The Klein-Gordon Equation is given as follows, using natural units ($\hbar \to1, c \to 1$):
$-\frac{\partial^2 \Psi}{{\partial t}^2} = -\nabla^2 \Psi + m^2 \Psi $
The way I tried to derive this is as follows, starting from the time-dependent Schrodinger equation:
$i\frac{\partial \Psi}{\partial t} = \hat{H} \Psi$,
where $\hat{H} = \frac{\hat{p}^2}{2m}$ and $\hat{p} = i\nabla$
Now, we are going to make the following substitution:
$\hat{H} = \frac{\hat{p}^2}{2m} \to \hat{H}^2 = \hat{p}^2+m^2$
Squaring the time-dependent Schrodinger equation gives:
$-\frac{\partial^2 \Psi}{{\partial t}^2} = \hat{H}^2 \Psi = (\hat{p}^2 + m^2) \Psi = (-\nabla^2 + m^2)\Psi = -\nabla^2\Psi + m^2\Psi $
as stated in the beginning.
However, is my derivation valid? For example, I squared the time-dependent Schrodinger equation, but this feels off to me, because I am not squaring the wave function but only the partial derivative and the $\hat{H}$ operator. Are there any flaws in this?
(I am not a masters student, so please avoid very complicated symbols).