1
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ 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. $\endgroup$
    – J. Murray
    Jan 3, 2021 at 21:00
  • 1
    $\begingroup$ 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. $\endgroup$
    – J. Murray
    Jan 3, 2021 at 21:05
  • $\begingroup$ Your last integrand contains $p$ and $k$ at the same time. $\endgroup$ Jan 11 at 16:33
  • $\begingroup$ @VladimirKalitvianski yh sorry that's a typo $\endgroup$ Jan 11 at 17:05

1 Answer 1

1
+100
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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 $\endgroup$ Jan 11 at 17:56
  • 1
    $\begingroup$ I don't really understand the linear independence argument $\endgroup$ Jan 11 at 17:57
  • 1
    $\begingroup$ @JoshuaPasa $e^{ipx}$ and $e^{-ipx}$ are linearly independent and therefore $Ae^{ipx}+Be^{-ipx}=0$ implies $A=B=0$. $\endgroup$ Jan 11 at 17:59
  • $\begingroup$ 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? $\endgroup$ Jan 11 at 18:10
  • 1
    $\begingroup$ @JoshuaPasa If you put $x=0$ you get $A+B=0$ and if first take a derivative and then put $x=0$ you get $A-B=0$. $\endgroup$ Jan 11 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.