# Atom in magnetic field modeled as simple harmonic oscillator

In a text I am using (Atomic Physics, by Christopher Foot), it states that an atom in a magnetic field can be modeled as a simple harmonic oscillator. I am assuming that there is periodic motion, possibly circular, which makes this a good candidate to be modeled as a simple harmonic oscillator.

The book states the following:

An atom in a magnetic field can be modelled as a simple harmonic oscillator. The restoring force on the electron is the same for displacements in all direction and the oscillator has the same resonant frequency $\omega_0$ for motion along the $x$-, $y$- and $z$- directions (when there is no magnetic field). In a magnetic field $\vec{B}$ the equation of motion for an electron with charge $-e$, position $\vec{r}$ and velocity $\vec{v} = \dot{\vec r}$ is $$m_e\frac{d\vec{v}}{dt} = -m_e \omega_0^2 \vec{r} - e \vec{v} \times \vec{B}$$

Does anyone have a good geometric interpretation of how how an atom in a magnetic field resembles the motion of a simple harmonic oscillator? Do I have the right idea in assuming that they are implying that in the absence of the magnetic field the electron can be modeled as if it is a simple harmonic oscillator $$m_e\frac{d\vec{v}}{dt} = -m_e \omega_0^2 \vec{r},$$ where the electron would be moving around as if on the surface of a sphere, and this is what is being modeled as a simple harmonic oscillator?

## 2 Answers

This text is presented in the book's Chapter 1.8 "Early atomic physics" / "The Zeeman Effect". So it is talking about history.

The paragraph's context is that, the structure of atom was still pretty much unknown, when in 1896 the (normal) Zeeman effect was observed (the spectral line splits in the presence of uniform magnetic field, with $\Delta\omega \propto eB/m$). In 1896 we didn't know there is a dense nucleus, not to mention quantum mechanics!

The model you mentioned was proposed by Lorentz trying to explain the effect. He assumed there is a force affecting the charged particles in the atom $\mathbf F = -k\mathbf r$. There is probably no good reason besides this field allows the electrons to oscillate in a constant frequency. But you may consider this compatible with the plum-pudding model, where the atom has a uniform sphere of positive charge, so $\mathbf F = q\mathbf E = -k\mathbf r$.

The rest of the chapter tries to show that the Zeeman effect and polarization of light can be explained using this classical assumption.

Ref: A. J. Kox. (1997) The discovery of the electron: II. The Zeeman effect. Eur. J. Phys. 18. pp139–144.

About the comment:

An electron oscillating parallel to $\mathbf B$ radiates an electromagnetic wave with linear polarization and angular frequency $\omega_0$. This $\pi$-component of the line is observed in all directions except along the magnetic field; in the special case of transverse observation the polarization of the $\pi$-component lies along $\hat{\mathbf e}_z$". — What is the $\pi$-component and what observation are they making?

The $\pi$- and $\sigma$-components are vector components of the polarization of the light wave. $\pi$- or $P$- means "parallel" (to $\mathbf B$, the external magnetic field), while $\sigma$- or $S$- means senkrecht, the German for "perpendicular".

In this context, we assume $\mathbf B = B\hat{\mathbf z}$, and put the atom at the origin $(0,0,0)$, and we detect the radiation intensity and polarization. The electron has 3 modes of oscillation:

• up-and-down along the $z$-axis with frequency $\omega$, which we call the $\pi$-component, and
• circularly clockwise/counter-clockwise on the $xy$-plane with frequencies $\omega\pm\Delta\omega$, which we call the $\sigma^{\pm}$-components.

Oscillating charge radiates spherical light wave. Thus “the $\pi$-component is observed in all directions (except along $\mathbf B$)”. And if we put the observer on the $xy$-plane (“transverse observation”), we will detect that the $\pi$-component has polarization along the $z$-axis.

• Thanks for your answer, I think I follow what you are saying. Could you please explain what is then meant by the line in the text "An electron oscillating parallel to $\vec{B}$ radiates an electromagnetic wave with linear polarization and angular frequency $\omega_0$. This $\pi$-component of the line is observed in all directions except along the magnetic field; in the special case of transverse observation the polarization of the $\pi$-component lies along $\hat{e}_z$". What is the $\pi$ component and what observation are they making? – user101311 Mar 9 '17 at 12:50
• @Moses Please check update. – kennytm Mar 9 '17 at 17:44
• Thanks a lot for your answer. So just to confirm, the $\pi$ and $\sigma$ components are the polarization vectors associated with the em radiation emmitted by an atom always in relation to a magnetic field? So $\pi$ and $\sigma$ polarization always requires as atom to be in a magnetic field and the resulting emission determines which waves are $\pi$ and which are $\sigma$? – user101311 Mar 17 '17 at 15:41
• @Moses The atom emits radiation after the electron is excited (e.g. by heating it). The magnetic field causes an emission line to split into three. – kennytm Mar 19 '17 at 21:04
• @Moses Yes the light is emitted in all directions. Unpolarized means every polarization direction is possible. – kennytm Mar 21 '17 at 13:52

Frankly, this model is a bit of a kludge: it does not come from anything deeper, and there are no fundamental reasons that underlie its validity. However, it is included in textbooks with some regularity because its conclusions coincide with the physics that we know does hold, and it is a simpler way to obtain those results using a friendlier toolbox.

In essence, though, the similarities that "an atom can be modelled as a simple harmonic oscillator" alludes to are barely more than "atoms radiate at a given frequency $\omega_0$ and harmonic oscillators do so too". The value of the model comes in that adding in the magnetic field gives you the correct form of the Zeeman effect (or one of them, anyway). However, it's not particularly fruitful to try and read more into the analogy.