I have a question in reading Polchinski's string theory vol I p 123, about the "old covariant quantization".

It is said

... $\langle 0;k | 0; k' \rangle = ( 2\pi)^D \delta^D (k-k') \tag{4.1.15}$ as follows from momentum conservation. The timelike excitation has a negative norm.

How to see "the timelike excitation has a negative norm"? Is Dirac delta function in the RHS of Eq. (4.1.15) either $0$ or $\infty$?

  • $\begingroup$ Dirac Delta function. Does that answer your question ? . $\endgroup$ – Abhimanyu Pallavi Sudhir Aug 8 '13 at 15:40
  • $\begingroup$ No... I didn't see the negative norm.... $\endgroup$ – user26143 Aug 8 '13 at 15:40
  • $\begingroup$ The definition of a time-like excitation, by the way, is to have a negative norm, i.e. $\|\psi\|^2<0$ . I don't exactly see the questionm here . $\endgroup$ – Abhimanyu Pallavi Sudhir Aug 8 '13 at 15:44
  • $\begingroup$ Do you mean Polchinski simply state the definition of time-like excitation here? I thought it should have some reasoning behind that.. $\endgroup$ – user26143 Aug 8 '13 at 15:49
  • $\begingroup$ Yes. That's what I mean. I see that @Prahar's answer is different . , but note, there's an $\eta^{00}=-1$ there, so it's essentially the same thing, . $\endgroup$ – Abhimanyu Pallavi Sudhir Aug 8 '13 at 15:50

We note $$ \left[ \alpha_m^0, \alpha_n^0 \right] = \eta^{00} \delta_{m+n,0} = - \delta_{m,-n} $$ A timelike excitation is $\alpha_{-n}^0 \left| 0; k \right>$. The norm of this state is \begin{equation} \begin{split} \left<0;k'\right| \alpha_{m}^0 \alpha_{-n}^0 \left|0;k\right> &= \left<0;k'\right| \left( \left[ \alpha_{m}^0 , \alpha_{-n}^0 \right] + \alpha_{-n}^0 \alpha_{m}^0 \right) \left|0;k\right> \\ &= - \delta_{m,n} \left<0;k'\right. \left|0;k\right> \\ &= - \delta_{m,n} (2\pi)^D \delta^D( k - k' ) < 0 \end{split} \end{equation} Thus a timelike excitation has negative norm.

  • 1
    $\begingroup$ Do you mean a timelike excitation is that, from $e \cdot \alpha_{-1} = e^{\mu} \cdot \alpha_{-1,\mu} $, we Lorentz transform it into $e^{\mu} \propto (1,0,0,0)$? $\endgroup$ – user26143 Aug 8 '13 at 16:05
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    $\begingroup$ Yes. That's what I mean. $\endgroup$ – Prahar Aug 8 '13 at 17:38

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