Consider a scalar field ϕ described by the Klein-Gordon Lagrangian density L=12∂μϕ∗∂μϕ−12m2ϕ∗ϕ.
As written in every graduate QM textbook, the corresponding conserved 4-current jμ=ϕ∗i↔∂μϕ gives non-positive-definite ρ=j0. If we are to interpret ϕ as a wave function of a relativistic particle, this is a big problem because we would want to interpret ρ as a probability density to find the particle.
The standard argument to save KG equation is that KG equation describes both particle and its antiparticle: jμ is actually the charge current rather than the particle current, and negative value of ρ just expresses the presence of antiparticle.
However, it seems that this negative probability density problem appears in QFT as well. After quantization, we get a (free) quantum field theory describing charged spin 0 particles. We normalize one particle states |k⟩=a†k|0⟩ relativistically:
⟨k|p⟩=(2π)32Ekδ3(→p−→k),Ek=√m2+→k2
Antiparticle states |ˉk⟩=b†k|0⟩ are similarly normalized.
Consider a localized wave packet of one particle |ψ⟩=∫d3k(2π)32Ekf(k)|k⟩, which is assumed to be normalized. The associated wave function is given by
ψ(x)=⟨0|ϕ(x)|ψ⟩=∫d3k(2π)32Ekf(k)e−ik⋅x
1=⟨ψ|ψ⟩=∫d3k(2π)32Ek|f(k)|2=∫d3xψ∗(x)i↔∂0ψ(x)
I want to get the probability distribution over space. The two possible choices are:
1) ρ(x)=|ψ(x)|2 : this does not have desired Lorentz-covariant properties and is not compatible with the normalization condition above either.
2) ρ(x)=ψ∗(x)i↔∂0ψ(x) : In non-relativistic limit, This reduces to 1) apart from the normalization factor. However, in general, this might be negative at some point x, even if we have only a particle from the outset, excluding antiparticles.
How should I interpret this result? Is it related to the fact that we cannot localize a particle with the length scale smaller than Compton wavelength ~ 1/m ? (Even so, I believe that, to reduce QFT into QM in some suitable limit, there should be something that reduces to the probability distribution over space when we average it over the length 1/m ... )
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