Example of analytic time evolution with a Pauli Hamiltonian

Question

I'm looking for any (non-trivial*) time-independent Hamiltonian expressed in the Pauli basis (with analytically known real coefficients), which unitary time evolves some analytic initial state to some analytic future state. That is, a system for which I can write down the time-evolved state without performing numerics, such as diagonalisation or matrix exponentiation.

*By non-trivial, I mean that the Hamiltonian is not identity, is not all pauli Z (and hence diagonal), and that it consists of multiple terms (since one-term matrices exponentiate trivially). For example, $H = 1 X Y + 2 Z X$ is sufficiently non-trivial. I'm also obviously looking for evolution $t \ne 0$. Furthermore, for the chosen initial state, time evolution shouldn't leave the state unchanged for all time (e.g. like a symmetric Hamiltonian on a uniform state).

This strange request comes from writing a unit test for software which numerically computes time-evolved states of Pauli Hamiltonians. I wish to write a unit test which performs no numerics of its own, and tests whether the software approximates the analytic solution. It's important my "reference" is analytic (and hence precise), since I'll be comparing it to several numerical techniques and must rank their accuracies.

Some ideas:

• If I knew the eigenspectrum of some Pauli Hamiltonian analytically, then I could evolve each eigenvector separately (as a simple oscillation in phase) and superpose them. Alas, I can't think of any system for which the Pauli coefficients and eigenvectors are simultaneously known.

• If I knew (analytically) the period of unitary time evolution of some Hamiltonian (e.g. the product of the eigenvalues, or through other means), then evolving to that period would produce my initial state. But the canonical physical systems with known periods tend to prescribe trivial Pauli Hamiltonians (e.g. the QHO prescribes only Pauli Z).

Any ideas?

Answer 1

Here's a somewhat uninteresting way. I pick some non-trivial Hamiltonian with yet undetermined coefficients,

$$ H = a \; X \, Y \, Z + b \; Y \, Z \, X + c \;Z \, X \, Y$$

and find its eigenvalues analytically, which are (with degeneracy)

$$ \lambda = \pm \sqrt{ a^2 + b^2 + c^2 } . $$

Since I know each eigenvector $|\phi\rangle$ evolves in time as $\exp( - i \lambda t ) | \phi \rangle $, and hence with period $\frac{2\pi}{\lambda}$, I can choose $a, b, c$ such that both periods are integers, and hence their product is a multiple of the system's total period. However in this simple example, since $|\lambda|$ are equal, I know already $\frac{2\pi}{\lambda}$ is the system's period. Hence, I constrain

$$ \lambda \equiv \frac{2 \pi}{n}, \;\; n \in \mathbb{Z} \backslash \{0\} $$ where for simplicity, integer $n$ can just be $\pm 1$. Finally, I can choose any $a, b, c$ such that

$$ 4 \pi^2 = a^2 + b^2 + c^2 $$

for example, $a = \pi \sqrt{2}$, $b = c = \pi$. Then we know

$$ \exp(- i H t) |\psi\rangle \;\big|_{t \,= \, 2\pi/\lambda \, = \, 1} = |\psi\rangle, \;\; \forall |\psi\rangle \in L_2. $$

Answer 2

Taken from the question/answer Deriving time-dependent Hamiltonian:

The state vector $$|\psi(t)\rangle=\begin{pmatrix} \cos(\Omega t) e^{i \omega_1 t} \\ \sin(\Omega t) e^{i \omega_2 t} \end{pmatrix}$$

is a solution for the time-dependent Hamiltonian $$H(t)=\begin{pmatrix} - \hbar \omega_1 & -i\hbar \Omega e^{ i(\omega_1 - \omega_2) t} \\ +i\hbar \Omega e^{-i(\omega_1 - \omega_2) t} & - \hbar \omega_2 \end{pmatrix}$$