## Exercise 2 Solution ### Step 1: Why the NS sector describes only bosonic states and the R sector only fermionic ones **Explanation:** - **NS (Neveu-Schwarz) Sector:** In the NS sector, the boundary conditions for the worldsheet fermions are **antiperiodic**. This means that the fermionic modes are half-integer moded (i.e., their mode numbers are \( r = n + 1/2 \), where \( n \) is an integer). The ground state in the NS sector is a scalar (bosonic), and the first excited state is created by a single fermionic oscillator, which still forms a bosonic representation due to the statistics from the vacuum structure. - **R (Ramond) Sector:** In the R sector, the boundary conditions for the worldsheet fermions are **periodic**. The fermionic modes are integer moded (\( r = n \)), and the ground state is degenerate due to the presence of zero modes. These zero modes satisfy a Clifford algebra and generate a representation of the spacetime gamma matrices, resulting in a ground state that is a spacetime spinor (fermion). **Summary for Step 1:** - NS sector → Bosonic states - R sector → Fermionic states --- ### Step 2: Why closed strings have four distinct sectors, and which correspond to bosonic/fermionic states **Explanation:** - **Closed String Sectors:** Closed strings have both left-moving and right-moving modes, and each can independently be in either the NS or R sector. This gives four possible combinations: - NS-NS (Left: NS, Right: NS) - NS-R (Left: NS, Right: R) - R-NS (Left: R, Right: NS) - R-R (Left: R, Right: R) - **Bosonic vs. Fermionic States:** - **NS-NS and R-R:** Both left and right sectors are either bosonic (NS) or both are fermionic (R). The combination of two bosonic sectors (NS-NS) or two fermionic sectors (R-R) results in an overall **bosonic state**, because a product of two fermions (spinors) can make either a scalar or higher-spin boson. - **NS-R and R-NS:** Here, one side is bosonic (NS) and the other is fermionic (R). The combination of a bosonic and a fermionic sector gives an overall **fermionic state**. **Summary for Step 2:** - NS-NS and R-R → Bosonic states - NS-R and R-NS → Fermionic states --- ## **Final Answer** 1. The NS sector describes only bosonic states because its ground state is a scalar, while the R sector describes only fermionic states because its ground state is a spacetime spinor due to the presence of fermionic zero modes. 2. Closed strings have four sectors (NS-NS, NS-R, R-NS, R-R) because both left- and right-movers can independently be in NS or R sectors. NS-NS and R-R correspond to bosonic states, while NS-R and R-NS correspond to fermionic states.

## Exercise 2 Solution ### Step 1: NS and R Sectors **Why the NS Sector Describes Bosonic States and the R Sector Describes Fermionic States** - **NS (Neveu-Schwarz) Sector:** - Boundary conditions for fermions are **antiperiodic**, leading to half-integer mode numbers \( r = n + 1/2 \). - The ground state is a scalar (bosonic), and excited states formed by adding fermionic oscillators remain bosonic due to vacuum structure. - **R (Ramond) Sector:** - Boundary conditions for fermions are **periodic**, allowing integer mode numbers \( r = n \). - Ground state is a degenerate spacetime spinor (fermion) arising from the presence of fermionic zero modes that satisfy Clifford algebra. **Summary for Step 1:** - NS sector → Bosonic states - R sector → Fermionic states --- ### Step 2: Closed Strings and Their Four Distinct Sectors **Why Closed Strings Have Four Distinct Sectors** - **Closed String Sectors:** Each closed string has left-moving and right-moving modes, with each mode able to be in either the NS or R sector. This results in: - NS-NS (Both Left and Right are NS) - NS-R (Left is NS, Right is R) - R-NS (Left is R, Right is NS) - R-R (Both Left and Right are R) - **Bosonic vs. Fermionic States:** - **NS-NS and R-R:** Both correspond to overall **bosonic states**. NS-NS consists of two bosonic states, while R-R, being a product of fermions, can yield bosonic excitations. - **NS-R and R-NS:** Each combination results in **fermionic states** since one side is bosonic and the other is fermionic. **Summary for Step 2:** - NS-NS and R-R → Bosonic states - NS-R and R-NS → Fermionic states --- ## Final Answer 1. The NS sector describes only bosonic states due to its scalar ground state, while the R sector describes only fermionic states because its ground state is a spacetime spinor from fermionic zero modes. 2. Closed strings have four sectors (NS-NS, NS-R, R-NS, R-R) because left- and right-moving modes can independently be in NS or R sectors. The NS-NS and R-R correspond to bosonic states, while NS-R and R-NS correspond to fermionic states.