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I learned that if we measure the spin angular momentum of an electron in one direction $J_z$, we get $\pm \frac{1}{2} \hbar$. But if we measure the magnitude of the angular momentum $\mathbf{J}^2$, we should get $\frac{3}{4} \hbar^2$. What experiment gives the latter result?

As @user1585635 notes, measuring $J_x$, $J_y$, and $J_z$ separately and summing their squares gives $\frac{3}{4} \hbar^2$. This is not what I'm looking for. First, if I measure $J_z$, measure angular momenta in three directions separately, and measure $J_z$ again, the two measurements of $J_z$ aren't guaranteed to be the same, since $J_z$, $J_x$, and $J_y$ don't commute. But $\mathbf{J}^2$ commutes with $J_z$. Second, when the spin is not $\frac{1}{2}$, say it's $j$, summing the squares of components of $\mathbf{J}$ gives 3ℏ²𝑗², where 𝐉² should be ℏ²𝑗(𝑗+1) doesn't always give $\hbar^2 j(j+1)$. (Thanks to @MichaelSeifert for pointing that out)

This question is not a duplicate of Why is orbital angular momentum quantized according to $I= \hbar \sqrt{\ell(\ell+1)}$?. That question is about how $\mathbf{J}^2$ is derived mathematically. Mine is about how it is confirmed experimentally.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/174018/2451 , physics.stackexchange.com/q/11197/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 9, 2021 at 5:33
  • $\begingroup$ So you are seeking examples of higher j multiplets whose degeneracy is split to $2j+1* different lines in a magnetic field? Have you leafed through an atomic physics book? $\endgroup$ Jan 9, 2021 at 13:32
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    $\begingroup$ @ChiralAnomaly Yes.I am asking how to measure $\mathbf{J}^2$ without affecting $J_z$. $\endgroup$
    – 30tah8uu
    Jan 9, 2021 at 23:47
  • $\begingroup$ @CosmasZachos I am not looking for examples of higher j. I only used higher j to show that measuring components of angular momentum is not what I want. $\endgroup$
    – 30tah8uu
    Jan 9, 2021 at 23:50
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    $\begingroup$ Note that if you measure $J_z$ for a spin-$j$ atom, you do not necessarily get $\pm j \hbar$; you can get any one of the options $+j \hbar, +(j-1) \hbar, ..., -(j-1)\hbar, -j\hbar$. The same goes for $J_x$ and $J_y$. This means that the result of the measurement $J_x^2 + J_y^2 + J_z^2$ will not necessarily be equal to $3 j^2 \hbar^2$, as you say in your second paragraph; any one of those three measurements could give you a value less than $j^2 \hbar^2$. $\endgroup$ Jan 11, 2021 at 15:40

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While it's not in the context of an atom, molecular rotations provide a particularly "clean" demonstration of the fact that $\mathbf{J}^2 = \hbar^2 j (j+1)$. In particular, it's not hard to show that the rotational KE of two point masses connected by a rigid rod is $$ H = \frac{J^2}{2 I} \quad \Rightarrow \quad E_j= \frac{\hbar^2 j(j+1)}{2 I}, $$ where $I$ is the moment of inertia about the center of mass. In particular, this means that transition energies between adjacent rotational energy states are $$ \Delta E = E_{j} - E_{j-1} = \frac{\hbar^2}{2I} j [j + 1 - (j-1)] = \frac{\hbar^2 j}{I}, $$ i.e., they are evenly spaced by energies of $\hbar^2/I$.

This even spacing has been confirmed by many experiments, and the values of $I$ inferred from these spectra are consistent with other measures of the molecular dimensions. Below is shown the absorption spectrum of carbon monoxide; the evenly spaced absorption lines are evident. While the large absorption lines of the molecules composed of carbon-12 and oxygen-16 (the most common isotopes of carbon and oxygen) are the most obvious features, we can also distinguish smaller absorption lines from molecules composed of carbon-12 and oxygen-18. Molecules composed of different isotopes will have the same bond lengths but different moments of inertia due to the the different masses of their constituent atoms; this results in a different spacing of the rotational lines.

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From Hollas, Modern Spectrosopy, 1992; taken from these lecture notes.

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  • $\begingroup$ I see the question as asking for an experiment that can be done every time. However, I see that this analysis only works when the electronic level that these rotational levels sit on has an OAM of zero. If the electron level has non-zero OAM, then the inferential analysis above wouldn't give the true $\hbar^2j(j+1)$ of the system. $\endgroup$
    – Dr. Nate
    Mar 29 at 18:05
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The Zeeman effect, discovered in the 19th century, was understood in quantum theory by Lande and Sommerfeld early on. Essentially, in a magnetic field, the energy levels of atoms are split by differences of magnetic moments, which are proportional to $J_z$, or the corresponding S and L components comprising J. This then allows to count the degenerate states so split of a system with fixed J (S, L, etc...) in a mixed population, graphically and dramatically. So at the most primitive level, the effect is a handle on degeneracy, $2j+1$.

This degeneracy, $2j+1$, is an easy counting observable, emerging as an eigenvalue of the formal operator $$ \sqrt{4\mathbf{J}^2/\hbar^2+1} , $$ the ready hands-on manifestation of the quantum number j, which bounds the eigenvalues' magnitude $|j_z|$ from above. Even if you had no idea about quantization, the discreteness of these eigenvalues would cue you in to the quantum nature of the lines' spacings, and the integer nature of all these quantities. In the 1920s, Wigner and his cohorts changed the face of physics fully understanding these regularities in terms of the theory of the rotation group (Their injection of theory was termed "Gruppenpest" --Group pestilence--by the experimental community at the time).

enter image description here

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    $\begingroup$ I suspect OP is looking for an experimental measurement that directly relates $\hbar^2 j(j+1)$ to an angular momentum. $\endgroup$ Jan 11, 2021 at 15:38
  • $\begingroup$ But that's just math (Wignerism), which he forfeits: I read the question as an inquiry on the "physical significance" of j, which must be the link to degeneracy. As for directness, we've lost the meaning of the term a long time ago... $\endgroup$ Jan 11, 2021 at 15:42
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I suppose we measure spin component $S_z$ by measuring the shift in energy spectrum when we add a magnetic field, which gives an extra term $B_z S_z$ in the hamiltonian.

Similarly, to measure $S^2$ we probably need a term that relates it to some energy. The one I thought of is Fine Structure correction of the hydrogen spectrum, in which the energy is different for different values of total angular momentum $j$. But $\vec{J}=\vec{L}+\vec{S}$ so the eigenvalues j depends on $S^2$ being $\frac34 \hbar^2$.

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It is true that $J_x$, $J_y$ and $J_z$ does not commute.

But $J_z$ does commute with $\textbf{J}^2=J_x^2+J_y^2+J_z^2$. So there can be an eigenstate of both $J_z$ and $\textbf{J}^2$, where you can simultaneously measure both the z component of the angular momentum and the magnitude of the angular momentum.

If the possible eigenstates of $J_z$ each has an eigenvalue of from $-l\hbar$ to $+l\hbar$, each eigenstate will also be an eigenstate of $\textbf{J}^2$ with eigenvalue of $l(l+1)\hbar^2$.

For example, if the possible eigenvalues of $J_z$ are $-\hbar,0,+\hbar$, each of these three eigenstates of $J_z$ will also be an eigenstate of $\textbf{J}^2$ with eigenvalues $2\hbar^2$.

Note that the way we measure the magnitude of angular momentum is not by measuring the individual $J_x$, $J_y$ and $J_z$ components and then summing them up. That would be the classical way of measuring things.

In QM, we need to find the eigenvalues of the operator $\textbf{J}^2=J_x^2+J_y^2+J_z^2$ and then take the square root of its eigenvalues to find the magnitude of angular momentum we are able to measure physically.

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    $\begingroup$ My question is how it can be confirmed experimentally. I edited it in case I was not clear enough. $\endgroup$
    – 30tah8uu
    Jan 9, 2021 at 12:20

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