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
 
 
 
2015.12.12 Saturday (c) All rights reserved by ACHK