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Figure 1.16: FANOUT with the Toffoli gate, with the second bit being the input to the FANOUT (and the other two bits standard ancilla states), and the output from the FANOUT appearing on the second and third bits. enter image description here Source: Quantum Computation and Quantum Information: 10th Anniversary Edition (Fig 1.16, p.30) by Michael A. Nielsen & Isaac L. Chuang

Why does the output from FANOUT appear on the second and third bits? How do we extract and use this output? I learned FANOUT with the following definition and diagram. enter image description here Thus, is the value of $a$ output from our circuit with Toffoli gate the value of $N$ in the above diagram? Isn't $a$ either 0 or 1? Then does it mean the value for fan-out can only be 0 and 1? I didn't think so as it's there are clearly more than 1 gate from fan-out in the above diagram.

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The second diagram is a classical circuit where the output of one gate is being used to drive multiple other gates. In classical digital hardware, if you increase the number if gates being driven too much, the driving gate will become too loaded and won't work as per specification. The maximum number of gates which can be driven in the Fanout of the driving gate in this case.

In the quantum circuit (fig 1), the sense in which the authors use the term Fanout seems to be in analogy with (but a bit different from) the classical case. In this circuit, it is desired to create a circuit which can replicate the state of one qubit ($|a>$) from one quantum wire to 2 quantum wires. Consider the case $a = 0$. Then the control of the CCNOT is inactive and the third wire retains its state $|0>$. In the case $a=1$, the third wire's state gets flipped to $|1>$. Combine both these cases, and you can see that the output state on the third wire is the same as input state on second wire, $a$. (Note that section 1.4.1 involves only classical computations on a quantum computer, so there is no possibility of a being in a superposition. Also, no operation on the second quantum wire and no effects like phase kickbacks (since we are doing classical computation via the QC) means that the second wire retains its state $a$).

$N$ from the second diagram should correspond to the number of wires in the first diagram on which the quantum state is being replicated (2 in this case).

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