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May 29, 2021 at 23:53 answer added Joseph Van Name timeline score: 1
Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Mar 23, 2015 at 23:49 comment added Suvrit In other words, if $A=X \otimes Y$ is tensor product of density matrices, then $\text{tr}_2(A)=X$ and $\text{tr}_1(A)=Y$ (this is the marginalizing out that Carlo mentioned).
Mar 23, 2015 at 13:26 comment added Igor Khavkine Partially dualized, $f\in \mathrm{Hom}(X,Y)\otimes (U\otimes U^*)$. Partial trace is then the linear map $\mathrm{Hom}(X,Y) \otimes (U\otimes U^*) \to \mathrm{Hom}(X,Y)$ given by $\mathrm{id} \otimes \delta$, where $\delta \in U^*\otimes U \cong (U\otimes U^*)^*$ is the dualized version of the identity map $U\to U$. I don't know if this is very geometric, but it's certainly an invariant description of the operation.
Mar 23, 2015 at 7:13 comment added Carlo Beenakker a physics interpretation is that the partial trace is the way one obtains marginal probability distributions in quantum mechanics: the density matrix $\rho$ describes the probability distribution $P_{X,U}$ of the combined systems $X$ and $U$ and by performing the partial trace over $U$ one obtains a new density matrix $\rho_X$ that describes the marginal distribution $P_X$ of system $X$ alone; this is not the geometric interpretation you are asking for, but I would think that if you have a "geometric interpretation" of the marginal distribution then you're done.
Mar 23, 2015 at 3:06 history asked David Spivak CC BY-SA 3.0