# The 4 solutions to the Dirac equation

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## Main Question or Discussion Point

Hello! I understand that the free Dirac equations has spinors as solutions, of dimension 4, and one can't discard the negative energy solutions (as one needs a complete basis to span the Hilbert space of solutions), and these negative energy particles are interpreted as positive energy antiparticles. Now, if one adds an EM field, the Dirac equation becomes: $$\gamma^\mu(\partial_\mu-ieA_\mu)\psi+im\psi = 0$$ This equation still requires 4 linearly independent solutions, but I am not sure how to think of them in this case. One can't pretend that 2 of them are particles and 2 are anti-particles, because the charge sign appears explicitly here and at the same time, one can't discard 2 of the solutions. So how should I think of the 4 linearly independent solutions to this equations? Thank you!

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king vitamin
Gold Member
I understand that the free Dirac equations has spinors as solutions, of dimension 4, and one can't discard the negative energy solutions (as one needs a complete basis to span the Hilbert space of solutions), and these negative energy particles are interpreted as positive energy antiparticles.
I would disagree that you should interpret the negative energy states as positive energy antiparticles. Instead, the Dirac equation describes single particle/hole excitations on top of a fully-filled "Dirac sea" of positively charged antiparticles. The two positive energy solutions are states describing the spin up and down particle on top of this Dirac sea, while the two negative energy solutions correspond to removing a single spin up/down antiparticles from the filled Dirac sea. (The latter excitation is often called a "hole.") Thus, the charge of all four of these excitations is negative compared to the Dirac sea, and the fact that they couple to the electromagnetic field with the same charge is sensible.

If you're interested in how the Dirac equation is constructed from this "Dirac sea" of antiparticles, I wrote a post here with many of the details. It uses a some quantum field theory, but feel free to ask follow-up questions. This topic is rarely treated properly in QM courses.

• Spinnor
I would disagree that you should interpret the negative energy states as positive energy antiparticles. Instead, the Dirac equation describes single particle/hole excitations on top of a fully-filled "Dirac sea" of positively charged antiparticles. The two positive energy solutions are states describing the spin up and down particle on top of this Dirac sea, while the two negative energy solutions correspond to removing a single spin up/down antiparticles from the filled Dirac sea. (The latter excitation is often called a "hole.") Thus, the charge of all four of these excitations is negative compared to the Dirac sea, and the fact that they couple to the electromagnetic field with the same charge is sensible.

If you're interested in how the Dirac equation is constructed from this "Dirac sea" of antiparticles, I wrote a post here with many of the details. It uses a some quantum field theory, but feel free to ask follow-up questions. This topic is rarely treated properly in QM courses.
Thank you for this. However, I was wondering what do the other 2 solutions represent physically. In the case of the free equation you have actually 4 particles (spin up and down electron and positron) which you observe in the lab. However, in the case of an EM field, the equation gives you 4 linearly independent solutions WITH charge minus one, so you can't make the same interpretation. So what are these 4 "things", if you were to measure them in a lab? Are all of them physical, as in the case of the free equation? Also, I am not sure I want an explanation in terms of the Dirac sea. I could ask the same question for the W bosons and that explanation would fail, I think.

king vitamin
Gold Member
The extra solutions do not represent anything physical, because they arise from some unphysical assumptions made when constructing the approximate one-particle Dirac equation.

In relativistic quantum mechanics, particles can be created and destroyed, so modeling your system using a one-particle wave function with conserved probability is simply incorrect. As I showed in the link I gave above, you can still construct an approximate one-particle theory, but it requires you to consider excitations on top of this unphysical infinite-energy and charge "Dirac sea" of antiparticles. The "negative energy" solutions only have negative energy with respect to this infinite-energy sea, and they correspond to removing one of the antiparticles. These do not correspond to anything measured in a lab because labs do not have Dirac seas, so really only the positive-energy solutions are useful. But the positive and negative energy solutions mix over time (the approximation breaks down).

If you instead want to model the wave function of an antiparticle with charge $-e$, you would instead construct it as an excitation on top of a Dirac sea of particles, as I mentioned in the post.

I recommend you go through the post a little to understand this: https://www.physicsforums.com/threads/which-law-formula-proves-the-existence-of-antimatter.951989/#post-6030416. The Dirac equation is working in a pathological limit so you find some weird stuff like this. If you want a more accurate theory, you need to go to QFT.

• bhobba
The extra solutions do not represent anything physical, because they arise from some unphysical assumptions made when constructing the approximate one-particle Dirac equation.

In relativistic quantum mechanics, particles can be created and destroyed, so modeling your system using a one-particle wave function with conserved probability is simply incorrect.
This really helped a lot. Thank you! One more thing I am confused about a bit (I guessed it is more historical than physical): In the books I read, introducing Dirac spinors, it is mentioned again and again that for the free Dirac equations, one can't discard the negative energy solutions and they must represent physical states. Here is, for example, a quote from Modern Particle Physics by Mark Thomson: "In classical mechanics, the negative energy solutions can be dismissed as being unphysical. However, in quantum mechanics all solutions are required to form a complete set of states, and the negative energy solutions simply cannot be discarded." But based on what you said, the fact that one particle wavefunctions are just approximations to a QFT description, it could have been the case that the extra 2 degrees of freedom didn't represent anything physical. I guess my question is, if they were building an approximate theory, why do all these books emphasize that the other 2 degrees of freedom HAD TO BE physical? (They turned out to be, but was it mandatory to be the case, given that the theory was just an approximation?)

king vitamin
Gold Member
I agree with the Thomson quote you gave that you cannot discard these negative energy solutions because they are inherently part of the Hilbert space. This is actually part of the problem with the one-particle Dirac theory: for particular Hamiltonians and initial conditions, you can evolve from a purely electron-like state into one of these negative-energy "hole" states, so in some systems the approximation breaks down at longer time scales.

Answering your question more generally is harder - I'm not sure if I agree that these extra states needed to be physical without the specific argument given above. But I think a lot of books and courses teach the one-particle Dirac theory without properly cautioning on what the exact approximations are.

vanhees71
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