Physics, asked by nabhsjsjs8388, 1 year ago

Does the conserved quantity of the complex scalar field descend from a symmetry?

Answers

Answered by shubham85288
0
For a complex scalar field Φ, the field has the expansionΦ(x0,x)=∫d3p√2Ep(2π)3 [e−iEpx0+ip⋅xap+e+iEpx0−ip⋅xˉa∗p]

where Ep=√|p|2+m2 and ap, a∗p are the particle ladder operators, and ˉap, ˉa∗p are the antiparticle ladder operators.

With the canonical commutation relations (meaning [ap,a∗k]=[ˉap,ˉa∗k]=δ(3)(k−p), etc.), the Wightman function becomes the following:⟨0|Φ∗(x)Φ(y)|0⟩=∫d3p(2π)312Epe−iEp(x0−y0)+ip⋅(x−y)

In Weinberg's Volume I (Chapter 5.2), he evaluates this function by picking the separation to be space-like with (x−y)2=−(x0−y0)2+|x−y|2>0. This results in⟨0|Φ∗(x)Φ(y)|0⟩=m4π2√(x−y)2K1(m√(x−y)2)

where K1 is the 1st modified Bessel function of the second kind.

I have two questions:

1. How to compute this function for time-likeseparations? If (x−y)2<0 is it simply true that ⟨0|Φ∗(x)Φ(y)|0⟩=m4π2√−(x−y)2K1(m√−(x−y)2)?

2. Weinberg says that for spacelike separations (x−y)2>0, this function is symmetric under (x−y)↦(y−x). Meaning ⟨0|Φ∗(x)Φ(y)|0⟩=⟨0|Φ∗(y)Φ(x)|0⟩. I see that this is obvious from the formula involving the Bessel K1, but how does one see this from the integral representation ⟨0|Φ∗(x)Φ(y)|0⟩=∫d3p(2π)312Epe−iEp(x0−y0)+ip⋅(x−y)? To me this is not obvious from the integral representation for non-zero x0−y0...Is it still true that the Wightman function is symmetric under (x−y)↦(y−x) for timelike separations?

EDIT; a typo. I actually am interested in the Wightman function ⟨0|Φ∗(x)Φ(y)|0⟩ (note that ⟨0|Φ(x)Φ(y)|0⟩=0)

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