General solution to the dirac equation expressed as a Fourier transform

Question

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 that $$(\partial_{\mu}\partial^{\mu} + m^2)\int \frac{d^3k}{(2\pi)^3}\tilde{\phi}(k,t) e^{i \vec{k} \cdot \vec{x}}=0$$

Which is the same as

$$\int \frac{d^3k}{(2\pi)^3}(\partial_{\mu}\partial^{\mu} + m^2)\tilde{\phi}(k,t) e^{i \vec{k} \cdot \vec{x}}=0$$

From this we get a differential equation for $\tilde{\phi}$ and we get that $$\phi(\vec{x},t)=\int \frac{d^3 p}{(2\pi)^3 }(a(\vec{p})e^{-ip \cdot x} + a^{\dagger}(\vec{p})e^{-ip \cdot x})$$

For the Dirac field which equations of motion are given by

$$(i\gamma^\mu\partial_\mu - m)\psi(x,t) = 0$$

Can we arrive at the form:

$$\psi(\vec{x},t)= \sum_{s} \int \frac{d^3 p}{(2\pi)^3}(a_s(\vec{p})u_s(\vec{p})e^{-ip \cdot x} + b_s^{\dagger}(\vec{p}) v_s(\vec{p})e^{ip \cdot x})$$

Using the same technique that we used for the Klein Gordon equation where $$(i\gamma^\mu\partial_\mu - m)\int \frac{d^3 k}{(2\pi)^3}\tilde{\psi}(k,t)e^{-ik \cdot x}=0$$ Any help would be greatly appreciated, Thanks.

Answer 1

Yes. But this is very homework-like so I'll only sketch the logic. I hope other answers do the same.

Note that $(i\not\partial+m)(i\not\partial-m)=(\partial^2+m^2)$, almost by definition (this is the whole point of the Dirac equation). Therefore, if $\psi$ satisfies $$ (i\not\partial-m)\psi=0\tag1 $$ it also satisfies $$ (\partial^2+m^2)\psi=0\tag2 $$ Therefore, you can use your own derivation to write $$ \psi_\alpha(\vec{x},t)=\int \frac{d^3 p}{(2\pi)^3 }(a_\alpha(\vec{p})e^{-ip \cdot x} + a_\alpha^{\dagger}(\vec{p})e^{-ip \cdot x})\tag3 $$ where $\alpha=1,2,3,4$ is a spinor index.

Now let us go back to the initial equation and check whether there are any further conditions (as we may have introduced fictitious solutions by manipulating the equation: the operator $(i\not\partial+m)$ is not invertible so multiplying by it can embiggen the space of solutions).

If we apply $(i\not\partial-m)$ to our solution $(3)$, we get $$ 0=(i\not\partial-m)\psi=\int \frac{d^3 p}{(2\pi)^3 }(\not p-m)a(\vec{p})e^{-ip \cdot x} + \text{c.c}\tag4 $$ and hence the spinor $a_\alpha$ must satisfy the algebraic equation $$ (\not p-m)a(\vec{p})\equiv0\tag5 $$ But this is easy to solve: let $u_s(\vec p)$, $s=1,2$, be the two linearly-independent vectors that solve this (there are two because this is the rank of the matrix $\not p-m$, as is easily checked by brute force). Then, the general solution to $(5)$ is $$ a(\vec p)=\sum_{s=1,2}u_s(\vec p)a_s(\vec p) $$ where $a_1,a_2$ are two $1\times 1$ annihilation operators (as opposed to $a_\alpha$, which is a $4\times1$ spinor).

Thus, finally, $$ \psi(\vec{x},t)=\sum_{s=1,2}\int \frac{d^3 p}{(2\pi)^3 }u_s(\vec p)a_\sigma(\vec p)e^{-ip \cdot x} + \text{c.c}\tag6 $$ as is well-known.