String Theory Demystified

CHAPTER 7 RNS Superstrings 133
In the second case, we have
and
Hence,
δψ
µ
µ
δψ
= µ
δψ
⎛ ⎞ −
⎜ ⎟
⎝ ⎠
α µ 0 µ 1 µ
ρ ∂ X ε = ρ ∂ X ε + ρ ∂ X ε
α
τ
σ
0 1 µ
= ( ρ ∂ + ρ ∂ ) X ε
τ σ
⎛ 0
=
⎝
⎜ i
−i⎞
0
0 ⎠
⎟ ∂ +⎛
⎝
⎜
⎞
0⎠
⎟
0
∂
⎡
⎢
⎣
τ
i
i ⎤ µ
⎥ X ε σ
⎦
⎛
=
⎝
⎜i(
∂ +∂ )
−i( ∂ −∂ ) ⎞
τ σ µ
X
⎠
⎟ ε
0
τ σ
⎛ 0 −2i∂ ⎞ − µ
=
X
⎝
⎜ i∂
⎠
⎟ ε
2 0
+
+
µ µ
δψ =−2∂ X ε
(7.11)
− − +
µ µ
δψ = 2∂ X ε
(7.12)
+ + −
Conserved Currents
At this point, we need to identify the conserved currents associated with the action
in Eq. (7.2). Before tackling supersymmetry, let’s review how to calculate a
conserved current by looking at momentum. You can practice in the Chapter Quiz
by looking at Lorentz transformations. Let’s start with a simple example to remind
ourselves of the method.
EXAMPLE 7.3
2
µ α
µ α
Consider the action S =−T/ 2∫
d σ ( ∂ X ∂ X −iψ ρ ∂ψ
)
α
µ
α µ
conserved current associated with translational invariance.
and fi nd the