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Suppose we have two qubits in the state $x|00\rangle+y|11\rangle $. What is the resulting state of the second qubit in that case? Use and to denote and respectively.

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closed as off-topic by TMM, Thomas, azimut, Cameron Buie, Vedran Šego Aug 26 '13 at 23:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is not about mathematics, within the scope defined in the help center." – TMM, Thomas, azimut, Vedran Šego
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Can you please introduce us briefly into this field of qubits? $\endgroup$ – Berci Feb 27 '13 at 12:38
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Given a pure state like yours, $| v \rangle=x|00\rangle+y|11\rangle$, with $|x|^2+|y|^2=1$, you don't get back another pure state after measuring the second qubit in general. You can trace out (Partial Trace) the first qubit from the density matrix $\rho=|v\rangle\langle v|$ and get $$ \rho_2=\operatorname{Tr}_1 \rho=\pmatrix {\color{red}{|x^2|}& \color{green}0\\0& \color{blue}{|y|^2} }, $$ which is in fact another pure state if $x=1$ or $y=1$. In case of an maximal entangled state, $x=y$, $\rho_2$ is the identity.

The operator $\operatorname{Tr}_1$ sums up the colored matrix elements of $\rho$: $$ \pmatrix{ \color{red}{|x|^2} &\color{green}0&0&x\bar y\\ 0&\color{blue}0&0&0\\ 0&0&\color{red}0&\color{green}0\\ \bar x y &0&0&\color{blue}{|y|^2}\\ } $$

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