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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.