

$$\langle \hat{x} \rangle=\int_{-\infty}^{\infty}\psi^*(x,t)\hat{x}\psi(x,t)$$ $$=\frac{1}{2\pi}\int_{-\infty}^{\infty}dx(\int_{-\infty}^{\infty}\phi(k,t)e^{ikx}dk)^*\hat{x}\int_{-\infty}^{\infty}dk'\phi(k',t)e^{ik'x}$$

Where $\phi(k,t)$ is the Fourier transform of $\psi$ (I included the $\omega t$ term in $\phi$)

$=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dkdk'\phi^*(k,t)\phi(k',t)\int_{-\infty}^{\infty}xe^{-i(k-k')x}dx$

The integral $\int_{-\infty}^{\infty}xe^{-i(k-k')x}dx=\frac{(i(k-k')x+1)}{(k-k')^2}e^{-i(k-k')x}\Bigg|_{-\infty}^{\infty}=???$ if $k\neq k'$ and it obviously equals $0$ when $k=k'$.

My understanding is that this limit diverges, but that doesn't really make sense since the expectation value does exist.

Answer 1

The last integral converges, not to a function but to a distribution: the fourier transform of a monomial (here $x$) is the $n$-th derivative of a Dirac, see Wikipedia.

You can get the intuition by starting from: $$ \int_{-\infty}^{+\infty} e^{-iux}du = 2\pi \delta(x) $$ Then: $$ \frac{d}{dx}\int_{-\infty}^{+\infty} e^{-iux}du =\int_{-\infty}^{+\infty} -iu e^{-iux}du = 2\pi \frac{d}{dx} \delta(x) = 2\pi \delta^{(1)} (x) $$ So your last integral would evaluate to : $$ \int_{-\infty}^{+\infty} x e^{-i\left(k-k'\right)x}dx=i 2\pi \delta^{(1)} (k-k') $$

This distribution behaves as the (minus) derivative operator when you apply it to a function (meaning integrating the distribution times the function). You can re-derive this by integrating by parts $\delta^{(1)} (x) f(x)$.

Another nice way to understand this is the representation of the $\hat{x}$ operator in $k$-space as the derivative operator: $$ \hat{x} = i \hat{\partial}_k $$

Then the expectation value in the momentum basis would be : $$ \langle \hat{x} \rangle = i \int_{-\infty}^{+\infty} \psi^*(k) \left(\partial_k \psi(k) \right) dk $$

Answer 2

It comes out to be the derivative of a Dirac delta function $$ \int x\exp(ikx) dx = -i\partial_k \int \exp(ikx) dx = -i\partial_k 2\pi\delta(k) $$ When you use the result inside the integral with the wave functions, integration by parts transfers the derivative to one of the wave functions.