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?
Best Answer
The equation for the expectation value of an operator:
$$ \langle\hat A\rangle = \int \psi^* \hat A \psi ~ d^3x $$
assumes that the wavefunction $\psi$ is normalised. If it is not normalised you need to use:
$$ \langle\hat A\rangle = \frac{\int \psi^* \hat A \psi ~ d^3x}{\int \psi^* \psi ~ d^3x} $$
The problem is that the infinite plane wave is not a physical state and cannot be normalised, so you can't simply calculate its kinetic energy. This isn't a problem because the infinite plane wave would describe an infinitely delocalised particle and in real life no particles are infinitely delocalised. A real particle would have a wavefunction that is a wave packet of some form.