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Basic questions about Spin3/2 formalism

  1. Jan 27, 2012 #1
    Hi there,

    I've been trying to understand the theory to descibe the delta(1232) resonance but I'm stuck... hope you guys could help me. After reading several papers, the following questions arose:

    1. How do I get (mathematically) the spinor respresentation (1/2,0)+(1,1/2)+(0,1/2)+(1/2,1)
    from the product of a vector and a spin-1/2 spinor?
    Why don't we use the (3/2, 0)-representation (I think this was alreay asked)?

    2. Problem with the lower spin components (s-1), (s-2), etc. corresponding to the (1/2,0)+(0,1/2) in the upper representation:
    How do I "see" them in the RS-Lagrangian?

    Is this problem exisiting in a Spin-1-theory: do Spin-0 d.o.f.s arise there?

    3. We remove the Spin 1/2 contributions by the constrain: γμψμ=0, where γ are the gamma matrices and ψ is the RS-field. What does this mean?

    4. Why the consistent(physical) vertex has to satisfy pμ[itex]\Gamma[/itex][itex]\mu[/itex]=0, with pμ being the 4-momentum?

    These questions remain unanswered to me.

    Thanks for reply!
    Last edited: Jan 27, 2012
  2. jcsd
  3. Jan 28, 2012 #2


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    PiepsNYC, The first thing to note is that the Rarita-Schwinger formalism is not the only way to go. There are three possible ways to describe higher spin particles:

    1) Directly in terms of spinors, i.e. quantities with explicit dotted and undotted indices. This is called the Fierz-Pauli method.
    2) As a single matrix quantity, obeying something that looks like the Dirac equation only with a set of larger matrices, β matrices in place of γ. (∂μβμ + m)ψ = 0 is known as the Bhabha equation.
    3) Rarita-Schwinger formalism, a hybrid approach using a Dirac-like quantity ψμ which is simultaneously a rank 4 matrix and a 4-vector.

    In all three approaches the main problem is to eliminate the extraneous degrees of freedom. A spin-3/2 particle should have 4 spin orientations for positive and negative energies each, a total of 8 degrees of freedom.

    Answer to question 1: Under the Lorentz group a 4-vector is the representation (1/2, 1/2). A Dirac spinor is (1/2, 0)⊕(0, 1/2). When you multiply these together, the first and second numbers combine like angular momenta. So since 1/2 ⊗ 1/2 = 1 ⊕ 0, we have

    (1/2, 1/2) ⊗ ((1/2, 0)⊕(0, 1/2)) = ((1, 1/2)⊕(0, 1/2))⊕((1/2, 0)⊕(1/2, 1))

    Count the components: 4 x 4 = 16 = 6 + 2 + 2 + 6. So there are 8 components to keep and 8 to eliminate.

    Now γμψμ is a Lorentz vector with 4 components, so setting γμψμ = 0 eliminates 4 of them. The other 4 comes from requiring ∂μψμ = 0. Imposing both of these conditions we are down to something that looks like a spin 3/2 particle.
  4. Jan 30, 2012 #3
    Thanks, Bill_K!
    As I understand a Spin 3/2 particle has 8 components for Spin 1/2, up and down, E>0 and E<0 (that would be 4) and 4 more for Spin 3/2. Is this true?
    So my question is: where do the other 8 come from? Is this just a mathematical issue?
    And shouldn't be the contributions of the spin 1/2 the ones we wish to eliminate (the 4 components mentioned before)? I am now confused about which components correspond to the Spin 1/2.
    To what would it correspond in the vector-spinor representation?

    How do I know which components I am eliminating by the constrains?
    ...:(........What characterizes a Spin 3/2 particle?

    The more I read, the less I understand....

    Thanks for reply!
  5. Jan 30, 2012 #4


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    PiepsNYC, It's a fact of life that the field quantities we have to work with (spinors, vectors, etc) do not correspond directly to one definite spin. Our only choice is to formulate with a field that contains the desired spin along with others, and then eliminate the ones we don't want.

    To count them again, a spin-3/2 particle will have 4 spin states with spin projection sz = +3/2, +1/2, -1/2 and -3/2. Similarly for E<0 (the anti-particle). Total of 8. But ψμ starts with 16.

    The first constraint γμψμ is an algebraic constraint that eliminates the (1/2, 0)⊕(0, 1/2) part entirely. Down to 12.

    The second constraint ∂μψμ = 0 is quite different, it's analogous to the Lorentz condition. As we know from QED this type of constraint cannot be imposed throughout a Feynman diagram, just on the external legs. So the remaining 4 'extra' components are present in all the calculations and need to be eliminated at the end.

    Hope that clears it up some.
    Last edited: Jan 30, 2012
  6. Feb 3, 2012 #5
    Thanks, Bill_K.
    So γμψμ=0 is imposed from the beginning, but how do I know that just the Spin 1/2 components are eliminated? Sorry but I got stuck with this question.
    I want (to try) to calculate just the equations of motion from the Lagrangian. Should this constrain be applied in any way?

    Thanks for reply!!
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