Is the density of states Lorentz invariant?

Question

This is something that has been confusing me. A system can have a multitude of quantum states, and the energy of each will change depending on the frame of reference. However, the number of states should stay the same in each reference frame, logically.
Given that the density of states is defined as $$frac{dN}{dE_text{s.e.}}$$ with $N$ being the number of states and $E_text{s.e.}$ being single state energy, I would therefore assume that the density of states is in fact not invariant.
However, my (undergraduate) particle physics course has included some derivations which depend on the density of states in fact being constant across reference frames.
I don't have the necessary background in relativistic quantum mechanics to delve into the derivations deeply with any understanding, so I'm not in a position to judge whether or not the derivations are being sloppy or whether the density of states is actually invariant. Any clarification on this issue from more senior physicists would be greatly appreciated.

mike stone · Answer

Integrals over the density of states are usually set up using  Lorentz Invariant Phase Space (LIPS) which  invoves the measure
$$
int frac{d^3p}{2E_p (2pi)^3}ldots
$$
Here $E_p=sqrt{|{bf p}|^2+m^2 }$. This is nice because the  combination of the $d^3p$ and the $1/2E_p$ remains invariant under Lorentz boosts. At the same time the normalization of plane-wave momentum eigenstates has to be
$$
langle {bf p}|{bf p}'rangle = 2E_p (2pi)^3 delta^3({bf p}-{bf p}').
$$
This might look odd and unnecessarily complicated  but makes sense if you think of the $|{bf p}rangle $ state as descibing  a beam of particles. After a boost, the number of particles in a region stays the same, but the length of the region is Lorentz contracted so the density goes up. This increase is accounted for by the increase in the $E_p$ factor.

Is the density of states Lorentz invariant?

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