# What's free evolution for a period T?

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.

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

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 $$H_a$$.

• Thank you for the detailed explanation! May 14, 2021 at 9:23