Introduction

For the Klein Gordon Field, the equations of motion are described by the equation

$$(partial_{mu}partial^{mu} + m^2)phi(vec{x},t)=0$$

Which when the field is expressed as a Fourier transform of the momentum we can get the solution that

$$phi(vec{x},t)=int frac{d^3 p}{(2pi)^3 2omega}(a(vec{p})exp{(-ip_{alpha} x^{alpha}}) + a^{dagger}(vec{p})exp{(ip_{alpha} x^{alpha}}))$$

For the Dirac field which equations of motion are given by

$$(igamma^mupartial_mu – m)psi(x,t) = 0$$

Can we arrive at the form:

$$psi(vec{x},t)= sum_{s} int frac{d^3 p}{(2pi)^3 2omega}(a_s(vec{p})u_s(vec{p})exp{(-ip_{alpha} x^{alpha}}) + b_s^{dagger}(vec{p}) v_s(vec{p})exp{(ip_{alpha} x^{alpha}}))$$

Using the same technique that we used for the Klein Gordon equation

Klein Gordon Field

For the Klein Gordon equation to begin with we need to express the state in a position basis using the momentum basis which we can do with a Lorentz invariant Fourier transform

$$phi(vec{x},t) = int frac{d^3 p}{(2pi)^3 2omega} phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}})$$

If we then substitute this expression into the Klein Gordon equation, we get

$$(partial_{mu}partial^{mu} + m^2) int frac{d^3 p}{(2pi)^3 2omega} phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) = 0$$

We can interchange the $(partial_{mu}partial^{mu} + m^2)$ operator with the integral due to its linearity which gives

$$int frac{d^3 p}{(2pi)^3 2omega} (partial_{mu}partial^{mu} + m^2) phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) = 0$$

Which is also equivalent to

$$int frac{d^3 p}{(2pi)^3 2omega} (partial_{mu}partial^{mu} (phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}})) + m^2 phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) ) = 0$$

If we use the product rule for differentiation we get

$$int frac{d^3 p}{(2pi)^3 2omega} (phi_p (vec{p},t) partial_{mu}partial^{mu} exp{(i vec{p} cdot vec{x}}) + exp{(i vec{p} cdot vec{x}}) partial_{mu}partial^{mu} phi_p (vec{p},t) + m^2 phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) ) = 0$$

If we expand out $partial_{mu}partial^{mu}$ we can see that

$$int frac{d^3 p}{(2pi)^3 2omega} (phi_p (vec{p},t) (partial_{t}partial^{t} – nabla^2) exp{(i vec{p} cdot vec{x}}) + exp{(i vec{p} cdot vec{x}}) (partial_{t}partial^{t} – nabla^2) phi_p (vec{p},t) + m^2 phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) ) = 0$$

For the first term, the time derivative does not contribute because the term we are differentiating is independent of time however $nabla^2 [exp{(i vec{p} cdot vec{x}})] = -|vec{p}|^2 exp{(i vec{p} cdot vec{x}})$. For the second term the only variable that is in common with the derivitive operator is the time derivitive therefore all together this becomes

$$int frac{d^3 p}{(2pi)^3 2omega} (phi_p (vec{p},t) |vec{p}|^2 exp{(i vec{p} cdot vec{x}}) + exp{(i vec{p} cdot vec{x}}) partial_{t}partial^{t} phi_p (vec{p},t) + m^2 phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) ) = 0$$

We can also write it like this

$$int frac{d^3 p}{(2pi)^3 2omega} (exp{(i vec{p} cdot vec{x}}) partial_{t}partial^{t} phi_p (vec{p},t) + (|vec{p}|^2 + m^2) phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) ) = 0$$

For this integration to result in $0$ this means that the integrand has to be $0$ and therefore

$$exp{(i vec{p} cdot vec{x}}) partial_{t}partial^{t} phi_p (vec{p},t) + (|vec{p}|^2 + m^2) phi_p (vec{p},t) exp{(i vec{p} cdot vec{x}}) = 0$$

We can divide through by common factors which leave us with

$$partial_{t}partial^{t} phi_p (vec{p},t) + (|vec{p}|^2 + m^2) phi_p (vec{p},t) = 0$$

Before we solve the differential equation notice that $omega = (|vec{p}|^2 + m^2)$ which means that

$$partial_{t}partial^{t} phi_p (vec{p},t) + omega phi_p (vec{p},t) = 0$$

This turns out to have the general solution that

$$phi_p (vec{p},t) = A exp{(-i omega t)} + B exp{(i omega t)}$$

Since the exponentials do not contain any momentum terms this must mean that the constants must depend on momentum because the function requires dependence on momentum.

$$phi_p (vec{p},t) = A(vec{p}) exp{(-i omega t)} + B(vec{p}) exp{(i omega t)}$$

Since this klein gordon field is real this must mean that $phi_p (vec{p},t)=phi^dagger_p (-vec{p},t)$. This means that if we use simultanious equations $A^dagger(-vec{p}) = B(vec{p})$ which implies

$$phi_p (vec{p},t) = A(vec{p}) exp{(-i omega t)} + A^dagger(-vec{p}) exp{(i omega t)}$$

We can now switch to the little $a$ notation

$$phi_p (vec{p},t) = a(vec{p}) exp{(-i omega t)} + a^dagger(-vec{p}) exp{(i omega t)}$$

We can now substitute this back into the fourier transform which gives us

$$phi(vec{x},t) = int frac{d^3 p}{(2pi)^3 2omega} (a(vec{p})exp{(-i omega t)}exp{(i vec{p} cdot vec{x}}) + a^dagger(-vec{p}) exp{(i omega t)exp{(i vec{p} cdot vec{x}})}) $$

Using a change of variables we can show that

$$phi(vec{x},t) = int frac{d^3 p}{(2pi)^3 2omega} (a(vec{p})exp{(-i omega t)}exp{(i vec{p} cdot vec{x}}) + a^dagger(vec{p}) exp{(i omega t)} exp{(-i vec{p} cdot vec{x})})$$

We can combine the exponentials and use the einstein summation notation to give

$$phi(vec{x},t) = int frac{d^3 p}{(2pi)^3 2omega} (a(vec{p})exp{(-i p_mu x^mu)} + a^dagger(vec{p}) exp{(i p_mu x^mu)})$$

Can we follow a similar approach for the Dirac Field?