Klein-Gordon Equation

Derivation

The Schrodinger equation is $$i\frac{\partial}{\partial t}\psi = H\psi,$$ where \(H\) is the Hamiltonian $$H=T+V=\frac{p^2}{2m}+V(\vec{x})$$ for a non-relativistic particle. In particular, for a free particle, the Hamiltonian is $$H=\frac{p^2}{2m}=\frac{\nabla^2}{2m}$$ To be consistent with special relativity, relativistic quantum theory must be Lorentz covariant. The relativistic energy of a free particle is $$E = \sqrt{p^2 + m^2}.$$ If we substitute this into our Schrodinger equation, we get $$i\frac{\partial}{\partial t}\psi=\sqrt{m^2-\nabla^2}\psi.$$ To get rid of the square root, we take square instead \begin{align} -\frac{\partial^2}{\partial t^2}\psi = (m^2-\nabla^2)\psi\\ (m^2+\Box^2)\psi = 0, \end{align} where \(\Box^2 \equiv \frac{\partial^2}{\partial t^2}-\nabla^2\) is the D’Alembertian operator. This is the Klein-Gordon equation for a free particle. The solutions are plane wave solutions $$\psi=\exp(-i\omega t + i\textbf{k}\cdot\textbf{x})=\exp(-ik\cdot x).$$

Negative Energy

This includes both positive and negative energy root $$H=\pm\sqrt{m^2+p^2}.$$ Negative energy solutions can be interpreted as the particle with positive energy but propagating backward in time $$\exp(-iEt)=\exp[-i(|E|)(-t)].$$ This is known as the Feynman-Stuckelberg Interpretation. Another interpretation can be obtained by minimal coupling. If the particle in the Klein-Gordon equation is coupled with E.M. fields, \(p^{\mu}\to p^{\mu}-qA^{\mu}\). By first quantization rule, \(\partial^\mu\to\partial^\mu + iqA^\mu\). Then, the Klein-Gordon equation becomes $$\left[(\partial^\mu + iqA^\mu)(\partial_\mu+iqA_\mu)+m^2\right]\phi_{-}=0,$$ where \(\phi_{-}\) is the negative energy solution of the Klein-Gordon equation. If we take complex conjugate on both side, $$\left[(\partial^\mu-iqA^\mu)(\partial_\mu-iqA_{\mu} + m^2)\right]\phi^*_{-}=0.$$ The complex conjugate of the negative energy solution is $$\phi^*_{-}=N\left[e^{i(\vec{p}\cdot\vec{x}+|E|t)}\right]^*=Ne^{-i(\vec{p}\cdot\vec{x}+|E|t)}=Ne^{i[(-\vec{p})\cdot\vec{x}-|E|t]}.$$ We obtained the usual phase \(e^{-i|E|t}\), with positive energy. The complex conjugate turned the sign of the charge of the particle \(+iqA_\mu\to-iqA_{\mu}\). But mass is still the same, \(m\). Thus, we say that the negative energy solution represents a particle of opposite charge but identical mass of the positive energy solution. We call them, antiparticles.

Current Conservation

\begin{align} (m^2+\Box^2)\psi &= 0\\ \psi^*(m^2+\Box^2)\psi - \psi((m^2+\Box^2)\psi)^* &= 0\\ \psi^*\left(\frac{\partial^2}{\partial t^2}-\nabla^2\right)\psi - \psi\left(\frac{\partial^2}{\partial t^2}-\nabla^2\right)\psi^* &= 0\\ \frac{\partial}{\partial t}\left[\psi^*\frac{\partial \psi}{\partial t}-\psi\frac{\partial \psi^*}{\partial t}\right] + \vec{\nabla}\cdot\left[-\psi^*\vec{\nabla}\psi+\psi\vec{\nabla}\psi^*\right]&=0.\\ \end{align} If we define the density as $$\rho\equiv\left[\psi^*\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^*}{\partial t}\right]$$ and the current density as $$\vec{J}\equiv \left[\psi^*\vec{\nabla}\psi-\psi\vec{\nabla}\psi^*\right],$$ we get the continuty equation $$\frac{\partial}{\partial t}\rho-\vec{\nabla}\cdot\vec{J}=0.$$ But \(\rho\) cannot be the probability density as it is not positive definite. But they can be charge density.