Rabi Oscillation

Rabi Oscillation

In quantum mechanics, Rabi Oscillation describes the oscillation within the two energy level in a two-level system, which is driven by an external electrical field. Let's delve deeper into it.

1. The problem begins when a two-energy system interacts with light

In the two-level system of an atom, the electron has two energy level. We refer to the lower level as \(\ket{g}\), and the upper level as \(\ket{e}\). The two level has the energy \(E_g\) and \(E_e\) respectively. We further define the Atom Frequency \(\Omega\) as \(\hbar \Omega = E_e - E_g\). The free hamiltonian could be written in the matrix form as:

\[H_0= \begin{pmatrix} E_e & \\ & E_g \end{pmatrix} = \begin{pmatrix} \frac{E_e+E_g}{2} + \frac 1 2 \hbar \Omega & \\ & \frac{E_e+E_g}{2} - \frac 1 2 \hbar \Omega \end{pmatrix}= \frac{E_e+E_g}{2} 1_2+\frac 1 2 \hbar \Omega \sigma_z \rightarrow H_0 = \frac 1 2 \hbar \Omega \sigma_z \]

where we have define the energy zero point exactly as the energy average of these two levels. Another way to understand it is we could simply define \(E_g=-\frac 1 2 \hbar \Omega\) and \(E_e=+\frac 1 2 \hbar \Omega\).

Next, a light is introduced to interact with this simple 'harmless' two-level energy. We model the light as single mode electric field:

\[\vec{E}(t)=\vec{E_0} \cos(\omega t + \varphi) \]

The interaction between the atom and the electric field could be characterized as \(V=-\vec{d}\cdot\vec{E}\).

Let's review the dipole moment of an atom. The dipole moment operator \(\vec{d}\) has the following form in the two-level system:

\[\vec{d}=e\vec{r}=e \begin{pmatrix} \braket{e|\vec{r}|e} & \braket{e|\vec{r}|g} \\ \braket{g|\vec{r}|e} & \braket{g|\vec{r}|g} \end{pmatrix} =\vec{d_{eg}}\ket{e}\bra{g} + \vec{d_{ge}}\ket{g}\bra{e} \]

where we has used the fact that diagonal terms are zero (\(\braket{e|\vec{r}|e}\)=0, \(\braket{g|\vec{r}|g}\)=0). This could be proved: 1. The operator \(\vec{r}\) has odd parity. 2. Assuming the potential in the atom solely depends on the distance \(r\) ( remembers \(\hat{\pi}|r|\hat{\pi}^\dagger=|r|\)), the electrons obeys parity invariant hamiltonian. 3. The system is non-degenerated, therefore the electron state ket has certain parity. 4. The parity selection tells us that when for a odd parity operator \(\vec{r}\), term \(\braket{a|\vec{r}|b}\) could have non-zero result only when \(\bra{a}\) and \(\ket{b}\) do not share the same parity.

To be simple, we assume \(\vec{d_{eg}} = \vec{d_{ge}} = \vec{\mu}\), which lead to:

\[\vec{d} = \vec{\mu} \sigma_x \]

Now, the Interaction Hamiltonian V could be expressed as:

\[\begin{equation} \begin{aligned} V&=-\vec{d}\cdot\vec{E} \\ &= -\vec{\mu} \cdot \vec{E}_0 \cos \left( \omega t + \varphi \right)\sigma_x \\ &= g \cos \left(\omega t + \varphi \right) \sigma_x \end{aligned} \end{equation} \]

, where the we have defined the coefficient g as the strength of the coupling between the atom and the electric field. Further, we will find that Rabi Frequency \(\Omega_R\) is \(\Omega_g = \frac{g}{2 \hbar}\), being the oscillation frequency between the upper and lower energy level when the frequencies of the electric field and the atom energy level are the same (\(\Omega = \omega\)).

For now, we obtained the complete hamiltonian that describe a two-level atom with frequency \(\Omega\) interacting with external electric field of frequency \(\omega\), as follow:

\[H = H_0 + V =\frac 1 2 \hbar \Omega \sigma_z + g \cos \left(\omega t + \varphi \right) \sigma_x \]

Our task to solve the Schrodinger equation (if possible):

\[i\hbar \frac{\partial}{\partial t} \Phi = H \Phi \]

Unfortunately, such hamiltonian is hard to crack.

2. Interaction Picture

It is natural to choose the interaction picture for solving such time-dependent perturbating hamiltonian. We set the free hamiltonian as

\[H_0 = \frac 1 2 \hbar \omega \sigma_z \]

Noting that in the interaction picture, we designed the free hamiltonian to be related to the field frequency \(\omega\). This aims to cancel the oscillating term (which frequency is also \(\omega\)) in the remaining hamiltonian.

Define the Detuned Frequency \(\Delta\) to be \(\Delta = \Omega - \omega\). The effective hamiltonian is

\[V^I = \exp(+\frac i \hbar H_0 t) V exp (-\frac i \hbar H_0 t) = \exp(+\frac i \hbar H_0 t) \left[ \frac 1 2 \hbar \Delta \sigma_z + g \cos (\omega t + \varphi) (\sigma^+ + \sigma^-)\right] exp (-\frac i \hbar H_0 t) \]

Using the Baker-Hausdorff lemma (\(e^{\xi A} B e^{-\xi A} = B + \xi [A,B] + \frac{\xi^2}{2!} [A,[A,B]] + \dots\)), the following equations could be obtained:

\[\begin{equation} \begin{aligned} \exp(i\frac 1 2 \omega \sigma_z t) \sigma_z \exp(-i\frac 1 2 \omega \sigma_z t) &= \sigma_z \\ \exp(i\frac 1 2 \omega \sigma_z t) \sigma_+ \exp(-i\frac 1 2 \omega \sigma_z t) &= \sigma_+ \exp(i\omega t) \\ \exp(i\frac 1 2 \omega \sigma_z t) \sigma_- \exp(-i\frac 1 2 \omega \sigma_z t) &= \sigma_- \exp(-i\omega t) \end{aligned} \end{equation} \]

Therefore, the effective hamiltonian under the interaction picture is:

\[H_I= V_I(t) = \frac 1 2 \hbar \Delta \sigma_z + g \cos (\omega t + \varphi) \left( \sigma_+ \exp(i\omega t) + \sigma_- \exp(-i\omega t) \right) \]

Not quite helping ... right?

3 Rotating Wave Approximation

In quantum optics, we often encounter a method called Rotation Wave Approximation (RWA), which refers to the removal of the fast oscillating terms.

\[\int e^{\pm in\theta} \text{d} \theta = \frac{e^{\pm in \theta}}{\pm in} \approx 0 \quad \text{(when n is large)} \]

We find that the \(\cos(\theta)\) term could be decomposed as \(\cos(\theta) = \frac 1 2 (e^{i\theta} + e^{-i\theta})\).

\[\begin{equation} \begin{aligned} H_I &= \frac 1 2 \hbar \Delta \sigma_z + g \cos (\omega t + \varphi) \left( \sigma_+ \exp(i\omega t) + \sigma_- \exp(-i\omega t) \right) \\ &= \frac 1 2 \hbar \Delta \sigma_z + \frac 1 2 g \left(\exp(i(\omega t + \varphi)) + \exp(-i(\omega t + \varphi) \right) \left( \sigma_+ \exp(i\omega t) + \sigma_- \exp(-i\omega t) \right) \\ &= \frac 1 2 \hbar \Delta \sigma_z + \frac 1 2 g \Bigl[\sigma_+ \Bigl(e^{i(2\omega t + \varphi)} + e^{-i\varphi}\Bigr) + \sigma_- \Bigl(e^{-i(2\omega t + \varphi)} + e^{i\varphi}\Bigr) \Bigr] \end{aligned} \end{equation} \]

According to the rotating wave approximation, the fast oscillation term with frequency \(2\omega\) could be canceled and only remain the constant part! Therefore, by applying rotating wave approximation, we obtain our effective hamiltonian as

\[H_I = \frac 1 2 \hbar \Delta \sigma_z + \frac 1 2 g \bigl( \sigma_+ e^{-i\varphi} + \sigma_- e^{i\varphi}\bigr) \]

Mysteriously enough, this hamiltonian could be solved easily.

4 Preparation

One final stop before the solution! We are going to prove one important identity:

\[\exp [ -i \vec{n} \cdot \vec{\sigma}] = \cos(|n|) 1_2 - i \sin(|n|)\Bigl(\frac{\vec{n}\cdot\vec{\sigma}}{|n|}\Bigr) \tag{1} \]

Proof:

The key property regarding Puali matrices is that for unit vector \(\hat{n}\), we have \(\left(\hat{n}\cdot\vec{\sigma}\right)^n=0\text{for n even} and 1\)

\[\left(\hat{n}\cdot\vec{\sigma}\right)^n= \left\{ \begin{aligned} &1 \quad & \text{for even n} \\ &\hat{n} \cdot \vec{\sigma} \quad & \text{for odd n} \end{aligned} \right. \]

Then,

\[\begin{equation} \begin{aligned} \exp[-i\vec{n}\cdot\vec{\sigma}] = & \Bigl[ 1 - \frac{|n|^2}{2!} (\hat{n}\cdot \vec{\sigma})^2 + \frac{|n|^4}{4!} (\hat{n} \cdot \vec{\sigma})^4 + \dots \Bigr] \\ & - i \Bigl[ |n| (\hat{n}\cdot\vec{\sigma}) - \frac{|n|^3}{3!} (\hat{n}\cdot\vec{\sigma})^3 + \dots \Bigr] \\ = & \cos (|n|) 1_2 - i sin(|n|) \Bigl(\frac{\vec{n} \cdot \vec{\sigma}}{|n|} \Bigr) \end{aligned} \end{equation} \]

Tada!

5 Solving the Schrodinger Equation - Resonance

We set the phase constant to be \(\varphi=0\). Otherwise, we could also make a unitary transformation (\(U_{12}=e^{i\varphi}\), \(U_{21}=e^{-i\varphi}\), a unitary matrix) to both the hamiltonian and the state ket. But that's not the big picture here.

Let's find out the evolution of the state under one additional intriguing condition: Resonance, \(\Omega = \omega\). In this context, the unduned term would would be zero \(\Delta = 0\). The Schrodinger equation is:

\[i \hbar \frac{\partial}{\partial t} \tilde{\Phi} = \frac 1 2 g \sigma_x \tilde{\Phi} \]

The solution is quite obvious:

\[\tilde{\Phi}= e^{-\frac i \hbar \frac 1 2 g \sigma_x t} \tilde{\Phi}(0) \]

By using Eq. (1), we have \(e^{- \frac i \hbar \frac 1 2 g \sigma_x t}=\cos |\frac{g}{2\hbar}| 1_2 -i\sin|\frac{g}{2\hbar}| \sigma_x\). Then:

\[\tilde{\Phi}(t) = \begin{pmatrix} \cos ( \frac{g}{2 \hbar} ) & -i \sin ( \frac{g}{2 \hbar} ) \\ -i \sin (\frac{g}{2 \hbar}) & \cos (\frac{g}{2 \hbar}) \end{pmatrix} \tilde{\Phi}(0) \]

Finally, we assume the initial state to be pure upper state \(\tilde{\Phi}(0)=\ket{e}\). Now we got the coefficient of the two level component at a time t for \(\tilde{\Phi}(t)\) as:

\[\tilde{\Phi}(t) = \begin{pmatrix} \cos ( \frac{g}{2 \hbar} ) & -i \sin ( \frac{g}{2 \hbar} ) \\ -i \sin (\frac{g}{2 \hbar}) & \cos (\frac{g}{2 \hbar}) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} =\begin{pmatrix} \cos ( \frac{g}{2 \hbar} ) \\ -i \sin ( \frac{g}{2 \hbar} ) \end{pmatrix} \]

The Rabi Frequency! Now we have proved that for an two-level atom interacting with a single mode electric field, the oscillation between two level is exactly the Rabi Frequency \(\Omega_g=\frac{g}{2\hbar}\).