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

CHAPTER 5. BRANES 63We should again define α 0 and ˜α 0 operators in such a way that we can “completethe sum” when we take the derivatives <strong>of</strong> the expansions. To this end, we define√2α ′ α0 25 = α ′K R− WR, (5.8)√2α′˜α 250 = α ′K R+ WR. (5.9)Now that we have these at our disposal, recall the (still unchanged) condition expressedin Eq. 2.99, as well as the defintion <strong>of</strong> L m in Eq. 2.97, and defineM 2 = −24∑µ=0p µ p µ , (5.10)i.e. the mass in the 25-dimensional non-compact space-time. Using all this, we find that0 = L 0 − 1,⎡= 1 ⎣ ( α 25 ) 224∑0 + α µ 02α 0µ +µ=0= ˜L 0 − 1,⎡= 1 )⎣ 25 224∑ (˜α 0 + ˜α µ 02˜α 0µ +µ=025∑+∞∑ν=0 n=125∑+∞∑ν=0 n=1α ν −nα nν⎤⎦ − 1,˜α ν −n˜α nν⎤⎦ − 1. (5.11)Remembering that for the closed bosonic string we defined α µ 0 = 1 2 l sp µ (but now withµ < 25) and definingthe above tells us thatN R =N L =25∑∑α−nα ν nν , (5.12)ν=0 n>025∑∑˜α −n˜α ν nν , (5.13)ν=0 n>012 α′ M 2 = ( α025 ) 2+ 2NR − 2 =(5.14))25 2 (˜α 0 + 2NL − 2. (5.15)Filling in Eqs. 5.8 and 5.9, and taking sum and difference allows us to conclude that[ (αN R − N L = −1 ′ )K 2 ( α ′ ) ]4α ′ R − WR K 2−R + WR= KW, (5.16)