# A Brief History¶

Unlike other Quantum Field Theory textbook, our introduction to the history will not be so long. In this section, we only plan to introduce the part that has direct connection to the motivation of Quantum Field Theory.

Note

The following context will call Geometric Unit System and Einstein Summation Convention. Review the link if you have never heard of them — they are important!

Assuming that you already have the knowledge of Quantum Mechanics. At least you are able to recognise the most famous equation shown below

$\mathrm{i}\frac{\partial}{\partial t}\psi = -\frac{1}{2m}\nabla^2\psi + V\psi$

It is called a wave equation. Why? If you take

$\psi = A\exp\{-\mathrm{i}\omega t + \mathrm{i}\vec k\cdot \vec x\}$

into the equation, you will find that it is a solution of the Schrödinger equation as long as the following holds

Recall that we have the following relationship in Quantum Mechanics

$\omega = E,\ \ \ \ \ k^i = p^i$

where $$E$$ is the energy and $$p^i$$ is the momentum with $$i=1,2,3$$. Now, we immediately find that the previous equation becomes

$E = \frac{\vec p^2}{m} + V$

which is nothing but the energy-momentum relationship in Newtonian classical mechanics.

Warning

Now we release symbol $$\psi$$. This indicates that $$\psi$$ in the following context will have a different meaning.

So now we are going to think: what if we want it to be relativistic? Assuming that you already have the knowledge of Relativity. In Special Theory of Relativity, the energy-momentum relationship becomes

$E^2 = \vec p^2 + m^2$

To achieve this result, simply reverse the previous steps. The above relationship should require

And this requires

$(\partial_\mu\partial^\mu - m^2)\varphi = 0$

where the signature of metric here is $$+2$$. This is the famous “Klein-Gordon equation” — the relativistic version of Schrödinger equation. The quotes here indicates that it still has some subtle difference with the real Klein-Gordon equation, as we are going to see next.

It seems that it is so easy to combine Quantum Mechanics and Special Theory of Relativity. Nonetheless, things are not so trivial. The Newtonian energy-momentum relation can guarantee that the energy of a free ($$V=0$$) particle is always positive. However, the relativistic version can not. The relativistic energy-momentum relation reveals that

$E = \pm\sqrt{\vec p^2+m^2}$

where we discover that there are both positive and negative solutions are possible. This issue is not paramount in classical theory, since the energy of a classical particle can only varies continuously, and there is a gap of $$2m$$ between the lowest positive energy $$E=m$$ (where $$\vec p=0$$) and the greatest negative value $$E=-m$$ (where $$\vec p=0$$). Thus, a classical particle can have either positive or negative energy — the only matter is to set an initial sign.

However, recall that quantised particles can jump between discrete energy levels. Hence, it is just a matter of probability for them to jump from a positive energy to a negative one and … fall all the way down to minus infinity. This then becomes a major difficulty of “Klein-Gordon equation”.

Also, let’s investigate the continuity equation. In Quantum Mechanics, there is

$\begin{split}\mathrm{i}\frac{\partial}{\partial t}\varphi = -\frac{1}{2m}\nabla^2\varphi + V\varphi\\ -\mathrm{i}\frac{\partial}{\partial t}\varphi^* = -\frac{1}{2m}\nabla^2\varphi^* + V\varphi^*\end{split}$

The first equation is the Schrödinger equation and the second equation is nothing other than the complex conjugate of the first one.

Now, we apply $$\varphi^*$$ to the first equation and $$\varphi$$ to the second one, and let the first one subtract the second

$\varphi^*\mathrm{i}\frac{\partial}{\partial t}\varphi + \varphi\mathrm{i}\frac{\partial}{\partial t}\varphi^* = \varphi\frac{1}{2m}\nabla^2\varphi^* - \varphi^*\frac{1}{2m}\nabla^2\varphi$

which, after a transform, becomes

$\mathrm{i}\frac{\partial}{\partial t}\varphi\varphi^* = \frac{1}{2m}\nabla(\varphi\nabla\varphi^* - \varphi^*\nabla\varphi)$

To gain the continuity equation, the only thing to do is to identify the probability density and probability current as

$\rho = \varphi\varphi^*,\ \ \ \ \ \vec{j} = {1\over2m\mathrm i}(\varphi\nabla\varphi^* - \varphi^*\nabla\varphi)$

The probability density is the square of the wave function, and will always be positive due to the property of square. Everything works fine here.

However, if you derive the continuity equation corresponding to the “Klein-Gordon equation”, you will find that the “probability density” now becomes

$\rho \sim \varphi^*\frac{\partial}{\partial t}\varphi - \varphi\frac{\partial}{\partial t}\varphi^*$

The minus sign indicates that it might be negative(!), which destroys the probabilistic interpretation of quantum theory. This is another major difficulty of “Klein-Gordon equation”.

To solve the difficulties, physicists decided to reinterpret symbol $$\varphi$$ here as classical field, instead of wave function. Then, similar to the Quantum Mechanics, path integral is used to quantise the classical field.

Now, if you can follow the pace, welcome to the next section.