Rabi cycle

A detailed mathematical description of the effect can be found on the page for the Rabi problem. For example, for a two-state atom (an atom in which an electron can either be in the excited or ground state) in an electromagnetic field with frequency tuned to the excitation energy, the probability of finding the atom in the excited state is found from the Bloch equations to be

${\displaystyle |c_{b}(t)|^{2}\propto \sin ^{2}(\omega t/2),}$ 

where ${\displaystyle \omega }$  is the Rabi frequency.

More generally, one can consider a system where the two levels under consideration are not energy eigenstates. Therefore, if the system is initialized in one of these levels, time evolution will make the population of each of the levels oscillate with some characteristic frequency, whose angular frequency^[1] is also known as the Rabi frequency. The state of a two-state quantum system can be represented as vectors of a two-dimensional complex Hilbert space, which means that every state vector ${\displaystyle |\psi \rangle }$  is represented by complex coordinates:

${\displaystyle |\psi \rangle ={\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}}=c_{1}{\begin{pmatrix}1\\0\end{pmatrix}}+c_{2}{\begin{pmatrix}0\\1\end{pmatrix}},}$ 

where ${\displaystyle c_{1}}$  and ${\displaystyle c_{2}}$  are the coordinates.^[2]

If the vectors are normalized, ${\displaystyle c_{1}}$  and ${\displaystyle c_{2}}$  are related by ${\displaystyle |c_{1}|^{2}+|c_{2}|^{2}=1}$ . The basis vectors will be represented as ${\displaystyle |0\rangle ={\begin{pmatrix}1\\0\end{pmatrix}}}$  and ${\displaystyle |1\rangle ={\begin{pmatrix}0\\1\end{pmatrix}}}$ .

All observable physical quantities associated with this systems are 2 × 2 Hermitian matrices, which means that the Hamiltonian of the system is also a similar matrix.

One can construct an oscillation experiment through the following steps:^[3]

1. Prepare the system in a fixed state; for example, ${\displaystyle |1\rangle }$ 
2. Let the state evolve freely, under a Hamiltonian H for time t
3. Find the probability ${\displaystyle P(t)}$ , that the state is in ${\displaystyle |1\rangle }$ 

If ${\displaystyle |1\rangle }$  is an eigenstate of H, ${\displaystyle P(t)=1}$  and there will be no oscillations. Also if the two states ${\displaystyle |0\rangle }$  and ${\displaystyle |1\rangle }$  are degenerate, every state including ${\displaystyle |1\rangle }$  is an eigenstate of H. As a result, there will be no oscillations.

On the other hand, if H has no degenerate eigenstates, and the initial state is not an eigenstate, then there will be oscillations. The most general form of the Hamiltonian of a two-state system is given

${\displaystyle \mathbf {H} ={\begin{pmatrix}a_{0}+a_{3}&a_{1}-ia_{2}\\a_{1}+ia_{2}&a_{0}-a_{3}\end{pmatrix}}}$ 

here, ${\displaystyle a_{0},a_{1},a_{2}}$  and ${\displaystyle a_{3}}$  are real numbers. This matrix can be decomposed as,

${\displaystyle \mathbf {H} =a_{0}\cdot \sigma _{0}+a_{1}\cdot \sigma _{1}+a_{2}\cdot \sigma _{2}+a_{3}\cdot \sigma _{3};}$ 

The matrix ${\displaystyle \sigma _{0}}$  is the 2 ${\displaystyle \times }$  2 identity matrix and the matrices ${\displaystyle \sigma _{k}\;(k=1,2,3)}$  are the Pauli matrices. This decomposition simplifies the analysis of the system especially in the time-independent case where the values of ${\displaystyle a_{0},a_{1},a_{2}}$  and ${\displaystyle a_{3}}$ are constants. Consider the case of a spin-1/2 particle in a magnetic field ${\displaystyle \mathbf {B} =B\mathbf {\hat {z}} }$ . The interaction Hamiltonian for this system is

${\displaystyle \mathbf {H} =-{\boldsymbol {\mu }}\cdot \mathbf {B} =-\gamma \mathbf {S} \cdot \mathbf {B} =-\gamma \ B\ S_{z}}$ , ${\displaystyle S_{z}={\frac {\hbar }{2}}\,\sigma _{3}={\frac {\hbar }{2}}{\begin{pmatrix}1&0\\0&-1\end{pmatrix}},}$ 

where ${\displaystyle \mu }$  is the magnitude of the particle's magnetic moment, ${\displaystyle \gamma }$  is the Gyromagnetic ratio and ${\displaystyle {\boldsymbol {\sigma }}}$  is the vector of Pauli matrices. Here the eigenstates of Hamiltonian are eigenstates of ${\displaystyle \sigma _{3}}$ , that is ${\displaystyle |0\rangle }$  and ${\displaystyle |1\rangle }$ , with corresponding eigenvalues of ${\displaystyle E_{+}={\frac {\hbar }{2}}\gamma B\ ,\ E_{-}=-{\frac {\hbar }{2}}\gamma B}$ . The probability that a system in the state ${\displaystyle |\psi \rangle }$  can be found in the arbitrary state ${\displaystyle |\phi \rangle }$  is given by ${\displaystyle {|\langle \phi |\psi \rangle |}^{2}}$ .

Let the system be prepared in state ${\displaystyle \left|+X\right\rangle }$  at time ${\displaystyle t=0}$ . Note that ${\displaystyle \left|+X\right\rangle }$  is an eigenstate of ${\displaystyle \sigma _{1}}$ :

${\displaystyle |\psi (0)\rangle ={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\1\end{pmatrix}}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\end{pmatrix}}+{\frac {1}{\sqrt {2}}}{\begin{pmatrix}0\\1\end{pmatrix}}.}$ 

Here the Hamiltonian is time independent. Thus by solving the stationary Schrödinger equation, the state after time t is given by $${\displaystyle \left|\psi (t)\right\rangle =\exp \left[{\frac {-i\mathbf {H} t}{\hbar }}\right]\left|\psi (0)\right\rangle ={\begin{pmatrix}\exp \left[{\tfrac {-iE_{+}t}{\hbar }}\right]&0\\0&\exp \left[{\tfrac {-iE_{-}t}{\hbar }}\right]\end{pmatrix}}|\psi (0)\rangle ,}$$  with total energy of the system ${\displaystyle E}$ . So the state after time t is given by:

${\displaystyle |\psi (t)\rangle =e^{\frac {-iE_{+}t}{\hbar }}{\frac {1}{\sqrt {2}}}|0\rangle +e^{\frac {-iE_{-}t}{\hbar }}{\frac {1}{\sqrt {2}}}|1\rangle }$ .

Now suppose the spin is measured in x-direction at time t. The probability of finding spin-up is given by:$${\displaystyle {\left|\langle +X|\psi (t)\rangle \right|}^{2}={\left|{\frac {\left\langle 0\right|+\left\langle 1\right|}{\sqrt {2}}}\left({{\frac {1}{\sqrt {2}}}\exp \left[{\frac {-iE_{+}t}{\hbar }}\right]\left|0\right\rangle +{\frac {1}{\sqrt {2}}}\exp \left[{\frac {-iE_{-}t}{\hbar }}\right]\left|1\right\rangle }\right)\right|}^{2}=\cos ^{2}\left({\frac {\omega t}{2}}\right),}$$ where ${\displaystyle \omega }$  is a characteristic angular frequency given by ${\displaystyle \omega ={\frac {E_{+}-E_{-}}{\hbar }}=\gamma B}$ , where it has been assumed that ${\displaystyle E_{-}\leq E_{+}}$ .^[4] So in this case the probability of finding spin-up in x-direction is oscillatory in time ${\displaystyle t}$  when the system's spin is initially in the ${\displaystyle \left|+X\right\rangle }$  direction. Similarly, if we measure the spin in the ${\displaystyle \left|+Z\right\rangle }$ -direction, the probability of measuring spin as ${\displaystyle {\tfrac {\hbar }{2}}}$  of the system is ${\displaystyle {\tfrac {1}{2}}}$ . In the degenerate case where ${\displaystyle E_{+}=E_{-}}$ , the characteristic frequency is 0 and there is no oscillation.

Notice that if a system is in an eigenstate of a given Hamiltonian, the system remains in that state.

This is true even for time dependent Hamiltonians. Taking for example ${\textstyle {\hat {H}}=-\gamma \ S_{z}B\sin(\omega t)}$ ; if the system's initial spin state is ${\displaystyle \left|+Y\right\rangle }$ , then the probability that a measurement of the spin in the y-direction results in ${\displaystyle +{\tfrac {\hbar }{2}}}$  at time ${\displaystyle t}$  is ${\textstyle {\left|\left\langle \,+Y|\psi (t)\right\rangle \right|}^{2}\,=\cos ^{2}\left({\frac {\gamma B}{2\omega }}\cos \left({\omega t}\right)\right)}$ .^[5]

Example of Rabi oscillation between two states in ionized hydrogen molecule.

An ionized hydrogen molecule is composed of two protons ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$ , and one electron. Because of their large masses, the two protons can be considered to be fixed. Let R be the distance between them and the ${\displaystyle |1\rangle }$  and ${\displaystyle |2\rangle }$  states where the electron is localised around ${\displaystyle P_{1}}$  or ${\displaystyle P_{2}}$ . Assume, at a certain time, the electron is localised about proton ${\displaystyle P_{1}}$ . According to the results from the previous section, we know that the electron will oscillate between the two protons with a frequency equal to the Bohr frequency associated with the two stationary states ${\displaystyle |E_{+}\rangle }$  and ${\displaystyle |E_{-}\rangle }$  of the molecule. This oscillation of the electron between the two states corresponds to an oscillation of the mean value of the electric dipole moment of the molecule. Thus when the molecule is not in a stationary state, an oscillating electric dipole moment can appear. Such an oscillating dipole moment can exchange energy with an electromagnetic wave of same frequency. Consequently, this frequency must appear in the absorption and emission spectrum of the ionized hydrogen molecule.

Consider a Hamiltonian of the form$${\displaystyle {\hat {H}}=E_{0}\cdot \sigma _{0}+W_{1}\cdot \sigma _{1}+W_{2}\cdot \sigma _{2}+\Delta \cdot \sigma _{3}={\begin{pmatrix}E_{0}+\Delta &W_{1}-iW_{2}\\W_{1}+iW_{2}&E_{0}-\Delta \end{pmatrix}}.}$$ The eigenvalues of this matrix are given by$${\displaystyle {\begin{aligned}\lambda _{+}&=E_{+}=E_{0}+{\sqrt {{\Delta }^{2}+{W_{1}}^{2}+{W_{2}}^{2}}}=E_{0}+{\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}\\\lambda _{-}&=E_{-}=E_{0}-{\sqrt {{\Delta }^{2}+{W_{1}}^{2}+{W_{2}}^{2}}}=E_{0}-{\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}},\end{aligned}}}$$ where ${\displaystyle \mathbf {W} =W_{1}+iW_{2}}$  and ${\displaystyle {\left\vert W\right\vert }^{2}={W_{1}}^{2}+{W_{2}}^{2}=WW^{*}}$ , so we can take ${\displaystyle \mathbf {W} ={\left\vert W\right\vert }e^{i\phi }}$ .

Now, eigenvectors for ${\displaystyle E_{+}}$  can be found from equation$${\displaystyle {\begin{pmatrix}E_{0}+\Delta &W_{1}-iW_{2}\\W_{1}+iW_{2}&E_{0}-\Delta \end{pmatrix}}{\begin{pmatrix}a\\b\end{pmatrix}}=E_{+}{\begin{pmatrix}a\\b\end{pmatrix}}.}$$ So$${\displaystyle b=-{\frac {a\left(E_{0}+\Delta -E_{+}\right)}{W_{1}-iW_{2}}}.}$$ Applying the normalization condition on the eigenvectors, ${\displaystyle {\left\vert a\right\vert }^{2}+{\left\vert b\right\vert }^{2}=1}$ . So$${\displaystyle {\left\vert a\right\vert }^{2}+{\left\vert a\right\vert }^{2}\left({\frac {\Delta }{\left\vert W\right\vert }}-{\frac {\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}{\left\vert W\right\vert }}\right)^{2}=1.}$$ Let ${\displaystyle \sin \theta ={\frac {\left\vert W\right\vert }{\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}}}$  and ${\displaystyle \cos \theta ={\frac {\Delta }{\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}}}$ . So ${\displaystyle \tan \theta ={\frac {\left\vert W\right\vert }{\Delta }}}$ .

So we get ${\textstyle {\left\vert a\right\vert }^{2}+{\left\vert a\right\vert }^{2}{\frac {({1-\cos \theta })^{2}}{\sin ^{2}\theta }}=1}$ . That is ${\displaystyle {\left\vert a\right\vert }^{2}=\cos ^{2}\left({\tfrac {\theta }{2}}\right)}$ , using the identity ${\textstyle \tan({\tfrac {\theta }{2}})={\tfrac {1-\cos(\theta )}{\sin(\theta )}}}$ .

The phase of ${\textstyle a}$  relative to ${\textstyle b}$  should be ${\textstyle -\phi }$ .

Choosing ${\textstyle a}$  to be real, the eigenvector for the eigenvalue ${\displaystyle E_{+}}$  is given by$${\displaystyle \left|E_{+}\right\rangle ={\begin{pmatrix}\cos \left({\tfrac {\theta }{2}}\right)\\e^{i\phi }\sin \left({\tfrac {\theta }{2}}\right)\end{pmatrix}}=\cos \left({\tfrac {\theta }{2}}\right)\left|0\right\rangle +e^{i\phi }\sin \left({\tfrac {\theta }{2}}\right)\left|1\right\rangle .}$$ Similarly, the eigenvector for eigenenergy ${\textstyle E_{-}}$  is$${\displaystyle \left|E_{-}\right\rangle =\sin \left({\tfrac {\theta }{2}}\right)\left|0\right\rangle -e^{i\phi }\cos \left({\tfrac {\theta }{2}}\right)\left|1\right\rangle .}$$ From these two equations, we can write$${\displaystyle {\begin{aligned}\left|0\right\rangle &=\cos \left({\tfrac {\theta }{2}}\right)\left|E_{+}\right\rangle +\sin \left({\tfrac {\theta }{2}}\right)\left|E_{-}\right\rangle \\\left|1\right\rangle &=e^{-\imath \phi }\sin \left({\tfrac {\theta }{2}}\right)\left|E_{+}\right\rangle -e^{-\imath \phi }\cos \left({\tfrac {\theta }{2}}\right)\left|E_{-}\right\rangle .\end{aligned}}}$$ Suppose the system starts in state ${\displaystyle |0\rangle }$  at time ${\textstyle t=0}$ ; that is,$${\displaystyle \left|\psi \left(0\right)\right\rangle =\left|0\right\rangle =\cos \left({\tfrac {\theta }{2}}\right)\left|E_{+}\right\rangle +\sin \left({\tfrac {\theta }{2}}\right)\left|E_{-}\right\rangle .}$$ For a time-independent Hamiltonian, after time t, the state evolves as$${\displaystyle \left|\psi \left(t\right)\right\rangle =e^{\frac {-i{\hat {H}}t}{\hbar }}\left|\psi \left(0\right)\right\rangle =\cos \left({\tfrac {\theta }{2}}\right)e^{\frac {-iE_{+}t}{\hbar }}\left|E_{+}\right\rangle +\sin \left({\tfrac {\theta }{2}}\right)e^{\frac {-iE_{-}t}{\hbar }}\left|E_{-}\right\rangle .}$$ If the system is in one of the eigenstates ${\displaystyle |E_{+}\rangle }$  or ${\displaystyle |E_{-}\rangle }$ , it will remain the same state. However, for a time-dependent Hamiltonian and a general initial state as shown above, the time evolution is non trivial. The resulting formula for the Rabi oscillation is valid because the state of the spin may be viewed in a reference frame that rotates along with the field.^[6]

The probability amplitude of finding the system at time t in the state ${\displaystyle |1\rangle }$  is given by ${\textstyle \left\langle \ 1|\psi (t)\right\rangle =e^{i\phi }\sin \left({\tfrac {\theta }{2}}\right)\cos \left({\tfrac {\theta }{2}}\right)\left(e^{\frac {-iE_{+}t}{\hbar }}-e^{\frac {-iE_{-}t}{\hbar }}\right)}$ .

Now the probability that a system in the state ${\displaystyle |\psi (t)\rangle }$  will be found to be in the state ${\textstyle |1\rangle }$  is given by$${\displaystyle {\begin{aligned}P_{0\to 1}(t)&={|\langle \ 1|\psi (t)\rangle |}^{2}\\&=e^{-\imath \phi }\sin \left({\frac {\theta }{2}}\right)\cos \left({\frac {\theta }{2}}\right)\left(e^{\frac {+\imath E_{+}t}{\hbar }}-e^{\frac {+\imath E_{-}t}{\hbar }}\right)e^{+\imath \phi }\sin \left({\frac {\theta }{2}}\right)\cos \left({\frac {\theta }{2}}\right)\left(e^{\frac {-\imath E_{+}t}{\hbar }}-e^{\frac {-\imath E_{-}t}{\hbar }}\right)\\&={\frac {\sin ^{2}{\theta }}{4}}\left(2-2\cos \left({\frac {\left(E_{+}-E_{-}\right)t}{\hbar }}\right)\right)\end{aligned}}}$$ This can be simplified to

$${\displaystyle P_{0\to 1}(t)=\sin ^{2}(\theta )\sin ^{2}\left({\frac {(E_{+}-E_{-})t}{2\hbar }}\right)={\frac {{\left\vert W\right\vert }^{2}}{\Delta ^{2}+{\left\vert W\right\vert }^{2}}}\sin ^{2}\left({\frac {(E_{+}-E_{-})t}{2\hbar }}\right)}$$

(1)

This shows that there is a finite probability of finding the system in state ${\displaystyle |1\rangle }$  when the system is originally in the state ${\displaystyle |0\rangle }$ . The probability is oscillatory with angular frequency ${\displaystyle \omega ={\frac {E_{+}-E_{-}}{2\hbar }}={\frac {\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}{\hbar }}}$ , which is simply unique Bohr frequency of the system and also called Rabi frequency. The formula (1) is known as Rabi formula. Now after time t the probability that the system in state ${\displaystyle |0\rangle }$  is given by ${\displaystyle {|\langle \ 0|\psi (t)\rangle |}^{2}=1-\sin ^{2}(\theta )\sin ^{2}\left({\frac {(E_{+}-E_{-})t}{2\hbar }}\right)}$ , which is also oscillatory.

These types of oscillations of two-level systems are called Rabi oscillations, which arise in many problems such as Neutrino oscillation, the ionized Hydrogen molecule, Quantum computing, Ammonia maser, etc.

The Rabi effect is important in quantum optics, magnetic resonance and quantum computing.

Any two-state quantum system can be used to model a qubit. Consider a spin-${\displaystyle {\tfrac {1}{2}}}$  system with magnetic moment ${\displaystyle {\boldsymbol {\mu }}}$  placed in a classical magnetic field ${\displaystyle {\boldsymbol {B}}=B_{0}\ {\hat {z}}+B_{1}\left(\cos {(\omega t)}\ {\hat {x}}-\sin {(\omega t)}\ {\hat {y}}\right)}$ . Let ${\displaystyle \gamma }$  be the gyromagnetic ratio for the system. The magnetic moment is thus ${\displaystyle {\boldsymbol {\mu }}={\frac {\hbar }{2}}\gamma {\boldsymbol {\sigma }}}$ . The Hamiltonian of this system is then given by ${\displaystyle \mathbf {H} =-{\boldsymbol {\mu }}\cdot \mathbf {B} =-{\frac {\hbar }{2}}\omega _{0}\sigma _{z}-{\frac {\hbar }{2}}\omega _{1}(\sigma _{x}\cos \omega t-\sigma _{y}\sin \omega t)}$  where ${\displaystyle \omega _{0}=\gamma B_{0}}$  and ${\displaystyle \omega _{1}=\gamma B_{1}}$ . One can find the eigenvalues and eigenvectors of this Hamiltonian by the above-mentioned procedure. Now, let the qubit be in state ${\displaystyle |0\rangle }$  at time ${\displaystyle t=0}$ . Then, at time ${\displaystyle t}$ , the probability of it being found in state ${\displaystyle |1\rangle }$  is given by ${\displaystyle P_{0\to 1}(t)=\left({\frac {\omega _{1}}{\Omega }}\right)^{2}\sin ^{2}\left({\frac {\Omega t}{2}}\right)}$  where ${\displaystyle \Omega ={\sqrt {(\omega -\omega _{0})^{2}+\omega _{1}^{2}}}}$ . This phenomenon is called Rabi oscillation. Thus, the qubit oscillates between the ${\displaystyle |0\rangle }$  and ${\displaystyle |1\rangle }$  states. The maximum amplitude for oscillation is achieved at ${\displaystyle \omega =\omega _{0}}$ , which is the condition for resonance. At resonance, the transition probability is given by ${\displaystyle P_{0\to 1}(t)=\sin ^{2}\left({\frac {\omega _{1}t}{2}}\right)}$ . To go from state ${\displaystyle |0\rangle }$  to state ${\displaystyle |1\rangle }$  it is sufficient to adjust the time ${\displaystyle t}$  during which the rotating field acts such that ${\displaystyle {\frac {\omega _{1}t}{2}}={\frac {\pi }{2}}}$  or ${\displaystyle t={\frac {\pi }{\omega _{1}}}}$ . This is called a ${\displaystyle \pi }$  pulse. If a time intermediate between 0 and ${\displaystyle {\frac {\pi }{\omega _{1}}}}$  is chosen, we obtain a superposition of ${\displaystyle |0\rangle }$  and ${\displaystyle |1\rangle }$ . In particular for ${\displaystyle t={\frac {\pi }{2\omega _{1}}}}$ , we have a ${\displaystyle {\frac {\pi }{2}}}$  pulse, which acts as: ${\displaystyle |0\rangle \to {\frac {|0\rangle +i|1\rangle }{\sqrt {2}}}}$ . This operation has crucial importance in quantum computing. The equations are essentially identical in the case of a two level atom in the field of a laser when the generally well satisfied rotating wave approximation is made. Then ${\displaystyle \hbar \omega _{0}}$  is the energy difference between the two atomic levels, ${\displaystyle \omega }$  is the frequency of laser wave and Rabi frequency ${\displaystyle \omega _{1}}$  is proportional to the product of the transition electric dipole moment of atom ${\displaystyle {\vec {d}}}$  and electric field ${\displaystyle {\vec {E}}}$  of the laser wave that is ${\displaystyle \omega _{1}\propto \hbar \ {\vec {d}}\cdot {\vec {E}}}$ . In summary, Rabi oscillations are the basic process used to manipulate qubits. These oscillations are obtained by exposing qubits to periodic electric or magnetic fields during suitably adjusted time intervals.^[7]

• Atomic coherence
• Bloch sphere
• Laser pumping
• Optical pumping
• Rabi problem
• Vacuum Rabi oscillation
• Neutral particle oscillation

• Quantum Mechanics Volume 1 by C. Cohen-Tannoudji, Bernard Diu, Frank Laloe, ISBN 9780471164333
• A Short Introduction to Quantum Information and Quantum Computation by Michel Le Bellac, ISBN 978-0521860567
• The Feynman Lectures on Physics, Volume III
• Modern Approach To Quantum Mechanics by John S Townsend, ISBN 9788130913148