Why normal Zeeman effect contains three components?

[misko]

Can someone explain to me why normal Zeeman effect splits spectral line into three components and not into 4, 5 or any other number?

 

[blue_leaf77]

Do you know the selection rule for the magnetic quantum number?

 

[misko]

Yes, it can be changed by +1, -1, 0.

 

[blue_leaf77]

In normal Zeeman effect, states with different $m$ split, accordingly the transition lines also split according to their value of $\Delta m$.

 

[misko]

I know that normal Zeeman effect happens when total spin of electrons in an atom is S=0.
This means that energy levels splitting in external magnetic field is done only on orbital angular momentum.

So for example P orbital will split into three sub-levels with slightly different energies. Now, it make sense to me that if there is transition from P orbital (L=1) to S orbital (L=0) there will be splitting to three components because P is split to three sub-levels and S orbital has no splitting.
But what about transition from say, D to P orbital? D orbital will be split into 5 sub-levels and P orbital will be split to 3 sub-levels so that gives many more combinations for transitions, even when selection rule for magnetic moment is taken into account.
So, (assuming total spin is zero), for D orbital we have J = 2,1,0,-1,-2 and for P orbital there is J=1,0,-1.
Possible transitions are then:
$D_2 \rightarrow P_1$
$D_1 \rightarrow P_1$
$D_0 \rightarrow P_1$

$D_1 \rightarrow P_0$
$D_0 \rightarrow P_0$
$D_{-1} \rightarrow P_0$

$D_0 \rightarrow P_{-1}$
$D_{-1} \rightarrow P_{-1}$
$D_{-2} \rightarrow P_{-1}$

So there are 9 lines here, each respecting the selection rule for magnetic quantum number.
What is wrong in this logic, where is my error in understanding?

 

[blue_leaf77]

The formula for the transition energy under Zeeman effect is
$$
\Delta E = \Delta E_0 + \mu_B B \Delta m
$$
where $\Delta E_0$ is the energy difference without magnetic field. As you see the energy difference in the presence of magnetic field only depends on $\Delta m$. In your example the lines $d_2 \to p_1$ and $d_1 \to p_0$ coincide.

 

[misko]

I think I am starting to get it now...

So in my example there are 9 transitions but they are grouped such that only 3 different lines are present, right?
If I group transitions that belong to the same line they would be like this:
line 1:
$D_2 \rightarrow P_1$
$D_1 \rightarrow P_0$
$D_0 \rightarrow P_{-1}$

line 2:
$D_1 \rightarrow P_1$
$D_0 \rightarrow P_0$
$D_{-1} \rightarrow P_{-1}$

line 3:
$D_0 \rightarrow P_1$
$D_{-1} \rightarrow P_0$
$D_{-2} \rightarrow P_{-1}$

Is this correct?

Btw, This formula you gave me, it is valid only for normal Zeeman effect right? I mean g-factor is 1 in the formula which applies to singlet states with total spin equal to zero.

 

[blue_leaf77]

Is this correct?

Yes.

This formula you gave me, it is valid only for normal Zeeman effect right? I mean g-factor is 1 in the formula which applies to singlet states with total spin equal to zero.

Yes.