Rabi cycle

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:

$|c_b(t)|^2 \propto \sin^2 (\omega t/2)$,

where $\omega$ 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 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 $\vert\psi\rangle$ is represented by two complex coordinates.

$|\psi\rangle = \begin{pmatrix} c_1 \\ c_2\end{pmatrix} = c_1\begin{pmatrix} 1 \\ 0\end{pmatrix} + c_2\begin{pmatrix} 0 \\ 1\end{pmatrix} ;$ where $c_1$ and $c_2$ are the coordinates.

If the vectors are normalized, $c_1$ and $c_2$ are related by ${|c_1|}^2+{|c_2|}^2 = 1$. The basis vectors will be represented as $|1\rangle = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ and $|2\rangle = \begin{pmatrix} 0 \\ 1\end{pmatrix}$

All observable physical quantities associated with this systems are 2 $\times$ 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:

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

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

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

If $|1\rangle$was an eigenstate of H, P(t)=1 and there are no oscillations. Also if two states are degenerate, every state including $|1\rangle$ is an eigenstate of H. As a result, there are no oscillations. So if H has no degenerate eigenstates, neither of which is $|1\rangle$, then there will be oscillations. The most general form of the Hamiltonian of a two-state system is given

$\mathbf{H} = \begin{pmatrix} a_0+a_3 & a_1-ia_2\\ a_1+ia_2 & a_0-a_3\end{pmatrix}$

here, $a_0,a_1, a_2$ and $a_3$ are real numbers. This matrix can be decomposed as,

$\mathbf{H} = a_0\cdot\sigma_0 + a_1\cdot\sigma_1 + a_2\cdot\sigma_2 + a_3\cdot\sigma_3 ;$

The matrix $\sigma_0$ is the 2 $\times$ 2 identity matrix and the matrices $\sigma_k (k = 1,2,3)$ are the Pauli matrices. This decomposition simplifies the analysis of the system especially in the time-independent case where the values of $a_0,a_1,a_2$and $a_3$are constants. Consider the case of a spin-1/2 particle in a magnetic field $\mathbf{B} = B\mathbf{\hat z}$. The interaction Hamiltonian for this system is

$H=-\boldsymbol{\mu}\cdot\mathbf{B}=-\gamma\mathbf{S}\cdot\mathbf{B}=-\gamma\ S_z B$.Where $S_z = \frac{\hbar}{2} \sigma _3 = \frac{\hbar}{2} \begin{bmatrix}1&0\\ 0&-1 \end{bmatrix}$

where $\mu$ is the magnitude of the particle's magnetic moment,$\gamma$ is Gyromagnetic ratio and $\boldsymbol{\sigma}$ is the vector of Pauli matrices. Here eigenstates of Hamiltonian are eigenstates of $\sigma_3$ that is $|1\rangle$ and $|2\rangle$. The probability that a system in the state $|\psi\rang$ will be found to be in the arbitrary state $|\phi\rangle$ is given by ${|\langle\phi|\psi\rangle|}^2$. Let system initially $t=0$ is in state $|+X\rangle$ that is eigen state of $\sigma_1$, $|\psi(0)\rang= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. That is $|\psi(0)\rang= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 0\end{pmatrix}+ \frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\end{pmatrix}$. Here Hamiltonian is time independent. So by solving time independent Schrödinger equation, we get state after time t is given by $|\psi(t)\rang=e^{\frac{-iEt}{\hbar}}|\psi(0)\rang$, where E is the total energy of system. So the state after time t is given by $|\psi(t)\rang=e^{\frac{-iE_+t}{\hbar}}\frac{1}{\sqrt{2}}|1\rangle + e^{\frac{-iE_-t}{\hbar}}\frac{1}{\sqrt{2}}|2\rangle$. Now suppose spin is measured in the x-direction at time t, the probability of finding spin-up is given by ${|\langle\ +X|\psi(t)\rangle|}^2={|\frac{1}{\sqrt{2}} (\langle\ 1|+\langle\ 2|)(e^{\frac{-iE_+t}{\hbar}}\frac{1}{\sqrt{2}}|1\rangle +e^{\frac{-iE_-t}{\hbar}}\frac{1}{\sqrt{2}}|2\rangle)|}^2= {\cos}^2(\frac{\omega t}{2})$ where $\omega$ is a characteristic angular frequency given by$\omega = \frac{E_- - E_+}{\hbar}=\gamma B$ where it has been assumed that $E_- \geq E_+$. So in this case probability of finding spin up state in X direction is oscillatory in time t when system is initially in Z direction. Similarly if we measure spin in Z direction then probability of finding $\frac{\hbar}{2}$ of the system is $\frac{1}{2}$.In the case $E_+ = E_-$, that is when the Hamiltonian is degenerate there is no oscillation. So we can conclude that if the eigenstate of an above given Hamiltonian represents the state of a system, then probability of the system being that state is not oscillatory, but if we find probability of finding the system in other state, it is oscillatory. This is true for even time dependent Hamiltonian. For example $H =-\gamma\ S_z B sin(\omega t)$, the probability that a measurement of system in Y direction at time t results in $\frac{+\hbar}{2}$ is ${|\langle\ +Y|\psi(t)\rangle|}^2={\cos}^2(\frac{\gamma B}{2\omega}\cos\omega t)$, where initial state is in $|+Y\rangle$.

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

Ionized hydrogen molecule is composed of two proton $P_1$ and $P_2$ and one electron. The two protons because of their large masses can be considered to be fixed. Let us call R be the distance between them and $|1\rangle$ and $|2\rangle$ the states where the electron is localised around $P_1$or$P_2$. Assume, at a certain time, the electron is localised about proton $P_1$. According to the results of previous section we know it will oscillate between the two protons with a frequency equal to the Bohr frequency associated with two stationary state $|E_+\rangle$ and $|E_-\rangle$ of 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 dipolemoment can exchange energy with an electromagnetic wave of same frequency. Consequently, this frequency must appear in the absorption and emission spectrum of Ionized hydrogen molecule.

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

Let us consider a Hamiltonian in the form $\mathbf{H} = E_0\cdot\sigma_0 + W_1\cdot\sigma_1 + W_2\cdot\sigma_2 +\Delta\cdot\sigma_3= \begin{pmatrix} E_0+\Delta & W_1-iW_2\\ W_1+iW_2 & E_0-\Delta\end{pmatrix}$.

The eigen values of this matrix are given by $\lambda_+ =E_+=E_0+ \sqrt{{\Delta}^2+{W_1}^2+{W_2}^2}=E_0+\sqrt{{\Delta}^2+ {\left\vert W \right\vert}^2}$ and $\lambda_- =E_-=E_0- \sqrt{{\Delta}^2+{W_1}^2+{W_2}^2}=E_0-\sqrt{{\Delta}^2+ {\left\vert W \right\vert}^2}$.Where $\mathbf{W}=W_1 + \imath W_2$ and ${\left\vert W \right\vert}^2={W_1}^2+{W_2}^2=WW^*$. so we can take $\mathbf{W}={\left\vert W \right\vert}e^{-\imath\phi}$.

Now eigen vector for $E_+$ can be found from equation :$\begin{pmatrix}E_0+\Delta & W_1-iW_2\\ W_1+iW_2 & E_0-\Delta\end{pmatrix} \begin{pmatrix}a\\b\\\end{pmatrix}=E_+\begin{pmatrix}a\\b\\\end{pmatrix}$.

So $b=-\frac{a(E_0+\Delta-E_+)}{W_1-\imath W_2}$.

Using normalisation condition of eigen vector,

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

Let $\sin\theta=\frac{\left\vert W \right\vert}{\sqrt{{\Delta}^2+ {\left\vert W \right\vert}^2}}$ and $\cos\theta=\frac{\Delta}{\sqrt{{\Delta}^2+ {\left\vert W \right\vert}^2}}$. so $\tan\theta=\frac{\left\vert W \right\vert}{\Delta}$.

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

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

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

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

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

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

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

Now a system is in one of the eigen states $|1\rang$ or $|2\rang$, 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 $|2\rang$ is given by $|\langle\ 2|\psi(t)\rangle|=e^{-\imath\phi}\sin(\theta/2)\cos(\theta/2)(e^{\frac{-\imath E_+ t}{\hbar}}-e^{\frac{-\imath E_- t}{\hbar}})$.

Now the probability that a system in the state $|\psi(t)\rang$ will be found to be in the arbitrary state $|2\rang$ is given by $P_1\to_2(t)={|\langle\ 2|\psi(t)\rangle|}^2=e^{+\imath\phi}\sin(\theta/2)\cos(\theta/2)(e^{\frac{+\imath E_+ t}{\hbar}}-e^{\frac{+\imath E_t}{\hbar}})e^{-\imath\phi}\sin(\theta/2)\cos(\theta/2)(e^{\frac{-\imath E_+ t}{\hbar}}-e^{\frac{i\imath E_t}{\hbar}})=\frac{\sin^2\theta}{4}(2-2\cos(\frac{(E_+-E_-)t}{\hbar}))$

By simplifying $P_1\to_2(t)=\sin^2(\theta)\sin^2(\frac{(E_+-E_-)t}{2\hbar})=\frac{{\left\vert W \right\vert}^2}{{\Delta}^2+{\left\vert W \right\vert}^2}\sin^2(\frac{(E_+-E_-)t}{2\hbar})$.........(1).

This shows that there is a finite probability of finding the system in state $|2\rang$ when the system is originally in the state $|1\rang$. The probability is oscillatory with angular frequency $\omega =\frac{E_+-E_-}{2\hbar}=\frac{\sqrt{{\Delta}^2+ {\left\vert W \right\vert}^2}}{\hbar}$, 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 $|1\rang$ is given by${|\langle\ +X|\psi(t)\rangle|}^2=1-\sin^2(\theta)\sin^2(\frac{(E_+-E_-)t}{2\hbar})$, 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 $\frac{1}{2}$ system with magnetic moment $\boldsymbol{\mu}$ placed in a classical magnetic field $\boldsymbol{B}=B_0\hat{z}+B_1(\cos\omega t \hat{x}-\sin\omega t \hat{y})$. Let $\gamma$ be the gyromagnetic ratio for the system. The magnetic moment is thus $\boldsymbol{\mu} = \frac{\hbar}{2} \gamma \boldsymbol{\sigma}$. The Hamiltonian of this system is then given by $\mathbf{H}=-\boldsymbol{\mu}\cdot\mathbf{B}= -\frac{\hbar}{2}\omega_0\sigma_z-\frac{\hbar}{2}\omega_1(\sigma_x\cos\omega t-\sigma_y\sin\omega t)$ where $\omega_0=\gamma B_0$ and $\omega_1=\gamma B_1$. One can find the eigenvalues and eigenvectors of this Hamiltonian by the above-mentioned procedure. Now, let the qubit be in state $|1\rang$ at time $t = 0$. Then, at time $t$, the probability of it being found in state $|2\rang$ is given by $P_{1\to2}(t)=\left(\frac{\omega_1}{\Omega}\right)^2\sin^2\left(\frac{\Omega t}{2}\right)$ where $\Omega=\sqrt{(\omega-\omega_0)^2+\omega_1^2}$. This phenomenon is called Rabi oscillation. Thus, the qubit oscillates between the $|1\rang$ and $|2\rang$ states. The maximum amplitude for oscillation is achieved at $\omega=\omega_0$, which is the condition for resonance. At resonance, the transition probability is given by $P_{1\to2}(t)=\sin^2\left(\frac{\omega_1 t}{2}\right)$. To go from state $|1\rang$ to state $|2\rang$ it is sufficient to adjust the time $t$ during which the rotating field acts such that $\frac{\omega_1 t}{2}=\frac{\pi}{2}$ or $t=\frac{\pi}{\omega_1}$. This is called a $\pi$ pulse. If a time intermediate between 0 and $\frac{\pi}{\omega_1}$ is chosen, we obtain a superposition of $|1\rang$ and $|2\rang$. In particular for $t=\frac{\pi}{2\omega_1}$, we have a $\frac{\pi}{2}$ pulse, which acts as: $|1\rang \to \frac{|1\rang+|2\rang}{\sqrt{2}}$. 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 $\hbar\omega_0$ is the energy difference between the two atomic levels, $\omega$ is the frequency of laser wave and Rabi frequency $\omega_1$ is proportional to the product of the transition electric dipole moment of atom $\vec{d}$and electric field $\vec{E}$ of the laser wave that is $\omega_1\propto \vec{d}\cdot\vec{E}\hbar$. 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.

See also

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

References

• 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 Vol 3 by Richard P. Feynman & R.B. Leighton, ISBN 978-8185015842
• Modern Approach To Quantum Mechanics by John S Townsend, ISBN 9788130913148

External links

• A Java applet that visualizes Rabi Cycles of two-state systems (laser driven)
• extended version of the applet. Includes electron phonon interaction