In what sense is $\sqrt{ {\bf p}^2c^2 +m^2 c^4}$ the Hamiltonian of special relativity? This Hamiltonian is used in the derivation of the Klein Gordon equation. How is this a Hamiltonian when it doesn't even have a position-dependent potential term? Is this the free-particle Hamiltonian? But a free particle in SR evolves exactly the same as a free particle in Newtonian mechanics, i.e. in a straight line. Why would the free-particle Hamiltonian be different in SR then?
Also, this Hamiltonian is the time component of the four-momentum, i.e. it's $m\frac{dt}{d\tau}$, where $\tau$ is proper time. What does this thing have to do with a Hamiltonian, which is supposed to be something that produces Hamilton's equations?
EDIT- ok let's say this is the Hamiltonian. Then,
$$\frac{dx}{dt}=\frac{d}{dp_x} \sqrt{p_x^2+p_y^2+p_z^2 +m^2}=p_x (p_x^2+p_y^2+p_z^2+m^2)^{\frac{-1}{2}}.\tag{1}$$
However, $\frac{dx}{dt}$ is clearly supposed to be $\frac{p_x}{m}$, assuming $p_x=mv_x$.
But it seems like I'm interpreting this Hamiltonian the wrong way. $p_x$ isn't $mv_x$. Instead it's $m\frac{dx}{d\tau}$
In that case, I got two questions:

*

*How come this $p_x$ still translates to $\frac{d}{dx}$ in the Klein-Gordon equation, even when it's not the usual momentum operator. It's the relativistic momentum operator


*I suppose this Hamiltonian predicts the equations of motion of the position co-ordinates $x$, $y$, $z$ and $t$, and the time parameter of this Hamiltonian is $\tau$. How come the time-parameter in the Klein Gordon equation is not $\tau$ (it's $t$ instead)?
 A: *

*The Hamiltonian
$$H~=~\sqrt{{\bf p}^2c^2+ m_0^2 c^4}\tag{A}$$
is the correct Legendre transform of the standard Lagrangian
$$L=-\frac{m_0c^2}{\gamma}, \qquad \gamma~:=~\frac{1}{\sqrt{1-\frac{\dot{\bf x}^2}{c^2}}}, \qquad  \dot{\bf x}~:=~\frac{d{\bf x}}{dt},  \tag{B}$$
for a relativistic point particle. Notice that $t$ is coordinate time, not proper time $\tau$.


*The formulations (A) and (B) are Lorentz covariant, although not manifestly Lorentz covariant, as time and space are treated differently. Concerning manifestly Lorentz covariant formulations, see e.g. this, this, this Phys.SE posts and links therein.


*OP's eq. (1) is correct as it is. Remembers that $H~=~\gamma m_0 c^2$, so that
$$ \dot{\bf x}~=~\frac{\partial H}{\partial {\bf p}}~=~\frac{{\bf p} c^2}{H}~=~\frac{{\bf p}}{\gamma m_0}. \tag{1}$$


*The formulas $$\hat{\bf x} ~=~{\bf x},  \qquad \hat{\bf p} ~=~ \frac{\hbar}{i}\nabla, \tag{C}$$ are just the Schrödinger representation, cf. e.g. this Phys.SE post.
A: If you're satisfied with $S=m\int d\tau$ being the action for a relativistic free particle, you can confirm that this is the correct Hamiltonian that you get when you perform a Legendre transformation. Or, you can verify that this Hamiltonian indeed produces the correct equations of motion when you write down the Hamilton's equations for this Hamiltonian.
As for it not containing a term dependent upon the position, the classical free-particle Hamiltonian doesn't contain it either. It simply reflects the fact that the canonical momentum is conserved. There are many reasons as to why you need a different Hamiltonian for a relativistic free particle even if the trajectories of relativistic and non-relativistic particles are both straight lines. The simplest one is that it's the Hamiltonian that gives you the relation between velocity and momentum in canonical formulation. This relation is obviously different in relativistic and non-relativistic cases. More broadly, why would you expect the non-relativistic Hamiltonian to work for a relativistic when the two systems have a completely different symmetry group?
A: For your edited questions:
We impose the condition
$$[x_i, p_j]=i\delta_{ij},\ \ E\leftrightarrow i\frac{d}{dt}$$
to quantize classical system. Here, the important point is $p_i$ is a canonical momentum of $x_i$. When we consider the Newtonian Lagrangean
$$H=\frac{mv_i^2}{2}+V(x),$$ the canonical momentum is $p_i^{NR}=mv_i$, and $[x_i, p^{NR}_j]=i\delta_{ij}$ is the quantizating condition. Thus we can identify  $p^{NR}_i\leftrightarrow \frac{1}{i}\frac{\partial}{\partial x_i}$.
On the other hand, in the case of relativistic mechanics, its canonical momentum is
$$p_i^R=m\gamma v_i,$$
so we have to impose the condition
$$[x_i, p^{R}_j]=i\delta_{ij}.$$
In this case, there is a correspondence $p_i^R\leftrightarrow \frac{1}{i}\frac{\partial}{\partial x_i} .$
The important point is that, following the canonical quantization formulation, the conjugate momentum is always interpreted as the derivative of the coordinates. Replacing the Newtonian momentum with a coordinate derivative is not proper quantization.
At this point, let’s revisit the quantization procedure for the energy $E\leftrightarrow i\frac{d}{dt}$. What is important in relativistic theory is that the theory becomes covariant under the Lorentz transformation. By no means does every time variable “$t$” that appears in relativity have to be replaced by proper time “$\tau$”. When considering the equations of motion, the time variables are replaced by proper time because otherwise the theory would not be covariant.
With these in mind, let's revisit the quantization law of energy. We already know the quantizing procedure for momentum:
$$p_i^R\leftrightarrow \frac{1}{i}\frac{\partial}{\partial x_i}. $$
On the other hand, four momentum must be covariant under the Lorentz transformation,
$$p^\mu=\begin{pmatrix}
E \\
p_i^R
\end{pmatrix}
\overset{quantize}{\longrightarrow} \begin{pmatrix}
? \\
-i\partial/\partial x_i
\end{pmatrix} $$
so under the appropriate quantization formulation, the above (?) term and $\partial/\partial x_i $ must form Lorentz covariant four vectors. (Because the left-hand side, $p^\mu$, is Lorentz covariant.)
We already know that $$\partial^\mu =\begin{pmatrix}
d/dt \\
-\partial/\partial x_i
\end{pmatrix} $$
is Lorentz covariant, so the usual quantizing procedure $E\leftrightarrow i\frac{d}{dt}$ is appropriate even in the sense of covariance. In short, relativistic quantizing procedure is summarized by following formula:
$$p^\mu\leftrightarrow i\partial^\mu.$$
Now, consider what would have happened if we had taken
$$E\leftrightarrow i\frac{d}{d\tau}$$as the quantization rule. In this case, the right hand side of
$$p^\mu=\begin{pmatrix}
E \\
p_i^R
\end{pmatrix}
\overset{???}{\longrightarrow} \begin{pmatrix}
i d/d\tau \\
-i\partial/\partial x_i\end{pmatrix}$$
does not preserve Lorentz covariance, so the covariance of the theory is lost. In this sense, too, the quantum counterpart of energy $E$ must be $id/dt$. It is never $id/d\tau$.
Finally, it should be pointed out that the energy $E$ is not Lorentz invariant quantity so there is no need to be $E\leftrightarrow i\frac{d}{d\tau}$.
A: P.A.M. Dirac's relativistic energy equation $ E^2 = (p_x^2 + p_y^2 + p_z^2)c^2 + m_0^2c^4 $ is the basis of the Hamiltonian of elementary particle such as the electron, yielding the Klein-Gordon equation. If we measure distances in light-seconds instead of in meters, c = 1 drops out of the equation, yielding the energy as a sum of four squares: $ E^2 = p_x^2 + p_y^2 + p_z^2 + m_0^2 $. The Klein-Gordon equation reduces then, after rearrangment, to $$ {\partial^2\Phi \over \partial t^2} - {\partial^2\Phi \over \partial x^2} - {\partial^2\Phi \over \partial y^2} - {\partial^2\Phi \over \partial z^2} = -{m_0^2\Phi \over {\hbar^2}},  $$ which is an eigenequation of the D'Alembert operator. This equation can be factored and solved using Dirac's trick. The massive particles are therefore the steady-state eigenfunctions of the D'Alembert operator, and their masses are the corresponding eigenvalues.
