# Corresponding particle-antiparticle solutions for Klein-Gordon equation

For free particle solutions in a box, the following 4 solutions are possible(Not all 4 are independent though) as $$\psi_+=A_+ \exp{\frac{i}{\hbar}(px-Et)}\\\psi_+^*=A_+^* \exp{\frac{-i}{\hbar}(px-Et)}\\\psi_-=A_- \exp{\frac{i}{\hbar}(px+Et)}\\\psi_-^*=A_-^* \exp{\frac{-i}{\hbar}(px+Et)}$$ where $$E>0$$. The properties of the solutions are as follows: $$\begin{matrix} Function &\psi_+ &\psi_+^* &\psi_- &\psi_-^* \\ Momentum & +p & -p & +p & -p\\ Energy & +E & -E &-E &+E \\ Charge & +e & -e & -e & +e\\ H_0^0 &E & E & E & E \end{matrix}$$ where $$H_0^0$$ is calculated from the energy-momentum tensor for the Klein-Gordon Field. Now, my question is, for which pairs of the solutions do we interpret it as particle antiparticle pair and I see that all $$+e$$ and $$-e$$ pairs have different energy value in terms of the sign. So how do we explain the negative energy business in the solutions? Also,there were online notes which said and I quote

Thus if φ is a negative-energy solution, we take it to represent an antiparticle with wave function φ ∗ (and hence positive energy, opposite charge and momentum).

where does this fit in this picture?

The Klein-Gordon (shortly K-G)-equation is a relativistic wave equation, and as such it has to be treated with the formalism of Quantum Field theory (QFT). First of all, one has to realize it is above all a wave equation, so its solutions are waves. These solutions do not describe particles, as it happened be the solutions of the Schrodinger equations. But the Schrodinger equation and its solution, the famous wave function, giving the probability amplitude of one particle are concepts of non-relativistic quantum mechanics which is a rather different approach from that of QFT. The wave function is a non-relativistic concept, which violates causality if one considers the finite propagation speed between cause and effect. Writing down a plane wave solution

$$\psi = A e^{-\frac{i}{\hbar}px} \equiv A e^{-\frac{i}{\hbar}( E t-\vec{p}\vec{x})}$$

it is not clear how many particles can actually be associated with this wave. As the phase of this solution is fixed, the particle number operator of such a wave can take on different values as phase and particle number as kind of conjugated variables (under some circumstances) , so if the phase is fixed, quasi nothing is known about the particle number. But nevertheless may be we can at least particle types associate with the type of wave. For doing so, we will further simplify the question in considering photons. The wave equation for photons is simply:

$$\Box A_\mu \equiv (\frac{\partial^2}{\partial t^2} -c^2 \nabla^2)A_\mu =0$$

If we solve this equation in Fourier space we get:

$$(\omega^2 -c^2\vec{k}^2)A_\mu(k)=0$$

and the solutions are (here the question of the number of degrees of freedom (according to gauge freedom) of the EM-field is not considered. $$\mu$$ is supposed to be a 4-vector index, but could also be considered as index for the 2 possible polarization states of the photon):

$$A_\mu(k) = \cal{A}_\mu e^{-ikx} = \cal{A}_\mu e^{-i(\omega t-\vec{k}\vec{x}) }$$ and $$A_\mu(k) = \cal{A}_\mu e^{+ikx}= \cal{A}_\mu e^{ i(\omega t-\vec{k}\vec{x})}$$.

These 2 solutions were already found before the advent of relativistic QM (i.e. Dirac and K-G equation) The reason for 2 solutions is the EM-field dispersion relation: $$\omega^2-c^2\vec{k}^2=0$$ (this is a quadratical equation after all) which has 2 solutions: $$\omega = \sqrt{c^2\vec{k}^2}$$ and $$\omega = -\sqrt{c^2\vec{k}^2}$$. But did the physicists interprete them as particles and anti-particles ?? Of course not !! The first solution is interpreted as in-going wave and the second solution as out-going wave. Upon later quantisation of the EM-field, the in-going wave solutions was interpreted as in-going photons whereas the out-going solution as out-going photons.

It's exactly the same for interpretation of the solutions of the K-G equation with the slight difference in the dispersion relation: $$\omega^2-c^2\vec{k}^2-\frac{(mc^2)^2}{\hbar^2} =0$$ giving : $$\omega = \sqrt{c^2\vec{k}^2 +\frac{(mc^2)^2}{\hbar^2}}$$ and $$\omega =-\sqrt{c^2\vec{k}^2 +\frac{(mc^2)^2}{\hbar^2}}$$, but the interpretation actually has not changed, these are in-going and out-going waves. But we actually don't know which type of particles could be associated to these in-going and out-going waves. This can only be done in the procedure of quantisation part of QFT. A distinction has to be made between real K-G fields and complex K-G fields. The latter have double number of freedoms, so a solution

$$\psi = A e^{ikx}$$

can actually be "out-going particle wave" or a "out-going anti-particle wave". The question of the particle type can only be answered in the formalism of QFT, if $$e^{ikx}$$ is together with particle creation operator $$a^\dagger$$, then it is a out-going particle wave, if $$e^{ikx}$$ is together with anti-particle creation operator $$b^\dagger$$, it is out-going anti-particle wave. In order to answer to the quote given, in relativistic QM (which actually tries to maintain the concepts of non-relativistic QM, in particular that of "wave function solutions" describing the probability amplitude of one particle; relativistic QM $$\neq$$QFT) the found solutions $$\psi = A e^{ikx}$$ are considered as "new" (although the EM-field already provided such solutions even with an interpretation) and were associated with anti-particles. Under this assumption that $$\psi = A e^{ikx}$$ actually describes a "out-going anti-particle wave", which upon complex conjugation turns into a $$\psi^{\ast} = A^{\ast} e^{-ikx}$$ in a "in-going anti-particle wave". This way this solution is comparable to the familiar standard particle solution $$\psi = A e^{-ikx}$$, with the only difference that it is now considered as a anti-particle in-going wave, because at the starting point $$\psi = A e^{ikx}$$ was associated with anti-particles. This is the way how relativistic QM handles the issue of so-called particle and anti-particle solutions. It is actually not fully satisfactory as assumptions had to be made which do not come out of the theory here, relativistic QM. However QFT gives these answers. In QFT $$\psi = A e^{\pm ikx}$$ are just wave equation solutions, and the particles respectively anti-particles are special one-particles states in the Fock space, something completely different. And both types of particle states have positve energy.

So to recap we have given the interpretation of 2 of the 4 solutions put in question. As the other 2 solutions still are missing an interpretation yet, consider that they are dependent on the other already explained solutions, they have to be considered as similar res. the same as the already explained solution they depend on.

• I think as I haven't still covered QFT, I had this silly doubt. So, in the light of Relativistic Quantum Mechanics(RQM), the negative energy solutions exists and we have no better way of interpreting them other than as anti-particles in RQM. But in QFT, we have some "fock space" interpretation as you say, in whose interpretation, we can actually have particles and anti-particles with positive energy and there, the negative energy solutions are just mathematical artifacts. I think you meant this, right? – Naman Agarwal Dec 19 '18 at 12:26
• @Naman Agarwal: I would not say that the negative energy solutions are mathematical artefacts, they exist and are useful. What changes is their interpretation. – Frederic Thomas Dec 19 '18 at 12:58
• So how are they interpreted in QFT?(In RQM, they are interpreted as anti-particles I think, right?) – Naman Agarwal Dec 19 '18 at 13:07
• The solutions of a wave equation are waves. And the wave's energy is $E\sim |A|^2$ which is always positive. Upon quantization the solutions become operators which only take on values when applied on Fock states. The energy of all (particle and anti-particle) Fock states is positive (or zero). – Frederic Thomas Dec 19 '18 at 13:16
• You know you why consider the wave having negative energy ? Because you still adopt the concept of Schrodinger's QM. In QFT this is no longer the case. I admit, it is very difficult to understand. Get acquainted with the Feyman-Stueckelberg(F-S) interpretation of anti-particles. Briefly it says: Let's take an in-going wave of negative 4-momentum $(-E,-\vec{p})$, it's $Ae^{ipx}$. According to F-S, it's like an outgoing wave of positive $(E,\vec{p})$. This what I've been saying : $Ae^{ipx}$ is an outgoing wave of positive energy. What changed ? Nothing except the interpretation of a result. – Frederic Thomas Dec 19 '18 at 14:03

All four solutions are independent free solutions - so not in a box. The energy is always positive, so it is the last row. The row you designate as E is actually the charge and is the same as that designated as q. The solutions with opposite charge are antimatter relative to one another. For each charge sign there are two solutions with opposite momentum. Good analysis.