i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf{r})\psi

In order to treat quantum mechanics in a Lorentz invariant framework, we must begin with the relativistic energy-momentum relationship:

E^2 = p^2c^2 + m_0^2c^4

Quantum mechanical operators for the energy and momentum can be defined as:

\begin{align*}

\hat{E}=-i\hbar\frac{\partial}{\partial t} \\

\hat{p}=-i\hbar\nabla

\end{align*}

For a plane wave solution of the form

`\psi(\mathbf{x},t)=A\exp(i\mathbf{k}\cdot\mathbf{x}-i\omega t)`

, these operators give the expected relationships for energy and momentum as their eigenvalues:

\begin{align*}

\hat{E}\psi = -i\hbar i\omega\psi = \hbar\omega\psi \\

\hat{p}\psi = -i\hbar i\mathbf{k} = \hbar\mathbf{k}

\end{align*}

These operators (squared) can be substituted directly into the relativistic energy-momentum relationship. From now on, I'm going to simplify matters by using natural units

`(\hbar=c=1)`

and writing the rest mass `m_0=m`

.

\begin{align*}

\hat{E}^2\psi = \hat{p}^2\psi+m^2\psi\\

\Rightarrow -\frac{\partial^2}{\partial t^2}\psi = -\nabla^2\psi + m^2\psi \\

\Rightarrow \left(-\frac{\partial^2}{\partial t^2} + \nabla^2 -m^2\right)\psi = 0 \\

\Rightarrow \left(\partial_\mu\partial^\mu + m^2\right)\psi = 0 \\

\Rightarrow \left( \Box +m^2\right)\psi(\mathbf{x},t) = 0

\end{align*}

Where

`\Box = \partial_\mu\partial^\mu = \frac{\partial^2}{\partial t^2}-\nabla^2`

.This is the Klein-Gordon Equation, and has plane-wave solutions of the form:

\begin{align*}

\psi(\mathbf{x},t)=N\exp(-iEt+i\mathbf{p}\cdot\mathbf{x}) \\

= N\exp(-ip\cdot x) \\

\mathrm{where~}p\cdot x = p_\mu x^\mu = Et - \mathbf{p}\cdot\mathbf{x}

\end{align*}

Since

`E^2=p^2+m^2`

, the energy solutions are:

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

i.e. there are positive and negative energy states associated with the Klein-Gordon equation. Before discussing the meaning of these, I'd like to first obtain the probability current for the Klein-Gordon equation. For the Schrödinger equation, the probability density is the square of the wavefunction,

`\rho = |\psi|^2 = \psi\psi^\dagger`

. If we take the Klein-Gordon equation as:

\frac{\partial^2\psi}{\partial t^2}-\nabla^2\psi + m^2\psi = 0

We can multiply by

`\psi^\star`

and subtract `\psi`

the complex conjugate of the Klein-Gordon equation. This results in:

\frac{\partial\rho}{\partial t}+\nabla\cdot\mathbf{j}=0

Where:

\begin{align*}

\rho=i\left[\psi^\star \frac{\partial\psi}{\partial t}-\left(\frac{\partial\psi^\star}{\partial t}\right)\psi\right] \\

\mathbf{j} = \frac{1}{i}\left[\psi^\star\nabla\psi - (\nabla\psi^\star)\psi\right]

\end{align*}

In four-vector notation, one can write the above as:

\partial_\mu j^\mu = 0

with

j^\mu = (\rho, \mathbf{j}) = i\left[\psi^\star\partial^\mu\psi - (\partial^\mu\psi^\star)\psi\right]

Since

`\psi`

is Lorentz invariant, and `\partial^\mu`

is a contravariant four-vector, `j^\mu`

is also contravariant.The spatial current j is identical to the Schrödinger current, but for the Klein-Gordon equation, the probability density contains time derivatives since K.G. is second-order in time derivatives. The probability density

`\rho`

is not, therefore, constrained to be positive definite. For plane wave solutions:

\begin{align*}

\psi=N\exp(-iEt-i\mathbf{p}\cdot\mathbf{x}) \\

\rho = 2|N|^2E \\

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

\Rightarrow \rho = \pm 2|N|^2\sqrt{\mathbf{p}^2+m^2}

\end{align*}

The probability density is therefore positive for positive-energy states and negative for negative energy states. The Klein-Gordon equation was abandoned for some time as a result of this, but the negative energy states may be interpreted as representing particles moving backwards in time, corresponding to anti-particles moving forwards in time! (The Feynman interpretation). The Klein-Gordon equation describes the relativistic quantum mechanics of (massless) spin-zero particles. (And no such fundamental particles exist within the Standard Model, but the neutral pion, composed of

`\frac{1}{\sqrt{2}}(u\bar{u}-d\bar{d})`

can be considered at low energies. In order to deal with fermions (spin-1/2 particles) we need to look at the Dirac equation, which I might write about another time.