I have been reading this Phys.SE answer in order to clarify my doubts. It seems to me that he claims that the postulates are the same no matter if it is QFT, QM or whatever. But some books tell us about the problems with the probability density in RQM and other issues so Im not quiet sure if they obey the same postulates of quantum mechanics.

How far can we extend the postulates of quantum mechanics (QM) to Relativistic quantum mechanics (RQM)?

I have read about the “postulates of quantum mechanics” in some non-relativistic QM textbooks but I have never heard about the “postulates of RQM”. It would be helpful if you can refer some paper or book about which postulate has to be modified or proving that in fact the postulates are the same.


2 Answers 2


Relativistic Quantum Mechanics arises as a sort of limit over the degrees of freedom of non-relativistic Quantum Mechanics. Without getting into technicalities, Quantum Mechanics deals with system with finitely many degrees of freedom (1D/2D/3D QHO, a system of finitely many QHOs, etc...). When you take the limit over the degrees of freedom you get into troubles, the main one being that von Neumann's uniqueness does not hold anymore, and one has to deal with a lot of inequivalent irreducible representations of the Weyl algebra (one can even exhibit a family of such representations that has the power of the continuum). To somehow fix this, one considers representations that are reminiscent of the theory with finitely many degrees of freedom: only those representations of the Heisenberg group for infinitely many degrees of freedom that integrates into Weyl form when restricted to any finite subset of degrees of freedom are considered as physically relevant). With this construction one naturally lands into the Fock representation.

In this sense, QM and RQM share a lot in common, the latter being based extensively on the former, but as you can see one has to require something more (like the postulate on the representations of the Heisenberg group). Furthermore, in a relativistic theory you also want to look at it as a dynamical system, where you have an action of the Poincaré group on the algebra of observables. This aspect introduces some more requirements (covariance, spectral condition, locality, ...) that are not considered in the non-relativistic case.

  • $\begingroup$ I am not sure why RQM arises as a sort of limit over the degrees of freedom. do you mean continuum degrees of freedom? $\endgroup$
    – Anthonny
    Jul 20, 2015 at 9:05
  • $\begingroup$ No I mean a system like countably many independent QHOs. For a system like this there is no unique irreducible representation, up to equivalence (see e.g. Gårding-Wightman, 1954). $\endgroup$
    – Phoenix87
    Jul 20, 2015 at 9:35

Relativistic quantum mechanics is quantum mechanics; therefore the interpretation is the same. In this context, the main difference (at least in my opinion) is that relativity allows (and it is an experimental fact as well) the creation and annihilation of particles.

To accommodate this possibility in the theory, a new framework has to be introduced. Naïvely, this new framework is to consider another type of Hilbert spaces (usually called Fock spaces), where the state carries information also on the possibility of having an arbitrary number of particles. The fashion in which these particles are created and destroyed is then described by the interaction.

The interpretation of the quantum nature is, however, unaffected. The state still has a probabilistic interpretation, the evolution is still linear and thus obeying Schrödinger equation (and there are still "problems" with the measurement procedure).


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