# 1.1.9. Physics Picture¶

## 1.1.9.1. Rabi oscillations¶

Hamiltonian of Rabi oscillation is

$\begin{split}H = -\frac{\omega_m}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} - A \cos(k t)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} .\end{split}$

The Hamiltonian we could solve is

$\begin{split}H &= -\frac{\omega_m}{2} \sigma_3 - \frac{A}{2} \begin{pmatrix}0 & e^{i k x} \\ e^{-i k x} & 0 \end{pmatrix} \\ & = -\frac{\omega_m}{2} \sigma_3 - \frac{A}{2} \cos(kx) \sigma_1 + \frac{A}{2} \sin (kx) \sigma_2 ,\end{split}$

which has a transition probability

$P(x) = \frac{\lvert A\rvert^2}{ \lvert A\rvert^2 + (k - \omega_m)^2 } \sin^2 \left( \sqrt{ \lvert A\rvert^2 + (k - \omega_m)^2 } x/2 \right).$

With two perturbations

$H' = -\frac{\omega_m}{2} \sigma_3 - \frac{1}{2} (A_1 \cos(k_1 x) + A_2 \cos(k_2 x)) \sigma_1 + \frac{1}{2}( A_1 \sin (k_1x) + A_2 \sin (k_2 x) ) \sigma_2.$

If $$k_1 \gg k_2$$, we can approximate by treating the slow rotating perturbation as a constant added to the energy gap, so that the new energy gap is shifted

$\omega_m' = \sqrt{ \omega_m^2 + A_2^2 },$

which could possibly shift the system out of resonance.

The best practice would be applying this to the different modes.

## 1.1.9.2. Oscillations and Modes¶

Using Jacobi-Anger expansion, for any system with Hamiltonian

$H = -\frac{\omega_m}{2} \sigma_3 + \frac{1}{2}\sum_n A_n \sin (k_n x) \cos 2\theta_m \sigma_3 - \frac{1}{2}\sum_n A_n \sin (k_n x) \sin 2\theta_m \sigma_1,$

we could rewrite the system into a composition of multiple Rabi oscillations

$\begin{split}H = -\frac{\omega_m}{2} \sigma_3 + \frac{1}{2} \sum_{n_1} \cdots \sum_{n_N} \begin{pmatrix} 0 & B_{n_1,\cdots,n_N} \Phi_{n_1,\cdots, n_N} e^{i \left( \sum_{a} n_a k_a \right)x} \\ B_{n_1,\cdots,n_N}^* \Phi_{n_1,\cdots, n_N}^* e^{-i \left( \sum_{a} n_a k_a \right)x} & 0 \end{pmatrix}.\end{split}$

For each mode, we have a Rabi oscillation

$\begin{split}H_{n_1,\cdots,n_N} = -\frac{\omega_m}{2} \sigma_3 + \frac{1}{2} \lvert B_{n_1,\cdots,n_N} \rvert \begin{pmatrix} 0 & e^{i \left( \sum_{a} n_a k_a \right)x} \\ e^{-i \left( \sum_{a} n_a k_a \right)x} & 0 \end{pmatrix},\end{split}$

where we have dropped $$\Phi_{n_1,\cdots, n_N}$$ and the possible sign and phase of $$B_{n_1,\cdots,n_N}$$ since these phase terms only determines the phase of the perturbation on xy plane.

To explain the interference, we explore the superposition of two modes,

$\begin{split}H \equiv -\frac{\omega_m}{2} \sigma_3 + \frac{1}{2} \lvert B_1 \rvert \begin{pmatrix} 0 & e^{i \left( \sum_{a} n_a k_a \right)x} \\ e^{-i \left( \sum_{a} n_a k_a \right)x} & 0 \end{pmatrix} + \frac{1}{2} \lvert B_2 \rvert \begin{pmatrix} 0 & e^{i \left( \sum_{a} n_a' k_a \right)x} \\ e^{-i \left( \sum_{a} n_a' k_a \right)x} & 0 \end{pmatrix},\end{split}$

which is composed of two Rabi oscillations. We choose the first mode to be the one close to resonance, i.e., $$\sum_a n_a k_a \sim \omega_m$$, while the second mode is far away from resonance.

For simplicity we use two perturbations, that is $$a=1,2$$. The Hamiltonian can be written as

$H = -\frac{\omega_m}{2} \sigma_3 + \frac{1}{2} ( \lvert B_1\rvert \cos(\phi_1 x) + \lvert B_1 \rvert \cos (\phi_2 x) )\sigma_1 - \frac{1}{2} ( \lvert B_1 \rvert \sin (\phi_1 x) + \lvert B_2 \rvert \sin(\phi_2x) ) \sigma_2,$

where we define $$\phi_1 = n_1 k_1 + n_2 k_2$$ and $$\phi_2 = n_1' k_1 + n_2' k_2$$. Using Pauli matrices are basis, this corresponds to a Hamilton vector

$\begin{split}\vec H = \begin{pmatrix} - \lvert B_1\rvert \cos(\phi_1 x) - \lvert B_2 \rvert \cos (\phi_2 x) \\ \lvert B_1 \rvert \sin (\phi_1 x) + \lvert B_2 \rvert \sin(\phi_2x) \\ \omega_m \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ \omega_m \end{pmatrix} + \begin{pmatrix} - \lvert B_1\rvert \cos(\phi_1 x) \\ \lvert B_1 \rvert \sin (\phi_1 x) \\ 0 \end{pmatrix} + \begin{pmatrix} - \lvert B_2 \rvert \cos (\phi_2 x) \\ \lvert B_2 \rvert \sin(\phi_2x) \\ 0 \end{pmatrix},\end{split}$

which has a z component and two rotating perturbations. We choose the system to be

$\begin{split}\phi_1 &\sim \omega_m \\ \phi_2 & \neq \omega_m.\end{split}$

We then have two different situations, $$\phi_2/\omega_m \gg 1$$ and $$\phi_2/\omega_m \ll 1$$.

### 1.1.9.2.1. Slow Perturbation¶

For $$\phi_2/\omega_m \ll 1$$, the second mode is a very slow rotating perturbation, which can be explained using the proposed theory. Fig. 1.36 Resonance interference of modes. The combination of the first mode and $$B_2=$$¶ Fig. 1.37 Compare with Rabi formula¶

As a test of the theory, we can calculate the ratio of each $$B_2$$, which depends on the modes, and the critical value $$B_2^C$$ which is the crtical value for the destruction of the resonance. Fig. 1.38 Ratio $$\lvert B_2\rvert/\lvert B_2^C \rvert$$¶

To summarize, in the modes view, resonance of some modes are destroyed by some certain modes.

## 1.1.9.3. Example of Full System¶

First we choose a system that is on resonance

$H_1 = -\frac{\omega_m}{2} \sigma_3 + \frac{\delta \lambda_1}{2} \cos 2\theta_m \sigma_3 - \frac{\delta \lambda_1}{2} \sin 2\theta_m \sigma_1,$

where $$\delta\lambda_1 = A_1 \sin (k_1 x)$$, where $$k_1 = \omega_m$$ and $$A_1 = 3.5\times 10^{-5}\omega_m$$. This sets the system to resonance.

Does the Diagonal Term Matter?

Removing the diagonal elements of the perturbation

$H_1' = -\frac{\omega_m}{2} \sigma_3 - \frac{\delta \lambda_1}{2} \sin 2\theta_m \sigma_1,$

will result in Fig. 1.40.

$\delta \lambda_2 = A_2 \sin (k_2 x),$

with

$\begin{split}A_2 &= 10^{-2},\\ k_2 &= 0.1.\end{split}$

Removing Diagonal Elements of Slow Perturbation

Removing the diagonal elements of slow perturbation

$H_2' = -\frac{\omega_m}{2} \sigma_3 + \frac{\delta \lambda_1 }{2} \cos 2\theta_m \sigma_3 - \frac{\delta \lambda_1 + \delta \lambda_2}{2} \sin 2\theta_m \sigma_1,$

gives us the result Fig. 1.42.

## 1.1.9.4. Explaination¶

Slow perturbation is slow and changes the energy gap of the system. Since the energy gap $$\omega_m$$ determines the resonance point, which is

$k_1 = \omega_m,$

adding the slow perturbation could increase $$\omega_m$$,

(1.27)$\begin{split}\omega_m' &= \sqrt{A_{2,\bot} ^2 + \omega_m^2} \\ & = \omega_m \sqrt{ \left(\frac{A_{2,\bot}}{\omega_m} \right)^2 + 1 } \\ & \approx \omega_m + \frac{A_{2,\bot}^2}{2\omega_m},\end{split}$

where $$A_{2,\bot}$$ is component perpendicular to z axis.

Only Perpendicular Component

In the calculation of the modified energy gap, we used only the perpendicular component of the new slow perturbation. This only holds for $$A_{2,\bot} \ll \omega_m$$.

PROOF

Shift The System Out of Resonance

Shift the system out of resonance, it is required that

$\lvert \omega_m' - k_1 \rvert \gtrsim \text{width of resonance} A_1.$

Width of resonance is basically determined by $$A_{1,\bot}$$. Apply equation (1.27), we can solve the condition to break the resonance,

$A_{2,\bot} \gtrsim \sqrt{2\omega_m A_{1,\bot}}.$

In our example, the condition becomes

$\begin{split}&A_2 \sin 2\theta_m \gtrsim \sqrt{2\omega_m A_1 \sin 2\theta_m} \\ \Rightarrow & A_2 \gtrsim \sqrt{2\omega_m A_1 \tan 2\theta_m/\cos 2\theta_m}.\end{split}$ Fig. 1.43 With $$A_2=\sqrt{2 A_1 \sin (2 \theta_m)}/ \cos ^2(2 \theta_m) =0.0190304\omega_m$$¶ Fig. 1.44 Compare to show destruction¶

Using Rabi formula the amplitudes are not matching the numerical calculations, Fig. 1.45.

As a reference, the Q values for each line are

$\begin{split}Q_1 & = \frac{\lvert k_1 - \sqrt{A_2 \sin^2(2\theta_m) + 1 } }{A_1\sin (2\theta_m)} = 1.11689, \\ Q_2 & = \frac{\lvert k_1 - \sqrt{A_2' \sin^2(2\theta_m) + 1 } }{A_1\sin (2\theta_m)} = 4.04469, \\ Q_3 & = \frac{\lvert k_1 - \sqrt{A_2'' \sin^2(2\theta_m) + 1 } }{A_1\sin (2\theta_m)} = 402.277.\end{split}$

However, the important question is whether the modified oscillation really Rabi oscillation. The answer is NO. Fig. 1.46 Is the oscillation with slow perturbation really Rabi oscillation? Upper panel: Theoretical and numerical calculation of original system; Lower panel: Theoretical and numerical calculation with slow perturbation added.¶

We can not predict the oscillation when we add in the new perturbation using the Rabi oscillation formula. That makes sense!

### 1.1.9.4.1. Introducing Another Component Perturbation¶

We add in the term that has two components,

$H = - \frac{\omega_m}{2} \sigma_3 + \frac{\delta \lambda(x)}{2} \cos 2\theta_m \sigma_3 - \frac{\delta \lambda(x)}{2} \sin 2\theta_m \sigma_1 + \frac{\delta \lambda(x)}{2} \sin 2\theta_m \sigma_2.$

### 1.1.9.4.2. Rotating Perturbation with Constant Strength¶

Construct a system with a mode at resonance and another rotating perturbation of constant length,

$H = - \frac{\omega_m}{2} \sigma_3 - \frac{1}{2} (A_1\cos(k_1x) + A_2 \sin(k_2x)) \sigma_1 + \frac{1}{2} ( A_1\sin (k_1 x) + A_2 \sin(k_2 x) ) \sigma_2,$

where we choose $$k_1\gg k_2$$.

The new $$\sigma_2$$ term is a rotating field with constant length, which makes sure the modified energy gap has a constant length rather than the slowly changing energy gap. Fig. 1.48 Reduction of transition amplitudes. Black dashed line: one perturbation at exact resonance; Green long dashed line: $$A_2=A_{2,\mathrm{Critical}}=0.0083666$$; Blue dotted line: $$A_2=0.01$$; Red line: $$A_2=0.02$$. The grid lines are the amplitude predicted using Rabi formula correspondingly.¶

## 1.1.9.5. Refs & Notes¶

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