# General solution to the dirac equation expressed as a Fourier transform

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.

• Hi Joshua. I didn't read through your question line-by-line, but it appears as though you derived an expression for the solution to the Klein-Gordon equation in fairly excruciating detail, and then asked if a similar path can be followed for the Dirac equation. Note that anybody who is in a position to answer this question already understands your derivation quite intimately, so approximately 95% of the body of your question is unnecessary. Jan 3, 2021 at 21:00
• Since the answer to your question is yes, my suggestion would be to try it and then ask a (succinctly worded) question if you run into conceptual difficulties which may arise when working with spinors. Otherwise, this is just a very lengthy request for the derivation of the corresponding solution to the Dirac equation, which can be found all over the internet. Jan 3, 2021 at 21:05
• Your last integrand contains $p$ and $k$ at the same time. Jan 11 at 16:33
• @VladimirKalitvianski yh sorry that's a typo Jan 11 at 17:05

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.

• Hi, I like the logic used here, but there's one step that I'm struggling to justify, why can you assume that each term of the expansion equals 0 when applying the Dirac equation when in general if the sum of terms equals 0 then each term doesn't necessarily have to be 0 Jan 11 at 17:56
• I don't really understand the linear independence argument Jan 11 at 17:57
• @JoshuaPasa $e^{ipx}$ and $e^{-ipx}$ are linearly independent and therefore $Ae^{ipx}+Be^{-ipx}=0$ implies $A=B=0$. Jan 11 at 17:59
• is it because if you expand them out as complex numbers you get that the imaginary and real part should be 0 and then it pops out from there? Jan 11 at 18:10
• @JoshuaPasa If you put $x=0$ you get $A+B=0$ and if first take a derivative and then put $x=0$ you get $A-B=0$. Jan 11 at 18:15