I'm trying to (understand and) solve this problem from Nielsen and Chuang's Quantum Computation and Quantum Information.

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I know the definition of Operation Elements: $\sum_{k} E_k \rho E_k^†$ with $E_k = \langle e_k|U|e_0 \rangle$ where $e_k$ form the basis for the environment. But I'm not quite sure how this definition may be used to solve the problem above.


So, let the system be $\rho$, and the environment $|0\rangle \langle 0|$.

The given operation (which you can check is unitary, and incidentally happens to be the CNOT operation), is applied on $\rho \otimes |0\rangle \langle 0|$.

So, you have: $$(P_0 \otimes I + P_1 \otimes X)(\rho \otimes |0\rangle \langle 0|)(P_0 \otimes I + P_1 \otimes X)$$

Now, you just expand it and multiply to get: $$P_0 \rho P_0 \otimes |0\rangle \langle 0| + P_0 \rho P_1 \otimes |0\rangle \langle 1| + P_1 \rho P_0 \otimes |1\rangle \langle 0| + P_1 \rho P_1 \otimes |1\rangle \langle 1| $$

When you trace out the environment, you're left with: $$P_0 \rho P_0 + P_1 \rho P_1$$

Thus the operator elements are $P_0$ and $P_1$.

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  • $\begingroup$ Awesome, thanks a lot. So, in order to find operation elements, I need to use $U(\rho \otimes |0_E\rangle \langle0_E|) U^\dagger$. $\endgroup$ – Bashir Jul 14 '19 at 20:24
  • $\begingroup$ @Bashir, yes, and then trace out the environment. $\endgroup$ – Mahathi Vempati Jul 15 '19 at 5:35
  • $\begingroup$ Yes, exactly. Thanks. $\endgroup$ – Bashir Jul 15 '19 at 22:29

As an alternative to the other answer (which is perfectly valid), we can use the formalism specified in the question. We're given a $U=P_0\otimes I+P_1\otimes X$. We know that the environment starts in state $|e_0\rangle=|0\rangle$. We need to establish an orthonormal basis for the environment, $|e_k\rangle$. In this case, it's straightforward as the environment is only two-dimensional: $$ |e_0\rangle=|0\rangle\qquad |e_1\rangle=|1\rangle. $$ Now we can apply the formula $E_k=(I\otimes \langle e_k|)U(I\otimes |e_0\rangle)$. Note that I've added in some identity elements to make it clearer that the environment basis only applies to the second subsystem. Now, $$ U(I\otimes |e_0\rangle)=P_0\otimes|0\rangle+P_1\otimes|1\rangle, $$ and so we have $$ E_0=(I\otimes\langle 0|)(P_0\otimes|0\rangle+P_1\otimes|1\rangle)=P_0, $$ and $E_1=P_1$.

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  • $\begingroup$ Thanks for the answer using a different approach. $\endgroup$ – Bashir Jul 15 '19 at 22:27

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