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This question, posed in a problem sheet that I have been asked to do, has stumped me. I really don't know what to do here. Any help would be greatly appreciated.

I know that the magnetic moment of an electron orbiting a nucleus is given by $$\mu=-\frac e{2m}L.$$

I also know that the angular momentum quantum number is zero so therefore so is $m_l$. I don't know which one is more relevant since $J_z=\hbar m_l$. How would one incorporate spin into this?

Qmechanic
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RobChem
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1 Answers1

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When you have orbital angular momentum and spin angular momentum for an electron then the resulting magnetic moment $\mu_j$ is $$\vec \mu_j=\vec \mu_l+\vec \mu_s= -\frac{e}{2m_e}(\vec l+g\vec s)=-\frac{\mu_b}{\hbar}(\vec l+g\vec s),$$ where $\mu_b=\frac{e\hbar}{2m_e}$ is Bohr's magneton, $l$ denotes the orbital angular momentum, $s$ the spin angular momentum, $e$ the electron charge and $m_e$ the electron mass. $g$ is the Landé factor which can be measured in experiment and calculated. The value of the Landé factor for an electron is $g\approx2$.

In your question the electron is in an s-state, so $\vec l$ is zero and the total magnetic moment is

$$\vec \mu_j=-g\frac{\mu_b}{\hbar}\vec s=-\frac{e}{m_e}\vec s$$

You can calculate the z-component and the magnetude of $\vec\mu_j$ by considering the z-component and the magnitude of $\vec s$:

$$|\vec s|=\sqrt{3/4}\hbar, \quad s_z=\pm1/2\hbar$$

Anticipating the next question from your sheet:

What if $l\neq0$, is there a nice, clean expression for $\mu_j$?

The problem is that this magnetic momentum vector $\vec\mu$ is not parallel to the total angular momentum vector $$\vec j=\vec l+\vec s.$$ Due to an interaction between the electron spin and it's orbital motion the vector $\vec s$ is not fixed in space. This then means that $\vec\mu_j$ is also not fixed in space. However, you can calculate the average value of $\vec\mu_j$ by projecting it on the vector $\vec{j}$, this is why it says "averaged according to..." in your problem sheet. If you do this then you will find that the average value is

$$\langle\mu_j \rangle = g_j\frac{\mu_b}{\hbar}|\vec j|=g_j\frac{\mu_b}{\hbar}\sqrt{j(j+1).} $$

$g_j$ is the Landé factor from your problem sheet. Since you consider an electron, the gyromagnetic factor for it's motion ($g_l$) is one (for the neutron it's zero) and $g_s$ is two, so you need the second formula.

In the formula for $g_j$ and $\mu_l$ you have the values for s,l and j. When you calculate $g_j$ you have always to consider those values for a specific state. So for a 2p electron $l=1$, $s=1/2$ and $$j=l\pm s=1/2 \quad \text{or} \quad 3/2.$$ So, for each of those set of values (1,1/2,3/2 and 1,1/2,1/2) you calculate $\langle \mu_j\rangle$.

I hope this helps!

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  • Why did you use $l$ rather than $ m_l$ ? – RobChem Apr 04 '15 at 16:53
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    @RobChem $l$ in my answer is the angular momentum quantum number and $\vec l$ the angular momentum vector. $m_l$ on the other hand is only the $z$-component of $\vec l$. You need $m_l$ when your electron is in an external magnetic field and you want to figure out the different values the energy can take. Since for any given value of $l$ the number $m_l$ can take on different values the energy depends on $m_l$. (See zeldredge's answer to your other question.) Similarly, for any value $j$, the total angular momentum number, there are different values of $m_j$, it's z-component. –  Apr 04 '15 at 17:19
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    So to make things more clear: the numbers $s$,$l$ and $j$ are the quantum numbers associated with the length of the according vectors, while the numbers $m_s$,$m_l$ and $m_j$ are the quantum numbers associated with the z-component of the according vectors. –  Apr 04 '15 at 17:25
  • Thank you very much. Could you please explain how you got the value for $|\vec s|$ and $s_z$ please? Also, what's the best book/website/you tube video(s) to help me learn these fundamentals of QM (at a first year chemistry undergraduate level)? – RobChem Apr 04 '15 at 18:50
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    @RobChem You're welcome! The values for $|\vec s|$ and $s_z$ are obtained from the same eigenvalue equations as for the angular momentum operator $L$, except that you change l for s: $\hat{S}^2 | n s m_s \rangle = \hbar^2 s( s + 1) | n s m_s \rangle$ and $\hat{S}_z | n s m_s \rangle = m_s \hbar | n s m_s \rangle$. You can do this because these relations are obtained from operator relations and are thus true for any angular momentum (including $\vec j$). The length of $\vec s$ is then obtained by taking the square root of the first equation and setting $s=1/2$. –  Apr 04 '15 at 19:24
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    You then have $|\vec s|=\sqrt{1/2*3/2}\hbar$. The z component is obtained by setting in the second equation $m_s=\pm 1/2$. Hmmm I think you should look at Griffiths "Introduction to quantum mechanics" https://archive.org/details/IntroductionToQuantumMechanics_718 But if I were you I would go to your library, borrow a few books and try which suits you best. –  Apr 04 '15 at 19:26
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    edit: you could also try to find these at your library: "Tipler: Physics" (contains a chapter on qm.) and "Halliday/Resnick:Physics 2" (same). They're not as "fundamental" as griffiths, but they're good and usually contain a lot of text, which is always good if you want to understand something. ^^ –  Apr 04 '15 at 19:39
  • Thank you very much. The text from the link you gave me is very helpful so far. – RobChem Apr 04 '15 at 20:38