The zero-mode <Virasoro constraint> is
$$
-\alpha'M^2+\frac{|x_2-x_1|^2}{4\pi^2\alpha'}+N-1=0,
$$
where the <string level operator> is
$$
\boxed{N=\sum_{n=1}^\infty\alpha_{-n}\mathbin\cdot\alpha_n}.
$$
Since the <string tension> is
$$
\boxed{T=\frac1{2\pi\alpha'}},
$$
the mass formula is
$$
\boxed{M^2=T^2|x_2-x_1|^2+\frac1{\alpha'}(N-1)}.
$$

For a string beginning and ending on the same brane, the $N=1$ states are massless. Oscillators polarized tangentially give a gauge field on the $(p+1)$-dimensional worldvolume, while transverse polarizations give scalar fields describing fluctuations of the brane position. Strings joining two separated branes have the additional stretching mass $T|x_2-x_1|$ and are charged under the two endpoint gauge groups. When the two branes coincide, these off-diagonal states also become massless and <Coincident-D-brane gauge enhancement> enlarges $U(1)\times U(1)$ to $U(2)$. The bosonic open string also retains its $N=0$ tachyon.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 306 / 4 / 3 / Solution

{c}

When $n$ parallel D-branes coincide, oriented strings joining distinct branes become massless alongside strings whose ends lie on one brane. Their Chan-Paton labels fill the adjoint representation of $U(n)$, enhancing $U(1)^n$ to $U(n)$.


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 Incoming links: Coincident-D-brane gauge enhancement