Does the dipole moment of an atom modelled as a two level system depends on its frequency?

Question

Consider an atom modelled as a two level system :

$$H=\frac{\hbar \omega}{2} \sigma_z $$

$|0\rangle$ and $|1\rangle$ are the ground and excited states that span the Hilbert space.

In the Rabi oscillations study, we turn on an interaction Hamiltonian using a laser, and we define the Rabi Frequency as $\Omega=\frac{\vec{d}.\vec{E_0}}{\hbar}$ where $\vec{E_0}$ is the amplitude of the electric field of the laser, and $\vec{d}$ is the quantity :

$$\vec{d}=q\langle 0 | {\widehat{r}} | 1 \rangle $$

$\widehat{r}$ being the position operator (which is a vector operator) of the two level system.

My question is :

Do the components of the vector $\langle 0 | {\widehat{r}} | 1 \rangle$ actually depend on the frequency $\omega$ ?

If so, is there a resource where I can find this law ?

I guess the answer is true, because if I turn on a laser, it will polarize my atom. Thus $\langle 0 | {\widehat{r}} | 1 \rangle$ should depend on the strength of the laser. Then it may also depend on the frequency of the atom.

But how can I find the law (if I'm not wrong).

Answer 1

Your hamiltonian hides essential physics. Still, the question in the title can be answered. The frequency depends on the dipole moment, not the other way around.

Answer 2

Usually $\langle\psi_1 |\mathbf{r}|\psi_2\rangle$ does depend on time. The reason being is that you usually start in Schrodinger picture (wavefunctions depend on time, operators do not):

$i\hbar\partial_t|\psi\rangle =\left(H_0+H_1\right)|\psi\rangle$,

where $H_0$ is your default Hamiltonian, and $H_1$ is the interaction Hamiltonian. You then go into interaction picture (both wavefunction and operators depend on time):

$i\hbar\partial_t |\psi_i\rangle = V|\psi_i\rangle$

$V=\exp\left(\frac{t H_0}{i\hbar}\right)H_1\exp\left(-\frac{t H_0}{i\hbar}\right)$

Now, your operator $\mathbf{r}$ will arise as a result of decomposition of that interaction operator $V$. Lets say that $\mathbf{r}_s$ is the Schrodinger picture position operator (time-independent). Then the other operator, the one you will be working with, will be: $\mathbf{r}=\exp\left(\frac{t H_0}{i\hbar}\right)\mathbf{r}_s\exp\left(-\frac{t H_0}{i\hbar}\right)$.

Then your matrix element will be:

$\langle\psi_1 |\mathbf{r}|\psi_2\rangle=\langle\psi_1 |\exp\left(\frac{t H_0}{i\hbar}\right)\mathbf{r}_s\exp\left(-\frac{t H_0}{i\hbar}\right)|\psi_2\rangle=\langle\psi_1 |\exp\left(-i\omega_2 t\right)\mathbf{r}_s\exp\left(i\omega_1 t\right)|\psi_2\rangle$

The last step is due to the fact that one usually uses eigen-basis of the default Hamiltonian for the decomposition ($H_0|\psi_{1,2}\rangle=\hbar\omega_{1,2}|\psi_{1,2}\rangle$). Thus

$\langle\psi_1 |\mathbf{r}|\psi_2\rangle=\exp\left(-i\left(\omega_1-\omega_2\right) t\right)\langle\psi_1 |\mathbf{r}_s|\psi_2\rangle$

where $\langle\psi_1 |\mathbf{r}_s|\psi_2\rangle$ is time-independent.