I am currently studying a model of a quantum (atomic) clock. And in this paper, I came across the term "Free evolution for a period T":

  1. Free evolution for a period T where a phase difference of Φ accumulates between the oscillator and the qubits $\psi_{2} =\begin{pmatrix}1 & 0 \\0 & e^{-i\phi} \end{pmatrix}\psi_{1}$.

I was able to find references to this term in some other scientific works, but they also do not explain what it means. It would be great to see a simple explanation of this term.

  • 3
    $\begingroup$ It probably just means that you let the system evolve under whatever its natural Hamiltonian is, without attempting to control that evolution at all. $\endgroup$
    – DaftWullie
    May 13, 2021 at 6:31
  • $\begingroup$ The math and animations here are pretty useful. I suspect the $1$ in the top-right element of your matrix should be a $0$. $\endgroup$ May 13, 2021 at 16:08
  • $\begingroup$ @QuantumMechanic Thank you, I corrected this typo. $\endgroup$
    – alexhak
    May 13, 2021 at 19:37

1 Answer 1


The specific work you cite deals with Ramsey interferometry. In that case, there is a two-level atom with some states $|g\rangle$ and $|e\rangle$, which have different energies. Since it never matters where we set the $0$ of energy, people typically say that the ground state $|g\rangle$ has energy $0$ and the excited state has energy $\hbar \omega_a$, where $\hbar$ is the reduced Planck's constant and $\omega_a$ is a frequency ($a$ stands for atomic).

In this case, we can write the atomic Hamiltonian as $$H_a=\hbar\omega_a|e\rangle\langle e|.$$ This Hamiltonian has two energy eigenstates: $|g\rangle$ is an eigenstate with eigenvalue $E_g=0$, and $|e\rangle$ is an eigenstate with eigenvalue $E_e=\hbar\omega_a$, as might be anticipated. Thus, if the initial state of the system is $$|\psi(0)\rangle=c_g(0)|g\rangle+c_e(0)|e\rangle,$$ it evolves under the standard rules of quantum mechanics (the Schrödinger equation) to $$|\psi(t)\rangle=c_g(t)|g\rangle+c_e(t)|e\rangle,$$ where, as usual, $$c_g(t)=e^{-i E_g t/\hbar}c_g(0)=c_g(0)\quad\mathrm{and}\quad c_e(t)=e^{-i E_e t/\hbar}c_e(0)=e^{-i\omega_a t}c_e(0).$$ If this evolution happens for a period $t=T$, the coefficient $c_g(0)$ does not change while the coefficient $c_e(0)$ aquires a phase of $\omega_a T$. If we represent our state in the $\{|g\rangle,|e\rangle\}$-basis, as $$|\psi(t)\rangle= \begin{pmatrix}c_g(t)\\c_e(t)\end{pmatrix}=\begin{pmatrix}c_g(0)\\e^{-i\omega_a t}c_e(0)\end{pmatrix}=\begin{pmatrix}1&0\\0&e^{-i\omega_a t}\end{pmatrix}\begin{pmatrix}c_g(0)\\c_e(0)\end{pmatrix}=\begin{pmatrix}1&0\\0&e^{-i\omega_a t}\end{pmatrix}|\psi(0)\rangle,$$ the matrix you described is equivalent to the evolution of the state under the Hamiltonian $H_a$.

  • 1
    $\begingroup$ Thank you for the detailed explanation! $\endgroup$
    – alexhak
    May 14, 2021 at 9:23

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