Why does combining special relativity with quantum mechanics imply that particle number is not conserved?

I'm guessing that it as something to do with the fact that if the relativistic energy of a particle is large enough, then by the energy-momentum dispersion relation $E^{2}=m^{2}c^{4}+p^{2}c^{2}$, it is possible for additional particles to be created. This is the classical relativistic side - I'm guessing that the quantum side comes in considering the Heisenberg uncertainty relation. However, so far I have been unable to come up with a satisfactory answer in my mind. Any help would be much appreciated!


2 Answers 2


One way to think about this is to compare the free-particle relations $E=\frac{p^2}{2m}$ and $E^2=m^2+p^2$, where I'll write $c=\hbar=1$ for tidiness. Quantising, these respectively become the TDSE $\dot{\psi}=\frac{i}{2m}\nabla^2\psi$ and the KGE $\ddot{\psi}=\nabla^2\psi-m^2\psi$.

The non-relativistic TDSE gives $\partial_t\rho+\boldsymbol{\nabla}\cdot\mathbf{j}=0$ with $\rho=\psi^\ast\psi,\,\mathbf{j}=\frac{i}{2m}(\psi^\ast\boldsymbol{\nabla}\psi-\psi\boldsymbol{\nabla}\psi^\ast)$. A relativistic continuity equation would be $\partial_\mu j^\mu=0$, which with the obvious generalisation $j^\mu=\frac{i}{2m}(\psi^\ast\partial^\mu\psi-\psi\partial^\mu\psi^\ast)$ is true of any solution of the KGE. But $j^0$ can no longer be a probability density for one particle's location, because unlike the TDSE the KGE's solution set is closed under complex conjugation, which multiplies $j^\mu$ by $-1$, changing the sign of "probabilities" (as well as energies, as you can see by considering a plane-wave solution).

The obvious fix is to posit that integrating $j^0$ over space doesn't give $1$, but instead the number of particles minus the number of antiparticles. You can think of this as a charge conservation, where antiparticles have an opposite charge. When physicists first considered the problem, however, they didn't know about antimatter. Instead the hope was to come up with some reason the negative-energy solutions are unphysical. Paul Dirac tried to delete them with a first-order energy-momentum relation, because $E^2=m^2+p^2$ is consistent with changing the sign of $E$. But even the first-order Dirac equation still has these solutions; it's just that they're interpreted as antimatter. (The $4$-component Dirac spinor for electrons and positrons gives each $2$ components, and thereby explains their $S=\frac{1}{2}$ spin degeneracy.)

Clearly, if antimatter exists the conservation of particles minus antiparticles is consistent with pair production and annihilation. Whether a specific theory requires separate conservation of particles and antiparticles requires a fuller treatment in terms of ladder operators. For example, these lecture notes explain that, while in a free theory the particle number $N_c$ and antiparticle number $N_b$ are separately conserved and hence so is $N_c-N_b$, in an interacting theory only $N_c-N_b$ is conserved. (See in particular Tong's discussions of Eqs. (2.75), (3.5)-(3.7), (3.25).)

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    – ACuriousMind
    May 10, 2017 at 16:04

you said, "I'm guessing ..." ... and i say keep guessing because you've pretty much almost answered your own question. So take a QM particle jack in the box and apply energy uncertainty with the box size going to zero in the limit. Now you got enough energy to create a particle-antiparticle pair in the box ... what's the quantum state for jack and his new friends? ... oops ... exit QM & SR, enter QFT.

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    Nov 17 at 17:13

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