Atomic Physics |Anomalous Zeeman effect

In atomic physics, we observe the interaction of the magnetic field with a magnetic dipole moment. Its interpretation is complex when S’≠0. The Lorentz classical theory can’t explain the interaction theory and further the splitting of energy levels. But remember, along with the complex structure of the emission, external magnetic field strength compared to the internal magnetic field of the atom matters here. This is called in general the Anomalous Zeeman effect. This effect observed in atoms that have the only odd number of electrons like Hydrogen, Li, Sodium, etc. In the Normal Zeeman effect, the strength of the external magnetic field is more as compared to the internal magnetic field of the atom.

According to the vector atom model, Orbital angular momentum L, spin angular momentum and total angular momentum lies in the same plane. Quantization of these three can be understood by the concept of space quantization.  

Larmor Precession in Atomic Physics

In this physics topic, we will use the Larmor Precession term, if you know it then good. If you don’t know then please keep in mind this is the precession of a vector mentioned above for the atom (like spin, orbital and total angular momentum) about to the defined axis.

Further, this plane precesses about the J’. The reason for it is the Larmor precession of Spin angular momentum S and L in the presence of a magnetic field of an atom. This spin angular momentum interacts with the orbital angular momentum called spin-orbit interaction or LS coupling. The frequency of the precession depends upon the strength of the internal magnetic field of the atom.

Here important point is about the total magnetic moment μ which is not antiparallel to the J. This μ actually precess about the J with a precessional frequency which depends upon the strength of the magnetic field of the atom, as mentioned above for the L and S.      

Here, optically active electrons play the important role. What are the properties of these types of electrons? For it, consider the Alkali atoms (noble gas + 1electron; Li, Na, K, Rb, and Cs) in these atoms, p-subshell is filled and 1-e remains in an s-subshell. The optical excitations are dominated by the behavior of the s electrons. It means these s-subshell electrons absorb and emits the photon. Because of this reason these electrons are called the optically active electrons.

You know the strength of the internal magnetic field in the atom is of the order of 1 Tesla. So, this magnetic field acts on the optically active electron. For the Anomalous Zeeman effect, the strength of the external magnetic field should be less than 1 Tesla. This is the reason because of that magnetic moment doesn’t precess about the external magnetic field direction. Means LS coupling (spin-orbit interaction) is stronger.

Now the question is how to evaluate the potential energy of the orientations of the total magnetic moment μ in this internal magnetic field? The answer is the same as we did start in the Normal Zeeman effect. The potential energy is the interaction of magnetic dipole moment with the internal magnetic field B. In the presence of the external magnetic field total angular momentum precesses about the z-axis. This can be seen by the vector atom model

Anomalous Zeeman effect number 1
                                                                          Figure 1: Vector Atom Model


To evaluate the potential energy,

\dpi{120} \fn_jvn \large \Delta E=-\vec{\mu_{j}}.\vec{B}           (1)

\dpi{120} \fn_jvn \large \textrm{Total angular momentum of the atom is}\\ \vec{J}=\vec{L}+\vec{S} \\ Also\\ \vec{\mu_{J}}=\vec{\mu_{L}}+\vec{\mu_{s}}\\

The total magnetic moment of the atom is the vector sum of the orbital magnetic moment and spin magnetic moment, that is given in above equation.

The magnetic moment due to the orbital motion of an electron is given by 

\dpi{120} \fn_jvn \large \vec{\mu_{L}}=-\frac{\mu_{b}}{\hbar}\vec{L}

The magnetic moment due to the spin of an electron is given by 

\dpi{120} \fn_jvn \large \vec{\mu_{S}}=-\frac{2\mu_{b}}{\hbar}\vec{S}

So the total magnetic moment

\dpi{120} \fn_jvn \large \vec{\mu_{J}}=-\frac{\mu_{b}}{\hbar}\vec{L}+(-\frac{2\mu_{b}}{\hbar})\vec{S}\\

\dpi{120} \fn_jvn \large \vec{\mu_{J}}=-\frac{\mu_{b}}{\hbar}[\vec{L}+2\vec{S}]                                       (2)

From the above equation, you can see that μj is not antiparallel to the total angular momentum J.  This point you can observe in the above mention figure.  You know when the external magnetic field is weak then = coupling will be very strong.

Quantum mechanically we can write

(L+2S) = J   —-(3)

where g is a constant known as the Lande g Factor. Now taking the dot product with J

J.(L+2S)=g j2

The left-hand side can be defined as,




So L = S , taking the square of both sides,

\dpi{120} \large L^{2}=J^{2}+S^{2}-2 {\color{Red} J}.{\color{Red} S}\\ {\color{Red} J}.{\color{Red} S}=\frac{J^{2}+S^{2}-L^{2}}{2}

So the above equation can be written as,

J.(L+2S)=j2+J.S = gJ2

by putting the value of J.S  in left side, from the above equation, we get

\dpi{120} \fn_phv \large J^{2}+\vec{J}.\vec{S}=gJ^{2}\\ J^{2}+\frac{J^{2}+S^{2}-L^{2}}{2}=gJ^{2}\\

taking the quantum mechanically average on both sides, it will be

\dpi{120} \fn_phv \large J(J+1)\hbar^{2}+\frac{[J(J+1)+S(S+1)-L(L+1)]\hbar^{2}}{2}=gJ(J+1)\hbar^{2}

after simplification of it, we get,


\dpi{120} \fn_phv \large J(J+1)\hbar^{2}[1+\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}]=gJ(J+1)\hbar^{2}\\ g=1+\frac{[J(J+1)+S(S+1)-L(L+1)]}{2J(J+1)}

“g” is called “Lande g Factor” which is a dimensionless number. 


Now, from equations (1), (2) and (3);

The energy acquired by the atom=

\dpi{120} \fn_jvn \large \Delta E=-\vec{\mu_{j}}.\vec{B}

substituting the value of μj from eqn. (2) and then using eqn. (3);

\dpi{120} \fn_phv \large \Delta E=- \vec{\mu_{J}}.\vec{B}\\ = -[-\frac{\mu_{b}g}{\hbar}\vec{J}.\vec{B}]\\ =\frac{\mu_{b}g}{\hbar}\vec{J}.\vec{B}

From Fig. 1, as shown above, you can see the total angular momentum J precesses about Z-direction of the applied magnetic field B. The projection of J along the z-direction is Jz.

\dpi{120} \fn_phv \large \vec{J}.\vec{B}=JBCos(\vec{J}.\vec{B})\\ Cos(\vec{J}.\vec{B})=\frac{J_{z}}{J}\\ \vec{J}.\vec{B}=JB \frac{J_{z}}{J}

Now substituting this value,

\dpi{120} \fn_phv \large \Delta E= \frac{\mu_{b}g}{\hbar}JB\frac {J_{z}}{J}\\ \Delta E= \frac{\mu_{b}g}{\hbar}JB\frac {m_{j}\hbar}{J}\\ \Delta E= \mu_{b}g B m_{j}



Anomalous zeeman energy

So this extra energy would be added into the original energy, of different transition states, in the presence of an external magnetic field. In a simple way, If E1 and E2 are two different states represent the ground and excited states, in the absence of an external magnetic field, respectively. This delta E energy will be added in the E1 and E2 states when one will apply the external magnetic field.

Let Ei and Ef be the energies of the initial and final state in the presence of an external magnetic field. Then,

\dpi{120} \fn_phv \large E_{i}=E_{1}^{0}+\Delta E\\ E_{f}=E_{2}^{0}+\Delta E

Substituting the value of delta E, we obtain

\dpi{120} \fn_phv \large E_{i}=E_{1}^{0}+g_{i}\mu_{B}m_{j}^{i}B\\ E_{f}=E_{2}^{0}+g_{f}\mu_{B}m_{j}^{f}B

Now, subtracting Ei from Ef, we obtain

\dpi{120} \fn_phv \large E_{f}-E_{i}=(E_{2}^{0}-E_{1}^{0})+\mu_{B}B(g_{f}m_{j}^{f}-g_{i}m_{j}^{i})


\dpi{120} \fn_phv \large E_{f}-E_{i}=h\nu\\ E_{f}^{0}-E_{i}^{0}=h\nu_{0}

So the frequency of emitted radiation will be;

\dpi{120} \fn_phv \large \nu=\nu_{0}+\mu_{b}B(g_{f}m_{j}^{f}-g_{i}m_{j}^{i})

Selection rule for the transitions are;

\dpi{120} \fn_phv \large \Delta m_{j}=0, \pm 1; m_{j}=0 \Rightarrow m_{j}=0 \mathit{not}\mathit{allowed}

The Anomalous Zeeman effect includes all microscopic entities like orbital and spin angular momentum.