## Sunday, 27 April 2008

### Relativistic Quantum Mechanics: The Klein-Gordon Equation

The Schrödinger equation, used in non-relativistic quantum mechanics, is not Lorentz invariant. This can be seen upon inspection since it is first order in time derivatives, and second order in spatial derivatives. Special relativity must treat space and time on an equal footing, i.e. they must be of the same order. Here's the Schrödinger equation for reference:
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.