# Negative energy solutions in Dirac equation for hydrogen atom

Solving Dirac's equation for the hydrogen atom $$^1$$, $$(\boldsymbol{\alpha}\cdot\boldsymbol{p}+\beta\cdot m)\Psi=E\Psi$$ after some mathematical machinery we find the condition for energy eigenvalues $$\varepsilon^2=\frac{1}{1+\left(\frac{\alpha}{1+\gamma+n'}\right)^2}\tag{1}$$ where $$\varepsilon:=\frac{E}{m}\tag{A}$$ $$\lambda=j+\frac{1}{2}\qquad\gamma=-1+\sqrt{\left(j+\frac{1}{2}\right)^2-\alpha^2}\tag{B}$$ and $$n'$$ is a quantum number. Taking the square root in equation $$(1)$$ and using $$(A)$$ and $$(B)$$ only the positive energy solution is retained$$^2$$ $$E_{n'j}=\frac{m}{\sqrt{1+\left(\frac{\alpha}{\sqrt{\left(j+\frac{1}{2}\right)^2-\alpha^2}+n'}\right)^2}}.\tag{2}$$ Equation $$(2)$$ yields the correct non-relativistic result in the first terms of McLaurin expansion for $$\alpha$$. Is there any other argument to reject the negative energy solution of $$(2)$$ i.e.

$$E_{n'j}=\frac{-m}{\sqrt{1+\left(\frac{\alpha}{\sqrt{\left(j+\frac{1}{2}\right)^2-\alpha^2}+n'}\right)^2}}.\tag{2}$$ other than yielding the correct non-relativistic limit? The electron here is not a free particle, so I can't see any obvious reason to discard the negative energy as we would if it were.

$$^1$$J.J.Sakurai, J. Napolitano, Modern Quantum Mechanics. Third edition. Chapter 8, section 8.4.

$$^2$$See $$(8.149)$$ and $$(8.150)$$ of footnote 1.

Considering bound states of negative energy as result of the Dirac-equation means considering a relativistic one-particle theory of which we know that it is not possible. A relativistic quantum theory is always a multi-particle theory and can only be correctly be treated with second quantisation. So actually to deal with solutions of the Dirac-equation related with a bound state created by an electromagnetic field which lead to negative energies

$$(i\not\partial -e \not A -m)\psi =0$$

is to take the charge-conjugate of this Dirac-solution $$\psi^c =C\overline{\psi^T}$$ where $$C$$ is the charge conjugation matrix fulfilling $$C^{-1}\gamma^\mu C = -\gamma^{\mu T}$$

$$(i\not\partial +e \not A -m)\psi^c =0$$

and search solution of this equation (if the corresponding positive energy solution is already known, the conjugate solution can of course be immediately found by $$\psi^c =C\overline{\psi^T}$$).

$$I\,will\, in\, the\, following\, another\, approach\, to\, negative\, energy\, states:$$

Considering an electron with positive energy undergoing a transition into an electron of negative energy corresponds actually electron-positron annihilation with the emission of one virtual photon (which materializes "later" into another particle-anti-particle pair).

According to the Feyman-Stueckelberg interpretation the positron is considered as electron moving backwards in time.

A transition from positive electron energy to negative energy would lead from an electron moving forward in time into an electron moving backward in time. In order to take the in time backward-moving aspect into account means that the electron which undergoes such a transition runs into the past, i.e. in our real world with only forward-flowing time the "transited electron" -- the electron with finally negative energy -- must already have been there before the transition.

We observe the particle that must have already there as positron (but with positive energy).

Otherwise the electron cannot transit to negative energies. Without the presence of its backward in time moving branch right before the transition an electron will not undergo such a transition.

In other words an electron in a H-atom will never fall into negative energies if not already a positron is around.

Therefore for the study of a single electron bound in a H-atom according to the Dirac-equation we can completely forget of the negative energy states. The reaction of an electron (may be bound in a H-atom) colliding with a positron is another business because it is described by Feynman-diagrams whereas finding bound states consists of solving a differential equation system. And anyway it does not happen often that a positron comes along a H-atom to collide with its electron.

It is easy to see that the negative energy solutions would be unphysical. First of all, the proton is positively charged and the negative energy solutions would be a positron rather than an electron, and there should not be a positron bound to a proton.

If you are more mathematically minded, you could try to solve the original equations without the squaring step, and then there will not be the unphysical negative energy solutions coming in.