Why does applying the kinetic energy operator to a free particle result in a divergent integral?

The wavefunction of a free particle is just

$$\psi = Ae^{i(kx-\omega t)}$$

and when you plug this into the Schrodinger equation you get the dispersion relation

$$E = \frac{\hbar^2 k^2}{2m}$$

However, using the kinetic energy operator to get the expected value for the kinetic energy leads to a divergent integral

$$\hat{T} = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$$ $$\left< T \right> = \int_{-\infty}^{\infty}\psi^*\hat{T}\psi dx$$ $$\left< T \right> = \frac{A^2\hbar^2k^2}{2m} \cdot \int_{-\infty}^{\infty} dx$$

Why doesn't this approach work?

• You have not (and indeed cannot) normalised your wavefunction. For expectation values you need normalized wavefunctions. May 14 at 13:07

$$\langle\hat A\rangle = \int \psi^* \hat A \psi ~ d^3x$$
assumes that the wavefunction $$\psi$$ is normalised. If it is not normalised you need to use:
$$\langle\hat A\rangle = \frac{\int \psi^* \hat A \psi ~ d^3x}{\int \psi^* \psi ~ d^3x}$$