# Landé $g$-factor for $j=0$

I want to know how to calculate the Landé $g$-factor for the case $j=0$. The formula

$g_j = 1 + \frac{j(j+1) + s(s+1) - l(l+1)}{2j(j+1)}$

seems useless in this case. I've seen that $g_j=0$, because since total angular momentum is zero there is no interaction with the magnetic field. But I don't understand this because precisely the coupling between magnetic field and angular momentum depends on whether it is orbital ($g_l$) or intrinsic ($g_s$) angular momentum.

EDIT: What I mean in the last sentence is that, as far as I know, the Hamiltonian of the interaction is not proportional to the total angular momentum $\mathbf{J} = \mathbf{L}+\mathbf{S}$, but to a similar sum with weighs $g_l$ and $g_s$:

$H \propto \big( g_l \mathbf{L} + g_s \mathbf{S} \big) \cdot \mathbf{B} \approx \big( \mathbf{L} + 2 \mathbf{S} \big) \cdot \mathbf{B}$

• The claim that "there is no interaction with the magnetic field" is obviously incorrect (or, at least, nontrivial), as you point out. However, you cannot express this as $\boldsymbol \mu= g_j \mathbf J$ because the latter is zero, which would give a vanishing magnetic moment regardless of the gyromagnetic ratio. – Emilio Pisanty Dec 4 '18 at 9:30

## 3 Answers

Here's a useful idea. Revisit the derivation of the Lande g factor, but do not assume any particular value for $$g_l$$ or $$g_s$$; just keep them in the calculation throughout. I will present this whole calculation, and point out what happens when $$J=0$$.

We have the perturbing Hamiltonian $$H = -{\bf \mu} \cdot {\bf B} = g_l \mu_B B \hat{L}_z + g_s \mu_B B \hat{S}_z$$ so in first order perturbation theory the energy shift is $$\Delta E = \langle LSJM_J | g_l \mu_B B \hat{L}_z + g_s \mu_B B \hat{S}_z | L S J M_J \rangle$$ where I assumed the atomic state has well-defined $$L,S,J,M_J$$ (the so-called LS coupling approximation). Now it can be shown that $$\langle LSJM_J | \hat{L}_z | L S J M_J \rangle {\langle LSJM_J | \hat{\bf J} \cdot \hat{\bf J} | L S J M_J \rangle} = {\langle LSJM_J | \hat{\bf L} \cdot \hat{\bf J} | LSJM_J \rangle \langle LSJM_J | \hat{J}_z | LSJM_J \rangle} \;\;\;\;\; (1)$$ which you can prove either by using the vector model or by more formal methods. In the Zeeman effect calculation, the usual next step is to divide both sides of this formula by the second term on the left, i.e. $$\langle \hat{J}^2 \rangle$$, but in the case $$J=0$$ this would result in $$0/0$$ on the right hand side, thus a mathematically undefined quantity. In other words when $$J=0$$ we have to see that the above expression tells us nothing about $$\langle \hat{L}_z \rangle$$ and we have to go back to the beginning. The usual way to handle this is simply to say that we expect the answer to have the form $$\Delta E = g_J \mu_B B M_J$$ so in the case $$J=0$$, which implies the only value for $$M_J$$ is $$M_J=0$$, we will get $$\Delta E = 0$$ irrespective of what value $$g_J$$ may have. Since this amounts to "no Zeeman effect" we may as well agree to take $$g_J = 0$$. If this is the heart of your question, then skip to the last section below where I will discuss it some more.

The above will already suffice to answer your question at a preliminary level, but I will now complete the calculation for the case $$J \ne 0$$ in any case, so you can see that the above formula does allow one to do this.

The RHS of expression (1) has two expectation values. After using $${\bf J} = {\bf L} + {\bf S}$$ so $$({\bf J} - {\bf L})^2 = {\bf S}^2$$ the first evaluates to $$\frac{1}{2}(J(J+1)+L(L+1)-S(S+1)$$ and the second is $$M_J$$, so we obtain $$\langle \hat{L}_z \rangle = \frac{J(J+1)+L(L+1)-S(S+1)}{2J(J+1)} M_J.$$ The calculation for $$\hat{S}_z$$ goes similarly with $$L$$ and $$S$$ reversed. Upon adding the $$L$$ and $$S$$ parts, the overall result is $$\Delta E = g_J \mu_B B M_J$$ where $$g_J = g_L \frac{J(J+1)+L(L+1)-S(S+1)}{2J(J+1)} + g_S \frac{J(J+1)-L(L+1)+S(S+1)}{2J(J+1)}$$

You can now put in values for $$g_l$$ and $$g_s$$ if you like, and the above method remains accurate in more general situations where the sum of two angular momenta are involved.

But why is $$\Delta E \propto M_J$$ ?

Anyone still bugged by the thought that $$\bf L$$ and $$\bf S$$ (with their associated dipole moments) are 'hiding' inside $$\bf J$$ may not be satisfied with the claim that the energy shift goes to zero when $$J$$ is zero but $$L$$ and $$S$$ are not. In the above I based this claim on the fact that $$\Delta E \propto M_J$$ in the situation under consideration. The further information you need is that (1) the shift is zero only at first order in perturbation theory; at second and higher orders there is a shift; this shift is important in atomic clocks, for example; (2) we can formulate the general rule that a quantum system with no sense of direction in space cannot pick up an energy shift, at first order, from a Hamiltonian having the form of a scalar product of a vector property and a vector applied field. The maths here is connected to the Wigner-Eckart theorem. The physical insight is that if there were a first-order effect, how would it know whether to be a shift up or down? Zero total angular momentum amounts to saying the quantum system cannot tell one direction from another, but a scalar product would change sign when the direction of the applied field changed.

• Hmm ... down vote ... could you point out the mistake? Or is it too long? Don't like the notation? (Actually I am not in any serious doubt about its correctness since I have taught this exact subject for 30 years.) – Andrew Steane Dec 5 '18 at 22:11

Ok, I will check your reference. However, is difficult for me accept that there are many references books with wrong notation, at least 4 books which I have checked. Another example of this could be : https://books.google.es/books?id=4vY7DQAAQBAJ&printsec=frontcover&hl=es#v=onepage&q&f=false The second eq. of page 292 Anyway, I think you are older than me and of course more experienced, maybe I don't know which book is the best to read or at least a good and introductory reference book (ie.Bransden,but not really introductory), so I believe you.

• I went to google books but it didn't show page 292. I can only assume that a poor piece of reasoning got copied around a community that did not need to re-examine the derivation. Ultimately of course one prefers to follow the reasoning rather than trust one authority or another. I hope my comments and answer make it possible to do that. If this exchange has brought to light a mistake in textbooks then we should try to make sure it is noticed, and then we will have done everyone a favour. – Andrew Steane Dec 7 '18 at 17:00
• I pay attention to my reads, but sometimes one is not in position to put in doubt them, like me, or you, if you are learning about any subject and don't have any background. That's clear. – J0KerSpin Dec 10 '18 at 19:28

It makes no sense to calculate the Landé $g$-factor for $j=0$

Remember that the magnetic dipole moments are

$$\vec M_L = -g_L \mu_B \vec L$$ $$\vec M_S = -g_s \mu_B \vec S$$

Defining $$\vec M_J = \vec M_L + \vec M_S = -g_J \mu_B \vec J$$

Simplifying

$$g_J \vec J = g_L \vec L + g_S \vec S$$

So

$$g_J = g_L \frac {\vec L.\vec J}{\vec J^2} + g_S \frac {\vec S.\vec J}{\vec J^2}$$

Of course, if the total momentum $J$ is 0 you can take $g_0 = 0$

Hope that helps

J.

• But then what is the energy splitting when applying a magnetic field to a $J=0$ (but $L\neq 0$ and $S \neq 0$) state? – MBolin Dec 4 '17 at 20:59
• The Zeeman effect doesn't exist if J =0 , so, we don't care about the value of $g_j$ – J0KerSpin Dec 4 '17 at 21:17
• Have you understood it? – J0KerSpin Dec 5 '17 at 13:30
• No. Why it doesn't exist? – MBolin Dec 5 '17 at 14:20
• The Zeeman effect is due to the interaction between the magnetic energy levels of a system and an external magnetic field H, but, if the total momentum is zero the is no Zeeman effect, so it does not make sense to define a $g_J$ if $\vec J = 0$ – J0KerSpin Dec 5 '17 at 18:04