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

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.

• I recommend reading chapter 8 of "Modern Quantum Mechanics" by J.J. Sakurai. Commented Aug 7 at 13:26
• what correspondence principle did you use? Commented Aug 7 at 14:04
• @TheTiler what do you mean by correspondence? Commented Aug 7 at 14:17
• See en.wikipedia.org/wiki/Correspondence_principle The correspondence between classical values ​​and operators is $E\leftrightarrow i\hbar\frac{\partial}{\partial t}$, $\vec{p} \leftrightarrow -i\hbar \vec{\nabla}$ Commented Aug 7 at 14:27
• Already your equation does not depend on time...., in addition in relativity $E-mc^{2}=E_{k}=(\gamma -1)mc^{2}\approx E_{k_{classic}}$ Commented Aug 7 at 14:48

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