Electromagnetic Coupling

The coupling of scalar particles to an electromagnetic field is done using the gauge invariant approach explained in the previous section. Using the minimal coupling, $\hat{p}_\mu\rightarrow \hat{p}_\mu -e A_\mu$, and writing the Klein-Gordon equation so that all terms involving the electromagnetic potential appear on one side, we have

$$[\partial^\mu\partial_\mu + m^2]\phi(x) = -\hat{V} \phi(x) ,$$ (4.69)

where the potential operator $\hat{V}$ is

$$\hat{V} = ie\, ( \partial_\mu A^\mu + A^\mu \partial_\mu) - e^2 A^\mu A_\mu .$$ (4.70)

As the Klein-Gordon equation is second order, the coupling term has a quite complicated structure. It contains gradients $\partial_\mu$ and moreover is nonlinear in $A_\mu$ because of the quadratic last term.

Using minimal coupling, the conserved current becomes

$$j^\mu = \phi^*\left(i\hbar\partial^\mu - \frac{e}{c}A^\mu\ri... ...hi \left(i\hbar\partial^\mu + \frac{e}{c}A^\mu\right)\phi^* .$$ (4.71)

Multiplying by $e/2m$ to get the usual normalization gives the time and space components of the conserved current

$$\rho = \frac{i\hbar e}{2mc^2} \left(\phi^*\partial_t\phi - \phi\partial_t\phi^* -\frac{2e}{i\hbar} A_0\phi^*\phi\right) ,$$ (4.72)

$$\vec{j} = \frac{\hbar e}{2im} \left(\phi^*\vec{\nabla}\phi - \phi\vec{\nabla}\phi^* + \frac{2e}{i\hbar c} \vec{A}\phi^*\phi\right) .$$ (4.73)

Returning to our initial normalization and using natural units, we have

$$\fbox{$\displaystyle \rho = \phi^*(x)[i\stackrel{\leftrightarrow}{\partial}_0 -2eA^0(x)]\phi(x) $}\ .$$ (4.74)

For a stationary state, $\psi(\vec{x},t) = \psi(\vec{x}) e^{-iEt/\hbar}$, equation 4.72 gives

$$\rho = \frac{i\hbar e}{2mc^2} \left(-\frac{iE}{\hbar} -\frac{iE}{\hbar} -\frac{2e}{i\hbar}A_0 \right) \vert\psi(\vec{x})\vert^2$$ (4.75)

$$= e \frac{E-eA_0}{mc^2} \vert\psi(\vec{x})\vert^2 .$$ (4.76)

If $E>eA_0$, $\rho>0$ and the charge density has the same sign as $e$ of the particle. But if $E<eA_0$, $\rho<0$ and the charge density has the opposite sign as $e$ of the particle. In this case the field is strong and we would need to invoke field theory to show that particles are created in this case.