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I have a doubt on what kind of transformations can be applied to qubits. I understand that the transformations need to be reversible , but they also have to preserve the norm: that's why the transformations need to be unitary as $U^{\dagger}U=I$.

But then I was reading the paper "Quantum Mechanics Near closed Timelike Lines", by D.Deutsch. In it he has applied a transformation given by

$$|x\rangle |y\rangle \Rightarrow |x\rangle |y \vee f(x)\rangle$$

where $\vee$ stands for XOR and $x,y \in \{0,1\}^{n}$ and $f:\{0,1\}^{n} \rightarrow \{0,1\}^{n}$

Now I can see that that the transformation is reversible (apply twice and get the same thing) but it is not unitary. Am I missing something?

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    $\begingroup$ It is unitary. You are probably confused by the names "0" and "1" for the two states. Try to call them "a" and "b" instead, and think about what "XOR" and "f" actually means in terms of matrix operations. $\endgroup$ – Thomas Klimpel Sep 13 '14 at 21:46
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    $\begingroup$ And here: physics.stackexchange.com/questions/71949/… you have more information about how the XOR is implemented reversibly (note that |x> in front, that's the trick!) $\endgroup$ – Martin Sep 15 '14 at 8:02

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