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?

  • 5
    $\begingroup$ You have not (and indeed cannot) normalised your wavefunction. For expectation values you need normalized wavefunctions. $\endgroup$
    – mike stone
    Commented May 14, 2022 at 13:07

1 Answer 1


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.


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