# Rabi cycle

(Redirected from Rabi flopping)

In physics, the Rabi cycle (or Rabi flop) is the cyclic behaviour of a two-level quantum system in the presence of an oscillatory driving field. A great variety of physical processes belonging to the areas of quantum computing, condensed matter, atomic and molecular physics, and nuclear and particle physics can be conveniently studied in terms of two-level quantum mechanical systems, and exhibit Rabi flopping when coupled to an oscillatory driving field. The effect is important in quantum optics, magnetic resonance and quantum computing, and is named after Isidor Isaac Rabi.

A two-level system has two possible levels, and if they are not degenerate (i.e. equal energy), the system can become "excited" when it absorbs a quantum of energy. When an atom (or some other two-level system) is illuminated by a coherent beam of photons, it will cyclically absorb photons and re-emit them by stimulated emission. One such cycle is called a Rabi cycle and the inverse of its duration the Rabi frequency of the photon beam. The effect can be modeled using the Jaynes-Cummings model and the Bloch vector formalism.

## Mathematical treatment

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 every state vector ${\displaystyle \vert \psi \rangle }$ is represented by two 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 ${\displaystyle |1\rangle ={\begin{pmatrix}1\\0\end{pmatrix}}}$ and ${\displaystyle |2\rangle ={\begin{pmatrix}0\\1\end{pmatrix}}}$

All observable physical quantities associated with this systems are 2 ${\displaystyle \times }$ 2 Hermitian matrices, which means the Hamiltonian of the system is also a similar matrix.

## How to prepare an oscillation experiment in a quantum system

One can construct an oscillation experiment consisting of following steps:[3]

(1) Prepare the system in a fixed state say ${\displaystyle |1\rangle }$

(2) Let the state evolve freely, under a Hamiltonian H for time t

(3) Find the probability P(t), that the state is in${\displaystyle |1\rangle }$

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

On the other hand, if H has no degenerate eigenstates, neither of which is ${\displaystyle |1\rangle }$, 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

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

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 ${\displaystyle |1\rangle }$ and ${\displaystyle |2\rangle }$. 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 initially (at ${\displaystyle t=0}$) be in ${\displaystyle |+X\rangle }$ which 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 ${\displaystyle |\psi (t)\rangle =e^{\frac {-iEt}{\hbar }}|\psi (0)\rangle }$, with total energy of the system ${\displaystyle E}$. So the state after time t is given by:

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

Now suppose the spin is measured in x-direction at time t. The probability of finding spin-up is given by:

${\displaystyle {\left|\langle +X|\psi (t)\rangle \right|}^{2}={\left|{\frac {1}{\sqrt {2}}}(\langle \ 1|+\langle \ 2|)(e^{\frac {-iE_{+}t}{\hbar }}{\frac {1}{\sqrt {2}}}|1\rangle +e^{\frac {-iE_{-}t}{\hbar }}{\frac {1}{\sqrt {2}}}|2\rangle )\right|}^{2}={\cos }^{2}({\frac {\omega t}{2}})}$,

where ${\displaystyle \omega }$ is a characteristic angular frequency given by ${\displaystyle \omega ={\frac {E_{-}-E_{+}}{\hbar }}=\gamma B}$, where it has been assumed that ${\displaystyle E_{-}\geq E_{+}}$.[4] So in this case the probability of finding spin-up in x-direction is oscillatory in time ${\displaystyle t}$ when the system is initially in ${\displaystyle |+X\rangle }$ direction. Similarly, if we measure the spin in z-direction, the probability of finding ${\displaystyle {\frac {\hbar }{2}}}$ of the system is ${\displaystyle {\frac {1}{2}}}$. In case of ${\displaystyle E_{+}=E_{-}}$, that is when the Hamiltonian is degenerate, 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 ${\displaystyle H=-\gamma \ S_{z}B\sin(\omega t)}$, the probability that a measurement of the system in y-direction at time ${\displaystyle t}$ results in ${\displaystyle {\frac {+\hbar }{2}}}$ is ${\displaystyle {|\langle \ +Y|\psi (t)\rangle |}^{2}={\cos }^{2}({\frac {\gamma B}{2\omega }}\cos \omega t)}$, where the initial state is in ${\displaystyle |+Y\rangle }$.[5]

## Derivation of Rabi formula in a nonperturbative procedure by means of the Pauli matrices

Let us consider a Hamiltonian in the form ${\displaystyle \mathbf {H} =E_{0}\cdot \sigma _{0}+W_{1}\cdot \sigma _{1}+W_{2}\cdot \sigma _{2}+\Delta \cdot \sigma _{3}={\begin{pmatrix}E_{0}+\Delta &W_{1}-iW_{2}\\W_{1}+iW_{2}&E_{0}-\Delta \end{pmatrix}}}$.

The eigen values of this matrix are given by ${\displaystyle \lambda _{+}=E_{+}=E_{0}+{\sqrt {{\Delta }^{2}+{W_{1}}^{2}+{W_{2}}^{2}}}=E_{0}+{\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}}$ and ${\displaystyle \lambda _{-}=E_{-}=E_{0}-{\sqrt {{\Delta }^{2}+{W_{1}}^{2}+{W_{2}}^{2}}}=E_{0}-{\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}}$.Where ${\displaystyle \mathbf {W} =W_{1}+\imath W_{2}}$ and ${\displaystyle {\left\vert W\right\vert }^{2}={W_{1}}^{2}+{W_{2}}^{2}=WW^{*}}$. so we can take ${\displaystyle \mathbf {W} ={\left\vert W\right\vert }e^{-\imath \phi }}$.

Now eigen vector 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 ${\displaystyle b=-{\frac {a(E_{0}+\Delta -E_{+})}{W_{1}-\imath W_{2}}}}$.

Using normalisation condition of eigen vector,

${\displaystyle {\left\vert a\right\vert }^{2}+{\left\vert b\right\vert }^{2}=1}$.So ${\displaystyle {\left\vert a\right\vert }^{2}+{\left\vert a\right\vert }^{2}({\frac {\Delta }{\left\vert W\right\vert }}-{\frac {\sqrt {{\Delta }^{2}+{\left\vert W\right\vert }^{2}}}{\left\vert W\right\vert }})^{2}=1}$.

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 ${\displaystyle {\left\vert a\right\vert }^{2}+{\left\vert a\right\vert }^{2}{\frac {({1-\cos \theta })^{2}}{\sin ^{2}\theta }}=1}$. That is ${\displaystyle {\left\vert a\right\vert }^{2}=\cos ^{2}\theta /2}$. Taking arbitrary phase angle ${\displaystyle \phi }$,we can write ${\displaystyle a=\exp(\imath \phi /2)\cos \theta /2}$. Similarly ${\displaystyle b=\exp(-\imath \phi /2)\sin \theta /2}$.

So eigen vector for${\displaystyle E_{+}}$ eigen value is given by ${\displaystyle |E_{+}\rangle =e^{\imath \phi /2}{\begin{pmatrix}\cos \theta /2\\e^{-\imath \phi }\sin \theta /2\end{pmatrix}}}$.

As overall phase is immaterial so we can write ${\displaystyle |E_{+}\rangle ={\begin{pmatrix}\cos \theta /2\\e^{-\imath \phi }\sin \theta /2\end{pmatrix}}=\cos(\theta /2)|1\rangle +e^{-\imath \phi }\sin(\theta /2)|2\rangle }$.

Similarly we can find eigen vector for ${\displaystyle E_{-}}$ value and we get ${\displaystyle |E_{-}\rangle =-e^{\imath \phi }\sin(\theta /2)|1\rangle +\cos(\theta /2)|2\rangle }$.

From these two equations, we can write ${\displaystyle |1\rangle =\cos(\theta /2)|E_{+}\rangle -\sin(\theta /2)e^{-\imath \phi }|E_{-}\rangle }$ and ${\displaystyle |2\rangle =e^{\imath \phi }\sin(\theta /2)|E_{+}\rangle +\cos(\theta /2)|E_{-}\rangle }$.

Let at time t=0, system be in ${\displaystyle |1\rangle }$ That Is ${\displaystyle |\psi (0)\rangle =|1\rangle =\cos(\theta /2)|E_{+}\rangle -\sin(\theta /2)e^{-\imath \phi }|E_{-}\rangle }$.

State of system after time t is given by ${\displaystyle |\psi (t)\rangle =e^{\frac {-iEt}{\hbar }}|\psi (0)\rangle =\cos(\theta /2)e^{\frac {-\imath E_{+}t}{\hbar }}|E_{+}\rangle -\sin(\theta /2)e^{-\imath \phi }e^{\frac {-\imath E_{-}t}{\hbar }}|E_{-}\rangle }$.

Now a system is in one of the eigen states ${\displaystyle |1\rangle }$ or ${\displaystyle |2\rangle }$, it will remain the same state, however in a general state as shown above the time evolution is non trivial.

The probability amplitude of finding the system at time t in the state ${\displaystyle |2\rangle }$ is given by ${\displaystyle |\langle \ 2|\psi (t)\rangle |=e^{-\imath \phi }\sin(\theta /2)\cos(\theta /2)(e^{\frac {-\imath E_{+}t}{\hbar }}-e^{\frac {-\imath E_{-}t}{\hbar }})}$.

Now the probability that a system in the state ${\displaystyle |\psi (t)\rangle }$ will be found to be in the arbitrary state ${\displaystyle |2\rangle }$ is given by ${\displaystyle P_{1}\to _{2}(t)={|\langle \ 2|\psi (t)\rangle |}^{2}=e^{+\imath \phi }\sin(\theta /2)\cos(\theta /2)(e^{\frac {+\imath E_{+}t}{\hbar }}-e^{\frac {+\imath E_{t}}{\hbar }})e^{-\imath \phi }\sin(\theta /2)\cos(\theta /2)(e^{\frac {-\imath E_{+}t}{\hbar }}-e^{\frac {i\imath E_{t}}{\hbar }})={\frac {\sin ^{2}\theta }{4}}(2-2\cos({\frac {(E_{+}-E_{-})t}{\hbar }}))}$

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

This shows that there is a finite probability of finding the system in state ${\displaystyle |2\rangle }$ when the system is originally in the state ${\displaystyle |1\rangle }$. 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 ${\displaystyle |1\rangle }$ is given by${\displaystyle {|\langle \ +X|\psi (t)\rangle |}^{2}=1-\sin ^{2}(\theta )\sin ^{2}({\frac {(E_{+}-E_{-})t}{2\hbar }})}$, which is also oscillatory. This type of oscillations between two levels are called Rabi oscillation and seen in many problems such as Neutrino oscillation, ionized Hydrogen molecule, Quantum computing, Ammonia maser etc.

## Rabi oscillation in quantum computing

Any two state quantum system can be used to model a qubit. Consider a spin ${\displaystyle {\frac {1}{2}}}$ system with magnetic moment ${\displaystyle {\boldsymbol {\mu }}}$ placed in a classical magnetic field ${\displaystyle {\boldsymbol {B}}=B_{0}{\hat {z}}+B_{1}(\cos \omega t{\hat {x}}-\sin \omega t{\hat {y}})}$. 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 ${\displaystyle |1\rangle }$ at time ${\displaystyle t=0}$. Then, at time ${\displaystyle t}$, the probability of it being found in state ${\displaystyle |2\rangle }$ is given by ${\displaystyle P_{1\to 2}(t)=\left({\frac {\omega _{1}}{\Omega }}\right)^{2}\sin ^{2}\left({\frac {\Omega t}{2}}\right)}$ where ${\displaystyle \Omega ={\sqrt {(\omega -\omega _{0})^{2}+\omega _{1}^{2}}}}$. This phenomenon is called Rabi oscillation. Thus, the qubit oscillates between the ${\displaystyle |1\rangle }$ and ${\displaystyle |2\rangle }$ 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 ${\displaystyle P_{1\to 2}(t)=\sin ^{2}\left({\frac {\omega _{1}t}{2}}\right)}$. To go from state ${\displaystyle |1\rangle }$ to state ${\displaystyle |2\rangle }$ 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 ${\displaystyle |1\rangle }$ and ${\displaystyle |2\rangle }$. In particular for ${\displaystyle t={\frac {\pi }{2\omega _{1}}}}$, we have a ${\displaystyle {\frac {\pi }{2}}}$ pulse, which acts as: ${\displaystyle |1\rangle \to {\frac {|1\rangle +i|2\rangle }{\sqrt {2}}}}$. 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 ${\displaystyle \omega _{1}\propto {\vec {d}}\cdot {\vec {E}}\hbar }$. 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.[6]