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?

  • $\begingroup$ I guess my question is - is this result possible? If it is impossible then I have made an error in the calculation. $\endgroup$
    – user12101
    May 21, 2020 at 9:41
  • 1
    $\begingroup$ 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. $\endgroup$
    – Rammus
    May 21, 2020 at 9:53

1 Answer 1


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, $$\operatorname{Tr}(\rho_A)=\operatorname{Tr}(\operatorname{Tr}_B(\rho)) = (\operatorname{Tr}_A\otimes\operatorname{Tr}_B)\rho=\operatorname{Tr}(\rho)=1.$$


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