# Potential well: Why non-relativistic kinetic energy?

When studying a potential well, the energy is defined as that: $$E=\frac{\pi h^2 n^2}{2ma^2}$$ and then some books say $$E=\frac{p^2}{2m}$$. Why energy is just kinetic energy and we aren't considering relativistic energy? Is that because we are talking about non-relativistic quantum mechanic?

• Neither of your expressions for $E$ is the definition of energy... but rather a particular type of energy. $E_n=\frac{\pi h^2 n^2}{2ma^2}$ is the energy of a particular eigenstate of an infinite square well, treated quantum mechanically. $E_{nr,kin}=p^2/(2m)$ is (as you suggest) the non-relativistic kinetic energy. Commented Jun 9, 2020 at 16:18

Yes! the derivation of Energy of any system, as:

$$E=\frac{\pi h^2 n^2}{2ma^2}$$

is derived, ignoring the relativistic effect on the system. Now, considering the relativistic parameter, the energy of any system, in accord with QM is given by:

$$H = E = \sqrt {c^2 p.p + (mc^2)^2}$$

and in this case, it is suitable to call it Hamiltonian, as some specification provives the equality between Hamiltonian and Total Energy of the system.

And correspondingly: $$i\hbar \frac{\partial \psi}{\partial t} = \sqrt {c^2 p.p + (mc^2)^2} \psi$$

or,

$$E^2\psi = c^2p.p\psi + (mc^2)^2\psi$$ ($$i.e.$$ Klein–Gordon equation)

Gives the necessary information of the system, as wave-function, accompanying the total energy of the system.

• What are the eigenvalues of energy in a quantum-relativistic potential well? Commented Jun 20, 2020 at 20:47
• The term, $\sqrt{c^2p.p+(mc^2)^2}$ is the corresponding eigenvalue at that quantum state. Commented Jun 20, 2020 at 20:54