On Page 323 of Zwiebach 2ed, he says the state $$ (R_-,R_+): |R_a\rangle_L \otimes |R_{\bar b}\rangle_R \otimes |p^+, \vec{p}_T\rangle, $$ 'include the product of two $R$ ground states, they are "doubly" fermionic, and thus spacetime bosons.'

But just earlier he mentioned $|R_a\rangle$ are fermionic whereas $|R_{\bar b}\rangle$ are bosonic. $|R_a\rangle$ are defined as the eight massless states where you act an even number of fermionic creation operators on the vacuum $|0\rangle$ which is also fermionic; and $|R_{\bar b}\rangle$ are the eight massless states where you act an odd number of fermionic creation operators on the vacuum $|0\rangle$. He says thus it follows that $|R_{a}\rangle$ are fermions and $|R_{\bar b}\rangle$ are bosons.

But then how is this 'doubly fermionic'? We have a fermion cross a bosin, shouldn't it make the state a fermion? I trust everything he says but I'm having a hard time understanding the logic.


1 Answer 1


In the Ramond sector of the open string, the ground state is a degenerate multiplet consisting of 16 states that furnish an irreducible representation of the 8D Euclidean Clifford algebra. This implies that this multiplet, viewed as a 16-component spinor, transforms as a spinor of $\mathrm{Spin}(8)$ in spacetime.

Further analysis reveals that out of the initial 16 degenerate Ramond ground states, 8 of them, call them $|R_\alpha\rangle$, are worldsheet fermions while the other 8, $|R_{\bar\alpha}\rangle$ are worldsheet bosons. As it turns out, the 16-component spinor mentioned above is reducible to two 8-component spacetime spinors transforming in inequivalent representations of $\mathrm{Spin}(8)$, which correspond exactly to $|R_\alpha\rangle$ and $|R_{\bar\alpha}\rangle$. Since all the excited states are built by acting on the ground state with operators carrying Lorentz vector indices, all states in the Ramond sector are spacetime fermions, even though they may either be worldsheet bosons or worldsheet fermions.

The theory does not yet possess, among other things, spacetime supersymmetry - this is remedied with the GSO projection. However after the GSO projection, the Ramond sector is truncated depending on the eigenvalues of the states under the following projection operator: $$ G=\Gamma_{11}(-1)^{\sum_n d_{-n}^id_n^i}\equiv (-1)^F $$

There are two inequivalent choices of GSO projection: we can choose to preserve all states with a $G$-parity of $-1$ and discard the rest, or we could preserve all states with a $G$-parity of $+1$ and discard the rest. These correspond to the Ramond sector being truncated to $|R_\alpha\rangle$ and $|R_{\bar\alpha}\rangle$, respectively. In summary,

Ground state $|R_\alpha\rangle$ $|R_{\bar\alpha}\rangle$
Spacetime properties Spacetime fermion Spacetime fermion
$\mathrm{Spin}(8)$ representation in spacetime Spinor $\mathbf 8_s$ Conjugate Spinor $\mathbf 8_c$
Worldsheet properties Worldsheet fermion Worldsheet boson
$G$-parity $-1$ $1$

For the open string, you can take either projection since they both result in physically indistinguishable theories.

The closed string RR ground state is produced by tensoring together two such spinorial representations and so it is bosonic on the worldsheet. However, we can choose the $G$-parity (or equivalently, chirality: see the $\Gamma_{11}$ up front) of the Ramond ground state of the left-moving sector and right-moving sector independently, and the resulting theory is clearly different in each case. If both sectors have the same chirality, we obtain type IIB string theory, while choosing opposite chiralities constructs type IIA string theory.


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