# Can we express $\mathrm{tr}_A((A\otimes B)\rho_{AB})$ in terms of $A$, $B$, $\rho_A$ and $\rho_B$?

For a density matrix $$\rho_{AB}$$ and some operators $$A, B$$, is there a way to express

$$\text{Tr}_A((A\otimes B)\rho_{AB})$$

using the reduced states $$\rho_A$$ and $$\rho_B$$ and operators $$A$$ and $$B$$? Note that when $$A$$ is the identity operator, this is indeed possible and one has

$$\text{Tr}_A((I_A\otimes B)\rho_{AB})= B\rho_{B}$$

as shown here.

In general, the knowledge of the marginals $$\rho_A$$ and $$\rho_B$$ and the operators $$A$$ and $$B$$ is insufficient to compute $$\mathrm{tr}_A((A\otimes B)\rho_{AB})$$. Indeed, we can find two different density matrices $$\rho_{AB}$$ and $$\sigma_{AB}$$ with the same marginals for which

$$\mathrm{tr}_A((A\otimes B)\rho_{AB}) \ne \mathrm{tr}_A((A\otimes B)\sigma_{AB}).$$

One such counterexample uses the elements of the Bell basis $$|\beta_{ij}\rangle = X_A^iZ_A^j|\beta_{00}\rangle$$ with $$|\beta_{00}\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$$ and $$i,j=0, 1$$. In this case

\begin{align} \mathrm{tr}_A((A_A\otimes B_B)|\beta_{ij}\rangle\langle\beta_{ij}|) &= \mathrm{tr}_A\left[(A_A\otimes B_B)X_A^iZ_A^j|\beta_{00}\rangle\langle\beta_{00}|Z_A^jX_A^i\right] \\ &= \mathrm{tr}_A\left[(A_A\otimes B_B)Z_B^jX_B^i|\beta_{00}\rangle\langle\beta_{00}|X_B^iZ_B^j\right] \\ &= \mathrm{tr}_A\left[(I_A\otimes B_B)Z_B^jX_B^iA_B^T|\beta_{00}\rangle\langle\beta_{00}|X_B^iZ_B^j\right] \\ &= BZ^jX^iA^T\,\mathrm{tr}_A\left[|\beta_{00}\rangle\langle\beta_{00}\right]\,X^iZ^j \\ &= \frac12BZ^jX^iA^TX^iZ^j \end{align}\tag1

where we exploited a useful and easy to check property that $$(A\otimes I)|\beta_{00}\rangle = (I\otimes A^T)|\beta_{00}\rangle$$. We see that if $$A$$ and $$B$$ are non-singular, then for the expression $$(1)$$ to yield the same value for all $$i,j$$ we need $$A$$ to commute with $$X$$ and $$Z$$. However, any operator $$A$$ on $$\mathbb{C}^2$$ can be written as $$A=aI+bX+cZ+dXZ$$ and it is easy to see that if $$A$$ commutes with $$X$$ and $$Z$$ then $$b=c=d=0$$. Consequently, $$A=aI$$.

We conclude that if $$A\ne aI$$ then $$\mathrm{tr}_A((A\otimes B)\rho_{AB})$$ is not uniquely defined on the set of density matrices $$\rho_{AB}$$ with given marginals $$\rho_A$$ and $$\rho_B$$. Therefore, no general formula for $$\mathrm{tr}_A((A\otimes B)\rho_{AB})$$ in terms of $$A$$, $$B$$ and the marginals exists.

• Thank you. Sorry for the extra question but since it's so closely related, can I ask if there is a general expression for $\text{tr}_B((I_A\otimes B)\rho_{AB})$? Is the state obtained somehow proportional to $\rho_A$?
– JRT
Mar 16, 2021 at 8:55
• There is not. To see this, set $B=I$ in the proof above to obtain that $\mathrm{tr}_A((A\otimes I)\rho_{AB})$ is not unique for fixed $A$ and $B$ and among the density matrices with given marginals. Then interchange $A$ and $B$ to reach the conclusion. Mar 16, 2021 at 15:16

No - take $$\rho$$ one of the four Bell states, which all have the same marginals. Then the trace you give will evaluate to $$\mathrm{tr}[APBP]$$, with $$P$$ one of the four Paulis (including $$I$$), which is not only a function of $$A$$ and $$B$$ (as it depends on the Pauli, which cannot be inferred from the reduced states).