# Rotation on the surface of the Bloch sphere?

I have two questions.

1. Is rotation of a state by an angle $\theta$ about the $\hat{n}$ axis of the Bloch sphere the same as the rotation of the sphere about $\hat{n}$ axis by an angle $\theta$? I think so. So when we say "T is a rotation by pi/4 radians around the $\hat{z}$ axis on the bloch sphere". Do we mean rotation of state or rotation of sphere? I think they are the same.

2. Hadamard operation is just a rotation of the sphere about $\hat{y}$ axis by 90 degree, followed by rotation of the sphere about $\hat{x}$ axis by 180 degree. Why do we stress on that (rotation of sphere about $\hat{y}$)? As you know, $H = (X+Z)/\sqrt{2}$. Can I derive the relationship between $R_{H}, R_{X}, R_{Z}$ directly from this and how can I see the $R_{Y}$? I think rotation about y is a combination of x and z. Something to do with that. What is the representation of Hadamard operation as a rotation of a state by some angles about $\hat{x}$ axis and some angle aboyt $\hat{z}$ axis?

• So i think rotation of sphere is the same as rotation of state. But when do we prefer to say rotation of sphere instead? Oct 10 '15 at 5:58
• Hmm. I think we say rotation of sphere when we want to rotate state inside or on the sphere but not only rotating the surface state only, right? Oct 10 '15 at 6:01

1. "T is a rotation by $\pi/4$ radians around the ẑ axis on the bloch sphere" is usually taken to mean "rotation of state", meaning that T maps any state $|a\rangle$ to a state $|b\rangle$ that is rotated by $\pi/4$ around ẑ. This is the active view of a symmetry transformation: actively acting on the object of the transform. "Rotation of the sphere" would mean instead rotating the axes (the reference frame), while the vector representing the state "stays put", which is the passive view of the rotation: the map is obtained not by acting on the object, but by changing the point of view. The two are indeed equivalent, but for qubits, in general, it is the active "rotation of state" view that is considered.
2. In terms of axes $\hat{x}$ and $\hat{z}$, the Hadamard gate $H = \frac{1}{\sqrt{2}}(X + Z)$ can be viewed as a rotation of $\pi$ about axis $(\hat{x}+\hat{z})/\sqrt{2}$ (see "Quantum Hadamard Gate on Wikipedia"), because up to a phase factor it can be written as $$H = e^{i\frac{\pi}{2}} e^{-i\frac{\pi}{2}\left[\frac{1}{\sqrt{2}}(X + Z)\right]} = e^{i\frac{\pi}{2}} \left[ \cos\left(\frac{\pi}{2}\right) - i \sin\left(\frac{\pi}{2}\right)\frac{1}{\sqrt{2}}(X + Z)\right] = \frac{1}{\sqrt{2}}(X + Z)$$
Notice the use of $\frac{\pi}{2}$ as angle, not of $\pi$, although the rotation is referred to as "of angle $\pi$". The reason for this has to do with the fact that a "rotation of angle $\alpha$ around axis ${\bf n} = (n_x, n_y, n_z)$" is represented by a transformation $$R_{{\bf n}}(\alpha) = e^{-i\frac{\alpha}{2}(n_x X + n_y Y + n_z Z)}$$ In turn, this is defined so that if a ${\mathbb R}^3$ vector ${\bf a} = (a_x, a_y, a_z)$ transforms under a rotation of angle $\alpha$ around ${\bf n}$ into a vector ${\bf a}' = (a'_x, a'_y, a'_z)$, then under the same rotation the operator $a_x X + a_y Y + a_z Z$ transforms into $a'_x X + a'_y Y + a'_z Z$. This transformation is given by $$a'_x X + a'_y Y + a'_z Z = e^{i\frac{\alpha}{2}(n_x X + n_y Y + n_z Z)}(\alpha)\;(a_x X + a_y Y + a_z Z)\;e^{-i\frac{\alpha}{2}(n_x X + n_y Y + n_z Z)}$$
With this definition in mind, let us check the statement that the "Hadamard operation is just a rotation of the sphere about ŷ axis by 90 degree, followed by rotation of the sphere about x̂ axis by 180 degree". A rotation of $\frac{\pi}{2}$ around $\hat{y}$ is $$R_{y}\left(\frac{\pi}{2}\right) = e^{-i\frac{\pi}{4}Y} = \cos\left(\frac{\pi}{4}\right) - i \sin\left(\frac{\pi}{4}\right)Y = \frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 & -1 \\ 1 & 1\end{array} \right)$$ A rotation of $\pi$ around $\hat{x}$ is $$R_{x}\left(\pi\right) = e^{-i\frac{\pi}{2}X} = \cos\left(\frac{\pi}{2}\right) - i \sin\left(\frac{\pi}{2}\right)X = \left( \begin{array}{cc} 0 & -i \\ -i & 0\end{array} \right)$$ Then a rotation of $\frac{\pi}{2}$ around $\hat{y}$ followed by a rotation of $\pi$ around $\hat{x}$ produces $$R_{x}\left(\pi\right)R_{y}\left(\frac{\pi}{2}\right) = \left( \begin{array}{cc} 0 & -i \\ -i & 0\end{array} \right) \frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 & -1 \\ 1 & 1\end{array} \right) = -\frac{i}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \\ 1 & -1\end{array} \right)$$
This is indeed the Hadamard gate, up to the same phase factor of $e^{i\frac{\pi}{2}}$: $$H = e^{i\frac{\pi}{2}} R_{x}\left(\pi\right)R_{y}\left(\frac{\pi}{2}\right) = e^{i\frac{\pi}{2}} e^{-i\frac{\pi}{2}\left[\frac{1}{\sqrt{2}}(X + Z)\right]}$$
Since we obviously must have $$R_{x}\left(\pi\right)R_{y}\left(\frac{\pi}{2}\right) = R_{(\hat{x}+\hat{z})/\sqrt{2}}\left(\pi\right)$$ you are right to say that a $\hat{y}$ rotation must be some combination of $\hat{x}$ and $\hat{z}$ rotations. In this case the above just shows that $$R_{y}\left(\frac{\pi}{2}\right) = R_{x}\left(-\pi\right) R_{(\hat{x}+\hat{z})/\sqrt{2}}\left(\pi\right)$$