1

On one hand, the dipole moment is define as $$\vec{\mu} = q\vec{r},$$ where q is the charge and $\vec{r}$ is a position vector.

On the other hand, I know the transition dipole operator of a two level system can be expressed as $$ \vec{\mu} = \mu_{ge}|g\rangle \langle e| + \mu_{eg}|e\rangle \langle g|. $$

The interaction term between a two level system and the external electric field is usually written as $$ H_{\mathrm{int}} = -\vec{\mu}\cdot \vec{E}(t), $$ where $\vec{\mu}$ is the dipole moment operator and $\vec{E}(t)$ is the time-dependent electric field.

I want to know three exponents of $\vec{\mu}$ and their commutation relations, e.g., $\vec{\mu}_x=?$ and $[\vec{\mu}_x,\vec{\mu}_y]=?$.

But I don't know how to relate $\vec{\mu}=\mu_0\vec{r}$ with $\vec{\mu} = \mu_{ge}|g\rangle \langle e| + \mu_{eg}|e\rangle \langle g|$ .

Peace Wan
  • 111

1 Answers1

1

Let me first note that this form of the interaction is independent on the direction of the electric field - liek any scalar product.

The dipole moment is usually defined as $$ \vec{\mu} = \mu_0\vec{r}, $$ where $\vec{r}$ is just the position operator. One then needs to calculate the matrix elements of this operator between the states of interest.

Remark: There are a bit more details about how the dipole approximation comes about in my other post here.

Update Given a complete basis of states, $|n\rangle$ we can expand the arbitrary wave function as $$|\psi\rangle=\sum_n\psi_n|n\rangle, $$ and represent an arbitarry operator in terms of its matrix elements in this basis: $$ \hat{O}=\sum_{n,m}|n\rangle\langle n|\hat{O}|m\rangle\langle m | =\sum_{n,m}|n\rangle O_{nm}\langle m | $$ WHen dealing with the absorption of atoms/molecules one often restricst this basis to only two states (since others are energetically inaccessible): the ground state $|g\rangle$ and the excited state $|e\rangle$. One can than, e.g., express the dipole interaction as $$ -\vec{\mu}\cdot\vec{E} = -|e\rangle\langle e|\vec{\mu}\cdot\vec{E}|e\rangle \langle e| -|e\rangle\langle e|\vec{\mu}\cdot\vec{E}|g\rangle \langle g| -|g\rangle\langle g|\vec{\mu}\cdot\vec{E}|e\rangle \langle e| -|g\rangle\langle g|\vec{\mu}\cdot\vec{E}|g\rangle \langle g| $$ The diagonal elements are usually absorbed in the main two-level Hamiltonian $$ H= |e\rangle E_e \langle e| + |g\rangle E_g \langle g|, $$ and the quantization axis is often taken along the z_direction, leaving us with $$ -\vec{\mu}\cdot\vec{E} = -|e\rangle\langle e|\mu_z E_z|g\rangle \langle g| -|g\rangle\langle g|\mu_z E_z|e\rangle \langle e| = - \mu_0 E_z\left( |e\rangle\langle e|z|g\rangle \langle g| +|g\rangle\langle g|z|e\rangle \langle e|\right). $$

Roger V.
  • 68,984