# I Why is light polarised for the Zeeman effect?

1. Aug 31, 2016

### greswd

as seen in this diagram.

What's the underlying quantum explanation for it?

2. Aug 31, 2016

### vanhees71

3. Aug 31, 2016

### greswd

Unfortunately it explains very little

4. Aug 31, 2016

Staff Emeritus
It explains it all.

Do you agree that if I had a transition between two states or energy levels X and Y (neither an S state) from randomly aligned atoms, the light would be unpolarized?
Do you agree that if I had a transition between two states or energy levels X and Y (neither an S state) from perfectly aligned atoms, the light would be polarized?
Do you agree that the conditions for the Zeeman effect partially align the atoms?
Do you agree that a consequence of this is that light from atoms in this condition will be partially polarized?

5. Sep 1, 2016

### blue_leaf77

The transition rate in dipole approximation is proportional to $\mathbf e \cdot \mathbf r_{ab}$ where $\mathbf e$ is the unit vector along the polarization and $\mathbf r_{ab}$ is the matrix element of $\mathbf r$ between the states denoted by composite quantum numbers $a$ and $b$. Although this expression can be written in Cartesian components, it's more convenient in this case to write $\mathbf e \cdot \mathbf r_{ab}$ in spherical components in which it reads
$$\mathbf e \cdot \mathbf r_{ab} = e_1^*(\mathbf r_{ab})_1 + e_0^*(\mathbf r_{ab})_0 + e_{-1}^*(\mathbf r_{ab})_{-1}$$
where for a vector $\mathbf v$, its spherical components are
\begin{aligned} v_1 &= {-1\over\sqrt{2}}(v_x+iv_y) \\ v_0 &= v_z \\ v_{-1} &= {1\over\sqrt{2}}(v_x-iv_y) \end{aligned}
Now, for a transition with a given $\Delta m$, one terms in $\mathbf e \cdot \mathbf r_{ab}$ survives and the other two vanish. For example in the case $\Delta m = +1$, the term $e_1^*(\mathbf r_{ab})_1$ survive. Let's consider a photon emitted such that its propagation vector $\mathbf k$ is as shown in the figure below.

$\mathbf e_p$ and $\mathbf e_s$ are the two components of the polarization of the emitted photons. Thus $\mathbf k$, $\mathbf e_p$, and $\mathbf e_s$ are mutually perpendicular with $\mathbf e_s$ in the $xy$ plane. In this case, the Cartesian components of $\mathbf e_p$ and $\mathbf e_s$ are
$$(e_p)_x = \cos\theta\cos\phi; \hspace{1cm} (e_p)_y = \cos\theta\sin\phi; \hspace{1cm} (e_p)_z = -\sin\theta\\ (e_s)_x = -\sin\phi; \hspace{1cm} (e_s)_y = \cos\phi; \hspace{1cm} (e_s)_z = 0 \\$$
The spherical components of $\mathbf e_p$ and $\mathbf e_s$ that corresponds to the non-vanishing radiation are $(e_p)_1$ and $(e_s)_1$,
$$(e_p)_1 = {-1\over\sqrt{2}} (\cos\theta\cos\phi+i \cos\theta\sin\phi) = {-1\over\sqrt{2}} \cos\theta e^{i\phi} \\ (e_s)_1 = {-1\over\sqrt{2}} (-\sin\phi + i\cos\phi) = {i\over\sqrt{2}} e^{i\phi}$$
Now we can analyze the polarization state for a given propagation direction (just for reminder, we are considering the case of $\Delta m = +1$). In longitudinal direction, $\theta = 0$, and the emitted photon's polarization is circular because
$$\mathbf E = |\mathbf E| \left((e_p)_1\mathbf e_p + (e_s)_1\mathbf e_s\right) = |\mathbf E| \frac{1}{\sqrt{2}}e^{i\phi} \left(\mathbf e_p + i\mathbf e_s\right)$$
In transverse direction, $\theta = \pi/2$, the emitted photon's polarization is linear in the $xy$ plane (perpendicular to the magnetic field) because
$$\mathbf E = |\mathbf E| \left((e_p)_1\mathbf e_p + (e_s)_1\mathbf e_s\right) = |\mathbf E| \frac{1}{\sqrt{2}}e^{i\phi} i\mathbf e_s$$
($\mathbf e_s$ is in the $xy$ plane).
You can check for the other lines $\Delta m = 0,-1$ and they should coincide with your picture above.

Last edited: Sep 3, 2016
6. Sep 1, 2016

### greswd

Sorry, I can't see where the external magnetic field has been factored into this.

7. Sep 1, 2016

### blue_leaf77

The analysis above actually also applies when there is no magnetic field, i.e. when the atom is free. In the presence of uniform, weak magnetic field along the z direction, the perturbation is proportional to $B(L_z + 2S_z)$. This perturbation commutes with $L^2$ and $L_z$, therefore the selection rule is not affected in weak Zeeman effect. However, in Stark effect the perturbation does not commute with the angular momentum and the above analysis about light polarization does not apply.

8. Sep 2, 2016

### greswd

thanks blue leaf. do you have any maths on how the direction of the external magnetic field causes the transverse and longitudinal polarizations to align with it?

9. Sep 2, 2016

### blue_leaf77

The effect of magnetic field is implicit. The application of external magnetic field causes the levels differing in magnetic quantum numbers to split. This magnetic quantum number is that associated with the component of angular momentum in the field's direction. Despite this splitting, the state is still specified by the non-rotational quantum number and angular momenta quantum numbers $|\alpha,L,m_L,m_s\rangle$, although the spatial function of this state is different from the unperturbed atom. This means the selection rule in the presence of magnetic field is unchanged. This allows us to write $\mathbf e \cdot \mathbf r_{ab}$ as
$$\mathbf e \cdot \mathbf r_{ab} = e_1^*(\mathbf r_{ab})_1 + e_0^*(\mathbf r_{ab})_0 + e_{-1}^*(\mathbf r_{ab})_{-1}$$
using $z_{ab} = (\mathbf r_{ab})_0$ as the direction of magnetic field.
As I have said before, that the same analysis applies to free atoms. However, since in most cases one deals with a collection of atoms rather than a single atom, the emitted light is unpolarized because each atom has different "z-directions".

10. Sep 3, 2016

### greswd

Just one more question.
The polarisation of light and the alignment of light rays of different polarisations parallel and perpendicular to the B field are two direct consequences of the exact same underlying phenomenon?

11. Sep 3, 2016

### blue_leaf77

By "alignment of light rays" do you mean the angular distribution of light intensity such as in figure 2 in the pdf in post #2? The answer is yes.
For example for $\Delta m=1$, the transition rate is
$$W_{ab}(\theta,\phi) \propto |(e_p)_1\mathbf e_p + (e_s)_1\mathbf e_s|^2 = 1+\cos^2\theta$$
while for $\Delta m = 0$,
$$W_{ab}(\theta,\phi) \propto |(e_p)_0\mathbf e_p + (e_s)_0\mathbf e_s|^2 = \sin^2\theta$$
You can compare the above formula with figure 2 in that pdf.

12. Sep 3, 2016

### greswd

I meant the alignment parallel and perpendicular to the field as seen in my original diagram. Is it the same as what you've just mentioned above?

13. Sep 3, 2016

### blue_leaf77

Do you possibly assume that the photon is emitted only in the transverse and longitudinal direction with respect to the magnetic field? The emitted light has certain angular distribution depending on $\Delta m$ as discussed in the previous few posts, you can detect photons in e.g. $\theta = 60^o,70^o$ etc as long as there is no zero in the angular distribution.