Lectures on Quantum Field Theory. Ashok Das
Given the energy-momentum four vectors, we can construct the Lorentz scalar
The Einstein relation for a free particle (remember c = 1)
where m represents the rest mass of the particle, can now be seen as the Lorentz invariant condition
In other words, in this space, the energy and the momentum of a free particle must lie on a hyperbola satisfying the relation (1.31).
We already know that the coordinate representations of the energy and the momentum operators take the forms
We can combine these to write the coordinate representation for the energy-momentum four vector operator as
Finally, let us note that in the four dimensional space-time, we can construct two totally anti-symmetric fourth rank tensors ϵµνλρ, ϵµνλρ, the four dimensional contravariant and covariant Levi-Civita tensors respectively. We will choose the normalization ϵ0123 = 1 = −ϵ0123 so that
where ϵijk denotes the three dimensional Levi-Civita tensor with ϵ123 = 1. An anti-symmetric tensor such as ϵijk is then understood to denote
and so on. This completes the review of all the essential basic notation that we will be using in this book. We will introduce new notations as they arise in the context of our discussions.
1.3Klein-Gordon equation
With all these basics, we are now ready to write down the simplest of the relativistic equations. We recall that in the case of a non-relativistic particle, we start with the non-relativistic energy-momentum relation
and promote the dynamical variables (observables) to Hermitian operators to obtain the time-dependent Schrödinger equation (see (1.1))
Let us consider the simplest of relativistic systems, namely, a relativistic free particle of mass m. In this case, we have seen that the energy-momentum relation is none other than the Einstein relation (1.30), namely,
Thus, as before, promoting these to operators, we obtain the simplest relativistic quantum mechanical equation to be (see (1.33))
Setting ħ = 1 from now on for simplicity, the equation above takes the form
Since the operator in the parenthesis is a Lorentz scalar and since we assume the quantum mechanical wave function, ϕ(x, t), to be a scalar function, this equation is invariant under Lorentz transformations.
This equation, (1.40), is known as the Klein-Gordon equation and, for m = 0, or when the rest mass of the particle vanishes, it reduces to the wave equation (recall Maxwell’s equations). Like the wave equation, the Klein-Gordon equation also has plane wave solutions which are characteristic of free particle solutions. In fact, the functions
with kµ = (k0, k) are eigenfunctions of the energy-momentum operator, namely, using (1.33) (remember that ħ = 1) we obtain
so that ±kµ are the eigenvalues of the energy-momentum operator. (In fact, the eigenvalues should be ±ħkµ, but we have set ħ = 1.) This shows that the plane waves define a solution of the Klein-Gordon equation (1.39) or (1.40) provided
Thus, we see the first peculiarity of the Klein-Gordon equation (which is a relativistic equation), namely, that it allows for both positive and negative energy solutions. This basically arises from the fact that, for a relativistic particle (even a free one), the energy-momentum relation is given by the Einstein relation which is a quadratic relation in E, as opposed to the case of a non-relativistic particle, where the energy-momentum relation is linear in E. If we accept the Klein-Gordon equation as describing a free, relativistic, quantum mechanical particle of mass m, then, we will see shortly that the presence of the negative energy solutions would render the theory inconsistent.
To proceed further, let us note that the Klein-Gordon equation and its complex conjugate (remember that a quantum mechanical wave function is, in general, complex), namely,
would imply
Defining the probability current density four vector as
where
we note that equation (1.45) can be written as a continuity equation for the probability current, namely,
The probability current density,