QFT with negative mass

1. May 25, 2012

nonequilibrium

In the Klein-Gordon equation (spin 0), the mass dependence is (only) through $m^2$, whereas in the Dirac equation (spin 1/2) it's through $m$.

Does this mean that for spin 0 particles, we can just as well describe them as having negative mass without changing any of the physics (whereas for the spin 1/2 particle there would be a difference)?

Perhaps this is simply another way of saying that spin 0 particles are their own antiparticles (?), whereas spin 1/2 particles are not (in the Dirac sea picture)? However, that can't really be the case, since we also have $m^2$ in the Proca equation, which would mean that all spin 1 particles are their own antiparticles, which is not the case.

So what does these different types of mass-dependence signify?

2. May 25, 2012

dextercioby

Because the units must match, there's a power 2 in the mass term for 2-nd order field PDE's (spin 0,1,2) and a power 1 for the mass term in the 1-st order field PDE (spin 1/2,3/2).

So there's a connection between the number in the power of the mass and the degree of the PDE as one can see by the Dirac-Fierz-Pauli equations for a general spin s field.

3. May 25, 2012

nonequilibrium

But what is the physical implication of this observation?

4. May 25, 2012

Bill_K

mr. vodka, As the man says, it's just that m (or actually mc/ħ, the reciprocal of the Compton wavelength) has the dimensions of L-1, and appears as many times as the derivative appears. You can, if you like, describe spin zero particles by replacing the Klein-Gordon equation with a set of first order equations.

Negative mass is not, as you seem to be saying, associated with antiparticles. Under charge conjugation, the operation that interchanges particles and antiparticles, electric charge changes sign but mass does not. Antiparticles have positive mass no matter what their spin happens to be. The idea of a particle with negative mass is unphysical.