Anti-particles and spinors pop out of the Dirac equation very naturally, yet the Dirac equation is only a modified version of the Schrödinger equation which includes the relativistic energy-momentum equation in a mathematically sensible way. Therefore, special relativity leads to antiparticles and spin. However, our two main postulates for special relativity are light constancy and the principle of relativity and I don't understand how these would relate to the existence of anti-particles in any way.
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1$\begingroup$ More on Dirac equation & antimatter. $\endgroup$– Qmechanic ♦Commented Jun 21 at 15:18
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1$\begingroup$ Special relativistic Energy-mass-momentum relations allow for novel options such as particle-antiparticle creation/annihilation. $\endgroup$– Cosmas ZachosCommented Jun 21 at 15:21
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$\begingroup$ Where is a linear mapping in the non-rel QM, there is a square in the relativistic QM. That is needed because relativistic QM describes time and space as different colums of a matrix, thus things must use second derivate also on time and space coordinates. However, having a square, makes the negation similarly solution of the same equation. There is also some trickery, here we play with matrixes, and there is another "fun" of that there is a need to have an easy square root of a matrix. I think that was the first time in physics where I have got the impression that I read some RPG thing. $\endgroup$– peterhCommented Jun 22 at 13:22
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3 Answers
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Feynman provides a good summary in his Dirac Memorial Lecture. Basics are as follows:
Feynman considers the scattering of a particle by some potential, and then a second scattering which returns the particle to its original state. He calculates the QM probability amplitude for this overall process and finds that for it to be nonzero, the second scattering has to be outside the future light cone (spacelike separated) of the first.
If this is true, then there will exist an observer moving at some velocity relative to the frame in which the scatterings occur who will see the second event happening before the first. a world line drawn in his perspective for that particle will then have a kink in it, in which the scattered particle appears to be going backwards in time for a bit.
Then he assigns a negative charge to that particle and asserts that a negatively-charged particle going backwards in time is the same as a positively-charged particle moving forwards in time. He then identifies that particle as an antiparticle, and concludes that QM + relativity => antiparticles.
Each step in this line of reasoning strikes me as magic (why, for example, do the two scatterings have to be spacelike-separated for the amplitude to be nonzero???) but I can follow the steps. But in the second half of his Dirac Lecture (a video exists on youtube, BTW) he breezes effortlessly through an exposition of why integer and half-integer spin furnish different statistics by dealing with the example of the effect of particle exchange on the sign of the wavefunction, which I absolutely cannot fathom, and tops it all off with spacetime diagrams showing that the presence of a spectator particle affects the production of particle/antiparticle pairs. Can anyone recommend a different derivation of this that is less opaque?
answered Jun 21 at 16:10
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$\begingroup$ The explanation in Griffith's quantum mechanics textbook is a pretty straightforward calculation taking place explicitly in Minkowski space. $\endgroup$ Commented Jun 21 at 21:02
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$\begingroup$ Thank you for the wonderful recommended read. However, you are fundamentally misunderstanding what Feynman was saying. It is not that the amplitude is only non-zero outside of light cone. It is that the amplitude, which you expect to only be non-zero inside and on the light cone, cannot be non-zero outside the light cone. It is the same argument in basic QFT, where the naïve Green's function fails to respect locality. The rest of Feynman's argument is mostly Minkowski diagram; check "How to Teach Relativity to your Dog" about the alien from Mars going faster than light. $\endgroup$ Commented Jun 26 at 18:43
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$\begingroup$ @naturallyInconsistent, I am not surprised that I am misunderstanding this topic. it is right at the edge of my grasp. I will go back and re-read the Dirac Lecture. $\endgroup$ Commented Jun 27 at 5:24
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$\begingroup$ @nielsnielsen no prob about that. But both parts refer to QFT so you should read up on that. The relevant parts are 1) where we started talking about Green's functions and commutators in QFT, and 2) spin statistics theorem. In fact, both are really about spin-stats, because the reason why macrolocality and microlocality works in QFT is due to spin-stats. The reason why FTL -> antiparticles is best done by drawing the Minkowski diagram as per the pop sci book teaching SR to dog. Follow through on that, and it will be superbly obvious. $\endgroup$ Commented Jun 27 at 5:44
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How does special relativity lead to anti-particles?
The path is not direct and does not involve just special relativity. As you have already implied in your question, quantum mechanics is also involved.
Anti-particles and spinors pop out of the Dirac equation very naturally, yet the Dirac equation is only a modified version of the Schrödinger equation...
The historical path to the Dirac equation is via considerations of the inadequacies of the Klein-Gordon equation. Thus, I would think of the Dirac equation as a modified version of the Klein-Gordan equation, which has been modified in a very clever way in order to have a non-negative probability density. (At least in the context of "ordinary" quantum mechanics where we have not performed "second quantization.")
However, our two main postulates for special relativity are light constancy and the principle of relativity and I don't understand how these would relate to the existence of anti-particles in any way.
The thing that we want to take from special relativity is the relativistic energy-momentum relationship: $$ E^2 = c^2p^2 + m^2 c^4\;. $$
Below I will set $\hbar = c = 1$ for simplicity of notation.
A simplistic approach leads to the Klein-Gordon equation: $$ -\frac{\partial^2}{\partial t^2}\psi = -\nabla^2 \psi + m^2\psi\tag{A}\;. $$
But, if we are going to take Eq. (A) seriously, then the probability density $\rho$ and current density $\vec j$ have to be defined as: $$ \rho = \frac{i}{2m}\left(\psi^* \frac{\partial \psi}{\partial t} - \psi\frac{\partial \psi^*}{\partial t} \right)\tag{B} $$ and $$ \vec j = \frac{1}{2i}\left( \psi^*\vec\nabla\psi - \psi\vec\nabla\psi^* \right)\;. $$
Such that: $$ \frac{\partial \rho}{\partial t} = -\vec \nabla \cdot \vec j $$
Eq. (B) reduces to the correct non-relativistic expression, but is not tenable in general as a probability density since it can be negative.
The failure of Eq. (B) follows from the fact that Eq. (A) is second order in time and so both $\psi$ and $\partial\psi/\partial t$ can be specified arbitrarily. Thus, we abandon the KG equation. (But, spoiler alert, we later welcome the KG equation back into the fold via QFT.)
We seek a relativistic equation that is first order in time (for the reasons stated above). Due to the relationships imposed by special relativity on time and space, we also want the equation to be first order in space. And so we arrive at the equation: $$ \frac{\partial \psi}{\partial t} + \vec \alpha\cdot\vec\nabla \psi + m\beta\psi = 0\;.\tag{C} $$
I cannot provide a full derivation of the Dirac equation here, but rather provide a reference: Intermediate Quantum Mechanics by Bethe and Jackiw. Suffice it to say that the symbol $\psi$ must describe a multi-component object--it is not just a scalar function of space and time.
In fact, to satisfy all the requirements of special relativity and a have a linear first-order equation, in Eq. C the symbol $\psi$ must represent a four-component vector and the symbols $\vec\alpha$ and $\beta$ must represent 4-by-4 matrices. With the proper choices of $\vec\alpha$ and $\beta$ Eq. C is the Dirac equation.
The four components of $\psi$ represent the two spin states of the electron and the two spin states of the positron.
The non-negative probability density we desired is $$ \rho = \psi^\dagger \psi $$ and the current density is $$ \vec j = \psi^\dagger \vec \alpha \psi\;. $$
Thus, special relativity and quantum mechanics and the desire to achieve a non-negative probability density lead to the Dirac equation and its prediction of anti-particles.
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$\begingroup$ I don't agree that there is anything simplistic about the Klein-Gordon equation. Any non-interacting matter obeys it. The reason that the probability distribution is not positive definite is that it is actually the (Noether) charge distribution, divided by $e$. Only in the non-relativistic, Schrödinger, limit the charge distribution is of definite sign, and the it reduces to $e\psi^*\psi$. Inclusion of spin can be achieved by adding a term that is the relativistic generalisation of the Pauli interaction and using bispinors as wave function. $\endgroup$– my2ctsCommented Jun 21 at 17:05
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1$\begingroup$ When I write "simplistic" I mean the simplistic ad hoc substitution $E\to i\frac{\partial}{\partial t}$ and $\vec p\to -i\vec \nabla$. $\endgroup$– hftCommented Jun 21 at 17:33
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1$\begingroup$ In addition, if you want to supposed that the KG density is a "charge" density, then what do you propose to use for a probability density. In the context of this answer, we are doing ordinary quantum mechanics and we need a probability distribution otherwise we can not make predictions. For example, in ordinary quantum mechanics (not second quantized; not QFT) we need to write down expressions for, say, the expectation value of the position. How do you propose we write such a thing down without a probability distribution? $\endgroup$– hftCommented Jun 21 at 17:42
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1$\begingroup$ You are not "done" when you arrive at the KG equation, since there is no expression for probability density. That (and the fact that there are negative energy solutions, but those are also a problem with Dirac eq) are the major problems and are why KG can only be reasonably interpreted in the context of QFT, and not as a single-particle wave equation. $\endgroup$– hftCommented Jun 21 at 18:28
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1$\begingroup$ @my2cts The KG equation is only tolerable if you consider the absolutely free particle. Even in the non-interacting electron in a Coulomb background potential that we always use to consider the Hydrogen atom, the SR corrections that KG equation gives as improvements to the Schrödinger's equation version of the Hydrogen atom, is in the opposite sign compared to experiment (and to Dirac equation, obviously). And so no, you cannot retreat to assert that "Any wavefunction describing non-interacting particles obey the KG equation." because it is wrong to do so. $\endgroup$ Commented Jun 24 at 4:42
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Special relativity on its own does not. But you are combining it with quantum theory, so it isn't just that relativity alone holds all the weight. You need both.
answered Jun 21 at 14:55
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1$\begingroup$ You've just repeated what OP already said, and haven't answered the question. $\endgroup$ Commented Jun 22 at 0:26
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1$\begingroup$ @BlueRaja-DannyPflughoeft The OP is wondering how the postulates of SR lead to anti-particles. My answer is that they don't. You need to bring in more. I do enjoy the other answers, but I am still answering the question rather than complaining about other posts $\endgroup$ Commented Jun 22 at 1:58