# Basic questions about Spin3/2 formalism

1. Jan 27, 2012

### PiepsNYC

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$\Gamma$$\mu$=0, with pμ being the 4-momentum?

These questions remain unanswered to me.

Last edited: Jan 27, 2012
2. Jan 28, 2012

### Bill_K

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.

3. Jan 30, 2012

### PiepsNYC

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....

4. Jan 30, 2012

### Bill_K

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
5. Feb 3, 2012

### PiepsNYC

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?