# Getting particle/antiparticle solutions from the Dirac Equation

• A
peguerosdc
TL;DR Summary
How to motivate positive and negative energy solutions as a superposition of two solutions?
Hi!

I am studying Dirac's equation and I already have understood the derivation. Following Griffiths, from factoring Einstein's energy relation with the gamma matrices:

##
(\gamma^\mu p_\mu + m)(\gamma^\mu p_\mu - m) = 0
##

You take any of the two factors, apply quantization and you arrive to Dirac's equation. Griffiths takes for example the second one:

##
(\gamma^\mu p_\mu - m) \varphi = 0
##

Comparing this with Srednicki's, they look the same just that he renames ## \gamma^\mu p_\mu = i\not\!\partial ##, inverts signs and arrives to this expression for Dirac's equation (equation 37.23):

##
(-i\not\!\partial + m) \varphi = 0
##

Now, Srednicki motivates finding two solutions (one with +E and one with -E) by showing that the wave function must obey the Klein-Gordon equation, so he proposes the solution as (eq 37.27):

##
\varphi (x) = u(\mathbf p) e^{ipx} + v(\mathbf p) e^{-ipx}
##

where u is the positive energy solution and v the negative energy solution.

So far, so good. Now, where I need help is understanding the next step. He says:
Plugging eq. (37.27) into the eq. (37.23) we get eq (37.28):
$$(\not\!p + m)u(\mathbf p) e^{ipx} + (-\not\!p + m)v(\mathbf p) e^{-ipx} = 0$$
Thus we require (37.29):
$$(\not\! p + m)u(\mathbf p) = 0 \qquad (-\not\!p + m)v(\mathbf p) = 0$$
I don't understand why plugging eq. 37.27 into 37.23 yields equation 37.28. It looks like we are taking one different factor for each solution (that's why I introduced my question with Griffiths' derivation), but I don't understand the reasoning behind it.

Thanks!

Gold Member
I do not go into detail but the formula seems to say ##e^{ipx}## and ##e^{-ipx}## are linear independent.

peguerosdc
I do not go into detail but the formula seems to say ##e^{ipx}## and ##e^{-ipx}## are linear independent.
Thanks for the reply! Yes, I think that's why 37.28 implies 37.29, but my question is more how to get to equation 37.28.
Following Srednicki, if you just plug 37.27 into 37.23, you get:

\begin{align*} (-\not\!p + m) ( u(\mathbf p) e^{ipx} + v(\mathbf p) e^{-ipx} ) &= 0 \\ (-\not\!p + m)u(\mathbf p) e^{ipx} + (-\not\!p + m)v(\mathbf p) e^{-ipx} &= 0 \end{align*}

Which doesn't look like 37.28 as some signs are wrong. I was thinking about plugging 37.27 into the first equation in my post to get something like:
\begin{align*} (\not\!p + m)(\not\!p - m) \varphi &= 0 \\ (\not\!p + m)(\not\!p - m)u(\mathbf p) e^{ipx} + (\not\!p + m)(\not\!p - m)v(\mathbf p) e^{-ipx} &= 0 \end{align*}
But then I don't know how I could get 37.28 from that.

Homework Helper
Gold Member
2022 Award
Thanks for the reply! Yes, I think that's why 37.28 implies 37.29, but my question is more how to get to equation 37.28.
Following Srednicki, if you just plug 37.27 into 37.23, you get:

\begin{align*} (-\not\!p + m) ( u(\mathbf p) e^{ipx} + v(\mathbf p) e^{-ipx} ) &= 0 \\ (-\not\!p + m)u(\mathbf p) e^{ipx} + (-\not\!p + m)v(\mathbf p) e^{-ipx} &= 0 \end{align*}

Which doesn't look like 37.28 as some signs are wrong. I was thinking about plugging 37.27 into the first equation in my post to get something like:
\begin{align*} (\not\!p + m)(\not\!p - m) \varphi &= 0 \\ (\not\!p + m)(\not\!p - m)u(\mathbf p) e^{ipx} + (\not\!p + m)(\not\!p - m)v(\mathbf p) e^{-ipx} &= 0 \end{align*}
But then I don't know how I could get 37.28 from that.
As per the book, you must get a different factor when you differentiate ##e^{ipx}## and ##e^{-ipx}##. One must have a factor of ##-1## compared to the other.

• peguerosdc
peguerosdc
As per the book, you must get a different factor when you differentiate ##e^{ipx}## and ##e^{-ipx}##. One must have a factor of ##-1## compared to the other.
Oh, right! I knew this was most likely a dumb question. Thanks!

Homework Helper
Gold Member
2022 Award
Oh, right! I knew this was most likely a dumb question. Thanks!
Let's look at this anyway. We are looking for ##\varphi(x)## to satisfy:
$$(-i \not\!\partial + m)\varphi(x) = (-i \gamma^{\mu}\partial_{\mu}+m)\varphi(x) = (-i \gamma^{\mu}\frac{\partial}{\partial x^{\mu}}+m)\varphi(x) = 0$$ where
$$\varphi(x) = u(p)e^{ipx} + v(p)^{-ipx} = u(p)e^{ip_{\nu}x^{\nu}} + v(p)e^{-ip_{\nu}x^{\nu}}$$
Note that this is a different convention from Griffiths with ##u(p)## for antiparticles. In any case:
$$\frac{\partial}{\partial x^{\mu}}e^{ip_{\nu}x^{\nu}} = ip_{\mu}e^{ip_{\nu}x^{\nu}}= ip_{\mu}e^{ipx} \ \ \text{and} \ \ \frac{\partial}{\partial x^{\mu}}e^{-ip_{\nu}x^{\nu}} = -ip_{\mu}e^{-ip_{\nu}x^{\nu}} = -ip_{\mu}e^{-ipx}$$
So that:
$$(-i \not\!\partial + m)\varphi(x) = -i\gamma^{\mu}u(p) (ip_{\mu}e^{ipx}) -i\gamma^{\mu}v(p)(-ip_{\mu}e^{-ipx}) + mu(p)e^{ipx} + mv(p)e^{-ipx}$$ $$= (\gamma^{\mu}p_{\mu}+ m)u(p)e^{ipx} + (-\gamma^{\mu}p_{\mu}+ m)v(p)e^{-ipx}$$
Setting this to zero and using the linear independence of ##e^{ipx}, e^{-ipx}## gives:
$$(\gamma^{\mu}p_{\mu}+ m)u(p) = 0 \ \ \text{and} \ \ (-\gamma^{\mu}p_{\mu}+ m)v(p) = 0$$

• vanhees71
peguerosdc
Note that this is a different convention from Griffiths with ##u(p)## for antiparticles.
Yeah, I almost missed that and actually I think that's the convention followed in every book, so it's worth noting it for future readers that check this thread.

Thank you for showing the full derivation! Definitely helpful.

• PeroK