Resonances

\newcounter{question}[chapter] \renewcommand{\thequestion}{Q\thechapter-\arabic{question}}

Resonances and Two-Level Atoms

Introduction to Resonances and Precision Measurements

Classically, a resonance is a system where one or more variables change periodically such that when the system is no longer driven by an extended periodic process, the decay of the system's oscillation takes many (or at least several) oscillation periods. Equivalently, the response of the driven system as a function of drive frequency exhibits some form of peaked structure. The frequency corresponding to the maximum response of the system is called the resonance frequency ($f_{0}$) \begin{figure}

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\caption{A resonance in the time (left) and frequency (right) domains. The time domain picture shows the (decaying) oscillation in some variable of the undriven system, while the frequency-domain pictures shows the steady-state response amplitude of the driven system. The time units are chosen such that $f_{0}=\omega _{0}/2\pi =1$ while the quality factor is $Q=10$.}

\end{figure} The simplest resonance systems have a single resonance frequency, corresponding to the harmonic motion of the variable. Some form of dissipation (damping) results in a finite damping time ($\Delta t<\infty$) for the undriven system, and a finite width ($\Delta f>0$) of the resonance curve for the driven system. The quantities $\Delta {f}$ and $\Delta {t}$ are related by a Fourier transformation: any oscillation with time-varying amplitude must be made up of a superposition of different frequency components, giving the resonance curve a finite width in frequency space. Low dissipation results in a long decay time $\Delta t$, and a narrow frequency width $f_{0}$. The ratio of frequency width $\Delta f$ to resonance frequency $f_{0}$ is called the quality factor $Q$.

$Q={\frac {f_{0}}{\Delta f}}$

Atomic physics often deals with isolated atomic systems in a vacuum, providing resonances of high quality factor $Q$. For example, an optical transition, even in a room-temperature (ie Doppler-broadened) gas, will have a $Q$-factor of

Failed to parse (unknown function "\unit"): {\displaystyle \frac{f_0}{\Delta f} \sim \frac{\unit{10^{15}}{\hertz}}{\unit{10^9}{\hertz}} \sim 10^6 }

In contrast, high $Q$ factors in solids typically require cryogenic temperatures: typical $Q$ factors of mechanical or electrical systems are $\sim 10^{3}$ at room temperature and $\sim 10^{6}$ at Failed to parse (unknown function "\unit"): {\displaystyle \lesssim \unit{1}{\kelvin}} temperatures. Solid optical resonators are an exception to this: $Q\gtrsim 10^{8}$ has been achieved in the whispering gallery modes of spheres or cylinders of high-purity glass \cite{Armani2003}. \begin{figure}

File:Resonances-whispering-gallery

\caption{One of the whispering-gallery resonators described in \cite{Armani2003}. This particular resonator had a quality factor $Q=10^{8}$. The very high surface quality required for such low losses is achieved by having surface tension shape the rim of the disk after it has been temporarily liquefied by a laser pulse; the inset shows the resonator before this treatment.}

\end{figure} At the macroscopic scale, astronomical systems consisting of massive bodies travelling essentially undisturbed through the near-vacuum of space can also exhibit high $Q$ factors. A resonance is particularly useful for science and technology if it is reproducible by different realizations of the same system, and if it has a well-understood theoretical connection to fundamental constants. In atomic physics, quantum mechanics ensures both features: atoms of the same species are fundamentally identical (although the probing apparatus is not), and the quantum mechanical description of atomic structure is well understood and highly accurate, at least for atoms with not too many electrons. The resonance frequencies of hydrogen, and to a lesser degree helium, are directly tied via quantum mechanics to fundamental constants such as the Rydberg constant, Failed to parse (unknown function "\unit"): {\displaystyle R_\infty=\unit{1.0973731568525(73)\times10^7}{\reciprocal\metre}} \cite{codata2002}, or the fine structure constant $\alpha =1/137.03599911(46)$ \cite{codata2002}. In fact, the Rydberg constant is the most accurately known constant in all of physics, and most fundamental constants have been determined by atomic physics or optical techniques. Additional motivation for measuring fundamental constants with ever-increasing accuracy has been provided by the speculation (and claims of evidence for it \cite{Flambaum2004,Reinhold2006}), that the fundamental constants may not be so constant after all, but that their value is tied to the evolution (and therefore the age) of the universe. Furthermore unexpected resonances, or splittings of resonances, have in the past indicated the need for new theories or modifications of theories. For instance, "anomalous" effects in the Zeeman structure (magnetic-field dependance) of atomic lines led to the discovery of spin \cite{Uhlenbeck1926}. while Lamb's measurement of the splitting between the $^{2}S_{1/2}$ and $^{2}P_{1/2}$ states of atomic hydrogen, on the order of \unit{1000}{\mega\hertz} instead of 0 as predicted by the Dirac theory, fostered the development of quantum electrodynamics, the quantum theory of light.

Classical Resonances

A classical harmonic oscillator (HO) is a system described by the second order differential equation

${\ddot {q}}+\gamma {\dot {q}}+\omega _{0}^{2}q=0$

such as a mass on a spring, or a charge, voltage, or currant in an electric RLC resonant circuit. For weak damping ($\gamma ^{2}<4\omega _{0}^{2}$) the undriven system decays with an exponential envelope,

$e^{-{\frac {\gamma }{2}}t}e^{\pm {i\omega ^{\prime }t}}$

with $\omega ^{\prime }=\omega _{0}{\sqrt {1-\gamma ^{2}/4\omega _{0}^{2}}}$. In the cases we will consider, $\gamma ^{2}\ll {4\omega _{0}^{2}}$ so that $\omega ^{\prime }\approx \omega _{0}$. The decay rate constant for the amplitude is $\gamma /2$, while for the stored energy it is $\gamma$, i.e. the stored energy decays exponentially with a time constant $\tau =1/\gamma$. If the harmonic oscillator is driven at a frequency $\omega$ close to resonance, Failed to parse (unknown function "\lvert"): {\displaystyle \lvert\omega_0-\omega\rvert \ll\omega_0} , the amplitude response $q_{0}$ is a lorentzian:

$q_{0}={\frac {\text{const.}}{\omega _{0}-\omega +i{\frac {\gamma }{2}}}}={\frac {\text{const.}}{-\Delta +i{\frac {\gamma }{2}}}}$

where $\Delta =\omega -\omega _{0}$ is the detuning as shown on the right-hand side of Figure \ref{fig:example_resonances}. The full width at half maximum (FWHM) of a lorentzian curve is $\Delta \omega =\gamma$ leading to a quality factor

$Q={\frac {\omega _{0}}{\Delta \omega }}={\frac {\omega _{0}}{\gamma }}$

Note that $\omega _{0}$, $\Delta w$, and $\gamma$ are measured in angular frequency units \unit{\radian\per\second}, or \unit{\reciprocal\second} for short, and should not be confused with frequencies $f_{0}=\omega _{0}/2\pi$, $\Delta f=\Delta \omega /2\pi$ etc. When quoting values, we will often write explicitly Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_0=2\pi\times\unit{1}{\mega\hertz}} rather than Failed to parse (unknown function "\unit"): {\displaystyle \omega_0=\unit{6.28\times10^6}{\reciprocal\second}} to remind ourselves that the quantity in question is an angular frequency. We will never write Failed to parse (unknown function "\unit"): {\displaystyle \omega_0=\unit{6.28\times10^6}{Hz}} : although formally dimensionally consistent, this invites confusion with real (laboratory or Hertzian) frequencies. For brevity, we will often write or say "frequency" when we mean angular frequency, but will quote real frequencies as Failed to parse (unknown function "\unit"): {\displaystyle f_0=\unit{1}{\mega\hertz}} . Note that the natural unit for decay rate constants is that of an angular frequency, as in $e^{-i\omega _{0}t-\gamma t}$. Therefore we will express decay rates as inverse times in angular frequency units: Failed to parse (unknown function "\unit"): {\displaystyle \gamma=\unit{10^4}{\reciprocal\second}} for instance, but not Failed to parse (unknown function "\unit"): {\displaystyle \gamma=\unit{10^4}{\hertz}} , nor Failed to parse (unknown function "\unit"): {\displaystyle \gamma=2\pi\times\unit{1.6}{\kilo\hertz}} . Damping time constants are the inverse of decay rate constants, Failed to parse (unknown function "\unit"): {\displaystyle \tau=1/\gamma=\unit{100}{\micro\second}} .

Resonance Widths and Uncertainty Relations

From $\Delta \omega =\gamma$ it follows that damping time $\tau$ and the resonance linewidth $\Delta \omega$ of the driven system obey: $\Delta \omega \tau =1$ or, assuming $E=\hbar \omega$, $\Delta E\tau =\hbar$: a finite number of frequency components or energies ($\Delta \omega >0$) is necessary to synthesize a pulse of finite ($\tau <\infty$) width in time. \begin{figure}

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\caption{Measurement in a limited time $\Delta t$, modelled as an ideal monochromatic sine wave of frequency $\omega _{0}/2\pi =1$ gated by an observation window of finite width $\Delta t$}

\end{figure} This is quite similar to an uncertainty relation familiar from Fourier transformation theory, according to which the measurement of a frequency in a finite time $\Delta t$ (see Figure \ref{fig:gated_pulse}) results in a frequency spread $\Delta \omega$ obeying $\Delta \omega \Delta t\geq 1/2$. By using $E=\hbar \omega$ we can further connect this with the usual Heisenberg uncertainty relation $\Delta E\Delta t\geq \hbar /2$ \QA{ Does the Heisenberg uncertainty relation $\Delta E\Delta t\geq \hbar /2$, or rather the frequency uncertainty relation $\Delta \omega \Delta t\geq 1/2$ that follows from (classical) Fourier theory hold in classical systems? Can you measure the angular frequency of a classical harmonic oscillator in a time $\Delta t$ with an accuracy better than $\Delta \omega =1/2\Delta t$? }{ Yes: if you have a good model for the signal, for instance if you know that it is a pure sine wave, and if the signal to noise ratio (SNR) of the measurement is good enough, then you can fit the sine wave's frequency to much better accuracy than its spectral width $\Delta \omega$ (see Fig. \ref{fig:line_splitting}). As a general rule, the line center can be found to within an uncertainty $\delta \omega \approx \Delta \Omega /{\text{SNR}}$. This is known as splitting the line. Splitting the line by a factor of 100 is fairly straightforward, splitting it by a factor of a 1000 is a real challenge. As an extreme example of this, Caesium fountain clocks interrogate a \unit{9}{\giga\hertz} transition for a typical measurement time of \unit{1}{\second}, yielding $\Delta \omega /\omega \approx 10^{-10}$, but the best ones can nonetheless achieve relative accuracies of $10^{-16}$ \cite{Santarelli1999}. Of course, this sort of performance requires a very good understanding of the line shape and of systematics. } \begin{figure}

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\caption{Splitting the line in time and frequency domains. Given a sufficiently good model of the signal or line shape, and adequate signal-to-noise, one can find the center frequency $\omega _{0}$ of the underlying signal to much better than $1{\text{ cycle}}/\Delta t$ or the line width.}

\end{figure} \QU{ Can you measure the angular frequency of a quantum mechanical harmonic oscillator in a time $\Delta t$ to better than $\Delta \omega =1/2\Delta t$? } \QU{ Can you measure the angular frequency of a laser pulse lasting a time $\Delta t$ to better than $\Delta \omega =1/2\Delta t$?} If you answered \ref{q:quantum_line_splitting} and \ref{q:classical_line_splitting} differently, you may have a problem with \ref{q:laser_line_splitting}. Is a laser pulse quantum or classical? Clearly it is possible to beat $\Delta \omega =1/2\Delta t$ for a laser pulse. For instance, you can superimpose the laser pulse with a laser of fixed and known frequency on a photodiode, and fit the observed beatnote. The stronger the probe pulse, the better the SNR of the beatnote, and the more accurately you can extract the probe frequency. So how are we to reconcile quantum mechanics and classical mechanics, and what does the Heisenberg uncertainty really state? The Heisenberg uncertainty relation makes a statement about predicting the outcome of a single measurement on a single system. For a single photon, the limit $1/2\Delta t$ does indeed hold. However, nothing in quantum mechanics says that you cannot improve your knowledge about the average values of the quantities characterizing the system by repeated measurements on identical copies of it. In the laser example above, the frequency resolution improves with probe pulse strength because we are measuring many photons - for the purposes of this discussion we could have measured the frequencies of the photons sequentially. For independent measurement of $n$ uncorrelated photons the SNR is given by ${\sqrt {n}}$: the signal grows as $n$, while the noise is the shot noise of the photon number, which grows as ${\sqrt {n}}$. This SNR allows a frequency uncertainty of $1/2\Delta t{\sqrt {n}}$, known as the standard quantum limit. If we allow the use of entangled or correlated states of the photons, then the uncertainty can be as low as the Heisenberg limit $1/2\Delta tn$ \cite{Wineland1992,Wineland1994}. Let us now revisit \ref{q:quantum_line_splitting}. You may have heard that the quantum mechanical description of electromagnetism, quantum electrodynamics (QED), represents any single mode of the electromagnetic field by a harmonic oscillator of the same frequency. The ground state $n=0$ corresponds to no photons in the mode, the first excited state $n=1$ to one photon and so forth. This description is valid because photons to a very good approximation do not interact: the addition of a photon changes the system's energy always by the same increment. Since the many-photon pulse from \ref{q:laser_line_splitting} maps onto a quantum harmonic oscillator, it seems to follow that \ref{q:quantum_line_splitting} must also be answered in the affirmative. On the other hand, a harmonic oscillator is a single quantum system to which the Heisenberg limit must apply. The resolution of this apparent contradiction is that \ref{q:quantum_line_splitting} is asking about a measurement of the fundamental frequency of the harmonic oscillator, not the energy of a particular level. Therefore if you measure the energy of the n-th level to $\Delta E=\hbar /2\Delta t$, which can be done by exciting the $n$-th level from the ground state through a nonlinear process, then the uncertainty on the harmonic oscillator resonance frequency (that corresponds to the laser frequency in \ref{q:quantum_line_splitting}) is given by $\Delta \omega _{0}=\Delta E/n=1/n2\Delta t$. �If the interaction with the oscillator is restricted to classical fields or forces, then there is an additional uncertainty of ${\sqrt {n}}$ on which particular level of the oscillator is being measured and so we recover the standard quantum limit $1/2\Delta t{\sqrt {n}}$ mentioned previously.

Resonances and Two-Level Systems

Now, unlike in classical mechanics, most resonances studied in atomic physics are not harmonic oscillators, but two-level systems. Unlike harmonic oscillators, two-level systems show saturation. When a harmonic oscillator is driven longer or faster, higher and higher excited states are populated: the oscillator amplitude can become arbitrarily large. In contrast, the amplitude of oscillation in a two-level system is limited to one half in the appropriate dimensionless units.\footnote{An analogous dimensionless amplitude for the harmonic oscillator would be the amplitude measured in units of the oscillator ground state size.} Why, and under what conditions, do classical harmonic oscillator models of a two-level system work? A two-level system of energy difference $E$ can be approximated by a harmonic oscillator of frequency $\omega _{o}=E/\hbar$ when saturation effects in the two-level system are negligible, i.e. when the population of the second excited state in the harmonic oscillator problem is negligible, or equivalently when the population ratio of the first excited state to the the ground state is small: $P_{1}/P_{0}\ll 1$. This is the basis for the classical Lorentz model of the electron bound in the atom that describes many linear atomic properties (for instance the refractive index) very well. See "The origin of the refractive index" in chapter 31 of the Feynman Lectures on Physics \cite{Feynman1963}. When saturation comes into play, i.e. when the initial ground state is appreciably depleted, the harmonic oscillator ceases to be a good model. The limit on the oscillation amplitude in a two-level system suggests that a classical system with periodic evolution and a limit on the amplitude, namely rotation, could provide a better classical model of a two-level system. Indeed, the motion of a classical magnetic moment in a uniform field provides a model that captures almost all features of the quantum-mechanical two-level system, the exception beting the projection onto one of the two possible outcomes in a measurement. \section{Magnetic Resonance: The Classical Magnetic Moment in a Spatially Uniform Field}

Magnetic Moment in a Static Field

The interaction energy of a magnetic moment ${\mu }$ with a magnetic field ${\bf {B}}$ is given by

$W=-{\bf {\mu }}\cdot {\bf {B}}$

In a uniform field the force

${\bf {F}}=-{\bf {\nabla }}W=0$

vanishes, but the torque

${\bf {T}}={\bf {\mu }}\times {\bf {B}}$

does not. For an angular momentum ${\bf {L}}$ the equation of motion is then

${\dot {\bf {L}}}={\bf {T}}={\bf {\mu }}\times {\bf {B}}=\gamma {\bf {L}}\times {\bf {B}}$

where we have introduced the gyromagnetic ratio $\gamma$ as the proportionality constant between angular momentum and magnetic moment, as shown in Figure \ref{fig:static_precession}. \begin{figure}

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\caption{Precession of a the magnetic moment and associated angular momentum about a static field ${\bf {B}}$.}

\end{figure} This describes the precession of the magnetic moment about the magnetic field with angular frequency

$\Omega _{L}=-\gamma B$

$\Omega _{L}$ is known as the Larmor frequency. For electrons we have Failed to parse (unknown function "\unit"): {\displaystyle \gamma_e=2\pi\times\unit{2.8}{\mega\hertz\per G}} , for protons Failed to parse (unknown function "\unit"): {\displaystyle \gamma_e=2\pi\times\unit{4.2}{\kilo\hertz\per G}} .

An Alternative Solution: Rotating Coordinate System

A vector ${\bf {A}}$ rotating at constant angular velocity ${\bf {\Omega }}$ is described by

${\dot {\bf {A}}}={\bf {\Omega }}\times {\bf {A}}$

Then the rates of change of ${\bf {A}}$ measured in a coordinate system rotating at ${\bf {\Omega }}$ and in an inertial system are related by

${\dot {\bf {A}}}_{\text{in}}={\dot {\bf {A}}}_{\text{rot}}+{\bf {\Omega }}\times {\bf {A}}_{\text{in}}$

This follows immediately from the following facts: \begin{itemize}

* If ${\bf {A}}$ is constant in the rotating system then

${\dot {\bf {A}}}_{\text{in}}={\bf {\Omega }}\times {\bf {A}}_{\text{in}}$.

* If ${\bf {\Omega }}=0$ then

${\dot {\bf {A}}}_{\text{in}}={\dot {\bf {A}}}_{\text{rot}}$.

* Coordinate rotation is a linear transformation.

\end{itemize} This transformation is sometimes abbreviated as the schematic rule

$\left({\frac {d}{dt}}\right)_{\text{rot}}=\left({\frac {d}{dt}}\right)_{\text{in}}-{\bf {\Omega }}\times {\big (}{\big )}_{\text{in}}$

It follows that the angular momentum in a rotating frame obeys

${\dot {\bf {L}}}_{\text{rot}}={\dot {\bf {L}}}_{\text{in}}-{\bf {\Omega }}\times {\bf {L}}_{\text{in}}={\bf {L}}\times (\gamma {\bf {B}}+{\bf {\Omega }})$

If we choose ${\bf {\Omega }}={\bf {\Omega _{L}}}=-\gamma {\bf {B}}$, then ${\bf {L}}_{\text{rot}}$ is constant in the rotating frame. Often it is useful to think of a fictitious magnetic field ${\bf {B}}_{\text{fict}}={\bf {\Omega }}/\gamma$ that appears in a rotating frame.

Larmor's Theorem for a Charged Particle in a Magnetic Field

The vanishing of the torque on a magnetic moment when viewed in the correct rotating frame is reminiscent of Larmor's theorm for the motion of a charged particle in a magnetic field, which we now present. If the Lorentz force acts in an inertial frame,

${\bf {F}}_{\text{in}}=q{\bf {v}}_{\text{in}}\times {\bf {B}}$

then in the rotating frame, according to the rule \ref{eqn:rot_frame_transformation} we have \begin{align*}

 \dot{  ${\bf {r}}$ }_\text{rot}
 &= \dot{  ${\bf {r}}$ }_\text{in} -   ${\bf {\Omega }}$  \times   ${\bf {r}}$ _\text{in} \\
 \ddot{  ${\bf {r}}$ }_\text{rot}
 &= \ddot{  ${\bf {r}}$ }_\text{in} - 2   ${\bf {\Omega }}$  \times \dot{  ${\bf {r}}$ }_\text{in}
   +   ${\bf {\Omega }}$  \times \left(   ${\bf {\Omega }}$  \times   ${\bf {r}}$ _\text{in}
   \right)

\end{align*} resulting in a force ${\bf {F}}_{\text{rot}}=m{\ddot {\bf {r}}}_{\text{rot}}$ in the rotating frame given by \begin{align*}

   ${\bf {F}}$ _\text{rot}
 &= q   ${\bf {v}}$ _\text{in} \times   ${\bf {B}}$  - 2 m   ${\bf {\Omega }}$  \times
     ${\bf {v}}$ _\text{in} + m   ${\bf {\Omega }}$  \times \left(   ${\bf {\Omega }}$  \times
     ${\bf {r}}$ _\text{in} \right) \\
 &= q   ${\bf {v}}$ _\text{in} \times \left(   ${\bf {B}}$  + 2 \frac{m}{q}   ${\bf {\Omega }}$ 
   \right) + m   ${\bf {\Omega }}$  \times \left(   ${\bf {\Omega }}$  \times
     ${\bf {r}}$ _\text{in} \right)

\end{align*} where we have used $m{\ddot {r}}_{\text{in}}={\bf {F}}_{\text{in}}=q{\bf {v}}_{\text{in}}\times {\bf {B}}$. Choosing

$2{\frac {m}{q}}{\bf {\Omega }}=-{\bf {B}}=-B{\bf {\hat {e}}}_{z}$

yields

Failed to parse (unknown function "\vctr"): {\displaystyle {\bf{F}} _\text{rot} =\frac{q^2}{4m}B^2 {\bf{\hat e}} _z\times\left(\vctr{\hat e}_z\times {\bf{r}} _\text{in}\right)\approx 0 }

if the $B$ field is not too large. Thus the Lorentz force approximately disappears in the rotating frame. Note that while the vanishing of the force is approximate, the vanishing of the torque on a magnetic moment in the rotating frame is an exact result.

Rotating Magnetic Field on Resonance

\begin{figure}

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\caption{Field and moment vectors in the static and rotating frames for the case of resonant drive.}

\end{figure} Consider a magnetic moment ${\bf {\mu }}$ precessing about a field ${\bf {B}}=B_{0}{\bf {\hat {e}}}_{z}$ with ${\bf {\mu }}=(\mu ,\theta ,\phi =-\omega _{0}t)$ in spherical coordinates, where $\omega _{0}=-\gamma B_{0}$. Assume that we now apply an additional field ${\bf {B}}_{1}$, in the $xy$-plane rotating at $\omega _{0}$. To solve the resulting problem it is simplest to go into the rotating frame (Figure \ref{fig:rotating_frame}). Then ${\bf {B}}_{1}$ is stationary, say along ${\bf {\hat {e}}}_{x}$, and there is an additional fictitious field ${\bf {B}}_{\text{fict}}={\bf {\omega _{0}}}/\gamma =-{\bf {B}}_{0}$ which cancels the field ${\bf {B}}_{0}$. So in the rotating frame we are left just with the static field ${\bf {B}}_{1}$, and the magnetic moment precesses about ${\bf {B}}_{1}$ at the Rabi frequency

$\omega _{R}=\gamma B_{1}$

A magnetic moment initially along the $-{\bf {\hat {e}}}_{z}$ axis will point along the $+{\bf {\hat {e}}}_{z}$ axis at a time $T$ given by $\omega _{r}T=\pi$, while a magnetic moment parallel or antiparallel to applied magnetic field ${\bf {B}}_{1}$ does not precess in the rotating frame. \QU{ Assume the magnetic moment is initially pointing along the $-{\bf {\hat {e}}}_{z}$ axis. Assume that a rotating field $B_{1}\ll B_{0}$ is applied, but that it rotates at a frequency $\omega _{1}>\omega _{0}$, where $\omega _{0}$ is the Larmor frequency for the static field ${\bf {B}}=B_{0}{\bf {\hat {e}}}_{z}$. Compared to the on-resonant case, $\omega _{1}=\omega _{0}$, is the oscillation frequency of the magnetic moment. \begin{enumerate}

* larger
* the same
* smaller

\end{enumerate} } \QU{ Same question as \ref{q:rabi_freq_blue_detuned} but for $\omega _{1}<\omega _{0}$. }

Rotating Magnetic Field Off-Resonance

If the rotation frequency $\omega$ of ${\bf {B}}_{1}$ does not equal the Larmor frequency $\omega _{0}=\gamma B_{0}$ associated with the static field ${\bf {B}}_{0}$, then in the frame rotating with ${\bf {B}}_{1}$ at frequency $\omega$ the static field is not completely cancelled by the fictitious field $\omega /\gamma$, but a difference along ${\bf {\hat {e}}}_{z}$ remains, giving rise to a total effective field in the rotating frame

${\bf {B}}_{\text{eff}}=B_{1}{\bf {\hat {e}}}_{x}+\left(B_{0}-{\frac {\omega }{\gamma }}\right){\bf {\hat {e}}}_{z}$

The effective field is static, lies at an angle $\theta$ with the z

$\tan \theta ={\frac {B_{1}}{B_{0}-{\frac {\omega }{\gamma }}}}$

and is of magnitude

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lvertB_\text{eff}\rvert =\sqrt{B_1^2+\left(B_0-\frac{\omega}{\gamma}\right)^2} }

The magnetic moment precesses around it with an effective (sometimes called generalized) Rabi frequency

$\Omega _{R}=\gamma B_{\text{eff}}={\sqrt {(\omega _{0}-\omega )^{2}+\omega _{R}^{2}}}={\sqrt {\omega _{R}^{2}+\delta ^{2}}}$

where $\omega _{R}=\gamma B_{1}$ is the Rabi frequency associated with $B_{1}$, and $\delta =\omega -\omega _{0}$ is the detuning from resonance with the Larmor frequency $\omega _{0}=\gamma B_{0}$. \subsection{Geometrical Solution for the Classical Magnetic Moment in Static and Rotating Fields} \begin{figure}

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\caption{Geometrical relations for the spin in combined static and rotating magnetic fields, viewed in the frame co-rotating with the drive field ${\bf {B}}_{1}$. At lower right is a view looking straight down the ${\bf {B}}_{\text{eff}}$ axis.}

\end{figure} Referring to Figure \ref{fig:rotating_coord_construction}, we have \begin{align*}

 \phi
 &= \Omega_R t \\
 C
 &= \mu \sin \alpha \\
 A^2
 &= \mu^2 ( 1 - \cos \alpha )^2 + C^2 \\
 &= \mu^2 ( 1 - 2 \cos \alpha + \cos^2 \alpha + \sin^2 \alpha ) \\
 &= 2 \mu^2 (1 - \cos \alpha ) \\
 \Rightarrow \cos \alpha
 &= 1 - \frac{A^2}{2 \mu^2}

\end{align*} On the other hand \begin{align*}

 \frac{A}{2}
 &= \mu \sin \theta \sin \frac{\phi}{2} \\
 A
 &= 2 \mu \sin \theta \sin \frac{\Omega_R t}{2}

\end{align*} so that \begin{align}

 \cos \alpha
 &= 1 - \frac{4 \mu^2 \sin^2 \theta \sin^2 \frac{\Omega_R t}{2}}{2 \mu^2}
   \notag \\
 &= 1 - 2 \sin^2 \theta \sin^2 \frac{\Omega_R t}{2} \notag \\
 \mu_z(t)
 &= \mu \cos \alpha \notag \\
 &= \mu \left( 1 - 2 \frac{\omega_R^2}{\delta^2 + \omega_R^2} \sin^2
   \frac{\Omega_R t}{2} \right) \notag \\
 \mu_z(t)
 &= \mu \left( 1 - 2 \frac{\omega_R^2}{\Omega_R^2} \sin^2 \frac{\Omega_R t}{2}
   \right)
  

\end{align} With $\omega _{0}=\gamma B_{0}$ the Larmor frequency of the static field, $\delta =\omega -\omega _{0}$ the detuning, $\omega _{R}=\gamma B_{1}$ the resonant and $\Omega _{R}={\sqrt {\omega _{R}^{2}+\delta ^{2}}}$ the generalized Rabi frequencies. Note that the precession is faster, but the amplitude smaller for an off-resonant field than for the resonant case. The above result is also the correct quantum-mechanical result.

"Rapid" Adiabiatic Passage

Rapid adiabatic passage is a technique for inverting a spin by (slowly) sweeping the detuning of a drive field through resonance. "Slowly" means slowly compared to the Larmor frequency $\gamma B_{\text{eff}}$ about the effective static field in the rotating frame for all times. The physical picture is as follows. Assume the detuning is initially negative ($\delta <0$, Failed to parse (unknown function "\abs"): {\displaystyle \abs\delta\gg\omega_R} ). Since

$\tan \theta ={\frac {B_{1}}{B_{0}-{\frac {\omega }{\gamma }}}}={\frac {\omega _{R}}{\omega _{0}-\omega }}=-{\frac {\omega _{R}}{\delta }}$

the effective magnetic field initially points of a small angle relative to the ${\bf {\hat {e}}}_{z}$ axis. If the detuning is increased slowly compared to the Larmor frequency, the spin will continue to precess tightly around ${\bf {B}}_{\text{eff}}$, which for $\delta =0$ points along the x axis, and for $\delta \gg \omega _{R}$ along the $-{\bf {\hat {e}}}_{z}$ axis (see Figure \ref{fig:rapid_adiabatic_passage}). \begin{figure}

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\caption{Motion of the spin during rapid adiabatic passage, viewed in the frame rotating with ${\bf {B}}_{1}$. The spin's rapid precession locks it to the direction of ${\bf {B}}_{\text{eff}}$ and thus it is dragged through an angle $\pi$ as the frequency is swept through resonance.}

\end{figure} Thus the magnetic moment, starting out along ${\bf {B}}_{0}=B_{0}{\bf {\hat {e}}}_{z}$, ends up pointing along $-{\bf {B}}_{0}=B_{0}{\bf {\hat {e}}}_{z}$. Note that in the rotating frame Failed to parse (unknown function "\vctr"): {\displaystyle \vctr\mu} remains always (almost) parallel to the effective field ${\bf {B}}_{\text{eff}}$. A similar precess is used in magnetic traps for atoms, but there ${\bf {B}}_{\text{eff}}$ is a real, spatially dependent field constant in time. As the atom moves in this field, the fast precession of the magnetic moment about the local field keeps its direction locked to the local field, whose direction varies in the lab frame. Returning to rapid adiabatic passage, since the generalized Rabi frequency is smallest and equal to the resonant Rabi frequency $\omega _{R}$ at $\delta =0$, the adiabatic requirement is most severe there, i.e. near $\theta =\pi /2$. Near $\theta =\pi /2$ we have, with $B_{z}=B_{0}-1/\gamma \omega (t)$,

Failed to parse (unknown function "\lvert"): {\displaystyle \lvert\dot\theta\rvert =\frac{ \lvert\dot{B}_z\rvert }{B_1}=\frac{ \lvert\dot\omega\rvert }{\gamma B_1} =\frac{ \lvert\dot\omega\rvert }{\omega_R}\overset{!}{\ll}\omega_R }

where the exclamation point in ${\overset {!}{\ll }}$ indicates a requirement which we impose. Consequently, if the evolution is to be adiabatic, we must have Failed to parse (unknown function "\lvert"): {\displaystyle \lvert\dot{\omega}\rvert \ll\omega_R^2} . This means that the change $\Delta \omega$ of rotation frequency $\omega$ per Rabi period $T=2\pi /\omega _{R}$, Failed to parse (unknown function "\lvert"): {\displaystyle \lvert\Delta\omega\rvert = \lvert\dot\omega\rvert T= \lvert\dot\omega\rvert /\omega_R 2\pi} , must be small compared to the Rabi frequency $\omega _{R}$. The quantum mechanical treatment yields a probability for non-adiabatic transition (probability for the magnetic moment not following the magnetic field) given by

Failed to parse (unknown function "\lvert"): {\displaystyle P_\text{na}=exp\left(-\frac{\pi}{2}\frac{\omega^2_R}{ \lvert\dot\omega\rvert }\right) }

in agreement with the above qualitative discussion.

Quantized Spin in a Magnetic Field

Equation of Motion for the Expectation Value

For the system we have been considering, the Hamiltonian is

${\hat {H}}=-{\bf {\hat {\mu }}}\cdot {\bf {B}}_{0}=-\gamma {\bf {\hat {L}}}\cdot {\bf {B}}_{0}=-\gamma {\hat {L}}_{z}B_{0}{\text{ for }}{\bf {B}}=B_{0}{\bf {\hat {e}}}_{z}$

Recalling the Heisenberg equation of motion for any operator ${\hat {O}}$ is

${\frac {d}{dt}}{\hat {O}}={\frac {i}{\hbar }}\left[{\hat {H}},{\hat {O}}\right]+{\frac {\partial {\hat {O}}}{\partial t}}$

where the last term refers to operators with an explicit time dependence, we have in this instance

${\frac {d}{dt}}{\hat {\mu }}_{k}=\gamma {\frac {d}{dt}}{\hat {L}}_{k}={\frac {i\gamma }{\hbar }}\left[{\hat {H}},{\hat {L}}_{k}\right]=-{\frac {i\gamma ^{2}}{\hbar }}B_{0}\left[{\hat {L}}_{z},{\hat {L}}_{k}\right]$

Using $\left[{\hat {L}}_{k},{\hat {L}}_{l}\right]=i\hbar \epsilon _{klm}{\hat {L}}_{m}$ with the Levi-Civita symbol $\epsilon _{klm}$, we have \begin{align*}

 \frac{d}{dt} \hat{\mu}_x
 &= - \frac{i \gamma^2}{\hbar} B_0 i \hbar \hat{L}_y
 = \gamma^2 \hat{L}_y B_0
 = \hat{\mu}_y \gamma B_0 \\
 \frac{d}{dt} \hat{\mu}_y
 &= - \mu_x \gamma B_0 \\
 \frac{d}{dt} \hat{\mu}_z
 &= 0

\end{align*} or in short \begin{align}

 \frac{d}{dt}   ${\bf {\hat {\mu }}}$ 
 &= \gamma   ${\bf {\hat {\mu }}}$  \times   ${\bf {B}}$  \\
 \frac{d}{dt}   ${\bf {\hat {L}}}$ 
 &= \gamma   ${\bf {\hat {L}}}$  \times   ${\bf {B}}$ 

\end{align} These are just like the classical equations of motion \ref{eq:classical_precession_in_static_field}, but here they describe the precession of the operator for the magnetic moment ${\bf {\hat {\mu }}}$ or for the angular momentum ${\bf {\hat {L}}}$ about the magnetic field at the (Larmor) angular frequency $\omega _{L}=\gamma B_{0}$. Note that \begin{itemize} \item Just as in the classical model, these operator equations are exact; we have not neglected any higher order terms. \item Since the equations of motion hold for the operator, they must hold for the expectation value

Failed to parse (unknown function "\avg"): {\displaystyle \avg{\dot{ {\bf{\hat L}} }}=\gamma\avg{ {\bf{\hat L}} }\times {\bf{B}} }

\item We have not made use of any special relations for a spin-$1/2$ system, but just the general commutation relation for angular momentum. Therefore the result, precession about the magnetic field at the Larmor frequency, remains true for any value of angular momentum $l$. \item A spin-$1/2$ system has two energy levels, and the two-level problem with coupling between two levels can be mapped onto the problem for a spin in a magnetic field, for which we have developed a good classical intuition. \item If coupling between two or more angular momenta or spins within an atom results in an angular momentum $F>1/2$, the time evolution of this angular momentum in an external field is governed by the same physics as for the two-level system. This is true as long as the applied magnetic field is not large enough to break the coupling between the angular momenta; a situation known as the Zeeman regime. Note that if the coupled angular momenta have different gyromagnetic ratios, the gyromagnetic ratio for the composite angular momentum is different from those of the constituents. \item For large magnetic field the interaction of the individual constituents with the magnetic field dominates, and they precess separately about the magnetic field. This is the Paschen-Back regime. \item An even more interesting composite angular momentum arises when $N$ two-level atoms are coupled symmetrically to an external field. In this case we have an effective angular momentum $L=N/2$ for the symmetric coupling (see Figure \ref{fig:dicke_states}). \begin{figure}

Error creating thumbnail: Unable to save thumbnail to destination

\caption{Level structure diagram for $n$ two-level atoms in a basis of symmetric states \cite{Dicke1954}. The leftmost column corresponds to an effective spin-$n/2$ object. Other columns correspond to manifolds of symmetric states of the $n$ atoms with lower total effective angular momentum.}

\end{figure} Again the equation of motion for the composite angular momentum ${\bf {\hat {L}}}$ is a precession. This is the problem considered in Dicke's famous paper \cite{Dicke1954}, in which he shows that this collective precession can give rise to massively enhanced couplings to external fields ("superradiance") due to constructive interference between the individual atoms. \end{itemize}

The Two-Level System: Spin-\texorpdfstring{$1/2$}{1/2}

Let us now specialize to the two-level system and calculate the time evolution of the occupation probabilities for the two levels. \begin{figure}

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\caption{Equivalence of two-level system with spin-$1/2$. Note that for $\gamma >0$ the spin-up state (spin aligned with field) is the ground state. For an electron, with $\gamma <0$, spin-up is the excited state. Be careful, as both conventions are used in the literature.}

\end{figure} We have that

Failed to parse (syntax error): {\displaystyle \left<S_z\right> =\frac{\hbar}{2}\left(P_\uparrow-P_\downarrow\right) =\frac{\hbar}{2}\left(P_g-P_e\right) =\frac{\hbar}{2}\left(1-2P_e\right) }

where in the last equation we have used the fact that $P_{e}+P_{g}=1$. The signs are chosen for a spin with $\gamma >0$, such as a proton (Figure \ref{fig:two_level_spin_half}). For an electron, or any other spin with $\gamma <0$, the analysis would be the same but for the opposite sign of $S_{z}$ and the corresponding exchange of $\uparrow$ and $\downarrow$. If the system is initially in the ground state, $P_{g}(t=0)=1$ (or the spin along Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +\vctr{\hat e}_z} , $S_{z}(0)=+\hbar /2$), the expectation value obeys the classical equation of motion \ref{eq:classical_rabi_flopping}: \begin{align}

 P_e(t)
 &= \frac{1}{2} - \frac{1}{\hbar}   Failed to parse (syntax error): {\displaystyle \left<S_z\right>}
 
 = \frac{1}{2}
   - \frac{  Failed to parse (syntax error): {\displaystyle \left<S_z(0)\right>}
 }{\hbar}
     \left(1 - 2 \frac{\omega_R^2}{\Omega_R^2} \sin^2
     \frac{\Omega_R t}{2}\right) \notag \\
 P_e(t)
 &= \frac{\omega_R^2}{\Omega_R^2} sin^2 \left(\frac{\Omega_R t}{2}\right)
  

\end{align} Equation \ref{eqn:rabi_transition_probability} is the probability to find system in the excited state at time $t$ if it was in ground state at time $t=0$. Figure \ref{fig:rabi_signal} shows a real-world example of such an oscillation. \begin{figure}

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\caption{Rabi oscillation signal taken in the Vuleti\'{c} lab shortly after this topic was covered in lecture in 2008. The amplitude of the oscillations decays with time due to spatial variations in the strength of the drive field (and hence of the Rabi frequency), so that the different atoms drift out of phase with each other.}

\end{figure}

Matrix form of Hamiltonian

With the matrix representation \begin{align*}

   Failed to parse (syntax error): {\displaystyle \left|e\right>}
 
 &=   Failed to parse (syntax error): {\displaystyle \left|S_z = - \frac{1}{2}\right>}
 
 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
   Failed to parse (syntax error): {\displaystyle \left|g\right>}
 
 &=   Failed to parse (syntax error): {\displaystyle \left|S_z = + \frac{1}{2}\right>}
 
 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}

\end{align*} we can write the Hamiltonian $H_{0}$ associated with the static field ${\bf {B}}_{0}=B_{0}{\bf {\hat {e}}}_{z}$ as

${\hat {H}}_{0}=-{\bf {\hat {\mu }}}\cdot {\bf {B}}_{0}=-\gamma {\hat {S}}_{z}B_{0}=-\hbar \omega _{0}{\frac {{\hat {S}}_{z}}{\hbar }}={\frac {1}{2}}\hbar \omega _{0}{\begin{pmatrix}1&0\\0&-1\end{pmatrix}}={\frac {1}{2}}\hbar \omega _{0}{\hat {\sigma }}_{z}$

where $\omega _{0}=\gamma B_{0}$ is the Larmor frequency, and

${\hat {\sigma }}_{z}={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}$

is a Pauli spin matrix. The eigenstates are Failed to parse (syntax error): {\displaystyle \left|e\right> } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|g\right> } with eigenenergies $\pm \hbar \omega _{0}/2$. A spin initially along ${\bf {\hat {e}}}_{x}$, corresponding to

Failed to parse (syntax error): {\displaystyle \left|\psi(t=0)\right> =\frac{1}{\sqrt{2}}\left( \left|e\right> + \left|g\right> \right) =\frac{1}{\sqrt{2}}\left( \left|-\frac{1}{2}\right> + \left|+\frac{1}{2}\right> \right) }

evolves in time as

Failed to parse (syntax error): {\displaystyle \left|\psi(t)\right> =\frac{1}{\sqrt{2}} \left(e^{-i\omega_0 t / 2} \left|e\right> +e^{+i\omega_0 t / 2} \left|g\right> \right) =\frac{e^{-i\omega_0 t/2}}{\sqrt{2}}\left( \left|e\right> +e^{i\omega_0 t} \left|g\right> \right) }

which describes a precession with angluar frequency $\omega _{0}$. The field ${\bf {B}}_{1}$, rotating at $\omega$ in the $X-Y$ plane corresponds to \begin{align}

 \hat{H}_1
 &= -  ${\bf {\hat {\mu }}}$ \cdot  ${\bf {B}}$ _1(t)
   =-  ${\bf {\hat {\mu }}}$ \cdot\frac{\omega_R}{\gamma}
     \left(-  ${\bf {\hat {e}}}$ _x \cos\omega t-   ${\bf {\hat {e}}}$ _y \sin\omega t\right)
     \notag \\
 &= \omega_R\left(\hat{S}_x \cos\omega t + \hat{S}_y \sin\omega t\right) \notag
 \\
 &= \frac{\hbar\omega_R}{2}
   \left(\hat{\sigma}_x \cos\omega t + \hat{\sigma}_y \sin\omega t\right)
   \notag \\
 &= \frac{\hbar\omega_R}{2}
   \left(
     \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\cos\omega t
     + \begin{pmatrix} 0 & -i \\ i & 0     \end{pmatrix}\sin\omega t
   \right) \notag \\
 \hat{H}_1
 &= \frac{\hbar\omega_R}{2}
   \begin{pmatrix}
     0 & e^{-i\omega t} \\
     e^{i\omega t} & 0
   \end{pmatrix}

\end{align} where have used the Pauli spin matrices $\sigma _{x}$, $\sigma _{y}$. The full Hamiltonian is thus given by

${\hat {H}}={\frac {\hbar }{2}}{\begin{pmatrix}+\omega _{0}&\omega _{R}e^{-i\omega t}\\\omega _{R}e^{-i\omega t}&-\omega _{0}\end{pmatrix}}$

This is the famous "dressed atom" Hamiltonian in the so-called "rotating wave approximation". Its eigenstates and eigenvalues provide a very elegant, very intuitive solution to the two-state problem.

Solution of the Schr\"{o

Spin-\texorpdfstring{$1/2$}{1/2} in the Interaction Representation} The interaction representation consists of expanding the state Failed to parse (syntax error): {\displaystyle \left|\psi(t)\right> } in terms of the eigenstates Failed to parse (syntax error): {\displaystyle \left|e\right> } , Failed to parse (syntax error): {\displaystyle \left|g\right> } of the Hamiltonian $H_{0}$, including their known time dependence $e^{iH_{0}t/\hbar }$ due to $H_{0}$. That means we write here

Failed to parse (syntax error): {\displaystyle \left|\psi(t)\right> =a_e(t) \left|e\right> e^{-i\omega_0t/2}+a_g(t) \left|g\right> e^{i\omega_0t/2} }

Substituting this into the Schr\"{o}dinger equation

Failed to parse (syntax error): {\displaystyle i\hbar\frac{d}{dt} \left|\psi(t)\right> =\left(H_0+H_1\right) \left|\psi(t)\right> }

then results in the equations of motion for the coefficients \begin{align*} i\dot{a}_e&= \frac{\omega_R}{2}e^{i\left(\omega_0-\omega\right)t}a_g \\ i\dot{a}_g&= \frac{\omega_R}{2}e^{-i\left(\omega_0-\omega\right)t}a_e \end{align*} Where we have used the matrix form of the Hamiltonian, \ref{eq:dressed_atom_hamiltonian}. Introducing the detuning $\delta =\omega -\omega _{0}$, we have \begin{align*} i\dot{a}_g&= \frac{\omega_R}{2}e^{i\delta t}a_e \\ i\dot{a}_e&= \frac{\omega_R}{2}e^{-i\delta t}a_g \end{align*} The explicit time dependence can be eliminated by the sustitution \begin{align*} b_g&= e^{-i\delta t/2}a_g \\ b_e&= e^{i\delta t/2}a_e \end{align*} As you will show (or have shown) in the problem set, this leads to solutions for $b_{g,e}$ given by \begin{align*} b_g(t)&= e^{i\Omega_R t/2}B_1+e^{-i\Omega_R t/2}B_2 \\ b_e(t)&= \frac{\omega_R}{\delta-\Omega_R}e^{i\Omega t/2}B_1

       +\frac{\omega_R}{\delta+\Omega_R}e^{-i\Omega_R t/2}B_2

\end{align*} with two constants that are determined by the initial conditions. For $a_{g}(t=0)=1$ we find

Failed to parse (unknown function "\lverta"): {\displaystyle \lverta_e(t)\rvert ^2=\frac{\omega_R^2}{\Omega_R^2}\sin^2\frac{\Omega_R t}{2} }

as already derived from the fact that the expectation value for the magnetic moment obeys the classical equation.

Atomic Clocks and the Ramsey Method

When comparing the Hamiltonian for a spin-$1/2$ in a magnetic field to that of a two-level system with a coupling between the two levels characterized by the strength $\omega _{R}$ and frequency $\omega$, we see that the energy spacing between Failed to parse (syntax error): {\displaystyle \left|e\right> } and Failed to parse (syntax error): {\displaystyle \left|g\right> } corresponds to the Larmor frequency $\omega _{0}$ in the static field. This spacing can provide a frequency or time reference if perturbations affecting $\omega _{0}$ are sufficiently well controlled. For instance, the time unit "second" is defined via the transition frequency between two hyperfine states in the electronic ground state of the caesium atom, which is near \unit{9.2}{\giga\hertz} in the microwave domain. The task of an atomic clock is then to measure this frequency accurately by trying to tune a frequency source (the frequency $\omega$ of the rotating field $B_{1}$ in the spin picture) to the atomic frequency $\omega _{0}$. Equivalently, we want to find the frequency $\omega$ such that the detuning $\delta =\omega -\omega _{0}$ is equal to zero. Starting with an atom in Failed to parse (syntax error): {\displaystyle \left|g\right> } (spin along $+{\bf {\hat {e}}}_{z}$ for $\gamma >0$), we could try to find the resonance frequency by noting that according to

$P_{e}(t)={\frac {\omega _{R}^{2}}{\omega _{R}^{2}+\delta ^{2}}}sin^{2}\left({\sqrt {\omega _{R}^{2}+\delta ^{2}}}\,{\frac {t}{2}}\right)$

the population of the upper state is maximized for $\delta =0$ (i.e. the precession of the spin to the $-{\bf {\hat {e}}}_{z}$ direction is only complete on resonance). This is the so-called Rabi method. It suffers from a number of drawbacks. For one, the signal is only quadratic in the detuning $\delta$, i.e. the method is relatively insensitive near $\delta =0$. Furthermore, the optimum time $t$ depends on the strength $\omega _{R}$ of the coupling (i.e. the strength $B_{1}$ of the rotating field), so fluctuations in $\omega _{R}$ can be mistaken for changes in $\delta$. Finally the coupling by $\omega _{R}$ to other levels can lead to level shifts that are not intrinsic to the atom, but depend on the applied drive $\omega _{R}$ (Figure \ref{fig:rabi_third_level}). \begin{figure}

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\caption{The drive used for Rabi flopping within the Failed to parse (syntax error): {\displaystyle \left|e\right>} , Failed to parse (syntax error): {\displaystyle \left|g\right>} system can also off-resonantly couple one or both levels to other states, perturbing the transition frequency $\omega _{0}$.}

\end{figure} Norman Ramsey invented an alternative method (the so-called "separated oscillatory fields method", known for short as the "Ramsey method" \cite{Ramsey1949,Ramsey1950}, for which he received the Nobel prize), that fixes all of these problems. It leads to a signal that is linear rather than quadratic in the detuning $\delta$, does not require tuning the measurement time to match the applied field strength $\omega _{R}=\gamma B_{1}$, and, most importantly, eliminates level shifts due to $\omega _{R}$ altogether. The method is as follows. Instead of applying a pulse for a time t that corresponds to Rabi rotation of the spin by $\pi$ (called a $\pi$ pulse), the pulse is applied for half that time, corresponding to the Rabi rotation of the spin by $\pi /2$ into the $X-Y$ plane ($\pi /2$ pulse). Then the applied field $B_{1}$ is turned off and the system is left to precess in the static field $B_{0}$ (or at its natural frequency $\omega _{0}$) for a measurement time $T$. Finally, a second $\pi /2$ pulse, identical to the original one, is applied (see Figure \ref{fig:ramsey_sequence}). \begin{figure}

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\caption{Ramsey sequence}

\end{figure} The signal is the ${\bf {\hat {e}}}_{z}$ component of the spin after the second interaction. The signal after the second $\pi /2$ pulse is an oscillating signal in $T\delta$, depending on how much phase the spin has acquired relative to the local oscillator (the microwave signal generator at frequency $\omega$). Examples of such curves are shown in Figures \ref{fig:ramsey_signal} and \ref{fig:ramsey_vs_freq}. \begin{figure}

File:Resonances-Ramsey

\caption{Ramsey oscillation signal as a function of time taken in the Vuleti\'{c} lab in 2007. The drive field was deliberately detuned from resonance so that the oscillation at the detuning frequency would be visible.}

\end{figure} \begin{figure}

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\caption{Experimental data from Ramsey's original paper \cite{Ramsey1950}, showing the signal as a function of frequency. Note the narrow oscillation, whose width is set by the measurement time $T$, superimposed on the much broader background set up by the inhomogeneously broadened $\pi /2$ pulses.}

\end{figure} At the zero crossings we have maximum sensitivity of the signal with respect to frequency changes. Note that the signal as a function of $T$ looks similar to Rabi flopping. However, there the zero crossing measure the Rabi frequency, not $\delta =\omega -\omega _{0}$.

Decoherence Processes, Mixed States, Density Matrix

The purely Hamiltonian evolution described by the Schr\"{o}dinger Equation leaves the system in a pure state. However, often we have to deal with incoherent processes such as the uncontrolled loss or addition of atoms, or other perturbations that change the system evolution in an uncontrollable way. If the incoherent process is merely a (state-dependent) loss of atoms to a third state then we can describe the system by a Hamiltonian with complex eigenstates

$E_{g,e}\rightarrow E_{g,e}+i{\frac {\Gamma _{g,e}t}{2}}$

The decay of the norm Failed to parse (syntax error): {\displaystyle \left<\psi\vert\psi\right> } corresponds to decay out of the Failed to parse (syntax error): {\displaystyle \left|g\right> } , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|e\right> } system, as in figure \ref{fig:decay_out}. \begin{figure}

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\caption{Decay out of the Failed to parse (syntax error): {\displaystyle \left|e\right>} Failed to parse (syntax error): {\displaystyle \left|g\right>} system, which can be modelled by a non-Hermitian Hamiltonian, the imaginary part of the eigenvalue corresponding to a decay rate.}

\end{figure} However, other processes, such as spontaneous decay from Failed to parse (syntax error): {\displaystyle \left|e\right> } to Failed to parse (syntax error): {\displaystyle \left|g\right> } or loss of coherence (well-defined phase relationship) between Failed to parse (syntax error): {\displaystyle \left|e\right> } and Failed to parse (syntax error): {\displaystyle \left|g\right> } due to uncontrolled level shifts (e.g. collisions, uncontrolled B-field fluctuations) cannot be dealt with within the Hamiltonian approach and require the density matrix formalism.

Density Operator

The density operator formalism is necessary when the preparation process does not prepare a single quantum state Failed to parse (syntax error): {\displaystyle \left|\psi\right> } , but a statistical mixture of quantum states Failed to parse (syntax error): {\displaystyle \left|\psi_i\right> } with probabilities $P_{i}$. The density operator is the projection operator for that statistical mixture, or ensemble average.

Failed to parse (syntax error): {\displaystyle \hat{\rho}=\Sigma P_i \left|\psi_i\right> \left<\psi_i\right| =\overline{ \left|\psi\right> \left<\psi\right| } }

From the Schr\"{o}dinger equation it follows that the Hamiltonian evolution of the density operator is governed by

$i\hbar {\dot {\hat {\rho }}}=\left[{\hat {H}},{\hat {\rho }}\right]$

In the presence of damping and decay processes there are additional terms, and the time evolution is governed by a so-called Master equation. The expectation value of any operator is given by

Failed to parse (syntax error): {\displaystyle \left<\hat O\right> = Tr \left(\hat O\hat \rho\right)= Tr \left(\hat \rho\hat O\right) }

The expectation value of the unity operator must be 1, so $Tr\left({\hat {\rho }}\right)=1$ The expectation value of ${\hat {\rho }}$, $Tr\left({\hat {\rho }}^{2}\right)$, is unity for a pure state , and $<1$ for a mixed state. In any basis Failed to parse (syntax error): {\displaystyle \left\{ \left|n\right> \right\}} , ${\hat {\rho }}$ is specified in terms of its matrix elements

Failed to parse (syntax error): {\displaystyle \rho_{n^\prime n}= \left<n^\prime\right| \hat \rho \left|n\right> }

and the expectation value of any operator can be written as \begin{align*}

 Failed to parse (syntax error): {\displaystyle \left<\hat O\right>}
 

&= Tr \left(\hat O\hat\rho\right) =\Sigma_n Failed to parse (syntax error): {\displaystyle \left<n\right|} \hat O\hat \rho Failed to parse (syntax error): {\displaystyle \left|n\right>} =\Sigma_{n^\prime n} Failed to parse (syntax error): {\displaystyle \left<n\right|} \hat O Failed to parse (syntax error): {\displaystyle \left|n^\prime\right>} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left<n^\prime\right|} \hat \rho Failed to parse (syntax error): {\displaystyle \left|n\right>} \\

 Failed to parse (syntax error): {\displaystyle \left<\hat O\right>}
 &= \Sigma_{nn^\prime}O_{nn^\prime}\rho_{n^\prime n}

\text{ with } O_{nn^\prime}= Failed to parse (syntax error): {\displaystyle \left<n\right|} \hat O Failed to parse (syntax error): {\displaystyle \left|n^\prime\right>} \end{align*} The diagonal elements of the density matrix ($\rho _{n^{\prime }n}$) are called the probabilities since $\rho _{nn}$ is the probability to find the system in state $n$, while the off-diagonal elements are called coherences ($\rho _{n^{\prime }n}$ is the coherence between states $n^{\prime }$ and $n$). If we write $\rho _{n^{\prime }n}=Ae^{i\phi }$ with $A$, $\phi$ real then $\phi _{n^{\prime }n}$ is the phase between states $n^{\prime }$ and $n$, while $A\leq {\sqrt {\rho _{nn}\rho _{n^{\prime }n^{\prime }}}}$. The coherence is maximum when $A={\sqrt {\rho _{nn}\rho _{n^{\prime }n^{\prime }}}}$. Finally note that the density operator is hermitian, so that $\rho _{n^{\prime }n}={\rho _{n}n^{\prime }}^{\star }$, as we would expect for a meaningfully-defined coherence between two states.

Density Matrix for a Two-Level System and Bloch Vector

We now introduce a slightly different parametrization of the two-level Hamiltonian \ref{eq:dressed_atom_hamiltonian} and of the corresponding density matrix: \begin{align*} H&= \frac{\hbar}{2}

 \begin{pmatrix}
     \omega_0 & \omega_R e^{-i\omega t} \\
     \omega_R e^{i \omega t} & -\omega_0
 \end{pmatrix}

=\frac{\hbar}{2}

 \begin{pmatrix}
     \omega_0 & V_1-iV_2 \\
     V_1+iV_2 & -\omega_0
 \end{pmatrix} \\

&= \frac{\hbar}{2}

 \left(V_1\hat\sigma_x+V_2\hat\sigma_y+\omega_0\hat\sigma_z\right) \\

\rho&=

 \begin{pmatrix}
   \rho_{11} & \rho_{12} \\
   \rho_{21} & \rho_{22}
 \end{pmatrix}

=\frac{1}{2}

 \begin{pmatrix}
     r_0+r_3 & r_1-ir_2 \\
     r_1+ir_2 & r_0-r_3
 \end{pmatrix} \\

&= \frac{1}{2}

 \left(r_0\hat 1 + r_1\hat\sigma_x + r_2\hat\sigma_y + r_3 \hat\sigma_z\right)

\end{align*} In the problem set you will show (or have shown) that the Hamiltonian evolution of the density matrix

$i\hbar {\dot {\rho }}=\left[H_{1},\rho \right]$

implies the equation of motion

${\frac {d{\bf {r}}}{dt}}={\bf {\omega }}\times {\bf {r}}$

for the vectors ${\bf {r}}=\left(r_{1},r_{2},r_{3}\right)$ and ${\bf {\omega }}=\left(V_{1},V_{2},\omega _{0}\right)$ This slightly generalizes our previous result: for a pure state using the Heisenberg equations of motion we had shown that the spin-$1/2$ and the corresponding magnetic moment obey

${\frac {d{\bf {S}}}{dt}}={\bf {\omega }}\times {\bf {S}}$

but we have now generalized this result to mixed states, i.e. to ensemble averages that have less than the maximum possible magnetic moment. Note that Hamiltonian evolution does not change the purity of a state: a pure state always remains pure, no matter how violently it is rotated, and a mixed state retains its degree of mixedness ($Tr\rho ^{2}(t)={\text{const.}}$) Here we can check this explicitly by noting that

Failed to parse (unknown function "\lvert"): {\displaystyle Tr \rho^2 =\frac{1}{2}\left(r_0^2+r_1^2+r_2^2+r_3^2\right) =\frac{1}{2}\left(r_0^2+ \lvert {{\bf{r}} \rvert} ^2\right) }

Since ${\dot {\bf {r}}}\perp {\bf {r}}$, the length of ${\bf {r}}$ does not change, and hence $Tr\rho ^{2}$ is constant in time. \putbib