Suppose I have a Quantum circuit, which gives an output state $|\psi \rangle$ let's say. I wish to obtain the reduced density matrix by tracing out subsystem B, i.e. $\rho = |\psi \rangle \langle \psi |$, and : $$\rho_A = tr_B(\rho)$$. Finally, I wish to square $\rho_A$ and find its trace, i.e. $tr_A(\rho_A^2)$. I know there are methods in Qiskit to evaluate reduced density matrices and such. My question is: Is there a circuit formulation (without using pre-existing methods) that can do the following non-linear operation? For eg: any circuit that encodes information about the term $tr_A(\rho_A^2)$ in its output probabilities?I understand that I can reconstruct the state itself from the measurement outcomes, and then do classical computation to arrive at my result, but is there any better way to do it?
2 Answers
If you mean any kind of circuit, or more general quantum operation, that takes as input some $\rho$, and produces output probabilities equal to $\operatorname{tr}(\rho_A^2)$, or more generally a nonlinear function of $\rho$, then no, it's not possible.
That's easy to see: any unitary or non-unitary operation that is physically realisable is describable as a linear map applied to the input density matrix. And the function sending output states into output probabilities is also linear. Therefore it's not possible to ever have a nonlinear function of input density matrices into the output probabilities. In fact, any possible circuit or combination of a physical operation and any kind of measurement can be described concisely as some effective POVM applied to the input state, and therefore the output probabilities always have the form $p_b(\rho)=\operatorname{tr}(\mu_b\rho)$ for some "effective POVM" $(\mu_b)_b$. If you allow for linear post-processing of probabilities, then the same argument tells you you can get functions of the form $\operatorname{tr}(\mathcal O \rho)$ for some observable $\mathcal O$. Either way, these are linear in $\rho$.
Of course, if your input is multiple copies of the same state, say $\rho\otimes\rho$, then you can get things like $\operatorname{tr}(\rho_A^2)$, because those are linear in $\rho\otimes\rho$.
If you have two copies of $|\psi\rangle$, you simply perform the swap test on the "A" components.
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$\begingroup$ I understand, but doesn't the SWAP test give entanglement inside a state, and not in between states? What will be the interpretation here for the swap test only acting on the "A" components? @DaftWullie $\endgroup$ Commented Jun 19 at 11:59
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$\begingroup$ The swap test between two density matrices $\rho_A$ and $\rho_B$ basically evaluates $\text{Tr}(\rho_A\rho_B)$. So if I set $\rho_B=\rho_A$, it does exactly what you're asking for. $\endgroup$ Commented Jun 19 at 12:02