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What is the relationship between quantum spin and classical spin?

  1. Jan 21, 2005 #1
    What is the relationship between quantum spin and classical spin? I have seen this question asked many times but have never seen an answer that I was happy with.

    The most common answer I have seen is that there is no connection. The word "spin" has more to do with history than physics and there is no more connection here than there is between color forces between quarks and the color red.

    But then others have pointed out that spin, either quantum or classical is angular momentum.

    Can you convert classical spin to quantum spin? For example what would happen if you fed a black hole some polarized light?

    Can an object have both quantum spin and classical spin? Can a proton have classical spin for example? How about a proton attached to a neutron? A methane atom? A large protein? A baseball?

    Is classical spin quantized? It would seem that it must be but then you have two different kinds of quantum spin. . .
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  3. Jan 21, 2005 #2


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    Is going the other way a satisfactory answer? You can magnetize an object so all/most of the spins are aligned one way, then suspend it at rest from a thread, and flip the spins by reversing the magnetization. The object as a whole starts to rotate in order to conserve total angular momentum.

    Feynman refers to this experiment somewhere in his lectures.

    This is similar to the common General Physics demonstration in which a guy sitting on a turntable holds a spinning bicycle wheel and flips it over.
  4. Jan 21, 2005 #3
    Cool! Kinda what I'm looking for.

    Is conservation of quantum or classical spin being violated here? I'm guessing that flipping the spins of the atoms in the object requires applying a magnetic field which is flipping the spin on atoms outside the object in the opposite way. And I'm guessing that the classical spin of the object is matched by an opposite spin by whatever is applying the field. So both spins are conserved?

    What I'm looking for is a way to violate conservation of both quantum and classical spin in such a way that they obey a deeper conservation law.
  5. Jan 21, 2005 #4

    Tom Mattson

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    It's true that quantum mechanical spin has no classical analogue, and it is also true that quantum mechanical spins can add up to produce an observable rotation of a macroscopic object. It's called the Einstein-de Haas effect, and amazingly we have a recent discussion thread on it here. In fact, if you Google "Einstein de Hass effect", that thread is the first hit!
  6. Jan 22, 2005 #5
    There does not seem to be much information around on the Einstein de Hass effect. Is the conservation of quantum spin or classical spin violated by this effect?
  7. Jan 22, 2005 #6

    Tom Mattson

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    I know. :frown:

    I learned about it in my Quantum Mechanics II course. Basically, if you have a ferromagnet and you apply an external B field to align the spins, they will add up to produce a noticeable angular momentum, manifested by the rotation of the magnet.

    No. No exception to the conservation of angular momentum has ever been found.
  8. Jan 22, 2005 #7
    What about my example of dropping polarized particles into a black hole?

    Remember I'm not trying to violate conservation of total angular momentum. I'm trying to convert quantum spin to classical spin so that they each seem to be violated individually but are conserved when you look at both.

    A black hole breaks the symmetry between matter and antimatter. In theory anyway. I wonder if it would do the same for quantum and classical spin.
  9. Jan 22, 2005 #8
    I also find myself intensely interested in this question. I have some questions to ask:

    1. Since the electrons are not actually "orbiting" the nucleus, how can they have "orbital momentum" that adds up to actual angular momentum of the object they make up?

    2. Since the electrons only make up a vanishingly small portion of the mass of an atom, yet are very much further from the center of mass than the nucleus, what is the distribution of angular momentum within the atom like?

    3. The fact that BECs can be created hints that the angular momentum of an atom must overall be zero, but doesn't this experiment belie that?

    4. What are the implications of the spins of the individual electrons, and of the nucleus or of its constituent nucleons?
  10. Jan 22, 2005 #9


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    The electrons don't have trajectories, and their angular momenta about any two axes in space obey the uncertainty principle (the operators to find them don't commute), but given these caveats the angular momenta are quantum observables.

    I'll let somebody else handle this one; I would only be guessing.

    Is this a misunderstanding? Bosons aren't restricted to zero spin, but to integral spin. The spins of the subatomic particles in an atom add up algebraically, and if they come out an integer the atom is bosonic and theoretically can form a BEC. If they come out an integer plus 1/2, the atom is fermionic, and single atoms like that can't form a BEC. There has been recent progress in forming correlated pairs of fermionic atoms which act like bosons and can themselves form BECs.

    I think I answered this at question 3, but if not, just say so.
  11. Jan 22, 2005 #10
    Hmmm, I thought it was quite clear by my stating that they are not actually orbiting the nucleus that I understood they could not have a trajectory. Also, I am not asking about their rotational angular momentum, which I thought was clear from my use of the term "orbital momentum."

    Let me be very clear: I am asking specifically since they do not have a trajectory how they can have orbital angular momentum, which appears in an explanation above (unless I misinterpreted what someone else said).

    Thanks for clarifying my thinking on this issue; you were correct, I confused "zero" and "integer" spin, which is easy enough to do, zero being an integer. :blushing:

    Now of course the issue of the orbital angular momentum (as opposed to spin angular momentum) becomes even more important.

    OK, perhaps I was overly concise. What are the implications for the overall spin of the atom of the spin angular momenta of the individual electrons, and of the nucleus and/or of its constituent nucleons, as opposed to the implications for the overall spin of the atom of the orbital angular momenta of the electrons?
  12. Jan 22, 2005 #11


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    To quote my pal Marlon,QM and group theory.Their Hamiltonian (both at classical and at QM level) is rotation invariant.Applying Noether's theorem (quantum case),one gets an operator which describes angular momentum for a quantum system rotation invariant...

    Could u brake this one into more parts???I simply don't get it... :confused:

  13. Jan 22, 2005 #12

    Hans de Vries

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    You're probably interested to see the two video's here:


    - Photon spin angular momentum transferred to a microscopic particle.

    - Orbital angular momentum transferred from a Laguerre-Gaussian
    (LG) light beam to a microscopic particle.

    Regards, Hans.

    PS. have a look here also: http://www.physics.gla.ac.uk/Optics/projects/AM/
  14. Jan 22, 2005 #13


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    That's an interesting question.It's solvable (and i hope u can understand why) only for the Schroedinger's version of H-atom...I invite u to compute the distribution of angular momentum (expressed by the average values for the quantum observables) for that "vanishingly small portion (...)" for an arbitrary quantum state labeled:
    [tex] |n,l,m\rangle [/tex]

    for the (orbital) angular momentum operators...Use the orthonormalization for the eigenstates and the answer will be immediate...

  15. Jan 23, 2005 #14
    OK, that makes sense. What do the results look like? Do we get a quantized spin out of this, as we do for spin angular momentum? How is the result related to spin angular momentum quantum numbers?

    OK, we have three elements here:
    1. The orbital angular momentum of the electrons, individually or as a whole. That is, the class of Hamiltonians you described above. Hopefully this can be represented on some basis that will allow us to take it into account with the spin angular momenta below.

    2. The spin angular momentum of the electrons, which since it is quantized will come out as some whole (or half-integer) number.

    3. The spin angular momentum of the nucleus, which since it is made up of neutrons and protons will come out as some whole (or half-integer) number.

    Once we add these three elements together (assuming of course that we can state the electron orbital angular momentum in a basis that allows it to be combined with the spin angular momenta of the electrons and nucleons/nucleus), we can describe the spin of the atom.

    If there is no basis in which we can describe the orbital and spin angular momenta so that they can be combined, then we need to account for the fact that atoms can form BECs without reference to their orbital angular momentum (since only spin would be left) and show what effect the orbital angular momentum might have on their behavior and that of the BEC.
  16. Jan 23, 2005 #15
    Unless I misunderstand, this may be the basis (from the last post) in which we can combine the orbital and spin angular momenta, but I cannot for the life of me see how it yields the Hamiltonian you spoke of in the last post, and I also cannot see how it relates either of these angular momenta to the spin angular momentum of the nucleus. Which is really kind of what I'm asking.

    I should also point out that it has to be clear that the angular momentum is a function not only of the mass but of the radius; thus, I was implying that the electrons, though light, have a large radius compared to the nucleus, and the nucleus, though small in radius, is very heavy, and asking if these two cancel or nearly cancel one another out.
  17. Jan 23, 2005 #16


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    What spin??What has spin gotta do with ORBITAL angular momentum...???There's absilutely no connection whatsoever.For an electron found in a eigenstate of [itex] \hat{L}_{z} [/itex],one cannot tell anything about its spin state...

    WE HAVE 4:Two types for angular momentum for the nucleus (seen as a whole,i.e.its orbital & spin angular momentum are found by composing/ading the spins and orbital angular momenta of the particles which compose it)) and 2 for the electrons which form the shells+subshells...
    The "spin of the atom" is described only by 2 elements of the 4 mentioned above.I hope u see which ones... :rolleyes:

    I believe Clebsch-Gordan's theorem comes up with a base ...

  18. Jan 23, 2005 #17
    Gee, Dex, I don't know what spin ANGULAR MOMENTUM might have to do with orbital ANGULAR MOMENTUM. I was hoping you might tell me.
  19. Jan 23, 2005 #18
    Look, here is how spin and orbital momentum were introduced in QM...

    I am sure you know the famous Stern and Gerlach experiment. What happened there was a beam of silver atoms passing through an inhomogeneous magnetic field B, aligned in the z-direction. Now theory predicts that de force acting on the silver atoms is equal to the magnetic moment in the z-direction multiplyed by the first derivative of B with respect to x ,y and z. This immediately explains why the B-field cannot be homogenous because this force would be zero (dB/d(x,y,z) =0). In this case the magnetic moment of the atoms would just precess around the z-axis with a frequency that is called the Larmor-precession. Now, as you know from elementary electromagnetics, a little circular current is equivalent to a magnetic dipole. This implies that each atom is a magnetic dipole because it can be seen as a nucleus with an electrical circular current coming from the orbiting electrons. Please, let's not get into the discussion whether or not electrons actually orbit the nucleus because it is useless. I am giving you the justification for the reason why atoms can be seen as magnetic dipoles. If you have many atoms, you have many dipoles and if you add them all up you get the magnetization. This magnetization M is directly connected to the angular momentum L because of the above stated reasons (duality of circular current coming from the electrons and a magnetic dipole). In the Stern and Gerlach experiment, the force is thus also equal to the z-component of L multiplyed by the first derivatives of B with respect to x,y and z. These yield the three force components.

    Now back to the experiment: Bohr proved that L was quantized using the results of the Franck-Hertz experiment and the famous Ballmer-spectral lines of a hydrogen atom. He found out that L=n * hbar where n needed to be an integer. Now, in our experiment, the magnetic moments (you know, the dipoles, or in other words the silver atoms) were randomly directed. Therefore, when passing through the B-field it was expected that the beam would deflect in many ways all symmetric around the z-axis. The reason for this is the fact that L_z varied between M and -M (the minimal and maximal magnetization value of the silver-atoms). However this was not observed. We observed that the bean had split up in only TWO subbeams yielding two dots on the detector screen. Basically this meant that L_z only had TWO possible values ofcourse. In QM it was proven that L² was proportional to l(l+1) and that L_z had 2l+1 possible values. l is the orbital quantumnumber. So we have from Stern and Gerlach that 2l+1 = 2 or that l = 1/2...But this s a big no no because it contradicts the fact that l needs to be an integer. This was the first result that suggested another quantumnumber was necessary. The experimental verification for the spin was the Zeemann-effect. You know, the doubling of energylevels when placed into an external magnetic field. It was back in 1924 (Stern and Gerlach was in 1921) that both Goudsmit and Uhlenbeck proposed the existence of the socalled intrinsic spin quantum number. This number had nothing to do with the current-dipole duality but it was a property of each atom on itself. All this was implemented in a theory and the final result was the total angular momentum J = LA + sa...I am sure you know of this. For silver l = 0 and s = 1/2 so that j=1/2 and therefore the degeneracy 2 (you know , of LA_z in stern and gerlach) = 2j+1 is RESPECTED for and integer l. All this thanks to spin s.

    The best experimental verification of spin is the very accurate determination of the gyromagnetic factor g.

  20. Jan 23, 2005 #19
    Marlon, an excellent and detailed explanation. Thank you very much. I appreciate the effort.
  21. Jan 23, 2005 #20

    Hans de Vries

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    My favorite book on this is: "The Story of Spin", by QED nobel prize
    winner Sin-Itiro Tomonaga.

    Regards, Hans.
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