Getting particle/antiparticle solutions from the Dirac Equation

In summary: Thus, you get 37.28.In summary, the derivation of Dirac's equation involves factoring Einstein's energy relation with the gamma matrices and using quantization to arrive at the equation. Srednicki takes a different approach by renaming one of the factors and inverting signs, which leads to a different expression for Dirac's equation. When finding solutions for this equation, it is necessary to consider both positive and negative energy solutions, as shown in equation 37.27. Plugging this solution into the original equation yields 37.28, which in turn leads to the conditions in equation 37.29. The difference between the two solutions is due to the fact that different factors are obtained when differentiating the terms
  • #1
peguerosdc
28
7
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!
 
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  • #2
I do not go into detail but the formula seems to say ##e^{ipx}## and ##e^{-ipx}## are linear independent.
 
  • #3
anuttarasammyak said:
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.
 
  • #4
peguerosdc said:
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.
 
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  • #5
PeroK said:
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!
 
  • #6
peguerosdc said:
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:
$$(\gamma^{\mu}p_{\mu}+ m)u(p) = 0 \ \ \text{and} \ \ (-\gamma^{\mu}p_{\mu}+ m)v(p) = 0$$
 
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  • #7
PeroK said:
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.
 
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Related to Getting particle/antiparticle solutions from the Dirac Equation

1. What is the Dirac Equation and why is it important in particle physics?

The Dirac Equation is a mathematical equation that describes the behavior of fermions, which are particles with half-integer spin. It was developed by physicist Paul Dirac in the 1920s and is important in particle physics because it accurately describes the behavior of fundamental particles such as electrons, protons, and neutrons.

2. How does the Dirac Equation relate to the concept of antiparticles?

The Dirac Equation predicts the existence of antiparticles, which have the same mass as their corresponding particles but opposite charge. This is because the equation has both positive and negative energy solutions, and the negative energy solutions correspond to antiparticles.

3. Can the Dirac Equation be used to obtain solutions for both particles and antiparticles?

Yes, the Dirac Equation can be used to obtain solutions for both particles and antiparticles. The positive energy solutions correspond to particles, while the negative energy solutions correspond to antiparticles.

4. What is the significance of obtaining particle/antiparticle solutions from the Dirac Equation?

The ability to obtain particle/antiparticle solutions from the Dirac Equation is significant because it provides a theoretical framework for understanding the behavior of fundamental particles and their interactions. It also allows for the prediction of new particles and the study of their properties.

5. Are there any practical applications of the Dirac Equation in modern technology?

Yes, the Dirac Equation has practical applications in modern technology. For example, it is used in the development of quantum computers, which utilize the principles of quantum mechanics to perform calculations much faster than classical computers. The Dirac Equation is also used in medical imaging techniques such as positron emission tomography (PET) scans.

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