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An issue with the vector model of spin and its operators

  1. Sep 7, 2012 #1
    I cant seem to reconcile a part of the vector model of spin and some of its operators.

    to quote wiki just above the "Bohr model" section:


    "2.The magnitude of the vectors must be constant (for a specified state corresponding to the quantum number),
    so the two indeterminate components of each of the vectors must be confined to a circle,
    in such a way that the measurable and un-measurable components (at an instant of time)
    allow the magnitudes to be constructed correctly, for all possible indeterminate components."

    I would assume pythagorean theorem holds true and "the two indeterminate components (S_x & S_y)
    of each of the vectors must be confined to a circle" of radius r:

    magnitude = sqrt(s(s+1))

    r = sqrt(s(s+1) - m^2)

    however the operators are defined as:

    S+ = sqrt(s(s+1) - m(m+1))
    S- = sqrt(s(s+1) - m(m-1))


    I must be missing something obvious. Please help.

    A visual of spin 5/2 with some notes:

    http://dl.dropbox.com/u/13155084/SPIN/SPIN-5-2-ladder-crop.png [Broken]
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 8, 2012 #2


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    Jeremy, It's unusual to see someone taking the vector model seriously - like driving a model T! :smile:
    Be sure to understand that its only a model. In quantum mechanics angular momentum is not really represented this way, as precessing vectors.

    If you look more closely at the definitions of S+ and S-, you'll see that they are stepping operators, that is they act to change the value of m up or down by one:

    S+ |m> = √(s(s+1) - m(m+1)) |m+1>
    S- |m> = √(s(s+1) - m(m-1)) |m-1>

    so you can't just treat them as magnitudes.
  4. Sep 10, 2012 #3
    Thanks Bill for you insight.

    I am very new to this vector/spinor model, hence my unusual view.

    I have heard that there is some controversy about all these
    quantum and non-quantum visualizations of spin from Stan Sykora. He maintains the
    high-resolution NMR spectra simulation dll in Mnova software and added some comments on
    http://oeis.org/A003991 that peaked my interest on the subject.

    I am trying to fully understand the concept of these spin operators because they are
    eerily close to a root system I've been working on involving prime number distribution.

    We already know a connection between number theory and quantum mechanics comes from
    the discovery that the spacing's between consecutive zeros of the zeta function also appear to
    behave statistically like the spacing's between consecutive eigenvalue's of large random matrices
    which physicists use to obtain estimates of the average spacing between consecutive energy levels
    of heavy atomic nuclei and other complex quantum systems.

    So these ladder operators act to change the value of m up or down by one essentially changing the
    observed value on the Z axis. Is it not correct to say that once S_z is observed, the un-measurable components
    must lie somewhere along a circle of radius sqrt(s(s+1) - m^2)?
    Last edited: Sep 10, 2012
  5. Sep 11, 2012 #4


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    Jeremy, In the vector model, Sx and Sy could take on any continuous value along the circumference of a circle. The modern viewpoint is different - Sx and Sy (and any other spin projection) can only take integer or half-integer values. They can't be measured simultaneously, and so for a state in which Sz has been determined there will be some probability that Sx takes on each of its allowed discrete values.
  6. Sep 11, 2012 #5
    Thank you again. I assumed the values of Sx and Sy would be quantized as well. It seems logical seeing as the Z axis is actually an arbitrary direction usually determined by an external magnetic field. Sz is just used for convention, we could substitute Sx and Sy just as easily. Is there a formula for the probability of these other allowed discrete values?

    Also, I still can't understand the ladder operators fully. They have been described as the relative intensities of the states with the equivalence S+ = Sx + i Sy and S- = Sx - i Sy, but the geometry doesn't seem to fit with the eigenvalues. Can you shed some light on this for me?
    Last edited: Sep 11, 2012
  7. Sep 15, 2012 #6

    If you think of the possible spin states as a function of the magnitude of spin then:

    |S| = magitude
    S+-=|x +- iy| = magnitude

    applying the ladder opperators on the S_x states
    one could come up with something like this model.

    Notice the X-axis:
    https://dl.dropbox.com/u/13155084/SPIN/SPIN-5-2-ladder%20XYZ.png [Broken]
    One link to the fact that you cannot observe the other projections is because
    they do not fall on integer or half-integer values.
    Is this a proper view of the ladder operators of the un-observable states?
    Last edited by a moderator: May 6, 2017
  8. Sep 21, 2012 #7
    Here is a 3D model of possible S_x and S_y states based upon the ladder operators.
    You will need the Flash plugin to view. This runs through all Spin states up to s=75/2.
    http://dl.dropbox.com/u/13155084/SPIN/index.html [Broken]
    Last edited by a moderator: May 6, 2017
  9. Sep 21, 2012 #8
    Can't say as I understand your Flash model, but it sure does look cool, Jeremy. Feel free to explain it a bit more. Also, since you mentioned spin 5/2...
    ... here is a purely numerological observation that may or may not be of interest to you in relation to Stan Sykora's comments here at http://oeis.org/A003991 (the ones that piqued your interest):

    [itex]S[/itex]=[itex]\frac{\hbar}{2}[/itex][itex]\sqrt{n(n+2)}[/itex] = [itex]\frac{\hbar}{2}[/itex][itex]\sqrt{5(5+2)}[/itex] == [itex]\frac{\hbar}{2}[/itex][itex]\sqrt{5 + 8 + 9 + 8 + 5}[/itex] = [itex]\frac{\hbar}{2}[/itex][itex]\sqrt{35}[/itex] for n = 5

    In other words, for the special case of n = 5, then 35 is tetrahedral as per Sykora's example (sum of relative intensities of Spin 5/2 transition states) and also one less than a square, meaning that it follows form n(n + 2) as per the above formula for spin. The maths work out in this instance because any tetrahedral number is equal to the product of 3 consecutive integers divided by 6. Therefore, 5*6*7/6 = 35, the fifth tetrahedral number, and because the 6's cancel it is also one less than a square (=5*7).

    - AC
    Last edited by a moderator: May 6, 2017
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