Why the quantum entanglement doesn't break quantum cryptography If the Alice sends to Bob an q-bit $\alpha\left|\mathbf{0}\right> + \beta\left|\mathbf{1}\right>$ in quantum key exchange (such as BB84) it is assumed that Eve cannot copy the state (due to no cloning theorem) and attempts to read the state requires guessing the base which would allow to detect interception.
However if the Eve put $\left|\mathbf{1}\right>$ through C-Not gate than the state would be:
$$\phi_0 = (\alpha\left|\mathbf{0}\right> + \beta\left|\mathbf{1}\right>)\otimes\left|\mathbf{1}\right>$$
$$\phi_1 = \alpha\left|\mathbf{00}\right> + \beta\left|\mathbf{11}\right>$$
Now Eve can store her q-bit somewhere and send the 'original' one to Bob. Since the base disclosure is on public channel she can measure her q-bit in correct base (the state would be collapsed but it shouldn't matter).
Why then quantum entanglement doesn't break quantum cryptography?
Edit: To clarify what I understand (I haven't read the text about density matrices).
Alice sends to Bob two qubits - $\left|\mathbf{0}\right>$ and $\left|\mathbf{+}\right>$. Let's assume they are not discarded then in first case after Eve entangle it's qubit with sent qubit the state in case one:
$$\phi_{\left|\mathbf{0}\right>} = \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)\left(\begin{array}{r} 0 \\ 1 \\ 0 \\ 0\end{array}\right) = \left(\begin{array}{r} 1 \\ 0 \\ 0 \\ 0\end{array}\right) = \left|\mathbf{00}\right>$$
Now the only possible value read by Bob is $\mathbf{0}$ assuming he guessed base. Eve can then overhear the disclosure of bases and measure the reduced state on previously stored entangled qubit.
In second case:
$$\phi_{\left|\mathbf{+}\right>} = \left(\begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}{r} 0 \\ 1 \\ 0 \\ 1\end{array}\right) = \frac{1}{\sqrt{2}}\left(\begin{array}{r} 1 \\ 0 \\ 0 \\ 1\end{array}\right) = \left|\mathbf{++}\right>$$
Assuming that the Bob guessed the base correctly the Eve would also measured $\mathbf{+}$ in similar way as previously. If Bob didn't guessed it correctly Eve can discard particular qubit.
Now if quantum cryptography works:


*

*For some reasons the entanglement is forbidden (if yes - what is the difference between sending qubit and processing it inside quantum computer?)

*Reduction of state by Bob does not happen in a way I think (possibly due to not knowing the density matrix).

 A: Since Bob can't receive information from Eve (which would mean Eve giving their game away), Bob effectively receives the qubits in the reduced state
$$\rho_\text{Bob}=\text{Tr}_\text{Eve}\left[\left(\alpha|\mathbf{00}\rangle+\beta|\mathbf{11}\rangle\right)\left(\alpha^\ast\langle\mathbf{00}|+\beta^\ast\langle\mathbf{11}|\right)\right]
=|\alpha|^2|\mathbf{0}\rangle\langle\mathbf{0}|+|\beta|^2|\mathbf{1}\rangle\langle\mathbf{1}|.$$
Notice that the intervention of Eve has completely killed the off-diagonal elements in this density matrix. For the case of $\alpha=\pm\beta$ this state is the completely mixed state, and Bob can extract no information from it. Thus, even if Alice sends $|+\rangle$, Bob might measure $|-\rangle$, and when they compare results they will find that an unacceptable fractionof their measurements don't match.
A: After reading your clarification, I think the problem is that you cannot have a quantum map that takes
$$ | 0 \rangle \rightarrow |00 \rangle \qquad \mathrm{and} \qquad
 | + \rangle \rightarrow |++ \rangle .$$
Isn't this cloning? 
The state $|++\rangle$ is $\frac{1}{2}(1,1,1,1)$. 
