Schrödinger equation on the Bloch sphere

Question

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}

Answer 1

Unitary evolution gives a circle, if you evolve on a 2-dimensional sphere. Unitary implies a rotation, in any dimensions, and in 3 dimensions every rotation is about an axis.

Answer 2

I think I get it. The self-adjoint matrix ${\cal H}$ can be written as a linear combination of $I$ and the Pauli matrices: $$ {\cal H} = aI + b_x \sigma_x + b_y \sigma_y + b_z \sigma_z $$ with real coefficients $a$, $b_x$, $b_y$, $b_z$ (these four matrices are orthogonal and therefore form a basis, and it is easy to see that the coefficients are real when ${\cal H}$ is self-adjoint, by getting them with scalar products). Hence $$ U_t = e^{-i t {\cal H}} = e^{-i t a I} e^{-it(b_x \sigma_x + b_y \sigma_y + b_z \sigma_z)}, $$ therefore, up to a phase factor, $U_t$ is equal to $$ U'_t = e^{-it(b_x \sigma_x + b_y \sigma_y + b_z \sigma_z)}. $$ Set $\lambda = \sqrt{b_x^2+b_y^2+b_z^2}$, $n_x=b_x/\lambda$, $n_y=b_y/\lambda$, $n_z=b_z/\lambda$ so that $$ U'_t = e^{-i\lambda t(n_x \sigma_x + n_y \sigma_y + n_z \sigma_z)}. $$ It is well-known in quantum mechanics that $U'_t$ induces a rotation on the Bloch sphere, namely, the rotation of angle $2\lambda t$ around the unit axis $(n_x, n_y, n_z)$ (this is the content of Exercise 4.6 in Nielsen & Chuang's book). Thus the rotation axis as well as the angle are straightforwardly given by the linear combination of Pauli matrices giving ${\cal H}$.