# What is the rotation matrix corresponding to a point on the Bloch sphere?

A qubit is given in the following form:

$$\left|\psi\right\rangle = \cos\left(\dfrac{\theta}{2}\right)\left|0\right\rangle + e^{i\phi}\sin\left(\dfrac{\theta}{2}\right)\left|1\right\rangle$$.

Let's us start at $$\left|0\right\rangle$$ and rotate about the $$x$$-axis $$180^{\circ}$$ (we should end up at $$\left|1\right\rangle$$). Mathematically, it could be shown easily:

Let $$\theta = 180^{\circ}$$ and $$\phi = 0^{\circ}$$:

$$\left|\psi\right\rangle = \cos\left(\dfrac{180}{2}\right)\left|0\right\rangle + e^{i(0)}\sin\left(\dfrac{180}{2}\right)\left|1\right\rangle\\ \left|\psi\right\rangle = \cos\left(90\right)\left|0\right\rangle + \sin\left(90\right)\left|1\right\rangle\\ \left|\psi\right\rangle = \left|1\right\rangle$$

Now, let's use the rotation matrix instead. The matrix is given as: $$R_x(\theta) \equiv e^{-i \theta \mathbb{X}/2} = \cos(\theta/2)\mathbb{I} -i\sin(\theta/2)\mathbb{X} = \begin{bmatrix} \cos(\theta/2) & -i\sin(\theta/2) \\ -i\sin(\theta/2) & \cos(\theta/2)\end{bmatrix}$$, where $$\mathbb{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and $$\mathbb{X} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$.

Using $$R_x(\theta)$$, we get

$$R_x(180) = \begin{bmatrix} \cos(180/2) & -i\sin(180/2) \\ -i\sin(180/2) & \cos(180/2)\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\ R_x(180) = \begin{bmatrix}0 & -i\\ -i & 0\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\ R_x(180) = \begin{bmatrix}0\\-i\end{bmatrix}\\ R_x(180) = -i\begin{bmatrix}0\\1\end{bmatrix}.$$

Of course, I feel that I am missing something. The vector obtained is correct but with a phase shift of $$-i$$.

Also, I am wondering why it is okay to let $$\phi = 0$$ (if it is not correct, then what should be the value?).

Lastly, I would like to know why the rotation matrix only have $$\theta$$ but not $$\phi$$.

• You could use a $R_y$ gate instead Oct 3, 2021 at 17:36

Be careful with your choice of notation. You're using $$(\theta,\phi)$$ to describe the input state, and you're using $$\theta$$ as the angle of rotation. These two are different $$\theta$$s.

Now $$\theta=\pi$$ and $$\phi=0$$ simply because you chose your initial state to be $$|0\rangle$$. (Actually, $$\phi$$ could be arbitrary, so you pick it to be 0 for simplicity.)

It perhaps helps to think about a picture of the Bloch sphere. An arbitrary pure state (on the surface of the sphere) requires two parameters to describe it, $$(\theta,\phi)$$. An arbitrary rotation requires three parameters - an axis (which is two parameters, entirely equivalent to the $$(\theta,\phi)$$ of the pure state), and an angle of rotation about that axis. Now, in your example, you have selected a fixed axes, $$X$$, and the $$\theta$$ you're using describes the angle of rotation about that axis. See, it's really incomparable to the other $$\theta$$ you're using.

Finally, you are correct that the $$R_x$$ operation gives you the answer that you want only up to a global phase factor. But global phase factors make no difference, and can be neglected.

Also, the -i you see outside the state vector is part of global phase. Those are not considered phase shifts (only relative phase shifts are). And since $$R_x$$ does not introduce any phase shift, $$\phi$$ is not in its rotation matrix.

PS: As @DaftWullie pointed out, axis needs two parameters.

• I appreciate your feedback! Can you please elaborate further on the difference between the input state and angle of rotation. Also, you mentioned that "An arbitrary rotation requires three parameters - an axis, and an angle of rotation about that axis". What is the third parameter? Aug 14, 2019 at 7:19