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1

Your example matrix $A$ isn't unitary. The right-most column has a length of 2 instead of 1, and the left-most column is only perpendicular to the right-most column if $a=-1$. You also seem to have computed $\rho_{\text{new}} = A \rho$ instead of $\rho_{\text{new}} = A \rho A^\dagger$. You can tell your $\rho_{\text{new}}$ must be wrong at a glance because ...


1

I was puzzled by this same thing when I took QFT classes several years back. After thinking about it, the reason is so trivial as to not merit an explanation in the literature, especially Peskin and Schroeder. Look at th LHS of your first equation: $\begin{align}\sum_{s,s'}\bar{v}^{s'}_a(p_2)\gamma^\mu_{ab}u^s_b(p_1)\bar{u}^{s}_c(p_1)\gamma^\nu_{cd}v^s_d(...


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An $A_{ab}B_{bc}$ yields a $C_{ac}$. Contracting all indices, but the outer ones of your expression yields a $[(\not{p}_2-m)\gamma^\mu(\not{p}_1+m)\gamma^\nu]_{dd}$. Now executing the $dd$ contraction is just the trace.


2

$G^\dagger G$ and $G G^\dagger$ are unitary equivalent in the finite dimensional case : There exists the polar decomposition $G = R U$ where $R \geq 0$ and $U$ unitary. Then $G^\dagger G = U^\dagger R^2 U = U^\dagger (G G^\dagger) U$. But from this it follows that $G^\dagger G$ and $G G^\dagger$ have the same eigenvalues. In the infinite dimensional case ...


4

This follows from the cyclicity of the trace, i.e. the property that $$\mathrm{Tr}(AB)=\mathrm{Tr}(BA),$$ which extends to the cyclic permutation $\mathrm{Tr}(ABC\cdots XYZ)=\mathrm{Tr}(BC\cdots XYZA)$ for larger products. Thus, if you expand $f$ in its Taylor series, you get \begin{align} \mathrm{Tr}\left(f(G^\dagger G)\right) & = \mathrm{Tr}\left(\sum_{...



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