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

CHAPTER 5. BRANES 65<strong>Open</strong> stringsNext, we turn ourselves to the case <strong>of</strong> the open string. We consider an open string obeyingNeumann boundary conditions in all 25 spatial dimensions, for which the expansionwas given by Eq. 2.41, which we refresh:X µ (σ, τ) = x µ + l 2 sp µ τ + il s∑n≠01n αµ ne −inτ cos (nσ).Again compactifying the X 25 dimension, this time we do not need to include any specialextra terms as in the closed string case. It remains true however that the p 25 momentumcomponent becomes quantized, so that p 25 = nR −1 with n ∈. Further recalling thatwe defined α µ 0 = l sp µ , we find for the expansion <strong>of</strong> X 25 :∑X 25 (σ, τ) = x 25 + l s α0 25 1τ + il sn α25 n e −inτ cos (nσ). (5.22)Although we did not do so before because it did not bare any interesting features withit at the time, we will now also expand this into left- and right-movers. We may do so,as we can consider the oscillations <strong>of</strong> an open string to be superpositions <strong>of</strong> left- andright-movers that combine into standing waves. We will do so, because <strong>of</strong> course, wewould like to be able to formulate the theory using a “dual coordinate” ˜X 25 , just as inthe closed string case. This expansion amounts ton≠0XR 25 (τ − σ) = x25 − ˜x 25+ 1 2 2 l sα0 25 (τ − σ) + i l ∑s 12 n α25 n e −in(τ−σ) , (5.23)n≠0∑XL 25 (τ + σ) = x25 + ˜x 25+ 1 2 2 l sα0 25 (τ + σ) + i l s2n≠01n α25 n e −in(τ+σ) , (5.24)where again ˜x 25 is an as-<strong>of</strong>-yet arbitrary constant that cancels in the sum. The biggestdifference with the closed string case, is that this time there is only one set <strong>of</strong> modes (i.e.,there are no ˜α’s). T-dualizing by sending R −→ ˜R = α ′ R −1 will again send X R −→ −X Rand X L −→ X L , and so we define the dual coordinate ˜X 25 (σ, τ) = X L − X R , whichexplicitely becomes∑˜X 25 (σ, τ) = ˜x 25 + l s α0 25 1σ + l sn α25 n e −inτ sin(nσ). (5.25)We see that the zero-mode gets multiplied by σ instead <strong>of</strong> τ, as was the case for X 25 ,and that the cos (nσ) got exchanged for −i sin(nσ). This hints at something that turnsout to be true, namelyn≠0∂ σ X 25 = ∂ τ ˜X25 , (5.26)∂ τ X 25 = ∂ σ ˜X25 , (5.27)