Is $\hat{H}^2 = c^2\left(\hat{p}^2+m^2c^2\right)$ valid quantum physics?

Question

From Schrödinger equation we know that $\hat{H}=\hat{E} $ for non-relativistic systems.

For relativistic scenarios we use the usual $\hat{E}^2 = c^2\left(\hat{p}^2+m^2c^2 \right)$. From this we can derive Klein-Gordon and Dirac equations.

Nonetheless, what would happen if we replace $\hat{E}$ with $\hat{H}$? Is that valid physics? The immediate result would be (for a free particle):

$$\left(\frac{\hbar^2}{4m^2c^2}\nabla^4 + \nabla^2 - \frac{m^2c^2}{\hbar^2}\right)\psi = 0$$

Where $\nabla^4$ is the biharmonic operator. I guess this equation does not apply as we would be comparing something classical to something relativistic, but I am curious to know if this is the case and what would the equation represent if not.

Answer 1

We start off with the relativistic square-rooted energy-momentum relation:

$$ \hat{H} = c\sqrt{ m^2 c^2+ \hat{p}^2}$$

For non-relativistic velocities $\hat{p}$ is small, so one can develop the right side of the square-rooted equation:

$$ \hat{H} = mc^2\sqrt{ 1 + \frac{\hat{p}^2}{m^2c^2}} \approx mc^2\left(1 + \frac{\hat{p}^2}{2 m^2c^2}\right) = mc^2 + \frac{\hat{p}^2}{2m}$$ So when you plug in for $\hat{H}$ on the left side not $\frac{\hat{p}^2}{2m}$, but $mc^2 + \frac{\hat{p}^2}{2m}$ because $\hat{H}$, as it is basically relativistic, also includes the energy at rest, then the result turns out to be a trivial identity.

Otherwise, plugging without precaution in a relativistic equation a non-relativistic relation, leads to contradictions.