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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:

enter image description here

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 $(\theta, \varphi)$.

enter image description here

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}

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  • $\begingroup$ @BenMcKay $\psi(t)$ is a unit vector in $\mathbb{C}^2$. Up to a phase factor, one can represent a unit vector $\psi \in \mathbb{C}^2$ on the Bloch sphere (which is nothing but the unit sphere, or the Riemann sphere). This is done in such a way that the stereographic projection of $\psi$ coincides with the stereographic projection of its representation on the unit sphere. $\endgroup$ – Stéphane Laurent Jul 4 '16 at 10:33
  • $\begingroup$ @BenMcKay, I added some explanations in the "Clarification" paragraph. $\endgroup$ – Stéphane Laurent Jul 4 '16 at 11:26
  • $\begingroup$ Can one explain me why this question deserves a downvote ? $\endgroup$ – Stéphane Laurent Jul 4 '16 at 19:03
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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.

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  • $\begingroup$ Thank you. Do you have a reference for this claim ? How can we get the rotation axis from $\mathcal{H}$ ? Also note that the equation is for unit vectors $\psi(t) \in \mathbb{C}^2$, it is not directly defined on the unit sphere. The sphere is just a representation of the unit vectors in $\mathbb{C}^2$ up to a phase factor (as say in the comment below my OP). The relation $\psi(t) = U(t) \psi(0)$ holds in $\mathbb{C}^2$ but does not hold on the sphere. $\endgroup$ – Stéphane Laurent Jul 4 '16 at 10:38
  • $\begingroup$ (continuation) The map sending $\psi$ to its representation on the sphere is not linear, so I'm not sure it's straightforward that there's a corresponding unitary evolution on the sphere (I could be wrong cause I'm not good in this area). $\endgroup$ – Stéphane Laurent Jul 4 '16 at 10:43
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    $\begingroup$ The relevant map here is the one taking $SU(2) \to SO(3)$, taking a unitary evolution up to rescaling to a rotation of the sphere. This is clear from the description in Wikipedia, which says that the map from states to points of the sphere is the Hopf fibration, which is well known to be equivariant for the map $SU(2) \to SO(3)$. $\endgroup$ – Ben McKay Jul 4 '16 at 11:08
  • $\begingroup$ The wiki about the Bloch sphere ? Ok, I gonna take a look. $\endgroup$ – Stéphane Laurent Jul 4 '16 at 11:27
  • $\begingroup$ But I'm rather ignorant in this area. I never saw the word "equivariant" before. I would be happy if you could provide a minimal set of details to guide me, just to follow the reasoning. $\endgroup$ – Stéphane Laurent Jul 4 '16 at 11:31
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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 ${\cal H}$.

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