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Measuring angular momentum in atomic systems

by pellis
Tags: angular, atomic, measuring, momentum, systems
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pellis
#1
Oct30-08, 07:21 AM
P: 16
Perhaps rather an elementary question, but I can't find a clear answer in my textbooks:

Would I be right in thinking that we never actually measure directly the angular momentum of atomic systems, but rather: using the results of QM calculations about structure, and knowledge of the selection rules, we *infer* it from spectral transitions?
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jtbell
#2
Oct30-08, 07:53 AM
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What would you consider to be a "direct measurement" of angular momentum?

The Einstein - de Haas effect demonstrates that the microscopic angular momentum of electrons in a metal contributes to the object's total macroscopic angular momentum. Briefly (and probably oversimplified), you start with an object that's not rotating, then flip the spins of the electrons, and observe that the object starts to rotate macroscopically in order to maintain the same total angular momentum.
pellis
#3
Oct30-08, 08:32 AM
P: 16
You've identified the origin of my question, as I could not think of how an experiment could observe angular momentum in atomic systems directly.

Clearly, the Einstein-de Haas effect does demonstrate a way to do this - and spin angular momentum at that (which is what I'm pursuing).

However, I'm still left wondering how angular momentum is identified from atomic spectroscopy. I doubt that early work on atomic spectra relied on the Einstein-de Haas effect.

Can you, or anyone else, tell me how it's conventionally done?

ZapperZ
#4
Oct30-08, 09:00 AM
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Measuring angular momentum in atomic systems

Er.. the angular momentum of an atom is the origin of magnetism in solids! One doesn't need any "spectroscopy" studies to get that.

Zz.
pellis
#5
Oct30-08, 09:26 AM
P: 16
OK - let me put it a bit more precisely:

I'm asking about the orbital and spin angular momenta of electrons in, for example, low density gaseous states.

Textbooks glibly mention electrons being in states |n,l,m,s>

But how, observationally, do we come to know what m(l) and m(s) are for the states between which we observe spectral lines?

Suppose I excite some sodium vapour, and as Wikipedia states:

http://en.wikipedia.org/wiki/Sodium

"One notable atomic spectral line of sodium vapor is the so-called D-line, which may be observed directly as the sodium flame-test line (see Applications) and also the major light output of low-pressure sodium lamps (these produce an unnatural yellow, rather than the peach-colored glow of high pressure lamps). The D-line is one of the classified Fraunhofer lines observed in the visible spectrum of the sun's electromagnetic radiation. Sodium vapor in the upper layers of the sun creates a dark line in the emitted spectrum of electromagnetic radiation by absorbing visible light in a band of wavelengths around 589.5 nm. This wavelength corresponds to transitions in atomic sodium in which the valence-electron transitions from a 3p to 3s electronic state. Closer examination of the visible spectrum of atomic sodium reveals that the D-line actually consists of two lines called the D1 and D2 lines at 589.6 nm and 589.0 nm, respectively. This fine structure results from a spin-orbit interaction of the valence electron in the 3p electronic state. The spin-orbit interaction couples the spin angular momentum and orbital angular momentum of a 3p electron to form two states that are respectively notated as 3p(2p0,1/2) and 3p(2p0,3/2) in the LS coupling scheme. The 3s state of the electron gives rise to a single state which is notated as 3s(2S1 / 2) in the LS coupling scheme. The D1-line results from an electronic transition between 3s(2S1 / 2) lower state and 3p(2p0,1/2) upper state. The D2-line results from an electronic transition between 3s(2S1 / 2) lower state and 3p(2p0,3/2) upper state. Even closer examination of the visible spectrum of atomic sodium would reveal that the D-line actually consists of a lot more than two lines. These lines are associated with hyperfine structure of the 3p upper states and 3s lower states. Many different transitions involving visible light near 589.5 nm may occur between the different upper and lower hyperfine levels.[8][9]"
(see original in Wiki to see the term symbols displayed correctly).

Now, how precisely do we come to be able to state that a transition is between any of the above two states - ie to identify the states' various quantum numbers including the angular momenta?

As posed in my original question - the only way I can see this being achieved is if one first *calculates* the structure of the spectrum and thus the associated n,l,m,s, values, and then one assigns the observed spectral lines to those theoretically identified states. So one never actually observes the m(l) and m(s) values, but as mentioned in the first post, one infers them.

Or is there some other way to do this?
jtbell
#6
Oct30-08, 11:05 AM
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One way (there are probably others) to associate spin and orbital angular momentum quantum numbers of initial and final states with particular spectral lines is via the Zeeman effect. When you apply an external magnetic field, the energy levels of the different spin states shift and/or split by amounts that depend on the angular momentum quantum numbers, and on the strength of the magnetic field.
pellis
#7
Oct30-08, 03:35 PM
P: 16
Yes, what you say is correct, but I think I'm failing to make explicit the point of my question.

It seems to me that all the answers I've received come from hindsight, as is the case with textbooks.

But how do we know *from the outset* what the quantum numbers are (which lines correspond to transitions from the lowest values), and what the units of spin and orbital angular momentum are initially, unless we have an atomic model of some sort to start with.

For example: If you look at Ch1, Vol 1 of P.W. Atkins' "Molecular Quantum Mechanics", he outlines how Balmer, Rydberg and Ritz worked out some regularities in spectral lines which led Bohr to propose a model for hydrogen, based on a number of assumptions, including that:

"The stationary states are to be determined by the condition that the ratio of the total energy of the electron to its frequency of rotation shall be an integral multiple of h/2. For circular orbits this is equivalent to the restriction of the angular momentum of the electron to integral multiples of h/(2pi)"

...and the calculation based on (all) the postulates yields the electron's energies in the hydrogen atom as:

E(n) = - mu*e^4/(8n^2h^2eta^2) (mu being a reduced mass)

...where n = 1,2,3... is the first quantum number. And the result agrees well with experiment (as far as early observations went).

So, it appears that even at the outset, the unit of angular momentum is fed into the model, and not itself observed. The first quantum number is identified by a model, and I suspect that the *ranges* of possible values of l and m(l) drop out of the spherical harmonics as solutions to the Schrodinger equation - and that these provide the original basis for *interpretation* of the observations, rather than direct measurement of orbital angular momentum. And m(s) emerges from the doublet structure of spectral lines but still refers to the *calculated* unit h/(2pi), again rather than being measured directly.

I think I'm convincing myself that my original point was true, but it would be good to know if I'm wrong.

Thanks for the stimulus of your contributions.
jtbell
#8
Oct30-08, 07:09 PM
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Bohr's atomic model has been superseded for over eighty years by modern quantum mechanics. The quantum numbers for orbital angular momentum arise directly from the solution of the Schrödinger equation.

For spin, I think you have to go further to the relativistic Dirac equation, and assume that the magnitude of the spin angular momentum has a certain value; but after that, the mathematics of addition of quantum-mechanical angular momentum determine everything else. (Someone with more expertise than I in atomic physics is welcome to correct me on this.)


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