It seems that when trying to identify the physical degrees of freedom for the string some authors$^1$ use: $$ q^-=\frac{1}{\ell}\int_0^{\ell} X^-(\tau,\sigma)d\sigma$$

Then, the commutation relation is $$[q-,p^+]=-i$$

Other authors$^2$ choose the constant part of $X^-=x_0^- + f(\tau,\sigma)$ so the commutation relation is $$[x_0^- ,p^+]=-i$$

You can think that this is just a free choice, however everybody uses the same hamiltonian $$H=2\alpha' p^+p^-$$ Of course, $p^-$ is an integral but my point is that the same $H$ cannot work for both cases.

Because in the second case the equation of motion for $x_0^-$ leads to $$\dot{x_0^-} =0$$ which is expected($x_0^-$ is constant as pointed out in $^2$).

But in the first case it would be $q^-=$constant which is wrong, or there must be something I am missing here.

In short, my questions are:

Why this is happening?

Why some people use $[q-,p^+]=-i$ and others use $[x_0^- ,p^+]=-i$?

RELATION BETWEEN $q^-$ and $x^-_0$

I am going to use eq. (12.98) in Zwiebach's book. $$ X^-(\tau,\sigma)=x^- _0 +\sqrt{2\alpha'}\alpha^-_0 \tau + oscillators$$ Since the oscillators will not contribute anything, integration leads $$ q^-=\frac{1}{\ell}\int_0^{\ell} X^-(\tau,\sigma)d\sigma=x^- _0 +\sqrt{2\alpha'}\alpha^-_0 \ \tau$$

$^1$ Sundermeyer; Constrained Dynamics, Lecture Notes in Physics; page 220; also see this post.

also see Ralph Blumenhagen, Dieter Lust and Stefan Theisen; Basic Concepts of String Theory; pages 24 and 44.

$^2$ Zwiebach; A first course in String Theory, 2nd edition; 2009; page 238

  • $\begingroup$ To me, the way you've defined $q^-$, you're just going to get $x_0^-$ back. The oscillators won't contribute anything. How do you get your equations of motion? $\endgroup$ Jun 12, 2015 at 1:58
  • $\begingroup$ $i\dot{E}=[E,H]$ $E=q^-$ or $E=x_0^-$ $\endgroup$
    – Anthonny
    Jun 12, 2015 at 3:14
  • $\begingroup$ If you want a relation between them, it would be $q^-=x_0^- +constant \tau$. Correct me if I am wrong. $\endgroup$
    – Anthonny
    Jun 12, 2015 at 20:33
  • $\begingroup$ Write down the expression for $X^-$ and explicitly integrate it. See what you get. $\endgroup$ Jun 12, 2015 at 20:51
  • $\begingroup$ @leastaction I saw what I got. Integration has been added to the question. what else? what is the idea? $\endgroup$
    – Anthonny
    Jun 14, 2015 at 5:14

1 Answer 1


Comments to the question (v3):

  1. The $X^-$ coordinate has (a part from a zero mode) been integrated out in the light-cone (LC) formalism. The above mentioned LC Hamiltonian cannot fully address questions about the $X^-$ coordinate.

  2. To get the well-known expansion of $X^-$ as a sum of zero and oscillator modes including the sought-for $\alpha^-_0$ mode term, one has to go back to the full string formulation prior to the LC reduction, cf. e.g. my Phys.SE answer here. By doing this, one may show that the $\alpha^-_0$ mode term indeed appears in the expansion of $X^-$ as it should.


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