1
$\begingroup$

I know that a magnetic dipole moment is given by $$\mu=\frac {-e}{2m}I$$ and that the z component of angular momentum is $$m_j\hbar.$$ However, I have also seen that angular momentum $I$ is given by $$I=\hbar\sqrt{l(l+1)}.$$ Is this also the component in the z direction or is it different? Are both expressions for angular momentum right? Which one do I use to answer the question?

(This question was asked in a problem sheet that I have been asked to do over the vacation. See below for context of question). enter image description here

$\endgroup$
1
$\begingroup$

The first equation you have is for spin. Let's look at it in a operator form

To properly treat the upcoming discussion, quantum electrodynamics is introduced to describe the eigenstates of spin. The eigenstates of spin has a direct correlation with spin. Formally we depict it as the following

$$ μ_B =-g \frac {eℏ}{2m} \frac { \boldsymbol {S}}{ℏ} $$

where μ is the magnetic moment operator, g is the g factor, ℏ is plank’s constant over 2π, m is the mass of the particle and S is the spin operator.


For orbital magnetic moment (your question) you use:

$$\mu =\hbar\sqrt{l(l+1)} \boldsymbol {\mu_B} $$

Recall:

s: l=0 p: l=1 d: l=2 f: l=3

So plugging in:

$$\mu =\hbar\sqrt{1(1+1)} \boldsymbol {\mu_B} $$ $$\mu =\hbar\sqrt{2} \boldsymbol {\mu_B}$$

(Units are $ \boldsymbol {\mu_B}$)

$\endgroup$
  • $\begingroup$ How are you defining $\mu_b$? Because, I understand it to be $\frac{e\hbar}{2m}$. If that is the case, shouldn't $\mu=\sqrt 2 \mu_b$, rather than $\mu=\hbar \sqrt 2 \mu_b$ $\endgroup$ – RobChem Apr 5 '15 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.