polarized trace

As you said, one way would be to specify a matrix representation for the spinors but it restrains the study. Another way lies on using the helicity projection operators to define a particular polarization. This method leads to traces solvable by applying $gamma$-matrices identities, so this is a pretty convenient way.

The helicity projection operators

We note the longitudinal spin component $sigma_{textbf{p}} = frac{mathbf{sigma}mathbf{.p}}{|mathbf{p}|}$ and we define the helicity projection operators by $$ Pi^{pm}(mathbf{p}) = frac{1}{2}(1 pm sigma_mathbf{p}) $$ Those operators project the spinors over the parallel (or antiparallel) spin to the direction of motion, as shown in the following identities (where the index r stands for the spin states $uparrow$ and $downarrow$). $$ Pi^+(mathbf{p})u_r(mathbf{p}) = delta_{ruparrow}u_r(mathbf{p})=u_uparrow(mathbf{p})$$ $$ Pi^-(mathbf{p})u_r(mathbf{p}) = delta_{rdownarrow}u_r(mathbf{p})=u_downarrow(mathbf{p})$$ $$ Pi^+(mathbf{p})v_r(mathbf{p}) = delta_{rdownarrow}v_r(mathbf{p})=v_downarrow(mathbf{p})$$ $$ Pi^-(mathbf{p})v_r(mathbf{p}) = delta_{ruparrow}v_r(mathbf{p})=v_uparrow(mathbf{p})$$

The idea is to use those operators in order to replace the specific spinor spin states by a sum over all states (where the Kronecker delta keep track of the specific state) so that you still have a sum over all states which leads to a trace when one applies the completeness relations. I will give an example to illustrate how it works. I don't really understand where the trace you get in your question comes from, so I assume you obtained it squaring the unpolarized Feynman amplitude. If it the case I hope my example will help you, otherwise this is irrelevant...

Let consider that we have a Feynman amplitude $mathcal{M} = bar{u}_downarrow(mathbf{p'})Gamma u_uparrow(mathbf{p})$ given for a particular polarization state (I have chosen a flip of spin state to fit your example) where $Gamma$ is a $4x4$ matrix built up out of $gamma$-matrices. We introduce $tilde{Gamma} = gamma^0Gamma^daggergamma^0$ which enables us to write the square amplitude as $$|mathcal{M}|^2 = (bar{u}_downarrow(mathbf{p'})Gamma u_uparrow(mathbf{p}))(bar{u}_uparrow(mathbf{p})tilde{Gamma} u_downarrow(mathbf{p'}))$$ We are free to rewrite this expression introducing the helicity projection operators (they will not contribute since the positive (negative) operator acts on the positive (negative) spin state): $$|mathcal{M}|^2 = (bar{u}_downarrow(mathbf{p'})GammaPi^+(mathbf{p}) u_uparrow(mathbf{p}))(bar{u}_uparrow(mathbf{p})tilde{Gamma}Pi^-(mathbf{p'}) u_downarrow(mathbf{p'}))$$

We can now sum over the spins because thanks to the helicity projection operators only one of the four terms is nonzero.

$$|mathcal{M}|^2 = sum_{r = uparrow downarrow}sum_{s = uparrow downarrow}(bar{u}_s(mathbf{p'})GammaPi^+(mathbf{p}) u_r(mathbf{p}))(bar{u}_r(mathbf{p})tilde{Gamma}Pi^-(mathbf{p'}) u_s(mathbf{p'}))$$

We can now write this expression as a trace. For that we should write down the spinors indices, move $u_s(mathbf{p'})$ to the left and use the completeness relations. Doing that we obtain :

$$|mathcal{M}|^2 = Tr[(not{p'} +m)GammaPi^+(mathbf{p})(not{p} +m)tilde{Gamma}Pi^-(mathbf{p'})]$$

This expression can now be computed using $gamma$-matrices identities and trace technology. By the way, in the relativistic limit $Egg m$ the helicity projection operators simplify to

$$ Pi^{pm}(mathbf{p}) = frac{1}{2}(1 pm gamma^5) $$

which leads to a considerable simplification during the computation of the trace.

As a conclusion, for a definite polarization the idea is to compute the trace as usual projecting over the relevant spin states via the helicity operators. I hope this answers your question, but I might be wrong when I try to understand where your trace comes from.