Rabi cycle explained

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 optical 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 is one that has two possible energy levels. These two levels are a ground state with lower energy and an excited state with higher energy. If the energy levels are not degenerate (i.e. not having equal energies), the system can absorb a quantum of energy and transition from the ground state to the "excited" state. 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 is the Rabi frequency of the system. The effect can be modeled using the Jaynes–Cummings model and the Bloch vector formalism.

Mathematical description

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

2
|c
b(t)|

\propto\sin2(\omegat/2),

where

\omega

is the Rabi frequency.

|\psi\rangle

is represented by complex coordinates:

|\psi\rangle=\begin{pmatrix}c1\c2\end{pmatrix}=c1\begin{pmatrix}1\ 0\end{pmatrix}+c2\begin{pmatrix}0\ 1\end{pmatrix},

where

c1

and

c2

are the coordinates.[2]

If the vectors are normalized,

c1

and

c2

are related by
2
|c
1|

+

2
|c
2|

=1

. The basis vectors will be represented as

|0\rangle=\begin{pmatrix}1\ 0\end{pmatrix}

and

|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.

Procedure

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

  1. Prepare the system in a fixed state; for example,

|1\rangle

  1. Let the state evolve freely, under a Hamiltonian H for time t
  2. Find the probability

P(t)

, that the state is in

|1\rangle

If

|1\rangle

is an eigenstate of H,

P(t)=1

and there will be no oscillations. Also if the two states

|0\rangle

and

|1\rangle

are degenerate, every state including

|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

H=\begin{pmatrix}a0+a3&a1-ia2\a1+ia2&a0-a3\end{pmatrix}

here,

a0,a1,a2

and

a3

are real numbers. This matrix can be decomposed as,

H=a0 ⋅ \sigma0+a1 ⋅ \sigma1+a2 ⋅ \sigma2+a3 ⋅ \sigma3;

The matrix

\sigma0

is the 2

x

2 identity matrix and the matrices

\sigmak(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

a0,a1,a2

and

a3

are constants. Consider the case of a spin-1/2 particle in a magnetic field

B=B\hatz

. The interaction Hamiltonian for this system is

H=-\boldsymbol{\mu}B=-\gammaSB=-\gammaBSz

,

Sz=

\hbar
2

\sigma3=

\hbar
2

\begin{pmatrix}1&0\ 0&-1\end{pmatrix},

where

\mu

is the magnitude of the particle's magnetic moment,

\gamma

is the Gyromagnetic ratio and

\boldsymbol{\sigma}

is the vector of Pauli matrices. Here the eigenstates of Hamiltonian are eigenstates of

\sigma3

, that is

|0\rangle

and

|1\rangle

, with corresponding eigenvalues of

E+=

\hbar
2

\gammaB,E-=-

\hbar
2

\gammaB

. The probability that a system in the state

|\psi\rang

can be found in the arbitrary state

|\phi\rangle

is given by

{|\langle\phi|\psi\rangle|}2

.

Let the system be prepared in state

\left|+X\right\rangle

at time

t=0

. Note that

\left|+X\right\rangle

is an eigenstate of

\sigma1

:

|\psi(0)\rang=

1
\sqrt{2
}\begin 1 \\ 1 \end= \frac\begin 1 \\ 0\end+ \frac\begin0\\1\end.

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

E

. So the state after time t is given by:
-iE+t
\hbar
|\psi(t)\rang=e
1
\sqrt{2
}|0\rangle + e^\frac|1\rangle .

Now suppose the spin is measured in x-direction at time t. The probability of finding spin-up is given by:^2= ^2= \cos^2\left(\frac \right),where

\omega

is a characteristic angular frequency given by

\omega=

E+-E-
\hbar

=\gammaB

, where it has been assumed that

E-\leqE+

.[4] So in this case the probability of finding spin-up in x-direction is oscillatory in time

t

when the system's spin is initially in the

\left|+X\right\rangle

direction. Similarly, if we measure the spin in the

\left|+Z\right\rangle

-direction, the probability of measuring spin as

\tfrac{\hbar}{2}

of the system is

\tfrac{1}{2}

. In the degenerate case where

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 \hat = -\gamma\ S_z B \sin(\omega t); if the system's initial spin state is

\left|+Y\right\rangle

, then the probability that a measurement of the spin in the y-direction results in

+\tfrac{\hbar}{2}

at time

t

is ^2 \,= \cos^2 \left(\frac \cos \left(\right) \right).[5]

Derivation using nonperturbative procedure by means of the Pauli matrices

Consider a Hamiltonian of the form \hat = E_0\cdot\sigma_0 +W_1\cdot\sigma_1 +W_2\cdot\sigma_2 +\Delta\cdot\sigma_3 =\beginE_0 + \Delta & W_1 - iW_2 \\W_1 + iW_2 & E_0 - \Delta\end.The eigenvalues of this matrix are given by\begin\lambda_+ &= E_+ = E_0 + \sqrt = E_0 + \sqrt \\\lambda_- &= E_- = E_0 - \sqrt = E_0 - \sqrt,\endwhere

W=W1+iW2

and

{\left\vertW\right\vert}2=

2
{W
1}

+

2
{W
2}

=WW*

, so we can take

W={\left\vertW\right\vert}ei

.

Now, eigenvectors for

E+

can be found from equation\begin E_0 + \Delta & W_1 - i W_2 \\ W_1 + i W_2 & E_0 - \Delta \end \begin a \\ b \end = E_+ \begin a \\ b \end.So b = -\frac . Applying the normalization condition on the eigenvectors,

{\left\verta\right\vert}2+{\left\vertb\right\vert}2=1

. So^2 + ^2\left(\frac - \frac\right)^2 = 1 . Let
\sin\theta=\left\vertW\right\vert
\sqrt{{\Delta

2+{\left\vertW\right\vert}2}}

and

\cos\theta=

\Delta
\sqrt{{\Delta

2+{\left\vertW\right\vert}2}}

. So

\tan\theta=

\left\vertW\right\vert
\Delta
.

So we get ^2+^2\frac=1. That is

{\left\verta\right\vert}2=\cos2\left(\tfrac{\theta}{2}\right)

, using the identity \tan(\tfrac) = \tfrac.

The phase of a relative to b should be -\phi.

Choosing a to be real, the eigenvector for the eigenvalue

E+

is given by\left|E_+\right\rang = \begin\cos \left(\tfrac\right) \\e^\sin\left(\tfrac\right)\end= \cos \left(\tfrac\right) \left|0\right\rang + e^ \sin \left(\tfrac\right) \left|1\right\rang.Similarly, the eigenvector for eigenenergy E_- is\left|E_-\right\rang =\sin \left(\tfrac\right) \left|0\right\rang - e^ \cos \left(\tfrac\right) \left|1\right\rang.From these two equations, we can write\begin\left|0\right\rang &=\cos \left(\tfrac\right) \left|E_+\right\rang + \sin \left(\tfrac\right) \left|E_-\right\rang\\\left|1\right\rang &=e^ \sin \left(\tfrac\right) \left|E_+\right\rang - e^ \cos \left(\tfrac\right) \left|E_-\right\rang.\endSuppose the system starts in state

|0\rang

at time t = 0; that is,\left| \psi\left(0 \right) \right\rang =\left|0\right\rang =\cos \left(\tfrac\right) \left|E_+\right\rang + \sin \left(\tfrac\right) \left|E_-\right\rang.For a time-independent Hamiltonian, after time t, the state evolves as\left| \psi\left(t \right) \right\rang =e^ \left| \psi\left(0 \right) \right\rang =\cos \left(\tfrac\right) e^ \left|E_+\right\rang + \sin \left(\tfrac\right) e^ \left|E_-\right\rang.If the system is in one of the eigenstates

|E+\rang

or

|E-\rang

, 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

|1\rang

is given by \left \langle\ 1 | \psi(t) \right\rangle = e^ \sin \left(\tfrac\right) \cos\left(\tfrac\right)\left(e^-e^ \right).

Now the probability that a system in the state

|\psi(t)\rang

will be found to be in the state |1\rang is given by\beginP_(t) &=
\psi(t)\rangle
^2\\ &= e^ \sin\left(\frac\right) \cos\left(\frac\right)\left(e^-e^\right) e^ \sin\left(\frac\right)\cos\left(\frac\right)\left(e^ - e^ \right)\\&= \frac \left(2 - 2\cos\left(\frac \right) \right)\endThis can be simplified to

This shows that there is a finite probability of finding the system in state

|1\rang

when the system is originally in the state

|0\rang

. The probability is oscillatory with angular frequency

\omega=

E+-E-=
2\hbar
\sqrt{{\Delta
2+

{\left\vertW\right\vert}2}}{\hbar}

, which is simply unique Bohr frequency of the system and also called Rabi frequency. The formula is known as Rabi formula. Now after time t the probability that the system in state

|0\rang

is given by

{|\langle 0|\psi(t)\rangle|}2=1-\sin2(\theta)\sin

2\left((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.

In quantum computing

Any two-state quantum system can be used to model a qubit. Consider a spin-

\tfrac{1}{2}

system with magnetic moment

\boldsymbol{\mu}

placed in a classical magnetic field

\boldsymbol{B}= B0 \hat{z}+ B1\left(\cos{(\omegat)}\hat{x}-\sin{(\omegat)}\hat{y}\right)

. Let

\gamma

be the gyromagnetic ratio for the system. The magnetic moment is thus

\boldsymbol{\mu}=

\hbar
2

\gamma\boldsymbol{\sigma}

. The Hamiltonian of this system is then given by

H=-\boldsymbol{\mu}B=-

\hbar
2

\omega0\sigma

z-\hbar
2

\omega1(\sigmax\cos\omegat-\sigmay\sin\omegat)

where

\omega0=\gammaB0

and

\omega1=\gammaB1

. One can find the eigenvalues and eigenvectors of this Hamiltonian by the above-mentioned procedure. Now, let the qubit be in state

|0\rang

at time

t=0

. Then, at time

t

, the probability of it being found in state

|1\rang

is given by

P0\to1(t)=\left(

\omega1
\Omega

\right)2\sin

2\left(\Omegat
2

\right)

where
2}
\Omega=\sqrt{(\omega-\omega
1
. This phenomenon is called Rabi oscillation. Thus, the qubit oscillates between the

|0\rang

and

|1\rang

states. The maximum amplitude for oscillation is achieved at

\omega=\omega0

, which is the condition for resonance. At resonance, the transition probability is given by

P0\to1

2\left(\omega1t
2
(t)=\sin

\right)

. To go from state

|0\rang

to state

|1\rang

it is sufficient to adjust the time

t

during which the rotating field acts such that
\omega1t=
2
\pi
2
or
t=\pi
\omega1
. This is called a

\pi

pulse. If a time intermediate between 0 and
\pi
\omega1
is chosen, we obtain a superposition of

|0\rang

and

|1\rang

. In particular for
t=\pi
2\omega1
, we have a
\pi
2
pulse, which acts as:

|0\rang\to

|0\rang+i|1\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\omega0

is the energy difference between the two atomic levels,

\omega

is the frequency of laser wave and Rabi frequency

\omega1

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

\omega1\propto\hbar\vec{d}\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]

See also

References

Notes and References

  1. http://www.rp-photonics.com/rabi_oscillations.html Rabi oscillations, Rabi frequency, stimulated emission
  2. Book: Griffiths, David . Introduction to Quantum Mechanics . limited . 2nd . 2005 . 341.
  3. Web site: The physics of 2-state systems . Sourendu Gupta . Tata Institute of Fundamental Research . 27 August 2013.
  4. Griffiths, David (2012). Introduction to Quantum Mechanics (2nd ed.) p. 191.
  5. Griffiths, David (2012). Introduction to Quantum Mechanics (2nd ed.) p. 196
  6. Merlin. R.. Rabi oscillations, Floquet states, Fermi's golden rule, and all that: Insights from an exactly solvable two-level model . American Journal of Physics . 2021 . 89. 1 . 26–34. 10.1119/10.0001897 . 2021AmJPh..89...26M . 234321681 . free .
  7. A Short Introduction to Quantum Information and Quantum Computation by Michel Le Bellac,