The discussion centers on the interpretation of the Lorentz invariant quantity \(\bar{\psi}\psi\) in the context of fermion fields. Participants clarify that this quantity does not represent probability density due to its non-positive nature and instead transforms as a scalar under Lorentz boosts. The distinction between fermion field operators and wave functions is emphasized, with the probability density for a 1-particle wave function given by \(\psi^{\dagger}\psi\), where \(\psi\) is identified as a spinor, specifically a 4-component wave function.
PREREQUISITESPhysicists, particularly those specializing in quantum field theory, graduate students studying particle physics, and researchers interested in the mathematical foundations of fermion fields.
blue2script said:Hmm.. right. But then, what is the physical interpretation of the product above? And what is the probability density of a fermion field?
Fermion field is an operator, so it does not have a probabilistic interpretation. However, one should distinguish the field operator from the wave function which is a c-number function representing a quantum state. For a 1-particle wave function \psi, the probability density isblue2script said:Hmm.. right. But then, what is the physical interpretation of the product above? And what is the probability density of a fermion field?
Demystifier said:Fermion field is an operator, so it does not have a probabilistic interpretation. However, one should distinguish the field operator from the wave function which is a c-number function representing a quantum state. For a 1-particle wave function \psi, the probability density is
\psi^{\dagger}\psi