4

Consider the quantization axis ($B$-field direction) to be along $z$, and the field is small enough for the weak Zeeman effect to dominate. An atom has a ground state and an excited (degenerate) state with $l=1$, and in the presence of the magnetic field the degeneracy splits.

If I send in a laser along the $z$-axis, polarization parallel to the $x$-axis, then that light is a superposition of the following polarizations: $\langle \phi | = 1/(2\sqrt{2})( \sigma^+ + \sigma^-)$, where $\sigma^+$ drives the $m=+1$ and $\sigma^-$ drives the $m=-1$ transition and this is using the usual Jones matrices to describe polarizations.

But we know linearly polarized light also drives $m=0$ transitions.

What gives here? Which does it drive? All three? It cannot be that it depends on how I "consider" my polarization - i.e. as linear, or a superposition of two circular polarizations. What am I missing?

Corollary: suppose I send in light along the $x$-axis polarized along the $y$-axis. Then it is again a superposition of $\sigma^+$ and $\sigma^-$. Does anything change? i.e. does it drive all three?

Qmechanic
  • 220,844
user2022444
  • 63

1 Answers1

1

The answer to this question is contained within Jagerber48's answer to this question: $\pi, ~\sigma$ - atomic transitions with respect to a quantization axis .

As it relates to your question, you have presented a common point of confusion, namely:

But we know linearly polarized light also drives m=0 transitions.

This is not true. At least it isn't based on the way you defined "linearly polarized" and "$m=0$ transitions" ($\Delta m=0$ from now on).

Since $m$ is the projection of angular momentum onto a particular axis you need to define which axis you are projecting onto, and commonly use a subscript to note this (e.g. $m_x$ for the projection of angular momentum along $\hat{x}$). You have not explicitly stated this but your statement

$\sigma^+$ drives the $m=+1$ and $\sigma^-$ drives the $m=−1$ transition

means you must be talking about the projection along $z$, since this is only true for $m_z$ states.

Given these two definitions, light that is a superposition of $\sigma^\pm$ will drive a superposition of $\Delta m_z =\pm1$ transitions, and no $\Delta m_z =0$ transitions at all.

However, your idea that linear polarizations drive $\Delta m = 0$ transitions did not come from nowhere. There is a linear polarization which would drive $\Delta m_z = 0$ transitions, just not the one you described. Such a transition can only be driven by light that is linearly polarized along $\hat{z}$. Similarly, there are axes for which $\Delta m = 0$ transitions would be driven by your laser, just not the $z$ axis. Light that is linearly polarized along the $x$ axis* would exclusively drive $\Delta m_x = 0$ transitions. In fact, it would drive some degree of $\Delta m = 0$ for any axis that is not in the $yz$-plane.

So in summary, it does not matter if you describe your light as a superposition of $\sigma^+$ and $\sigma^-$ or a linear polarization along $\hat{x}$, as long as you understand how that relates to the angular momentum of an atom, and are clear about which component you are interested in.

*you have said both $x$-axis and $\frac{1}{2\sqrt{2}}(\sigma^+ + \sigma^-)$, I will note that with $\sigma^\pm$ defined with the convention I am familiar with, the latter would actually be polarized along $\hat{y}$. Linear $x$ polarization would be $\frac{1}{\sqrt 2}(-\sigma^+ +\sigma^-)$)

Alex
  • 66