For any linear operator $A$, the support of $A$ is the orthogonal complement of its kernel. Hence when we say, $supp(A)\subset supp(B)$, we have that for any vector $v$ in the kernel of $B$ i.e. $Bv = 0$, it also must be that $Av = 0$.

My question is if it is true (and if yes, how to show) that support containments for quantum states are preserved under partial trace. That is, if

$$\text{supp}(\rho_{AB})\subset \text{supp}(\sigma_{AB})$$

is it true that

$$\text{supp}(\rho_{A})\subset \text{supp}(\sigma_{A})$$

  • $\begingroup$ For quantum states, you can also use the following facts 1) The relative entropy is finite if and only if the support of the first argument is contained in the second 2) Data processing under partial trace cannot increase the relative entropy (so if it starts out finite, it stays finite). $\endgroup$ – rnva Mar 9 at 3:16

If $\rho_{AB}, \sigma_{AB} \ge 0$, i.e. they are positive semi-definite then yes (otherwise no).

To prove it observe that if $\text{supp}(\rho_{AB})\subset \text{supp}(\sigma_{AB})$ then we can write that $$ \sigma_{AB} = \epsilon \cdot \rho_{AB} + \Delta, $$ for some sufficiently small $\epsilon > 0$ and some operator $\Delta \ge 0$.

Partial trace is linear, so we can write $$ \sigma_A = \epsilon \cdot \rho_{A} + \text{Tr}_B(\Delta). $$ Since $\text{Tr}_B(\Delta) \ge 0$ we can deduce that $\text{supp}(\rho_{A})\subset \text{supp}(\sigma_{A})$.


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