Why Use the Non-Relativistic Momentum Operator in Relativistic Quantum Mechanics? In deriving the Klein Gordon equation one starts out with the relativistic energy relation $E^2 = p^2 + m^2$ and substitutes the quantum momentum operator that corresponds to non-relativistic QM, $\hat{p}= -i \frac {∂}{∂x}$ . I was wondering what justifies the use of this when this quantum operator was derived non-relativistically, but the $p$ in the $E^2$ equation is relativistic. I can't imagine how one would go about deriving a quantum operator corresponding to $\gamma mv$. 
 A: @knzhou is correct for the translation aspect, but it is also possible to derive this equality from canonical quantization.
It is easiest to see by considering the Polyakov action for a relativistic particle
$$S = \frac 12 \int d\tau\ \frac{1}{e} \dot x^\mu \dot x_\mu - e m^2$$
Then picking a time direction, we can define the momenta
$$p_\mu = \frac{\dot x_\mu}{e}$$
$$p_e = 0$$
With the Hamiltonian
\begin{eqnarray}
H &=& \frac 12 [p_\mu x^\mu(p) - \frac{1}{e} \dot x^\mu \dot x_\mu + e m^2]\\
&=& \frac e2( p^\mu p_\mu + m^2)\\
&=& \frac e2 (E - p^2 + m^2)
\end{eqnarray}
After some gauge fixing where we set $e = 1$, and some use of constraints, this is just equivalent to $E = p^2 + m^2$. Furthermore, we need to compute the Poisson brackets for the basic phase space quantities. This will be
\begin{eqnarray}
\left\{x^\mu, p_\nu \right\} = \delta^\mu_{\nu}
\end{eqnarray}
since $x$ and $p$ have the same definition as in non-relativistic mechanics. From there, we simply apply the Dirac quantization
$$\left\{x^\mu, p_\nu \right\} \to \frac{1}{i\hbar} [\hat x^\mu, \hat p_\nu]$$
or in other words, 
$$[\hat x^\mu, \hat p_\nu] = \delta^\mu_\nu i\hbar$$
via the Stone-von Neumann theorem, we know that the following are representations of such operators : 
\begin{eqnarray}
\hat x^\mu \psi &=& x^\mu \psi\\
\hat p_\nu \psi &=& -i\hbar \partial_\nu \psi
\end{eqnarray}
A: The operator
$ \hat{p}=-i \hbar \frac {∂}{∂x} \tag{1}$ 
is valid in both cases.
We end up with this operator my cascading from the math of what we consider to be a “well behaved” wave.
From De Broglie’s wave mechanics we could associate to a particle with momentum $p$ a wave with wave number $k$ linked to the momentum by
$p=\gamma mv=h/ \lambda=\hbar k \tag{2}$
Equation (2) is relativistic.
To this point no one knows (“knew?”) what exactly that wave or its actual form are. However, for its form we could postulate it. Starting from a plane wave
$$ψ=e^{i(kx+ω t)} \tag{3}$$
The derivative of this equation with respect to $x$ is
$$\frac {∂ }{∂x} ψ =ikψ \tag{4}$$
Defining the operator
$$\hat{k}=-i\frac {∂ }{∂x}  \tag{5}$$
one gets the Schrodinger equation for the operator $\hat{k}$, or
$$\hat{k}ψ =kψ \tag{6}$$
which is always valid, as long as we have the plane wave in mind when “deriving” our wave mechanics. Now, replacing $k$ by $\hat{k}$ in (2) and considering that that equation is valid in both situations, the operator (1) can be used in both cases for the same reasons.
PS: the operator (1) or (5) are not actually derived. They are postulated and checking their consequences with the experimental results will improve/reduce their merits.
A: The momentum operator is defined as the generator of translations, so we have
$$\hat{p} = - i \partial_x$$
in all cases, relativistic or not, in the same way that the Hamiltonian always generates time translations, $H |\psi \rangle = i \partial_t |\psi \rangle$.
The only thing that relativity changes is the relationship between $\hat{p}$ and the velocity, i.e. we have
$$\hat{p} = \gamma m \hat{v}$$
where the velocity operator is defined as 
$$\hat{v} = \frac{\partial H(p, q)}{\partial p}\bigg|_{p = \hat{p}, q = \hat{q}}$$
and gives the local velocity of the particle. So if we had instead tried to define $\hat{p} = m \hat{v}$ in all cases it would indeed be incompatible with relativity, but that's not what we're doing. 
A: You are absolutely right. It is necessary to replace the usual pulse with the momentum operator $p_{Nonrelativistic} \mapsto -i\hbar \triangledown$ in the expression for the coupling of the energy and momentum of the SRT, then we obtain a completely different equation M2. Which has no flaws in the Klein-Gordon equation.
 $$ \Delta \Psi -\frac{1}{\hbar^{2}}\left [ \frac{m^{4}c^{6}}{\left ( E-U\left ( \overrightarrow{r}\right ) \right )^{2}}-m^{2}c^{2} \right ]\Psi =0 $$
A: In the first place $p=\gamma m v$ is only valid when one ignores interactions. For a particle in an electromagnetic field, the momentum is $p=\gamma m v + eA$
In the second place one wouldn't confound an operator $\hat{p}$ with an eigenvalue $p$. Both are related by
$$\hat{p} \Psi = p \Psi$$
For free particles, the value of the momentum $p$ will be $p=\gamma m v$ or $p=m v$ depending of the state of the particle, but the operator is the same
$$\hat{p} =-i \hbar \frac {∂}{∂x}$$
