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The magnetic moment operator being $$\mu = -\frac{\mu_\text{B}}{\hbar}(L+2S),$$ a direct evaluation of its absolute value would be $$|\mu| = (\mu^2)^{1/2} = \frac{\mu_\text{B}}{\hbar}(L^2 + 4S^2 + 4L\cdot S)^{1/2}$$ whose expectation value is $$\langle|\mu|\rangle = \mu_\text{B} \left[ 2j(j+1) - l(l+1) +2s(s+1)\right]^{1/2}$$ in which we have used the equality $2L \cdot S = J^2 - L^2 - S^2$.

Now, a different approach could be to use the proportionality of $\mu$ and $J$ through the Landè g-factor $\mu = -g_J \frac{\mu_\text{B}}{\hbar}J$, which would give $$\langle|\mu|\rangle = (\mu^2)^{1/2} = \frac{\mu_\text{B}}{\hbar}g_J(J^2)^{1/2} = \mu_\text{B} g_J[j(j+1)]^{1/2} \\ = \mu_\text{B} \left[\frac{3}{2}+\frac{s(s+1)-l(l+1)}{2j(j+1)}\right][j(j+1)]^{1/2}$$ which is clearly different from the equation above. Putting some values for $l$, $s$, and $j$, one indeed obtains different results.

I tend to think that the first result is correct, but I cannot think about valid reasons to confute the second derivation.

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In general, $\hat{\boldsymbol{\mu}} \neq (-g_J \mu_\text{B}/\hbar) \hat{\mathbf{J}}$ as operators. Indeed, unless the g-factors of the individual angular momenta are the same, the two operators cannot be proportional since $\hat{\mathbf{J}} = \hat{\mathbf{L}} + \hat{\mathbf{S}}$. The equality is between their matrix elements with respect to the eigenstates of the $z$-component of the total angular momentum for a given $j$. So $$\langle j,m'_j| \hat{\boldsymbol{\mu}}|j,m_j\rangle = -g_J \frac{\mu_\text{B}}{\hbar} \langle j,m'_j|\hat{\mathbf{J}}|j,m_j\rangle$$ but this does not mean that $$\hat{\boldsymbol{\mu}} = -g_J \frac{\mu_\text{B}}{\hbar} \hat{\mathbf{J}}.$$

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  • $\begingroup$ I see. Not sure why, I have to check better the Wigner-Eckart theorem. Thank you $\endgroup$ Commented yesterday

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