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.
