Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Even in a non-relativistic theory Spinors can arise as irreducible representations of the rotation subgroup of the symmetries of the theory. Why do people then put so much emphasize on the role of Relativity and the Dirac's equation in the topic? Is it just for historical reasons?

share|cite|improve this question

2 Answers 2

The Dirac equation has three predictions which explain phenomena observed in the hydrogen atom energy levels.

  1. The gyromagnetic ratio of 2
  2. The spin orbit interation which includes the Thomas precession, please, see for example Gurtler and D. Hestenes
  3. The Darwin term (please see again, the last reference).

Now, Bargmann already in 1962 discovered the ray representations of the Galilean group (equivalently, the true representations of a central extension of the Galilean group now known as the Bargmann group) and described its spinor representations.

The fact is that it is possible to formulate Galilean invariant wave equations which give the correct gyromagnetic ratio, spin orbit interation and the Darwin term. please see for example FUSHCHYCH, NIKITIN,SALOGUB , however, these equations contain arbitrary parameters which must be fitted to the experiment.

The subject of Galilean invariant wave equations is still under research, please see for example a recent article by : Niederl and Nikitin .

In my opinion, one of the reasons that there are no free parameters in the relativistic case, which makes the relativistic theory especially elegant (in addition to its full consistency with experiment), is that the Lorentz group is simple in contrast with the Galilean group.

I don't know if it is possible to formualte a Galelian invariant theory fully consistent with the experiment in the nonrelativistic regime developed from first principles (as in the case of the Dirac equation) and without ad-hoc terms, but I think it is possible, because the Galilean group can be obtained as an Inonu-Wigner contarction of the Lorentz group. I think that at least, this way, one can reduce the number of free parameters.

share|cite|improve this answer
It's not a huge deal, but while I would agree that the g factor is a prediction of the Dirac equation, the Dirac equation isn't necessary in order to get it right. One can use the non-relativistic Schrodinger-Pauli equation. See pg78-79 in Sakuri's ' Advanced QM.' – DJBunk Jun 12 '12 at 18:29

From a practical standpoint it depends on your field and motivation whether relativity and the Dirac equation are necessary tools. If you are working with things like the Aharonov–Bohm effect or in fields like Quantum Chemistry then full-blown Quantum Field Theory is unnecessary and quantum mechanics will suffice. Single-particle wave functions (or linear combinations of them) will be all you need to do your calculations. If you really want to be more precise then you can add relativistic corrections to your Hamiltonian, but you still don't need the Dirac equation or full blown QFT yet. However, if you are a particle physicist and want to smash things together at large energies and see what comes out you are going to need more machinery. Because of relativity ($E = m c^2$, roughly) you are going to create new, heavier particles either as external states in the final product of the collision or you will see effects of intermediate states of these heavier particles. In short you need a multi-particle theory that takes relativity into account- QFT including the Dirac equation (well, Lagrangian actually) is indispensable in this case. For more information on the limitations of non-relativistic quantum mechanics and how QFT addresses these limitations for particle physics I would see the the end of a Quantum mechanics book (chapter 20 in Shankar for example) or the beginning of a QFT textbook (chapters 1 & 2 of Peskin).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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