On Polchinski, the mode expiation of an open string is given as $$X^i(\tau,\sigma)=x^i+\frac{p^i}{p^+}+i \sqrt{2\alpha'}\sum_{n\neq 0} \frac{1}{n}\alpha_n^ie^{-\frac{\pi i n c \tau}{\ell}}\cos\frac{\pi n \sigma}{\ell}\tag{1.3.22}$$ Giving the B.C. $\partial_a X^i=0$ at $\sigma = 0, \ell$.

However, In David Tong, he writes the general solution to the wave function(EOM) and says $$X^\mu = X_L^\mu(\sigma^+)+ X^\mu_R(\sigma^-)$$ and $$X_{L/R}^\mu=\frac{1}{2}x^\mu + \alpha' p^\mu \sigma^\pm + i \sqrt{\frac{\alpha'}{2}}\sum_{n\neq 0}\frac{1}{n}\alpha^\mu_n e^{-in\sigma^\pm}\tag{1.36}$$ then he impose the B.C. after mode expansion for to arrive Neumann,$\alpha_n^a=\bar \alpha_n^a$, and Dirichlet, $x^I=c^I, p^I=0,\alpha_n^a=-\bar \alpha_n^a$.

Is the mode expansion of David Tong is the most general one and once we impose the B.C. and apply lightcone gauge then we get the one on Polchinski? If that is true, there are 2 questions I have:

  1. I could not get the $\cos\frac{\pi n \sigma}{\ell}$ term when I try this.
  2. The mode expansion seems reasonable but where does the $\alpha'$ comes? I feel it comes from EOM but I am not sure.

1 Answer 1


1) $\sigma^\pm = \tau \pm \sigma$ so you get something of the form $$ \sum_n e^{-in\tau}\Big(e^{-in\sigma} + e^{in \sigma} \Big) $$ which gives you the cosine.

2) The $\alpha'$ (and $l$) is just a convenient normalisation.


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