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Isn't classical spin an instance of spin 1?

  1. Aug 4, 2009 #1
    isn't classical "spin" an instance of spin 1?

    A spin 1 field A(x) is a 3-component vector field which transforms under rotations by

    [tex]A(x)\rightarrow exp[i(S+L)\cdot \hat{n}\theta]A(x) = exp(iS\cdot \hat{n}\theta)A(R^{-1}(\theta)x)[/tex]

    where S are the three 3x3 matrices

    [tex]S_1=\left( \begin{array}{c1 ... cn}
    0 & 0 & 0 \\
    0 & 0 & i \\
    0 & -i & 0
    \end{array}\right) [/tex]

    [tex]S_2=\left( \begin{array}{c1 ... cn}
    0 & 0 & -i \\
    0 & 0 & 0 \\
    i & 0 & 0
    \end{array}\right) [/tex]

    [tex] S_3=\left( \begin{array}{c1 ... cn}
    0 & i & 0 \\
    -i & 0 & 0 \\
    0 & 0 & 0
    \end{array}\right)
    [/tex]

    This is certainly true in quantum theory. Multiplying S by hbar (and so replacing i with i/hbar in the transformation) gives the quantum spin operators for spin 1. But, as far as I can tell, there is nothing particularly "quantum" about this. The classical electromagnetic vector potential also transforms like this. Any vector field transforms like this.

    Furthermore (entering less certain ground now) a rotating rigid body with intrinsic angular momentum [tex]L_{spin}[/tex] located at position x is essentially a vector [tex]L_{spin}(x)[/tex] attached to the point x. It is similar to the field A(x), except that rather than a field it is a single vector attached to a particular x. And does it not also transform under rotations through the operation exp(iS\cdot \hat{n}\theta)[/tex]? So is it not then a spin 1 instance.

    In short, if spin 1 is vector, and since classical spin is a vector, is not "spin 1" in this sense? That is, their generators of rotations are the same representation of SO(3).

    The quantum case is not "quantum" because its generators of rotations do not commute. It is quantum because the structure of the theory requires eigenvectors of the spin matrices and we cannot have simultaneous eigenvectors of noncommuting matrices. The classical case has no restriction on its vectors.

    Am I stating this correctly?
     
  2. jcsd
  3. Aug 4, 2009 #2

    Demystifier

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    Re: isn't classical "spin" an instance of spin 1?

    You are correct that spin is a classical property of a classical field. However, what makes it quantum is the probabilistic interpretation. If you postulate that A(x,t) is nothing but an auxiliary quantity serving for calculating |A(x,t)|^2 (or something similar) which is interpreted as the probability of finding the PARTICLE at the position x at time t, then spin is no longer a property of a field, but a property of the pointlike particle. This does not have a classical analog.
     
  4. Aug 4, 2009 #3
    Re: isn't classical "spin" an instance of spin 1?

    I think your point about the probability is the same as my point about eigenstates, since the physical significance of a quantum state is in its interpretation as a probability amplitude.

    But what about the angular momentum of a rotating rigid body?

    I guess what I am asking is: what is the relationship between the generators of rotation and angular momentum for a classical system? In quantum theory it is clear: the generators are identical to the angular momenta observables.


    Edit:
    Don't bother replying yet. Let me review Noether's theorem first. I will check back in again after.
     
  5. Aug 4, 2009 #4

    Ben Niehoff

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    Gold Member

    Re: isn't classical "spin" an instance of spin 1?

    In classical mechanics, if G is some function of (q,p), then we can derive a one-parameter family of transformations on phase space via

    [tex]\frac{df}{d\lambda} = [f, G] + \frac{\partial f}{\partial \lambda}[/tex]

    where f is some function of interest, lambda is some parameter, and [f, G] is the Poisson bracket. G is then said to be a "generator", and this is called an "infinitesimal canonical transformation". If we take G to be the Hamiltonian H, and the parameter to be time, we recover the familiar

    [tex]\frac{df}{dt} = [f, H] + \frac{\partial f}{\partial t}[/tex]

    which shows us that the Hamiltonian is the generator of time evolution. However, we can take G to be some other function as well, such as

    [tex]L_z = x p_y - y p_x[/tex]

    and then we can write

    [tex]\frac{df}{d\theta} = [f, L_z] + \frac{\partial f}{\partial \theta}[/tex]

    which tells us how the function f transforms via a rotation by some angle theta (about the z axis). In this sense, L_z is the generator of rotations (about the z axis). If you work this out for some given function f, then you should recover the familiar transformation laws.

    Also, if you work out the Lie algebra for the classical angular momentum generators using Poisson brackets, you should find something very similar to the quantum mechanical algebra (mostly identical, except for factors of i and h-bar).

    To move from classical mechanics to quantum mechanics, we simply replace Poisson brackets with commutators, and insert factors of i and h-bar. The h-bar is merely a unit convention; and the factor of i is so that we can make our operators Hermitian instead of anti-Hermitian.

    So yes, spin-1 exists classically. Actually, all integer-numbered spins exist classically; they merely correspond to the rotation properties of vectors, 2nd-rank tensors, 3rd-rank tensors, etc.

    The half-integer spins come about because there are additional representations of SO(3) besides the ones represented by classical objects. These new representations have the same infinitesimal properties (i.e., the same algebra is satisfied by the operators), but the global properties are different: they are invariant under a [itex]4\pi[/itex] rotation, but not under a [itex]2\pi[/itex] rotation.
     
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