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November 7, 2013 303<br />

rank-2, spin-2 :<br />

rank-2, spin-1 :<br />

rank-3, spin-3 :<br />

(<br />

)<br />

1<br />

∇ µ α∇ ν β + ∇ µ β∇ ν α − 1 2<br />

2 ∇µν ∇ αβ ,<br />

1<br />

2<br />

1<br />

6<br />

(<br />

)<br />

∇ µ α∇ ν β − ∇ µ β∇ ν α ,<br />

(<br />

∇ µ α∇ ν β∇ ρ γ + ∇ µ β∇ ν γ∇ ρ α + ∇ µ γ∇ ν α∇ ρ β<br />

+ ∇ µ β∇ ν α∇ ρ γ + ∇ µ α∇ ν γ∇ ρ β + ∇ µ γ∇ ν β∇ ρ α<br />

− 1 (<br />

∇ µν ( ∇ ρ α∇ βγ + ∇ ρ β∇ γα + ∇ ρ )<br />

γ∇ αβ<br />

12<br />

+ ∇ νρ ( ∇ µ α∇ βγ + ∇ µ β∇ γα + ∇ µ )<br />

γ∇ αβ<br />

)<br />

+ ∇ ρµ ( ∇ ν α∇ βγ + ∇ ν β∇ γα + ∇ ν γ∇ αβ<br />

) ) (10.230)<br />

Compared to the massive case, some coefficients are different : -1/2 rather than<br />

-1/3 in the spin-2 case, and -1/12 instead of -1/15 for spin-3. This is due, of<br />

course, to the different traces of ∆ and ∇. The spin sum for the massless vector<br />

particle (rank-1, spin-1) is in fact that of the axial gauge discussed in Chapter<br />

6, with the gauge vector r chosen to be ¯p. Note that, whatever r µ , we can<br />

always move to the centre-of-mass frame of p µ and r µ , and in that frame we<br />

have precisely r µ = ¯p µ .<br />

10.11.7 Spin of the Kalb-Ramond state<br />

Concerning the Kalb-Ramond (KR) state, there may be some controversy. For<br />

a massless particle in this state, the spin along the axis of motion must, under<br />

measurement, always come out zero. It is not easy to see how such a particle can<br />

be distinguished from a scalar one. Indeed, in string theory where the KR state<br />

comes up naturally, it is considered to describe a (pseudo)scalar particle called<br />

the axion. In order to talk sensibly about the spin of the KR state it is useful<br />

to consider how it may be measured, for instance using fermions. We therefore<br />

consider the coupling of a rank-2, spin-1 state to fermions. The interaction<br />

vertex must have the properties that (a) it is an antisymmetric rank-2 tensor,<br />

and (b) it is current-conserving, in order to make sense in the massless limit.<br />

Denoting the two fermions by ψ and ψ the simplest choice appears to be<br />

ψ ɛ µνρσ p ρ (A + Bγ 5 ) γ σ ψ<br />

where p is the momentum of the antisymmetric tensor state, and A and B<br />

are constants. This interaction vertex vanishes trivially under the handlebar<br />

operation. For the process<br />

¯f(p 1 ) f(p 2 ) → f(p 3 ) ¯f(p 4 )<br />

by the exchange of a KR state of mass M, we then have the amplitude<br />

M = i¯h v(p 1 ) ɛ µνρσ p ρ (A + Bγ 5 ) γ σ u(p 2 )

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