Quantum Optics III Two-level system

In a calculation on a digital computer, information is processed in the form of so-called binary digits or bits, representing its smallest unit. Each bit can have the value 0 or 1, corresponding to two discrete states. Moving the processing of information itself into the quantum mechanical domain, one can identify a two-level system now called quantum bit, or qubit that is based on two distinguishable quantum states. Therefore, it is necessary to look on two-level system specifically.

Time-independent interaction in two-level system

Static aspect

The Hamiltonian is given by

$$H = H_0 + W$$

Taking the eigenstates $\{\ket{\varphi_1},\ket{\varphi_2}\}$ of $H_0$ with eigenenergy $E_1,E_2$ as the basis of state space. Suppose $W$ possesses purely non-diagonal matrix element, which means that $W$ only describes the coupling between two levels, $W_{11}=W_{22}=0$. The matrix representing $H$ is written

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

in which $W_{12} = W_{21}^*$. The diagonalization of matrix $H$ presents no problems. We find the eigenvalues

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

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

Introduce two parameters

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

The eigenvalues now are written

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

with eigenvectors (not normalized)

$$\begin{gathered} \ket{\mathrm{+}} = \frac{\Delta-\sqrt{\Delta^2 + |W_{12}|^2}}{W_{12}} \ket{\varphi_0} + \ket{\varphi_1} \\ \ket{\mathrm{-}} = \frac{\Delta+\sqrt{\Delta^2 + |W_{12}|^2}}{W_{12}} \ket{\varphi_0} + \ket{\varphi_1} \end{gathered}$$

Dynamic aspect

Assume that the system at time $t=0$ in the state $\ket{\psi(0)} = \ket{\varphi_1}$ . The probability amplitude of finding the system at time $t$ in the state $\ket{\varphi_2}$ is then written

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

This relation shows that the probability $\mathcal{P}_{12}(t)$ oscillates over time with a frequency of $\Omega_\text{rabi} = \sqrt{|W_{12}|^2+\Delta^2} / \hbar = (E_+ - E_-) / 2 \hbar$ and an amplitude of $\frac{|W_{12}|^2}{|W_{12}|^2+\Delta^2}$ . The generalized Rabi frequency $\Omega_\text{rabi}$ is the frequency at which the probability amplitudes of two atomic energy levels fluctuate in an oscillating electromagnetic field.

/* 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 $\ket{\varphi_2}$. */

Rabi Oscillation

In this section we shall calculate exactly the probability of a transition between two atomic states $\ket{g}$ and $\ket{e}$ driven by a quasi-resonant wave of angular frequency $\omega$ similar to the Bohr frequency of the transition $\omega_0 = (E_e - E_g) / \hbar$ .

Schrodinger Equation

We shall set to zero the energy $E_g$ of the lower state and denote by $\omega_0$ the atomic Bohr frequency

$$H_0 = \begin{bmatrix} 0 & 0 \\ 0 & \hbar \omega_0 \end{bmatrix}$$

We employ the electric dipole Hamiltonian of $W_{DE} = - \boldsymbol{D} \cdot \boldsymbol{E}$ , where $\boldsymbol{E} = \boldsymbol{E_0} \cos (\omega t + \varphi)$. We denote

$$W_{eg} = -\braket{e|\boldsymbol{D} \cdot \boldsymbol{E}_0 | g} = \hbar \Omega_1$$

in which $\Omega_1$ is called Rabi frequency. The complete Hamiltonian reads

$$H = \begin{bmatrix} 0 & \hbar \Omega_1 \cos (\omega t + \varphi) \\ \hbar \Omega_1 \cos (\omega t + \varphi) & \hbar \omega_0 \end{bmatrix}$$

Suppose $\ket{\psi(t)} = c_g(t) \ket{g} + c_e(t) \mathrm{e}^{-\mathrm{i}\omega_0 t} \ket{e}$ . Schrodinger equation reads

$$\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} t} \ket{\psi(t)} = H \ket{\psi(t)}$$

$$\mathrm{i}\frac{\mathrm{d} c_g}{\mathrm{d} t} = \left[ \frac{\Omega_1 \mathrm{e}^{\mathrm{i}\varphi}}{2} \mathrm{e}^{\mathrm{i}(\omega-\omega_0)t} + \frac{\Omega_1 \mathrm{e}^{-\mathrm{i}\varphi}}{2} \mathrm{e}^{-\mathrm{i}(\omega+\omega_0)t} \right] c_e$$

$$\mathrm{i}\frac{\mathrm{d} c_e}{\mathrm{d} t} = \left[ \frac{\Omega_1 \mathrm{e}^{-\mathrm{i}\varphi}}{2} \mathrm{e}^{-\mathrm{i}(\omega-\omega_0)t} + \frac{\Omega_1 \mathrm{e}^{\mathrm{i}\varphi}}{2} \mathrm{e}^{\mathrm{i}(\omega+\omega_0)t} \right] c_g$$

in which we decompose $\cos (\omega t + \varphi)$ into $\frac{1}{2} [\mathrm{e}^{\mathrm{i}(\omega t + \varphi)} + \mathrm{e}^{-\mathrm{i}(\omega t + \varphi)}]$ . Applying rotating wave approximation, $|\omega + \omega_0|$ terms are negligible, we obtain

$$\mathrm{i}\frac{\mathrm{d} c_g}{\mathrm{d} t} = \frac{\Omega_1 \mathrm{e}^{\mathrm{i}\varphi}}{2} \mathrm{e}^{\mathrm{i}(\omega-\omega_0)t}c_e$$

$$\mathrm{i}\frac{\mathrm{d} c_e}{\mathrm{d} t} = \frac{\Omega_1 \mathrm{e}^{-\mathrm{i}\varphi}}{2} \mathrm{e}^{-\mathrm{i}(\omega-\omega_0)t} c_g$$

Corollary : Under rotating wave approximation, the Hamiltonian could be seen as

$$H = \hbar \begin{bmatrix} 0 & \frac{\Omega_1}{2} \mathrm{e}^{\mathrm{i} (\omega t + \varphi)} \\ \frac{\Omega_1}{2} \mathrm{e}^{-\mathrm{i} (\omega t + \varphi) } & \omega_0 \end{bmatrix}$$

Hence we conventionally write the Hamiltonian as

$$H = \hbar \omega_0 |e\rangle\langle e| + \frac{\hbar \Omega_1}{2} \mathrm{e}^{\mathrm{i} (\omega t + \varphi)}|g\rangle\langle e| + \frac{\hbar \Omega_1}{2} \mathrm{e}^{-\mathrm{i} (\omega t + \varphi)}|e\rangle\langle g| = \hbar \omega_0 |e\rangle\langle e| + \frac{\hbar \Omega_1}{2} \left[\sigma \mathrm{e}^{\mathrm{i} (\omega t + \varphi)} + \sigma^\dagger\mathrm{e}^{-\mathrm{i} (\omega t + \varphi)} \right]$$

in which $\sigma:= |g\rangle\langle e|$ is the atomic lowering operator.

Define the detuning frequency from resonance $\delta := \omega - \omega_0$ and perform a transformation

$$\tilde{c}_g = c_g \mathrm{e}^{-\mathrm{i} \delta t/2} \qquad \tilde{c}_e = c_e \mathrm{e}^{\mathrm{i} \delta t/2}$$

The Schrodinger equations will become first order linear ODE with constant coefficients

$$\mathrm{i}\frac{\mathrm{d} \tilde{c}_g}{\mathrm{d} t} = \frac{\delta}{2} \tilde{c}_g + \frac{\Omega_1 \mathrm{e}^{\mathrm{i}\varphi}}{2} \tilde{c}_e$$

$$\mathrm{i}\frac{\mathrm{d} \tilde{c}_e}{\mathrm{d} t} = \frac{\Omega_1 \mathrm{e}^{-\mathrm{i}\varphi}}{2} \tilde{c}_g-\frac{\delta}{2} \tilde{c}_e$$

whose solution is the superposition of the eigenvector of coefficient matrix, multiplied by the exponential of the corresponding eigenvalues of the matrix. /* Hint: $\mathrm{i}\frac{\mathrm{d} \boldsymbol{x}}{\mathrm{d} t} = M \boldsymbol{x} \implies \boldsymbol{x}=\mathrm{e}^{-\mathrm{i}\lambda_1 t} \boldsymbol{v}_1+\mathrm{e}^{-\mathrm{i}\lambda_2 t} \boldsymbol{v}_2$ */ Therefore, it is easy to find the eigenvalues

$$\lambda_\pm = \pm\frac{1}{2} \Omega_\text{rabi} = \pm \frac{1}{2} \sqrt{\Omega_1^2 + \delta^2}$$

in which $\Omega_\text{rabi} := \sqrt{\Omega_1^2 + \delta^2}$ are defined as the generalized Rabi frequency. The eigenstates of them are called dressed states.

Assume that the system at time $t=0$ in the state $\ket{\psi(0)} = \ket{g}$ . We will find

$$\mathcal{P}_{g \to e} (t,t_0) = \frac{\Omega_1^2}{\Omega_1^2 + \delta^2} \sin^2 \frac{\Omega_\text{rabi}}{2} (t-t_0)$$

Rotating Frame Transformation

Back to Rabi Oscillation. The Hamiltonian is

$$H = \hbar \begin{bmatrix} 0 & \frac{\Omega_1}{2} \mathrm{e}^{\mathrm{i} (\omega t + \varphi)} \\ \frac{\Omega_1}{2} \mathrm{e}^{-\mathrm{i} (\omega t + \varphi) } & \omega_0 \end{bmatrix}$$

Rotate it by

$$U = \begin{bmatrix} 1 & 0 \\ 0 & \mathrm{e}^{\mathrm{i}\omega t} \end{bmatrix}$$

The transformed Hamiltonian $\tilde{H}(t) = UHU^\dagger+\mathrm{i}\hbar \dot{U} U^\dagger$ will be

$$\tilde{H} = \hbar \begin{bmatrix} 0 & \frac{\Omega_1}{2} \\ \frac{\Omega_1}{2} & \omega_0 - \omega \end{bmatrix} = \hbar \begin{bmatrix} 0 & \frac{\Omega_1}{2} \\ \frac{\Omega_1}{2} & -\delta \end{bmatrix}$$

We can then immediately obtain the Rabi frequency $\Omega_\text{rabi} = \frac{1}{2} \sqrt{\Omega_1^2 + \delta^2}$ using the result of time-independent two-level system.

Corollary The effect of the AC electric field of light on a two-level system whose energy interval is $\hbar \omega_0$, is equivalent to the effect of the DC electric field on a two-energy system whose energy interval is $- \hbar \delta = \hbar (\omega_0-\omega)$.

/* Note: $U$ can be written in non-matrix form $U = \mathrm{e}^{\mathrm{i}\omega t |e\rangle\langle e|}$ */

Proof

$$\begin{aligned} U &= \mathrm{e}^{\mathrm{i}\omega t |e\rangle\langle e|} = I+\sum_{n=1}^\infty \frac{(\mathrm{i}\omega t)^n}{n!} (|e\rangle\langle e|)^n = I+\sum_{n=1}^\infty \frac{(\mathrm{i}\omega t)^n}{n!} |e\rangle\langle e| \\ &=I+(\mathrm{e}^{\mathrm{i}\omega t}-1) |e\rangle\langle e| = |g\rangle\langle g| + \mathrm{e}^{\mathrm{i}\omega t}|e\rangle\langle e| \end{aligned}$$

AC Stark shift

In spectroscopy, AC Stark effect, is a dynamical Stark effect corresponding to the case when an oscillating electric field (e.g. laser) is tuned in resonance (or close) to the transition frequency of a given spectral line, and resulting in a change of the shape of the absorption/emission spectra of that spectral line. The AC Stark effect was discovered in 1955 by American physicists Stanley Autler and Charles Townes.

The old energy levels are given by

$$E_g = -\delta/2, \quad E_e = \delta/2$$

The new energy levels are given by

$$E_+ = \frac{1}{2} \sqrt{\Omega_1^2 + \delta^2}, \quad E_- = -\frac{1}{2} \sqrt{\Omega_1^2 + \delta^2}$$

For large detuning $\delta \gg \Omega_1$, we have

$$E_+ = \frac{\Omega_1^2}{4\delta}, \quad E_- = -\delta - \frac{\Omega_1^2}{4\delta}$$

$$\ket{\mathrm{+}} = \frac{-\dfrac{\Omega_1}{2\delta} \ket{g} + \ket{e}}{\sqrt{1+\dfrac{\Omega_1^2}{4\delta^2}}},\quad \ket{\mathrm{-}} = \frac{\dfrac{2\delta}{\Omega_1} \ket{g} + \ket{e}}{\sqrt{1+\dfrac{\Omega_1^2}{4\delta^2}}}$$

The effective Hamiltonian is given by

$$H = E_+ \ket{+}\bra{+} + E_- \ket{-}\bra{-}$$