Direction of rotation for single-qubit unitary operators

Question

The rotation operators for a single qubit are defined as $R_{v}(\theta) = e^{-i \theta X/2}$, with $v \in \{ X,Y,Z\}$. If we look at the direction of rotation of $R_v$ w.r.t. the positive eigenvalue, we will observe that it is counter-clockwise.

However, a rotation $e^{-ix}$ with $x \in \mathbb{R}$ happens in the clock-wise direction.

What is the intuitive reason behind this?

I find this particularly confusing, because e.g. a state $|\psi\rangle$ on the y-z axis with angle $\theta$ will rotate after the application of $R_x(\theta)$ to $|0\rangle$, not to some state $|\psi'\rangle$ with angle $2 \theta$.

Answer 1

This is because, while apparently similar, the "rotations" you are talking about refer to different phenomena.

A "rotation" $e^{-i x}$ refers to the fact that the point corresponding to its real and imaginary parts rotates with $x$ in $\mathbb{R}^2$. The direction of rotation is due to the choice of representing the imaginary unit as a unit vector in the positive vertical direction. If you were to represent complex vectors via the mapping $z\mapsto z_1 \binom10 + z_2 \binom{0}{-1}$, you'd get rotations in the opposite direction.

On the other hand, the "direction of rotation" of $e^{-i\theta X/2}$ is a bit different. These are not complex numbers, so do not admit a representation in the real plane $\mathbb{R}^2$. Rather, for each $\theta$, we have here a unitary matrix. The reason we can still think of these as "rotations" is that pairs of complex numbers defined up to multiplication with complex scalars (i.e. elements of the complex projective line $\mathbb{CP}^1$, i.e. qubit pure states) can be represented as elements in the sphere $S^2\subset\mathbb{R}^3$. This is due to the isomorphisms $\mathbb{CP}^1\simeq S^3/S^1\simeq S^2$. Details aside, the point is that elements of $\mathbb{CP}^1$ (kinda) amount to complex numbers $\mathbb C$ plus the point at infinity, and these can be represented as a sphere.

So, $2\times 2$ unitary matrices act naturally on pairs of complex numbers, as well as on pairs of complex numbers defined up to a phase. When these get represented as points in a sphere, this action can be pictured as a transformation of elements of a sphere, which then turns out to correspond to a rotation. To see what the direction of rotation comes from, let's break down these steps:

1. A state corresponds to a pair of complex numbers modulo multiplication by a scalar. Write it as some $[\alpha:\beta]\equiv(\alpha,\beta)\mathbb{C}\in\mathbb{CP}^1$ (the notation comes from projective geometry). Almost all of these can be represented as $[1:z]$ for some $z\in\mathbb{C}$. The remaining point becomes the point at infinity.

2. Via the stereographic projection, we map $[1:z]$ to $$\left( \frac{2z}{1+|z|^2}, \frac{1-|z|^2}{1+|z|^2}\right)\in S^2,$$ where in saying that this is an element of $S^2\subset\mathbb{R}^3$ I'm splitting the first component in its real and imaginary parts.

   To fix ideas, this mapping sends $$[1:0]\to (0,0,1), \qquad [1:\infty]\equiv [0:1]\to (0,0,-1),\\ [1:\pm 1]\to (\pm 1,0,0), \qquad [1:\pm i]\to (0,\pm1,0).$$ You can picture this mapping as taking the complex plane, and "bending it" downwards in a way that keeps the complex origin $(0,0)$ fixed, while everything else bends below it until it wraps into a sphere. The origin becomes the north pole of the sphere, while the point at infinity is sent to the south pole.

3. Now, a unitary operation $U$ maps $[1:z]$ to the complex ray corresponding to $U\binom{1}{z}$, i.e. $$U.[1:z] = [u_{11} +z u_{12}: u_{21} + z u_{22}] = \left[ 1 : \frac{u_{21} + z u_{22}}{u_{11} + z u_{12}} \right].$$ Let's focus for simplicity to $U_t\equiv e^{-i\theta X/2}$. If you visualise the direction this group of unitaries "pushes" each $z\in\mathbb{C}$, you get the following:

   

   Remember that this visualises the direction each element of the complex plane is sent. To get the corresponding movement in the sphere, you have to imagine this plane "wrapped down" into one.

Now, for the conclusion: the above reasoning leads us to conclude that the rotation induced by $e^{-i\theta X/2}$ is in the counter-clockwise direction with respect to the fixed point on the right. However, what would happen if we were to represent the complex plane as a sphere wrapping it the other way around? By this I mean mapping $(0,0)\in \mathbb{C}$ to the south pole, and the $\infty$ point to the north pole, rather than the other way around. We can absolutely do this, it won't change a thing, going for one option over the other is purely a matter of convention. But if we were to use this alternative wrapping of $\mathbb{C}\cup\{\infty\}$ into $S^2$, we'd get that $e^{-i\theta X/2}$ is represented as a rotation in the opposite direction.

In other words, I'd say the direction of the rotation is just a matter of convention.