Dealing in dimension $2$, I have tried to visualize the solution of the Schrödinger equation $$ \frac{{\mathrm{d}}\psi(t)}{{\mathrm{d}}t} = -{\mathrm{i}}{\mathcal{H}}\psi(t) $$ on the Bloch sphere, for an abritrary Hamiltonian $\mathcal{H}$ (a self-adjoint operator, possibly without physical meaning). Recall the solution is $$ \psi(t) = U_t\psi(0) \quad \text{with} \; U_t = {\mathrm{e}}^{-{\mathrm{i}}t{\mathcal{H}}}. $$ I always get a circle, like:

Is it true that the solution always gives a circle? How to prove it? It looks like the relation between $\psi(0)$ and $\psi(t)$ on the sphere is always a rotation with an angle $\alpha(t)$ depending on $t$ in a linear way. How to get the rotation axis and the angle $\alpha(t)$ from $\mathcal{H}$?

### Clarification.

It seems that some of you have knowledge about such unitary evolutions but don't know the representation on the Bloch sphere. Let me explain. The equation is defined for unit vectors $\psi(t) \in \mathbb{C}^2$. Up to a phase factor (a complex number having modulus $1$), a unit vector $\psi \in \mathbb{C}^2$ can be written $$ \psi "=" \begin{pmatrix} \cos \frac{\theta}{2} \\ e^{i\varphi}\sin\frac{\theta}{2}\end{pmatrix}. $$ This equality actually means that the two members define the same ray.

The representation of $\psi$ on the Bloch sphere (the unit sphere) is the vector with spherical coordinates $(\theta, \varphi)$.

This representation enjoys the following property: the stereographic projection $\xi\in \bar{\mathbb{C}}\cup\{\infty\}$ of the representation of $\psi=\begin{pmatrix} z_0 \\ z_1 \end{pmatrix}$ on the Bloch sphere is $\tfrac{z_1}{z_0}$. The point $\xi \in \bar{\mathbb{C}}\cup\{\infty\}$ is the usual representation of the ray defined by $\psi$ in $\bar{\mathbb{C}}\cup\{\infty\}$. Thus: \begin{multline} \textrm{Representation in $\bar{\mathbb{C}}\cup\{\infty\}$ of the ray defined by $\psi$} \\ = Stereographic(\textrm{representation of $\psi$ on the sphere}). \end{multline}