Physical interpretation of Lorentz invariant fermion field product?

[blue2script]

Hey all!
Just a very short question: May I interpret the Lorenz invariant quantity

\bar\psi\psi

as being the probability density of a fermion field? Thanks!
Blue2script

 

[Demystifier]

No, because it is not positive.

 

[blue2script]

Hmm.. right. But then, what is the physical interpretation of the product above? And what is the probability density of a fermion field?

 

[hardphysics]

Hmm.. right. But then, what is the physical interpretation of the product above? And what is the probability density of a fermion field?

as far as i know there is no propability interpretation of fermion fields only charge density.

 

[hardphysics]

the interpretation of \bar\psi\psi is that it transforms under lorentz boosts like a scalar.

 

[Demystifier]

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 is
\psi^{\dagger}\psi

 

[hardphysics]

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

you mean \psi^{*}\psi because \psi is a scalar ? but eitherway its okay to write it with a dagger.

 

[Demystifier]

I mean dagger because psi a spinor, i.e., a 4-component wave function.