- #1
ChrisVer
Gold Member
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Suppose that we have that:
[itex]\mathcal{M} = c \bar{u}^{s'}(p') u^s(p) \bar{u}^{r'}(k') u^r(k) [/itex]
For a fermion fermion scattering: [itex] f(k,r)+ f(p,s) \rightarrow f(k',r')+ f(p',s') [/itex]
Now if I want to calculate the polarization summed and averaged squared amplitude:
[itex]\frac{|c|^2}{4} \sum_{r,r',s,s'} \bar{u}^{s'}(p') u^s(p) \bar{u}^{r'}(k') u^r(k) \bar{u}^{r}(k) u^{r'}(k') \bar{u}^{s}(p) u^{s'}(p')[/itex]
In index form I think this can be written:
[itex]\frac{|c|^2}{4} \sum_{r,r',s,s'} \sum_{a,b,c,d} \bar{u}^{s'}(p')_a u^s(p)_a \bar{u}^{r'}(k')_b u^r(k)_b \bar{u}^{r}(k)_c u^{r'}(k')_c \bar{u}^{s}(p)_d u^{s'}(p')_d[/itex]
So I can move the spinors around, no? So that I will finally get:
[itex] \frac{|c|^2}{4} Tr \Big( [p+m][p'+m] \Big) Tr \Big( [k+m][k'+m] \Big) [/itex]?
where the momenta in the trace are in fact slashed...
[itex]\mathcal{M} = c \bar{u}^{s'}(p') u^s(p) \bar{u}^{r'}(k') u^r(k) [/itex]
For a fermion fermion scattering: [itex] f(k,r)+ f(p,s) \rightarrow f(k',r')+ f(p',s') [/itex]
Now if I want to calculate the polarization summed and averaged squared amplitude:
[itex]\frac{|c|^2}{4} \sum_{r,r',s,s'} \bar{u}^{s'}(p') u^s(p) \bar{u}^{r'}(k') u^r(k) \bar{u}^{r}(k) u^{r'}(k') \bar{u}^{s}(p) u^{s'}(p')[/itex]
In index form I think this can be written:
[itex]\frac{|c|^2}{4} \sum_{r,r',s,s'} \sum_{a,b,c,d} \bar{u}^{s'}(p')_a u^s(p)_a \bar{u}^{r'}(k')_b u^r(k)_b \bar{u}^{r}(k)_c u^{r'}(k')_c \bar{u}^{s}(p)_d u^{s'}(p')_d[/itex]
So I can move the spinors around, no? So that I will finally get:
[itex] \frac{|c|^2}{4} Tr \Big( [p+m][p'+m] \Big) Tr \Big( [k+m][k'+m] \Big) [/itex]?
where the momenta in the trace are in fact slashed...