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.

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

A solution to the equation can in principle be read off by comparing coefficients.

  • 1
    $\begingroup$ @user77146 Was this answer helpful? Would you like me to expand? $\endgroup$ – Nephente Sep 5 '14 at 13:02
  • $\begingroup$ @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.) $\endgroup$ – user120404 Feb 2 '16 at 18:05
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    $\begingroup$ @user120404 No, the partial trace isn't cylic anymore. The partial trace means we're interested in the evolution of the first qubit only, regarding it as an open quantum system. As it is, your last statement is not sensible because the lhs is just 1 while the rhs is a density matrix. There is an excellent book on open quantum systems by Breuer and Pettrucione. $\endgroup$ – Nephente Feb 2 '16 at 19:19

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