# Theoretical approach of the quantum CNOT gate

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

• 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. – Emilio Pisanty Mar 2 '18 at 21:29
• @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. – Alexandre Daubricourt Mar 3 '18 at 11:29
• The tensor product is now correct - you're just misreporting the two initial states. This is now answerable, though. – Emilio Pisanty Mar 3 '18 at 12:31