String Theory Demystified

CHAPTER 7 RNS Superstrings 131
SOLUTION
µ α
We can rewrite ψ ρ ∂ ψ in a more enlightening way by expanding out the sum
α µ
explicitly:
Now
So, the summation is
Hence,
µ α
µ 0 1
ψ ρ ∂ ψ = ψ ( ρ ∂ + ρ ∂ ) ψ
α µ
0 ⎛ 0
ρ ∂ = 0
⎝
⎜ i
1 0
ρ ∂ = 1
⎛
⎝
⎜ i
0
− ⎞
0 ⎠
⎟ ∂
i
⎞
0⎠
⎟ ∂
i
0 1 ⎛ 0 − ⎞ 0 ⎞
ρ ∂ + ρ ∂ =
0 1
τ σ
⎝
⎜ 0 ⎠
⎟ ∂ +⎛
⎝
⎜ 0⎠
⎟ ∂
i i
i
i
⎛ 0 −i( ∂ −∂ ⎞
τ
=
⎝
⎜ ∂ +∂
⎠
⎟ =
) σ ⎛ 0 −2i∂ ⎞ −
i(
) 0 ⎝
⎜ 2i∂0 ⎠
⎟
τ σ
µ α
µ 0 1
ψ ρ ∂ ψ = ψ ( ρ ∂ + ρ ∂ ) ψ
α µ
⎛ 0 −2i∂ ⎞
µ
−
= ψ
⎝
⎜ 2i∂0 ⎠
0
+
σ
τ
1
1
µ
µ
⎟ ψ µ
Now, we write out the components of the spinors. For simplicity, we suppress the
space-time index for a moment. First, note that
⎛ 0 −2i∂ ⎞ ⎛ 0 −2i∂ ⎞ ⎛ψ
⎞
−
− −
⎝
⎜ 2 ∂ 0 ⎠
⎟ ψ =
i
⎝
⎜ 2i∂0 ⎠
⎟
⎝
⎜ψ
⎠
⎟ =
⎛ −∂ ψ ⎞ − +
2i
⎝
⎜ ∂ ψ ⎠
⎟
µ † µ 0 Using ψ = ( ψ ) ρ , we have
+
⎛ 0 −2i∂ ⎞
µ
−
ψ ψ ψ ψ
µ
⎝
⎜ 2i∂0 ⎠
+
µ µ
⎟ = ( − + )
+
= ( ψ ψ − + )
⎛ 0
⎝
⎜ i
+
0 2
−i⎞
⎠
⎟ i −∂ ⎛ ψ
⎜
⎝ ∂ ψ + −
+
− + µ
⎞
⎟
⎠
+ −
⎛ 2∂
ψ ⎞ + −
⎜
⎝ 2∂ ψ − + µ ⎠ ⎟ µ
µ
= 2ψ ∂ ψ + 2ψ
∂ ψ
µ
µ µ µ
= 2(
ψ ⋅∂ ψ + ψ ⋅∂ ψ )
− + − + − +
µ
− + − + − + µ