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I used to think that CNOT flips amplitudes of the target qubit with the chance of $b^2$ where $b$ is the $|1\rangle$ amplitude of the control qubit. This yields that the final(after CNOT) state of the target qubit will be : $$ \sqrt{a_1^2 a_2^2 + b_1^2 b_2^2} \big|0 \big\rangle + \sqrt{a_1^2 b_2^2 + b_1^2 a_2^2} \big|1 \big\rangle $$ (the intuition is that with probability $a^2$ there is no flip, and with probability $b^2$ flip occurs) But in fact, after two sequential applies of such a formula on the target qubit, such a qubit will not end up at the initial state. So either the upper interpretation is wrong or I've blundered the math somewhere. If my interpretation is incorrect, could you please provide a correct one? If it is correct, help me please to derive it in probability terms and to understand how CNOT unitary matrix makes it work.

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For a single CNOT operation, the simplest approach is to think in terms of Boolean logic. Mathematically, this is usually represented by modular addition, which gives the action of CNOT as $$\vert A, B \rangle \rightarrow \vert A, A \oplus B \rangle,$$ where $\vert A, B \rangle$ is the tensor product of $\vert A \rangle$ and $\vert B \rangle$, and $\oplus$ is addition mod 2. If this seems foreign to you, just write out the $2 \times 2$ truth table and you'll see that there's nothing complicated going on here (except possibly some unfamiliar notation). It's also easy to see that $A \oplus A \oplus B = B$, giving the expected outcome after sequential application.

Another approach is to use the tools from linear algebra and represent CNOT as a unitary transformation. Since CNOT is a two-qubit operation, the dimension of the transformation is $2^2=4$, giving $$\vert A, B \rangle \rightarrow U_{CNOT}\vert A,B \rangle, \;\;\; U_{CNOT}=\begin{bmatrix} 1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0 \end{bmatrix}.$$ This is also well worth working this out by hand if there's any confusion as to why it works, noting that $U_{CNOT}{}^2=I$.

One of the problematic assumptions implicit in the OP equation is that $\vert A \rangle$ and $\vert B \rangle$ are separable after CNOT, which is not generally true. In other words, if $\vert A \rangle$ and $\vert B \rangle$ are entangled after CNOT, they can no longer be represented as a linear combination of pure states. At this point density matrices become relevant to avoid dealing with exponentially large vectors after a series of multi-qubit gates.

Edit in response to comment:

Taken to its logical end, your interpretation leads to a classical correlation between measurement outcomes of the two qubit states. Bell and CHSH experiments have shown beyond doubt that entangled qubit states are not classically correlated.

So at least in cases when CNOT creates entanglement, your interpretation is inconsistent with experiment. Unfortunately, after a century of debate on the subject, it's still not clear what the correct interpretation is (or even if such an interpretation exists). Even more unfortunately there are numerous plausible interpretations that are consistent with experiment.

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  • $\begingroup$ If after CNOT qubits |A⟩ and |B⟩ could be entangled then does it mean that (my)probability interpretation is not correct? Cause you certainly can not get it by simply swapping amplitudes of target qubit with arbitrary prob(prob of control qubit to be in 1 state). If so, what is a correct one? $\endgroup$ – quantumtweak Oct 11 '20 at 6:39
  • $\begingroup$ Your interpretation is not consistent with experiment, which is as close as physics gets to labeling an interpretation as incorrect. I edited my post to add clarification, as it's a bit unwieldy to fit into a comment. $\endgroup$ – Jonathan Trousdale Oct 11 '20 at 7:38
  • $\begingroup$ The problem is that even in case when CNOT is not generating an entangled state, my formula still does work only for the first time CNOT is invoked(second time applied, it does not reverse the state to the initial one). You mean that CNOT is just an operation that is doing (abab->abba) transform and there is no sense/intuition behind it, have I understood you correctly? As for boolean representation, I am not quite sure how xor is defined for the amplitude vector if it is not schematic. I am seeking concrete understanding here, so schematic representation can't help here I guess. $\endgroup$ – quantumtweak Oct 11 '20 at 10:48
  • $\begingroup$ Under the interpretation you stated there would be no general expectation that a second application of CNOT would return the circuit to its initial state. Your essentially trying to apply a bit flip with probability $p$ twice, which does not generally return an initial state. $\endgroup$ – Jonathan Trousdale Oct 11 '20 at 12:50
  • $\begingroup$ @quantumtweak Regarding your question, I'm afraid you have misunderstood me. There is both intuition and sense behind what's happening in the CNOT gate. Gaining that intuition begins by accepting and understanding the postulates of quantum mechanics. If you're seeking a concrete understanding, I strongly suggest you work through Nielsen and Chuang carefully. Speaking from personal experience, you will have the understanding you're looking for before you reach the end. $\endgroup$ – Jonathan Trousdale Oct 11 '20 at 13:08

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