# How does a rotation about the y-axis on a Bloch sphere affect the state of a qubit?

I am unsure how exactly it can be inferred that the following sequence of logic gates and rotation operators acting on a three-qubit state $$\lvert abc\rangle$$

$$R_{y,c}\left(\frac{\pi}{4}\right)CNOT(b,c)R_{y,c}\left(\frac{\pi}{4}\right)CNOT(a,c)R_{y,c}\left(-\frac{\pi}{4}\right)CNOT(b,c) R_{y,c}\left(-\frac{\pi}{4}\right)$$

(Where, for example, $$R_{y,c}(\frac{\pi}{4})$$ represents a $$\frac{\pi}{4}$$ rotation of qubit c about the y-axis and CNOT(b,c) represents the CNOT gate acting on the qubits b and c (b is the control and c is the target).)

Can represent that of a Toffoli gate but with relative phases.

My specific problem with understanding this sequence are the rotation operators. If I neglect the rotation operators and just apply the three CNOT logic gates for some state $$\lvert abc\rangle$$ (e.g. $$\lvert 110\rangle$$), it is apparent that the sequence of the three CNOT gates acts as a single Toffoli gate. But I'm confused by trying to understand what effect the rotation operators have on each c qubit. Do they cancel each other out? Do they impose a net rotation on some state $$\lvert abc\rangle$$?

I've tried to make sense of it all by envisioning rotations on a Bloch sphere but it hasn't helped. If anyone could provide an explanation as to what effect a rotation operator has on a qubit, that would be helpful. Thank you.

The key observation is that commuting $$X$$ through a $$Y$$ rotation changes the sign of the rotation angle

$$XR_y(\theta) = R_y(-\theta)X.$$

In order to understand how the circuit gives rise to a Toffoli-like gate, we consider three cases.

First, suppose that $$a$$ is in the state $$|0\rangle$$. Then $$CNOT(a, c)$$ acts as identity and the sequence simplifies as

$$R_{y,c}\left(\frac{\pi}{4}\right)CNOT(b,c)R_{y,c}\left(\frac{\pi}{4}\right)R_{y,c}\left(-\frac{\pi}{4}\right)CNOT(b,c) R_{y,c}\left(-\frac{\pi}{4}\right) = \\ R_{y,c}\left(\frac{\pi}{4}\right)CNOT(b,c)CNOT(b,c) R_{y,c}\left(-\frac{\pi}{4}\right) = \\ R_{y,c}\left(\frac{\pi}{4}\right)R_{y,c}\left(-\frac{\pi}{4}\right) = \\ I.$$

Next, suppose that qubits $$ab$$ are in the state $$|10\rangle$$. Then $$CNOT(a, c)$$ acts as Pauli $$X$$ on $$c$$ and each $$CNOT(b, c)$$ acts as identity and the sequence simplifies as

$$R_{y,c}\left(\frac{\pi}{4}\right)R_{y,c}\left(\frac{\pi}{4}\right)X_cR_{y,c}\left(-\frac{\pi}{4}\right) R_{y,c}\left(-\frac{\pi}{4}\right) = \\ R_{y,c}\left(\frac{\pi}{4}\right)R_{y,c}\left(\frac{\pi}{4}\right)R_{y,c}\left(\frac{\pi}{4}\right) R_{y,c}\left(\frac{\pi}{4}\right)X_c = \\ R_{y,c}(\pi) X_c = -iY_cX_c = Z_c.$$

Finally, suppose that qubits $$ab$$ are in the state $$|11\rangle$$. Then all three CNOTs act as Pauli $$X$$ on $$c$$ and the sequence simplifies as

$$R_{y,c}\left(\frac{\pi}{4}\right)X_cR_{y,c}\left(\frac{\pi}{4}\right)X_cR_{y,c}\left(-\frac{\pi}{4}\right)X_cR_{y,c}\left(-\frac{\pi}{4}\right) = \\ R_{y,c}\left(\frac{\pi}{4}\right)R_{y,c}\left(-\frac{\pi}{4}\right)X_cX_cR_{y,c}\left(-\frac{\pi}{4}\right)R_{y,c}\left(\frac{\pi}{4}\right)X_c = \\ R_{y,c}\left(\frac{\pi}{4}\right)R_{y,c}\left(-\frac{\pi}{4}\right)R_{y,c}\left(-\frac{\pi}{4}\right)R_{y,c}\left(\frac{\pi}{4}\right)X_c = X_c$$

We have shown that the action of the circuit on the computational basis is

$$\begin{array}{ccc} |000\rangle & \to & |000\rangle \\ & \dots & \\ |011\rangle & \to & |011\rangle \\ |100\rangle & \to & |100\rangle \\ |101\rangle & \to & -|101\rangle \\ |110\rangle & \to & |111\rangle \\ |111\rangle & \to & |110\rangle \end{array}$$

which neglecting the phase change on $$|101\rangle$$ is that of the Toffoli gate. Action of an operator on a basis completely determines the operator.

Note that it is not true that without the $$Y$$ rotations the circuit is equivalent to a Toffoli with phases. In that case, the two $$CNOT(b, c)$$ gates cancel and all we are left with is $$CNOT(a, c)$$.