# Magnetic moment of hydrogen atom in weak magnetic field from hyper fine structure

So I need some help with this question, which is formulated:

"Calculated the total magnetic moment of a hydrogen atom in the ground state in a weak magnetic field which arises from hyper fine splitting of the ground state. How many beams does hydrogen produce from a Stern-Gerlach analyzer with a weak magnetic field?"

I've calculated the magnetic moments from the hyper fine structure splitting to be

$$\mu_{1} = \mu_{B}, \mu_{2} = -\mu_{B}$$

but I don't really know how to get the total magnetic moment of hydrogen. My guess so far is to compute the vector sum of

• the electron's spin magnetic moment
• the electron's orbital magnetic moment
• the proton's spin magnetic moment
• the magnetic moment that arises from the hyper fine structure ($\mu_{1} \text{ and } \mu_{2}$),

but I have no idea where $\mu_{1} \text{ and } \mu_{2}$ "point". How do you "vectorize" them?

I'm also kind of confused as to how the magnetic moment of a particle is calculated. For example, the spin magnetic moment of a particle is

$$\mu_{S} = \frac{gq}{2m}S$$.

Since S is an operator, this also means that the magnetic moment, $\mu_{S}$, is an operator? This kind of confuses me. Is the "length" of $\mu_{S}$` given by the root of the eigenvalues of $\mu_{S}^2$, which would be $\frac{gq}{2m}\hbar\sqrt{s(s+1)}$?

Lastly, I'm guessing that to answer the question "How many beams does hydrogen produce from a Stern-Gerlach analyzer with a weak magnetic field?", you just count the number of different values you get for the total magnetic moment?

Thanks!

You perhaps missed the most important stipulation in this context, that the hydrogen atoms are in their ground state. That means, the electron is in the $1S$ state, and the azimuthal quantum number $l = 0$, which means no orbital angular momentum with this stipulation. Since $\vert {\vec \mu_L} \vert \propto \vert {\vec L} \vert$, that means no magnetic moment from the orbital angular momentum part, and hence, no magnetic interaction due to this either.

Thus, the only source of magnetic interaction possible in this case, is the spin angular momentum, and the consequent magnetic moment ${\vec \mu_S}$. To analyze this, we note that $s = \frac{1}{2}$ for an electron, and hence $\vert \vec S \vert$ is uni-valued, being equal to $\sqrt{3} \hbar/2$. With the magnitude determined, all one needs to find is the orientation of this $\vec S$, for which one has two possible scenarios: either $S_z = m_s \hbar$ with $m_s = +\frac{1}{2}$ ("up-spin"), or with $m_s = +\frac{1}{2}$ ("down-spin"). Since $\cos \theta = S_z/\vert {\vec S} \vert,$ this fixes the orientations of $\vec S$ with respect to the direction of the external magnetic field $\vec B$, to lie (on a cone) making an angle $54.7^{\circ}$, or $125.3^{\circ}$. Thus, there are two possible orientation angles for the spin angular momentum vector, with respect to the z-direction, or the direction of the external magnetic field $\vec B$.

Since ${\vec \mu_S} = - g_s {\vec S}$, where $g_s$ represents the spin gyromagnetic ratio for the electron, the anti-parallel relation between the two dictates that there are only two possible orientations of $\vec \mu_S$ w.r.t. $\vec B$. Upon passing through a Stern-Gerlach setup, there would be as many traces on the screen, as are the possible values of the product $(\vec \mu_S \cdot \vec B)$, which is 2.

On a sidenote, this question refers to the Phipps-Taylor implementation of the Stern-Gerlach experiment, carried out/published in 1927, which demonstrated the existence of "spin". If one ensures that the electrons in hydrogen atoms stay in the ground state only, which can done by cooling the atoms to a low temperature, the only source of magnetic interaction in the context can be the spin degrees of freedom of the electrons. (Of course, there is a caveat, there is also the magnetic interaction of protons in the nucleus with the external magnetic field, but nuclear magnetic moments are typically about 2000 times smaller than their electronic counterparts.)

(image source: paper cited above)

So, the two Stern-Gerlach traces in this case, invariably demonstrate the validity of the concept of "spin" as being real, and not just an abstraction.

• Thanks for the reply! :) I totally forgot that there's not orbital angular momentum since $l=0$, but I do know what role spin plays. My question now is: how does the magnetic momentum which arises from the hyper fine structure splitting play in? In an update to the assignment our professor gave us the matrix $$W = \begin{matrix} \mu_B B + A/4 & 0 & 0 & 0 \\ 0 & A/4 & 0 & \mu_B B\\ 0 & 0 & -\mu_B B + A/4 & 0 \\ 0 & \mu_B B & 0 & -3A/4 \end{matrix}$$ and said that the eigenvalues of this matrix are the energy corrections. – Trettman May 13 '18 at 12:50
• He also said that the magnetic moment for a weak magnetic field is given by $\mu = \partial E^1/\partial B$ as B → 0. This gave me $\mu_1 = \mu_B, \mu_2 = -\mu_B, \mu_3 = \mu_4 = 0$ . I don't really know how to calculate the total magnetic moment, with these values included. – Trettman May 13 '18 at 12:50