# Ramond sector zero modes

Problem 14.3b4

A First Course in String Theory

What are $\xi_1, \xi_2, \xi_3, \xi_4$ in Equation (14.44)?

p.315 “Ramond fermions are more complicated than NS fermions because the eight fermionic zero mode $d_0^I$ must be treated with care. It turns out that these eight operators can be organized by simple linear combinations into four creation operators and four annihilation operators. Let us call the four creation operators …”

Since there are 8 possible transverse directions, there are 8 possible $d_0^I$‘s, where $I = 2,3, ..., 9$.

What is the meaning of “… organized by simple linear combinations into four creation operators …”?

— Me@2015.11.01 03:53 AM

The $d_0^I$ operators are similar to but different from other $d_r^I$ operators. $d_0^I$‘s and $d_r^I$‘s are similar in the sense that they all follow Equation (14.43): $\{ d_m^I, d_n^J \} = \delta_{m+n, 0} \delta^{IJ}$

p.315 “Again, the negatively moded oscillators $d_{-1}^I, d_{-2}^I, d_{-3}^I, ...$, are creation operators, while the positively moded ones $d_{1}^I, d_{2}^I, d_{3}^I, ...$ are annihilation operators.” $d_0^I$‘s and $d_r^I$‘s are different in the sense that $d_0^I$‘s are neither creation nor annihilation operators.

(Based on the ideas from “Introduction to String Theory, A.N. Schellekens” and “A First Course in String Theory (Second Edition)” p.315:)

If we define $d_0 | 0 \rangle = 0$, $\{ d_0^I, d_0^J \} | 0 \rangle$ $= \left( d_0^I d_0^J + d_0^I d_0^J \right) | 0 \rangle$ $= 0$

which does not match the requirement of $\{ d_0^I, d_0^J \} = \delta^{IJ}$

So the definition $d_0 | 0 \rangle = 0$ does not work.

— Me@2015.11.12 11:30 AM

Instead, they are “organized by simple linear combinations into four creation operators”.

(Based on the idea from “Boundary Conformal Field Theory and the Worldsheet Approach to D-Branes, by Andreas Recknagel，Volker Schomerus” and “A First Course in String Theory (Second Edition)” p.315:)

Let $c_0^i = d_0^{i+1}$ $e_i = \frac{1}{\sqrt{2}} \left( c_0^{2i} - i c_0^{2i - 1} \right)$ $e_i^\dagger = \frac{1}{\sqrt{2}} \left( c_0^{2i} + i c_0^{2i - 1} \right)$.

Then $\left\{ e_i, e_j^\dagger \right\}$ $= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}$ $= \frac{1}{2} \delta^{ij} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2i} + i c_0^{2i - 1} \right) \right\}$

By p.315 Equation (14.43): $\{ d_0^I, d_0^J \} = \delta^{IJ}$

In other words, $\{ c_0^{I-1}, c_0^{J-1} \} = \delta^{I-1,J-1}$ $\{ c_0^{I}, c_0^{J} \} = \delta^{IJ}$ $\left\{ e_i, e_j^\dagger \right\}$ $= \frac{1}{2} \left\{ \left( c_0^{2i} - i c_0^{2i - 1} \right), \left( c_0^{2j} + i c_0^{2j - 1} \right) \right\}$ $= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} - \left\{ i c_0^{2i - 1}, i c_0^{2i - 1} \right\} \right]$ $= \frac{1}{2} \delta^{ij} \left[\left\{ c_0^{2i} , c_0^{2i} \right\} + \left\{ c_0^{2i - 1}, c_0^{2i - 1} \right\} \right]$ $= \frac{1}{2} \delta^{ij} \left[1 + 1 \right]$ $= \delta^{ij}$

This is compatible with the anti-commutator requirement for fermion creation and annihilation operators: $\{a^{\,}_i, a^\dagger_j\} = \delta_{i j}$

— Me@2015.11.13 11:14 PM