Quantum Optics II Two-level system

In classical computing, information is processed in the form of binary digits or bits, which represent the smallest unit of information. Each bit can take the value 0 or 1, corresponding to two discrete states. When information processing is extended into the quantum mechanical domain, a two-level system, known as a quantum bit or qubit, can be identified. A qubit is based on two distinguishable quantum states. Therefore, it is essential to focus specifically on two-level systems.

Two-level System

Basic

The Hamiltonian of a two-level system is^[1]

$$H = H_0 + W$$

Let $\{\ket{\varphi_1},\ket{\varphi_2}\}$ be the eigenstates of $H_0$ with eigenvalues $E_1,E_2$, which forms the basis of state space. Suppose that $W$ has purely non-diagonal matrix elements, which describes the interaction between the two levels. The matrix representation of $H$ is then

$$H = \begin{bmatrix} E_1 & W_{12} \\ W_{21} & E_2 \end{bmatrix}$$

where $W_{12} = W_{21}^*$. Diagonalize the matrix $H$, we can find the following eigenvalues

$$E_+ = \frac{1}{2} (E_1 + E_2) + \frac{1}{2} \sqrt{(E_1 - E_2 )^2 + 4 |W_{12}|^2}$$

$$E_- = \frac{1}{2} (E_1 + E_2) - \frac{1}{2} \sqrt{(E_1 - E_2 )^2 + 4 |W_{12}|^2}$$

Introducing two parameters

$$E_m = \frac{1}{2} (E_1 + E_2),\quad \Delta = \frac{1}{2}(E_1 - E_2)$$

The eigenvalues can be written as

$$E_+ = E_m + \sqrt{\Delta^2 + |W_{12}|^2},\quad E_- = E_m -\sqrt{\Delta^2 + |W_{12}|^2}$$

The corresponding eigenvectors (not normalized) are

$$\ket{\mathrm{+}} = \frac{\Delta-\sqrt{\Delta^2 + |W_{12}|^2}}{W_{12}} \ket{\varphi_0} + \ket{\varphi_1},\quad \ket{\mathrm{-}} = \frac{\Delta+\sqrt{\Delta^2 + |W_{12}|^2}}{W_{12}} \ket{\varphi_0} + \ket{\varphi_1}$$

Dynamics

Assume that the system was initially in the state $\ket{\psi(0)} = \ket{\varphi_1}$ at $t=0$ . The probability to find the system in the state $\ket{\varphi_2}$ at time $t$ is

$$\mathcal{P}_{12}(t) = \braket{\varphi_2|\psi(t)} =\frac{|W_{12}|^2}{|W_{12}|^2+\Delta^2} \sin^2 \left(\sqrt{|W_{12}|^2+\Delta^2} \frac{t}{\hbar}\right) = \frac{|W_{12}|^2}{|W_{12}|^2+\Delta^2} \sin^2 \left(\frac{E_+ - E_-}{2 \hbar} t\right)$$

This expression shows that the probability $\mathcal{P}_{12}(t)$ oscillates over time with a frequency of $\Omega = \sqrt{|W_{12}|^2+\Delta^2} / \hbar = (E_+ - E_-) / 2 \hbar$ and an amplitude of $\frac{|W_{12}|^2}{|W_{12}|^2+\Delta^2}$.

/* Note: If $E_1 = E_2$, which means $\Delta = 0$, the amplitute of oscilation will be $1$. There exists some certain times when all particles transmit from $\ket{\varphi_1}$ to $\ket{\varphi_2}$. */

Dipole Interaction

Atom Field Interaction

We begin our treatment with a general description of the atom-field interaction^[2]. We will assume the field is monochromatic and linear polarized with angular frequency $\omega$

$$\boldsymbol{E} = E_0 \cos \omega t \,\hat{\mathbf{e}}$$

Here, $\hat{e}$ is the unit polarization vector of the field.

/* Note: We are ignoring the spatial dependence of the field, only writing down the field at the location of the atom. This is appropriate in the dipole approximation or long-wavelength approximation, where we assume that the wavelength of the field is much longer than the size of the atom, so that we can neglect any variations of the field over the extent of the atom. */

The atomic free-evolution Hamiltonian is given by

$$H_\text{atom} = \frac{1}{2} \hbar \omega_0 \sigma_z$$

The atom-field interaction Hamiltonian is

$$H_\text{int} = - \boldsymbol{D} \cdot \boldsymbol{E}$$

where $\boldsymbol{D} = -e \boldsymbol{R}$ is the electric dipole moment of an atom, $\boldsymbol{R}$ is the position vector pointing from the nuclei to the electron. Using parity, we can easily prove that $\braket{g|\boldsymbol{D}|g} = \braket{e|\boldsymbol{D}|e} = 0$. Therefore, in the two dimensional state space, the dipole moment operator can be decomposed into

$$\boldsymbol{D} = \braket{g|\boldsymbol{D}|e} \ket{g}\bra{e} + \braket{e|\boldsymbol{D}|g} \ket{e}\bra{g}$$

We can choose the phase of the dipole matrix element $\braket{g|\boldsymbol{D}|e}$ such that it is real (using Clebsch-Gordan), in which case the expression we can write the dipole moment operator as

$$\boldsymbol{D} = \braket{g|\boldsymbol{D}|e} (\sigma^\dagger + \sigma)$$

The atom-field interaction Hamiltonian becomes

$$H_\text{int} = -\braket{g|\boldsymbol{D}\cdot \hat{\mathbf{e}} |e}E_0 (\sigma^\dagger + \sigma) \cos \omega t$$

Defining Rabi frequency $\Omega_0 = -\braket{g|\boldsymbol{D}\cdot \hat{\mathbf{e}} |e}E_0/\hbar$, the Hamiltonian now becomes

$$H_\text{int} = \hbar \Omega_0 \cos\omega t\, (\sigma^\dagger + \sigma) = \frac{\hbar \Omega_0}{2} (\sigma^\dagger + \sigma)(\mathrm{e}^{\mathrm{i}\omega t} + \mathrm{e}^{-\mathrm{i}\omega t})$$

The total Hamiltonian is

$$H = H_\text{atom} + H_\text{int} = \frac{1}{2} \hbar \omega_0 \sigma_z + \frac{\hbar \Omega_0}{2} (\mathrm{e}^{\mathrm{i}\omega t} + \mathrm{e}^{-\mathrm{i}\omega t}) (\sigma^\dagger + \sigma)$$

where $\sigma^\dagger$ is the raising operator $\ket{e}\bra{g}$ and $\sigma$ is the lowering operator $\ket{g}\bra{e}$, $\sigma_x = \sigma^\dagger + \sigma$.

Rabi Oscillation

The Hamiltonian of a two-level system is

$$H = \frac{1}{2} \hbar \omega_0 \sigma_z + \frac{1}{2}\hbar \Omega_0 (\mathrm{e}^{\mathrm{i}\omega t} + \mathrm{e}^{-\mathrm{i}\omega t}) (\sigma^\dagger + \sigma)$$

Performing the following rotating frame transformation

$$U=\exp \left( \frac{\mathrm{i}}{2} \omega t \sigma_z \right)$$

The Hamiltonian in the rotating frame will become

$$\tilde{H} = UHU^\dagger+\mathrm{i}\hbar \dot{U} U^\dagger = -\frac{1}{2}\hbar\delta \sigma_z + \frac{1}{2} \hbar \Omega_0 \sigma^\dagger (1+ \mathrm{e}^{\mathrm{i} 2\omega t}) + \frac{1}{2} \hbar \Omega_0 \sigma (1+ \mathrm{e}^{-\mathrm{i} 2\omega t})$$

where $\delta := \omega - \omega_0$ is the detuning of laser frequency. Since $\mathrm{e}^{\mathrm{i} 2\omega t}$ and $\mathrm{e}^{-\mathrm{i} 2\omega t}$ are rapid oscillating terms, we can omit them. This is called rotating-wave approximation (RWA). RWA requires $|\omega_0 - \omega| \ll \omega_0 +\omega$. In this case, the Hamiltonian becomes

$$\tilde{H}_\text{RWA} = -\frac{1}{2}\hbar\delta \sigma_z + \frac{1}{2} \hbar \Omega_0 (\sigma^\dagger + \sigma)$$

We can actually make the rotating-wave approximation from the very beginning. The electric field and the dipole moment can be decomposed into the positive-frequency part and the negative frequency part:

$$\boldsymbol{E} = \boldsymbol{E}^+ +\boldsymbol{E}^- = \frac{E_0}{2} \mathrm{e}^{-\mathrm{i}\omega t}\,\hat{\mathbf{e}} + \frac{E_0}{2} \mathrm{e}^{\mathrm{i}\omega t}\,\hat{\mathbf{e}}$$

$$\boldsymbol{D} = \boldsymbol{D}^+ + \boldsymbol{D}^- = \braket{g|\boldsymbol{D}|e} \sigma + \braket{g|\boldsymbol{D}|e} \sigma^\dagger$$

Only $\boldsymbol{E}^+ \cdot \boldsymbol{D}^-$ and $\boldsymbol{E}^- \cdot \boldsymbol{D}^+$ are on-resonant, which means the effective terms in the Hamiltonian remain

$$H_\text{RWA} = \frac{1}{2} \hbar \omega_0 \sigma_z + \frac{1}{2}\hbar \Omega_0 (\sigma^\dagger \mathrm{e}^{-\mathrm{i}\omega t}+\sigma\mathrm{e}^{\mathrm{i}\omega t})$$

If we directly perform rotating frame transformation on $H_\text{RWA}$, we will immediately get

$$\tilde{H}_\text{RWA} = -\frac{1}{2}\hbar\delta \sigma_z + \frac{1}{2} \hbar \Omega_0 (\sigma^\dagger + \sigma)$$

We find that the Hamiltonian in the rotating frame $\tilde{H}$ is time-independent, which enables us to use time evolution operator to depict the evolved state at any time

$$\tilde{H} = -\frac{1}{2}\hbar\delta \sigma_z + \frac{1}{2} \hbar \Omega_0 (\sigma^\dagger + \sigma) = \begin{bmatrix} -\frac{1}{2}\hbar \delta & \frac{1}{2}\hbar \Omega_0 \\ \frac{1}{2}\hbar \Omega_0 & \frac{1}{2}\hbar \delta \end{bmatrix}$$

$$\tilde{U} = \exp\left( - \frac{\mathrm{i}}{\hbar} \tilde{H} t\right) = \begin{bmatrix} \cos\frac{\Omega t}{2} + \mathrm{i} \frac{\delta}{\Omega} \sin\frac{\Omega t}{2} & -\mathrm{i} \sin\frac{\Omega t}{2} \frac{\Omega_0}{\Omega} \\ -\mathrm{i} \sin\frac{\Omega t}{2} \frac{\Omega_0}{\Omega} & \cos\frac{\Omega t}{2} - \mathrm{i} \frac{\delta}{\Omega} \sin\frac{\Omega t}{2} \end{bmatrix}$$

For an atom initialized in the ground state $\ket{g}$, the final state in the rotating frame is

$$\ket{\tilde{\psi}(t)} = \tilde{U} \ket{\tilde{\psi}(0)} = \tilde{U} \ket{g}$$

switching back to the lab frame (Schrödinger picture)

$$\ket{\psi(t)} = \mathrm{e}^{-\frac{\mathrm{i}}{2}\omega t\sigma_z} \ket{\tilde{\psi}(t)} = \mathrm{e}^{-\frac{\mathrm{i}}{2}\omega t\sigma_z} \tilde{U} \ket{g} = \begin{bmatrix} -\mathrm{i} \sin\frac{\Omega t}{2} \frac{\Omega_0}{\Omega} \mathrm{e}^{-\frac{\mathrm{i}}{2}\omega t} \\ \mathrm{e}^{\frac{\mathrm{i}}{2}\omega t} \left( \cos\frac{\Omega t}{2} - \mathrm{i} \frac{\delta}{\Omega} \sin\frac{\Omega t}{2} \right) \end{bmatrix}$$

where $\Omega = \sqrt{\Omega_0^2 + \delta^2}$ is called the generalized Rabi frequency. The probability to find the atom in the excited state at time $t$ is

$$|\langle e| \psi(t)\rangle|^2 = \frac{\Omega_0^2}{\Omega^2} \sin^2\frac{\Omega t}{2} = \frac{\Omega_0^2}{\Omega_0^2+\Delta^2} \sin^2\frac{\sqrt{\Omega_0^2+\Delta^2} t}{2}$$

Dressed State

As long as we continuously shining the laser on the atom, the atom and the electric field can keep interacting and form a stable coupled system. The atom-field as a whole is now fully described by $\tilde{H}$. By diagonalizing $\tilde{H}$, we can find the new eigenenergies and eigenstates, which are so-called dressed states.

Still start with

$$\tilde{H} = -\frac{1}{2}\hbar\delta \sigma_z + \frac{1}{2} \hbar \Omega_0 \sigma_x = \hbar \begin{bmatrix} -\delta/2 & \Omega_0/2 \\ \Omega_0/2 & \delta/2 \end{bmatrix}$$

Diagonalize it we get

$$E_+ = \frac{\hbar}{2}\sqrt{\Omega_0^2+\delta^2}= \frac{\hbar\Omega}{2},\quad\ket{+}=\begin{bmatrix} \cos(\theta/2)\\ \sin(\theta/2) \end{bmatrix}$$

$$E_- = -\frac{\hbar}{2}\sqrt{\Omega_0^2+\delta^2}= -\frac{\hbar\Omega}{2} ,\quad \ket{-}=\begin{bmatrix} -\sin(\theta/2) \\ \cos(\theta/2) \end{bmatrix}$$

in which $\tan \theta = -\Omega_0/\delta ,0\le\theta < \pi$ is called the Stückelberg angle, $\Omega = \sqrt{\Omega_0^2+\delta^2}$ is the generalized Rabi frequency.

/* Note: The sign convention and domain of Stückelberg angle are important! */

Free-evolution

When there is no external light field, the energy levels of the ground state and the excited state (in the rotating frame) are

$$E_e = -\frac{\hbar\delta}{2},\quad E_g = \frac{\hbar\delta}{2}$$

which coincides with the classical occasion: angular frequency of rotation is changed and part of energy is removed.

On-resonance

When the laser is on-resonant, $\delta = 0$. The energy level of dressed states are

$$E_+ = \frac{1}{2}\hbar\Omega_0,\quad E_- =-\frac{1}{2}\hbar\Omega_0$$

Meanwhile, $\theta = \pi/2$, thus

$$\ket{+} = \frac{1}{\sqrt{2}} (\ket{e}+\ket{g}),\quad \ket{-} = \frac{1}{\sqrt{2}} (\ket{e}-\ket{g})$$

Far-off-resonance

For large detuning $|\delta| \gg \Omega_0$, we have^[3]

• $\delta < 0$, red detuned, $\theta \to 0$, $\ket{+} = \ket{e}$ and $\ket{-}=\ket{g}$, thus

  $$E_+ = E_e \approx -\frac{\hbar \delta}{2} - \frac{\hbar \Omega_0^2}{4\delta},\quad E_- =E_g \approx \frac{\hbar\delta}{2} + \frac{\hbar \Omega_0^2}{4\delta}$$

• $\delta >0$, blue detuned, $\theta \to \pi$, $\ket{+} = \ket{g}$ and $\ket{-}= -\ket{e}$, thus

  $$E_+ = E_g \approx \frac{\hbar\delta}{2} + \frac{\hbar \Omega_0^2}{4\delta},\quad E_- = E_e \approx -\frac{\hbar \delta}{2} - \frac{\hbar \Omega_0^2}{4\delta}$$

We can conclude that, when the detuning reverses its sign, although the mapping between $\ket{+},\ket{-}$ and $\ket{e},\ket{g}$ is swapped, the expressions for the corrected energy level of the ground state and the excited state remain unchanged.

Moreover, we find a unified energy shift formula

$$\Delta E_g = \frac{\hbar \Omega_0^2}{4\delta}$$

This is crucial for building an optical dipole trap.

Autler-Townes Doublet

Consider a three-level atom with bare states $\ket{0},\ket{1}$ and $\ket{2}$. State $\ket{1}, \ket{2}$ are strongly driven by a near-resonant laser^[4]. As a result, the energy of the upper level $\ket{2}$ is pulled down by the light field, and the atom and field as a whole form two new dressed states $\ket{+}, \ket{-}$ with energy splitting $\Omega$, replacing the old energy levels $\ket{1}$ and $\ket{2}$.

Now we apply a second weaker laser to probe the transition between the state $\ket{0}$ and the state $\ket{1}$. As the frequency of the probe laser is swept through resonance with the bare transition frequency $\omega_0$, the absorption spectrum obtained exhibits two peaks separated by the generalized Rabi frequency $\Omega$ associated with transitions between the bare states $\ket{1}$ and $\ket{2}$, as shown in the following figure. The positions of these peaks differ from the bare transition frequency $\omega_0$ by $\pm\Omega/2$. In other words, the weaker laser actually detected the transition of $\ket{0}\leftrightarrow\ket{+}$ and $\ket{0}\leftrightarrow\ket{-}$!

In 1955, Autler and Townes showed the spectrum discussed above utilizing a microwave transition of the OCS molecule^[5]. The corresponding doublet is called the Autler-Townes doublet, or the dynamic Stark splitting.

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1. Cohen-Tannoudji C, Diu B, Franck Laloë. Quantum Mechanics, Volume 1: Basic Concepts, Tools, and Applications[M]. 2. Wiley VCH, 2019: 411-422. ↩︎

2. Daniel A. Steck, Quantum and Atom Optics, available online at http://steck.us/teaching (revision 0.16.2, 15 November 2024). ↩︎

3. Grynberg, G., Aspect, A., & Fabre, C. (2010). Introduction to Quantum Optics: From the Semi-classical Approach to Quantized Light. Cambridge: Cambridge University Press. ↩︎

4. Barnett S M, Radmore P M. Methods in Theoretical Quantum Optics[M]. Oxford University Press, 2002 :182-221. ↩︎

5. Autler, S. H., & Townes, C. H. (1955). Stark Effect in Rapidly Varying Fields. Phys. Rev., 100(2), 703–722. https://doi.org/10.1103/PhysRev.100.703 ↩︎