# Can an ensemble of pure states give probability less than 1?

I am calculating the reduced density matrix of a bipartite system, I ended up getting that it was the sum of two density matrices of pure states each with a probability $$1/3$$. My understanding was that the coefficients in this ensemble should add up to $$1$$. But perhaps this is only for density matrices of whole systems and not for a reduced density matrix? So have I made an error in the calculation (I don't think I have) or the result I got possible?

• I guess my question is - is this result possible? If it is impossible then I have made an error in the calculation.
– user12101
May 21, 2020 at 9:41
• If you have a normalised bipartite state $\rho_{AB}$ and you calculate the reduced state $\rho_A = \mathrm{Tr}_A[\rho_{AB}]$, then the reduced state should also be normalised. (Partial trace is a TPCPM). Are all the density matrices in your ensemble and the original density matrix normalized? As glS says, it is probably best if you were to post your calculation. May 21, 2020 at 9:53

Given $$\rho\equiv\sum_{ijk\ell}\rho_{ij,k\ell}|ij\rangle\!\langle k\ell|$$, the reduced state is $$\rho_A = \sum_{ik} \left(\sum_j \rho_{ij,kj} \right)|i\rangle\!\langle k|.$$ Therefore, $$\operatorname{Tr}(\rho_A)\equiv \sum_i (\rho_A)_{ii} = \sum_{ijk}\rho_{ij,ij}=1.$$
Equivalently, $$\operatorname{Tr}(\rho_A)=\operatorname{Tr}(\operatorname{Tr}_B(\rho)) = (\operatorname{Tr}_A\otimes\operatorname{Tr}_B)\rho=\operatorname{Tr}(\rho)=1.$$