# Classical free particles (i.e. Klein-Gordon)

*TL;DR – The extended Hamiltonian for a classical free particle is the classical version of the Klein-Gordon equation.*

After having seen the classical version of antiparticles we will see the classical version of the Klein-Gordon equation. This is the essentially the extended Hamiltonian for the free particle.

Let’s look at the details.

1. Extended Hamiltonian for the free particle

The extended Hamiltonian for a free particle is:

\begin{equation}

\begin{aligned}

\mathcal{H}(x^i,t,p_j,E) = \frac{1}{2m} \left(p^2 – \left(\frac{E^2}{c^2}\right)\right) + \frac{1}{2} mc^2\\

\end{aligned}

\label{ExtendedHamiltonFree}

\end{equation}

Remember that, for valid physical states, we have $\mathcal{H}=0$. Therefore:

\begin{equation}

\begin{aligned}

\left(p^2 – \left(\frac{E^2}{c^2}\right)\right) = – m^2c^2\\

\end{aligned}

\label{FourMomentum}

\end{equation}

If you studied special relativity, you recognized the norm of the four-momentum. Moreover:

\begin{equation}

\begin{aligned}

\frac{E^2}{c^2} &= p^2 + m^2c^2\\

E &= \pm c \sqrt{p^2 + m^2c^2} \\

\end{aligned}

\label{HamiltonFree}

\end{equation}

This is the energy as a function of momentum: it the relativistic Hamiltonian for a free particle. But in this case the energy can both be positive or negative. Note that:

\begin{equation}

\begin{aligned}

\frac{dt}{ds} = – \frac{\partial \mathcal{H}}{\partial E} = \frac{E}{mc^2}

\end{aligned}

\label{TimeEvolution}

\end{equation}

So if the energy is negative, then the parametrization is anti-aligned to the time: $t$ decreases as $s$ increases. As we saw in the previous post, these correspond to anti-particle states.

So, not only the extended Hamiltonian recovers relativistic free-particles, it also recovers anti-particles and it is has much nicer form than the relativistic Hamiltonian (i.e. it is quadratic and does not have any square root).

If you use the standard classical to quantum substitutions $\left( p_i, -E \right)$ to $\left(-\imath \hbar \frac{\partial}{\partial x^i}, -\imath \hbar \frac{\partial}{\partial t} \right)$ we have:

\begin{equation}

\begin{aligned}

\mathcal{H} = &\frac{1}{2m} \left(- \hbar^2 \nabla^2 + \hbar^2 \left(\frac{\partial^2}{c^2\partial t^2}\right)\right) + \frac{1}{2} mc^2 = 0\\

&\frac{1}{c^2}\frac{\partial^2}{\partial t^2}- \nabla^2 + \frac{m^2 c^2}{\hbar^2} = 0\\

\end{aligned}

\label{KleinGordon}

\end{equation}

If you studied quantum field theory, that should look familiar: if we add a $\psi$ it’s the Klein-Gordon equation. (Note: the mass term is positive as we are using the relativity metric convention $(-,+,+,+)$ instead of the particle physics metric convention $(+, -, -, -)$).

So why should we add $\psi$? There is good reason to do this, even in the classical setting.

Suppose you have a phase space distribution $\rho(x^i,t,p_j,E)$. This would correspond to a distribution of matter over space, time, momentum and energy. Consider $\mathcal{H} \rho$, the product between the extended Hamiltonian and the distribution. As we saw, $\mathcal{H}=0$ is a constraint between energy and the other variables so that only states that satisfy that equation are physical. Therefore $\rho$ can only be different than zero for those states where $\mathcal{H}$ is zero. But this means that $\mathcal{H} \rho = 0$ everywhere. In other words: $\mathcal{H} \rho = 0$ can be used as a constraint to find physically meaningful distributions over the extended phase space.

Now, in the quantum case states are always ensembles: they are always distributions. Therefore we use the constraint $\mathcal{H}\psi = 0$. Which, in the case of the free particle, corresponds to the Klein-Gordon equation.

The extended Hamiltonian for a free particle has a much nicer form than the relativistic Hamiltonian one finds in standard textbooks, and it ties in more with quantum concepts. The Klein-Gordon equation is the extended Hamiltonian for a free particle acting on the distribution. The only thing it does is constrain the state so that the norm of the four-momentum is $-m^2c^2$.