02.06.2013 Views

String Theory Demystified

String Theory Demystified

String Theory Demystified

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

CHAPTER 12 Heterotic <strong>String</strong> <strong>Theory</strong> 217<br />

The usual formulas for the modal expansions apply<br />

x p i<br />

X<br />

n e<br />

i i<br />

∞ i<br />

i<br />

n<br />

( τ − σ) = + ( τ −σ)<br />

+ ∑<br />

2 2 2 n=<br />

1<br />

X<br />

i<br />

α −2in( τ−σ) i i<br />

i<br />

x p i n in<br />

n n e<br />

∞ α − 2 ( τ+ σ)<br />

( ) + ∑<br />

2 2 2 = 1<br />

( τ + σ) = + τ + σ<br />

x p i<br />

X<br />

n e<br />

I I<br />

∞ I<br />

I<br />

n<br />

( τ + σ) = + ( τ + σ)<br />

+ ∑<br />

2 2 2 n=<br />

1<br />

∞<br />

a −2in( τ−σ) n<br />

n=−∞<br />

a<br />

S ( τ − σ)<br />

= ∑ S e<br />

<br />

α − 2in(<br />

τ+ σ)<br />

Using this approach, we write down two number operators. The number operator<br />

for the right-moving sector is<br />

∞<br />

⎛ i i 1 − ⎞<br />

N = ∑ n n + nS n Sn<br />

⎝<br />

⎜α−<br />

α<br />

− Γ<br />

⎠<br />

⎟<br />

2<br />

n=<br />

1<br />

For the left-moving sector we have<br />

∞<br />

N<br />

i i I I<br />

= ∑( α−nαn+ α−nαn) n=<br />

1<br />

The mass can be written in terms of the canonical momentum p I as<br />

1<br />

1<br />

4<br />

1<br />

16<br />

2<br />

m N N p<br />

2<br />

I<br />

= + − + ∑(<br />

)<br />

Now, let’s see how compactifi cation affects these results. It is easiest to<br />

understand by stepping back to compactifi cation of one dimension as we did in<br />

Chap. 8. Kaku gives a nice example which we restate here. Take a single-dimensional<br />

theory and let<br />

x = x+2π R<br />

Consider a fi eld defi ned on this space φ( x ) . Since the coordinate is periodic, the<br />

fi eld must be also<br />

I=<br />

1<br />

φ( x) = φ( x+2πR) 2

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!