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I want to give an expression for the mass formula of an open string with has Neumannn condtion in $m$ directions and Dirichlet in $n$ directions $X^{i}(\sigma ^{1}=0)=x_{0}^{i}, X^{i}(\sigma ^{1}=\pi)=x_{1}^{i}$. If I use the condition $L_{0}=0$ in order to obtain the mass taking into account eq (4.18): enter image description here

I obtain this for the mass squared:

enter image description here

Where is there the dependence with the concrete boundary conditions of my example?

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The problem is that equation (4.18) is not true for the open string with DD boundary conditions, only for NN. In your case you have boundary conditions such that the open string ends on two different branes. Therefore you need the oscillator expansion for the open string with DD boundary conditions

\begin{equation} X^\mu \left(\sigma,\tau\right) =x_0^\mu+\frac{1}{l}\left(x_1^\mu-x_0^\mu\right)\sigma+\sqrt{2\alpha '}\sum\limits_{n\neq 0}\frac{1}{n} \alpha_n^\mu e^{-\frac{i\pi}{l}n\tau} \sin \left(\frac{\pi n \sigma}{l}\right) \end{equation} such that when writing down the expansion \begin{equation} \partial_\pm X^\mu= \pm \frac{\pi}{l}\sqrt{\frac{\alpha '}{2}}\sum_\limits{n=-\infty}^{n=\infty}\alpha_n^\mu e^{\frac{-i\pi n}{l}(\tau\pm\sigma)} \end{equation} you have that \begin{equation} \alpha_0^\mu = \frac{1}{\sqrt{2\alpha '}}\frac{1}{\pi} \left(x_1^\mu-x_0^\mu\right). \end{equation} Therefore you get an additional contribution in the referenced calculation due to tension of the string stretching from one brane to the other of the form

\begin{equation} \frac{1}{2}a_0^2|_{\text{DD-direction}} = \frac{1}{(\alpha')^2} \sum_i\left(\frac{x_1^i-x_0^i}{2\pi}\right)^2 =\left(T\Delta x^i\right)^2. \end{equation}

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