Two Level Atom Algebra

Definitions and meaning of the operators§

Two level atom by definition only have state $\ket{g}$ and $\ket{e}$, which correspond to the ground state and excited state. Thus the atomic ladder operators ($\sigma^-$ lowering and $\sigma^+$ raising) can be defined as

$$\hat{\sigma}^- = \ket{g}\bra{e}$$ $$\hat{\sigma}^+ = \ket{e}\bra{g}$$

And the population projectors

$$\hat{\sigma}_e = \ket{e}\bra{e}$$ $$\hat{\sigma}_g = \ket{g}\bra{g}$$

With consistency to the pauli matricies,

$$\hat{\sigma}_z = \hat{\sigma}_e - \hat{\sigma}_g$$ $$\hat{\sigma}_x = \sigma^+ + \sigma^-$$ $$\hat{\sigma}_y = -i(\sigma^+ - \sigma^-)$$

Intepretation§

If we have a state $\ket{\psi}=g\ket{g}+e\ket{e}$, let's find some intepretation to these operators

With normalisation condition

$$g^2+e^2=1$$

Population Inversion§

With the physical meaning of $\ket{e}$ and $\ket{g}$, we can call $\hat{\sigma}_z$ as population inversion

When applied to a state

$$\hat{\sigma}_z\ket{\psi}=-g\ket{g}+e\ket{e}$$

Which flips the phase of the ground state

When taken the expectaiton value.

$$\braket{\sigma_z}=e-g$$

Which measures the populations of the state, is it more likely to be on excited state or ground state.

Dipole Moment§

$$\hat{d}=-e\hat{r}$$

For a simple free atom, the distribution of position for ground state and the first exicted state are shperical syemmtereic. Thus their expectation value is 0

$$\bra{e}\hat{r}\ket{e}=0\qquad\bra{g}\hat{r}\ket{g}=0$$

Thus in this basis

$$\hat{d}=\hat{I}\hat{d}\hat{I}=\sum_{i,j\in\{e,g\}}\ket{i}\bra{i}\hat{d}\ket{j}\bra{j}$$

Uisng the known facts about expecatation values, and forcing it to be hermitian as it should be a physically measurable value.

$$\hat{d}=d_{eg}(\ket{e}\bra{g}+\ket{g}\bra{e}))=d_{eg}\hat{\sigma}_x$$

Acting on the state

$$\hat{\sigma}_x\ket{\psi}=e\ket{g}+g\ket{e}$$

The expectation value

$$\braket{\sigma_x}=eg^*+ge^*$$

Here we see that for non superposition state for now we will call pure, there will be no dipole, as said, the atom is spherical symetrric. But for any superposition of state, the dipole is non 0 due to the symetry breaking due to superpositions. With more math that is not relavent to the topic, we can find that

$$\braket{x}=x_{eg}\cos(\omega_{eg}t)$$

where

$$\omega_{eg}=\frac{E_e-E_g}{\hbar}$$

The oscilating expectation value of position is why there will be non zero dipole.

** And for $\hat{\sigma}_y$ it is the same meaning for $\hat{\sigma}_x$ just measured in a differnet direction.

Relation between oprators§

We know that $\hat{\sigma}_e$ should be a projector to state $\ket{e}$, well another way of obtaining it is by lowering then raising.

$$\hat{\sigma}_e=\hat{\sigma}^+\hat{\sigma}^-$$

Similarly,

$$\hat{\sigma}_g=\hat{\sigma}^-\hat{\sigma}^+$$

The rest of the relation is standard pauli matrix algebra.

Commutation§

Given the above, obviously

$$[\hat{\sigma}^+,\hat{\sigma}^-]=\hat{\sigma}_z$$

Also

$$[\hat{\sigma}^+\hat{\sigma}^-,\hat{\sigma}^-]=-\hat{\sigma}^-$$