# Does an electron have a frequency (and hence an energy)?

The formulation is provocative, the question is similar to the question here. There I can follow the question, but not the answers, which for me imply that an electron in a momentum eigenstate does not have a certain energy.

In other words again: in Wikipedia and many textbooks you can find the dispertion relation $$\omega (k)=\frac{\hbar k^2}{2m}$$ which is a consequence of the Schrödinger equation $$-{\frac {\hbar ^2}{2m}}\nabla ^{2}\ \psi (r,t)=i\hbar {\frac {\partial }{\partial t}}\psi (r,t)$$ for free particles if you set $$p=\hbar k,\quad E=\hbar \omega$$ On the other hand the relativistic dispersion relation is $$\omega (k)=\sqrt{ k^2 c^2 + \frac{1}{\hbar^2} m^2 c^4 }$$ The non-relativistic approximation with $$k^2 c^2 \ll \frac{1}{\hbar^2} m^2 c^4$$ then is , because $$\sqrt{1+x} = 1 + \frac{1}{2} x + O(x^2)$$ , $$\omega (k) \simeq \frac{mc^2}{\hbar} + \frac{\hbar k^2}{2m}$$ which differs from the first equation by a term also known as Compton frequency.

So my question is: Is the first equation a wrong approximation? Or a different formulation would be: is it true that there is no experiment that can detect an electron's frequency (hence energy)? Only wavelengths can be observed like in the double slit experiment? Or in a more philosophical way: Does an electron have a frequency if there is no experiment that can detect it?

If the first equation is wrong, shouldn't the usual Schrödinger equation in position base read $$i\hbar {\frac {\partial }{\partial t}}\Psi (\mathbf {r} ,t)=\left[{\frac {-\hbar ^{2}}{2\mu }}\nabla ^{2}+V(\mathbf {r} ,t)+mc^2\right]\Psi (\mathbf {r} ,t)$$ yielding the correct non-relativistic limit for the frequency when setting $$\lambda(p) = \frac{h}{p}, \quad \omega(p) = \frac{p^2}{2m\hbar} + \frac{mc^2}{\hbar}$$ rather than $$i\hbar {\frac {\partial }{\partial t}}\Psi (\mathbf {r} ,t)=\left[{\frac {-\hbar ^{2}}{2m}}\nabla ^{2}+V(\mathbf {r} ,t)\right]\Psi (\mathbf {r} ,t)$$ The last equation has solutions with negative energies when introducing an attracting force $$V(r)$$ like in the hydrogen problem. With $$mc^2$$ you get purely positive energy eigenvalues which better fits to a relationship of energy and measurable information, the latter one being positive by definition and limited by the dimension of the Hilbert space, which itself is positive by definition.

• I don't really understand your question, you give a formulation for finding the energy and frequency of an electron and ask if it exists. – Ismasou Aug 6 '18 at 21:21
• See the linked question and there the comments "So you can just redefine energy by removing this value and take energy levels by referring to the constant term as the 0-energy level" and "A constant energy term will affect nothing since only energy differences are physically meaningful." They argue that energy has no physical meaning. But I say that it has a meaning, and the usual forms of the non-relativistic Schrödinger equation and dispertion relation are not correct. – Harald Rieder Aug 7 '18 at 12:09