DBI Analysis of Open String Bound States on Non-compact D-branes

CHAPTER 5. BRANES 62but if the string is wound a number <strong>of</strong> times then simply stating that X 25 (σ + π, τ) =X 25 (σ, τ) would identify the wrong points!The next thing to do is <strong>of</strong> course to take a look at the mode expansion <strong>of</strong> our string,and see if anything has changed. The first thing to note is that nothing changes in the24 spatial dimensions that were left uncompactified, and hence the mode expansion inthese dimensions also does not change. In the 25 th dimension however, for exactly thesame reason mentioned just earlier, we should add an extra term to account for thewinding <strong>of</strong> the string. The second thing to note is that compactifying does not changeanything to the oscillator modes. And the third and final thing to note is that due tothe compactification, the momentum in the 25 th dimension becomes quantized. In orderto see this, recall e.g. the case <strong>of</strong> Kaluza-Klein compactification <strong>of</strong> a free massless scalarfield ψ (x µ , y) living in a D +1-dimensional Minkowski space <strong>of</strong> which one dimension, y,is compactified, satisfying the massless Klein-Gordon equation( + ∂2y)ψ = 0, (5.2)where = ∂ µ ∂ µ with µ ∈ {0, . . .,D − 1}. Given that y is periodic, so is the Fourier expansionin this dimension, and hence as a general solution to the Klein-Gordon equationwe findψ (x µ , y) =+∞∑n=−∞e i n R y ψ k (x µ ), (5.3)in which R is the radius <strong>of</strong> the compactified dimension, and e i n R y = e ipyy , implying thatp y = n R . (5.4)In the case <strong>of</strong> the string, the same argument holds (recall that the X µ ’s are scalar fields),but with this distinction that we definep 25 = K R , (5.5)where we have defined the Kaluza-Klein excitation number K ∈. Taking all this intoaccount, and starting from the generic solutionX 25 (σ, τ) = x 25 + 2α ′ p 25 τ + 2RWσ + . . ., (5.6)splitting in left- and right-movers leaves us withXR 25 (τ − σ) = 1 (x 25 − ˜x 25) +(α ′K )2R − WR (τ − σ) + oscillators,XL 25 (τ + σ) = 1 (x 25 + ˜x 25) +2(α ′K R + WR )(τ + σ) + oscillators, (5.7)where ˜x 25 is some arbitrary constant that cancels in the sum, which will prove itselfvery useful in a second.