# How can I solve an equation involving partial trace?

I am unable to find the solution to the following equation:

Tr$_{2}[U(|\psi\rangle \langle\psi|\otimes \rho)U^{\dagger}]=\rho$

Here $\psi$ is state vector representing a qubit and $\rho$ state of second qubit(the partial trace is over its subspace).

Also $U$ is a unitary operator given by $\sum |x \vee y\rangle |y\rangle \langle x| \langle y|$ where $x,y \in \{0,1\}$

$\vee$ stands for bit XOR and $\otimes$ for tensor product and $U$ operates on joint space of both the qubits.

• What exactly are you struggling with? – Nephente Sep 4 '14 at 6:57
• Deutsch theory regarding closed timelike curves – sashas Sep 4 '14 at 7:09
• Nah :-) I meant specifically with the calculation. I imagine you have trouble evaluating the lhs? – Nephente Sep 4 '14 at 7:21
• yes how to get $\rho$ on the other side, yes can't resolve the lhs – sashas Sep 4 '14 at 7:54

Hint: Start by representing $\psi$ and $\rho$ in the basis $\{ \vert x\rangle,\vert y\rangle\}$. Shouldn't be too difficult to calculate the action of $U$ once you've done that. If you don't know how to take the partial trace, post the intermediate result and ask back.
These steps will produce an equation like $$A_{xy}(\psi,\rho)\vert x\rangle\langle y\vert = \rho = \rho_{xy}\vert x\rangle\langle y\vert$$
The notation on the lhs. indicates ,that the coefficients $A$ will in general depend on both $\rho$ and $\psi$.
• @Nephente Can't we use the cyclic property of the trace to pair the unitary operator conjugates into the identity, then throw out the $\rho$ of the second system to get $Tr(|\psi \rangle \langle \psi |) = \rho$ ? (Be merciful if it's nonsense. I am new to this.) – user120404 Feb 2 '16 at 18:05