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Let's $ |\psi \rangle $ be $ \sqrt{1} \begin{pmatrix} \frac{2}{8} \\\ \frac{6}{8} \end{pmatrix} $, $ |\phi \rangle $ $ = \sqrt{1} \begin{pmatrix} \frac{3}{4} \\\ \frac{1}{4} \end{pmatrix} $,

and $ |\omega \rangle $ = |$ \psi \rangle $ $ \otimes $ |$ \phi \rangle $ = $ \begin{pmatrix} \sqrt{\frac{2}{8}} \begin{pmatrix} \sqrt{\frac{3}{4}} \\\ \sqrt{\frac{1}{4}} \end{pmatrix} \\\ \sqrt{\frac{6}{8}} \begin{pmatrix} \sqrt{\frac{3}{4}} \\\ \sqrt{\frac{1}{4}} \end{pmatrix} \end{pmatrix} = $ $ \begin{pmatrix} \frac{\sqrt{3}}{4} \\\ \frac{1}{4} \\\ \frac{3}{4} \\\ \frac{\sqrt{3}}{4} \end{pmatrix} $ be our 2-qubit quantum state.

Here is my question:
If a do a CNOT on $ \omega $:
$ |\omega_{NOT} \rangle = CNOT(\omega) = $ $ \begin{pmatrix} \frac{\sqrt{3}}{4} \\\ \frac{1}{4} \\\ \frac{\sqrt{3}}{4} \\\ \frac{3}{4} \end{pmatrix} $

Now, $ \omega $ being the "old" system, and $ \omega_{NOT} $ our new system: $ \delta_{\omega} \neq \delta_{\omega_{NOT}} $ where $ \begin{pmatrix} \alpha \\\ \beta \\\ \delta \\\ \gamma \end{pmatrix} $

The chances of |$ \omega \rangle $ being either |$ 00 \rangle $ or |$ 01 \rangle $ are the same as $ \omega_{NOT} $. But the chances of being |$ 10 \rangle $ or |$ 11 \rangle $ have changed.
But theses last two chances should only change if the control bit ($ \psi $) is $ = $ |$ 1 \rangle $ thus the control has $ \frac{6}{8} $ chances to be |$ 1 \rangle $

I Would like to know what i am misconceiving.
Thanks

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  • $\begingroup$ The tensor product you've quoted is wrong (it's not the tensor product of the vectors you quote above it), and the line starting $\alpha |\frac{\sqrt{3}}{4}\rangle+\cdots$ is plain word salad - it doesn't mean anything (you seem to be confusing coefficients with basis-vector labels). That makes it impossible to even get near the CNOT parts of your question. $\endgroup$ – Emilio Pisanty Mar 2 '18 at 21:29
  • $\begingroup$ @EmilioPisanty Sorry for the "salad" x).. I really don't see where the tensor product is wrong.. :/ I added one more step to make it clearer. $\endgroup$ – Alexandre Daubricourt Mar 3 '18 at 11:29
  • $\begingroup$ The tensor product is now correct - you're just misreporting the two initial states. This is now answerable, though. $\endgroup$ – Emilio Pisanty Mar 3 '18 at 12:31

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