We know that nuclear magnetic moment can be expressed in terms of the expected value for nuclear spin as:

$$\langle\mu\rangle =[g_lj+(g_s-g_l)\langle s_z\rangle]\frac{\mu_N}{\hbar}$$

(Cf. Krane), where $\vec{j}$ is the total angular momentum, $\vec{l}+\vec{s}$.

How does the expected $\langle s_z\rangle$ value relate to the $\vec{j}$-component of spin, $\langle s_j\rangle$? Krane mentions that only that value is needed, given that it remains constant.

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    $\begingroup$ It would be helpful if you give more more details about your source (so far define just as "Krane"). I'm puzzled by "$\vec{j}$-component of spin $\langle s_j \rangle$" -- what do you mean? $\endgroup$ – Slaviks Mar 12 '12 at 20:15
  • $\begingroup$ The source is Kenneth Krane's book 'Introductory Nuclear Physics'. $\langle s_j \rangle$ is the expected value for the $\vec{j}$-component of the spin vector, $\vec{s}$. We assume a fixed $z$ axis around which $\vec{j}$ -the sum of angular and spin momentum- rotates. In the book, it is said that if we want to measure $\langle s_z \rangle$, it suffices to know $\langle s_j \rangle$. Why? $\endgroup$ – Sebgr Mar 12 '12 at 20:40
  • $\begingroup$ I have not access to the book now, but I remember that In Krane's book the terminology is a bit sloppy. He calls spin to total angular momentum, can this be the source of your doubt? $\endgroup$ – DaniH Mar 12 '12 at 20:47
  • $\begingroup$ I don't think so, total angular momentum is clearly defined as the sum of angular momentum and spin. $\endgroup$ – Sebgr Mar 12 '12 at 20:52
  • $\begingroup$ OK, I found the book. It seems that he´s applying the Wigner-Eckart theorem although it does not say it explicitly. Also, if I remember correctly there was an error in one of the formulas 5.9 although I do not remember in which one. I will try to elaborate later... $\endgroup$ – DaniH Mar 13 '12 at 20:36

From the magnetic moment $$\mathbf{\mu}=\mathbf{\mu_L}+ \mathbf{\mu_S}=(g_l \mathbf{L}+g_s\mathbf{S})\frac{\mu_N}{\hbar}\tag1$$ take the scalar product with $\bf{J}$ $$\mathbf{\mu_J} ·\mathbf{J}=(1/2(g_l +g_s)\mathbf J^2+1/2(g_l -g_s)(\mathbf L^2-\mathbf S^2))\frac{\mu_N}{\hbar} \tag2$$ so with commutator relation $$\mu=(1/2(g_l +g_s)j+1/2(g_l -g_s)\frac{(l-s)(l+s+1)}{j+1}) \mu_N \tag3$$

since $s=1/2$ and $j= l\pm1/2$ you end with two possible values for $\mu$ $$\mu=(jg_l-1/2(g_l -g_s)) \mu_N \quad\text{for }j= l+1/2\\ \mu=(jg_l+(g_l -g_s)\frac{j}{2j+1}) \mu_N \quad\text{for }j= l-1/2 \tag4$$

this is your $⟨s⟩$ projection in the $j$ direction.

For $s\neq 1/2$ equation, $(3)$ still holds. But for $s>l$ the factor must be changed to the general form $$\frac{l(l+1)- s (s+1)}{j+1}.$$

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  • $\begingroup$ Nuclei usually have more than one nucleon in them, so you cannot assume s=1/2. The spin operator S in question is presumably S_1 + S_2 + ..., i.e. the sum of the spin operators for each nucleon. $\endgroup$ – Steve Byrnes Mar 17 '12 at 12:37
  • $\begingroup$ Edited answer. I was thinking on an odd unpaired nucleus. However without proper quantum chromodynamics all classical answers are approximations. I am not a specialist on particle physics. $\endgroup$ – Alex1167623 Mar 17 '12 at 22:54

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