# Poisson brackets of bosonic string oscillators

I'm reading string theory books and I'm stuck at the moment when we consider a Hamiltonian version of the classical string. Namely I don't understand how to derive the Poisson brackets for string oscillators. Here is the question in details.

From the equations of motion following from the Polyakov action in conformal gauge $$S = \int \mathcal L d\sigma d\tau = \frac12 \int ( \dot X^2 - X'{}^2 ) d\sigma d\tau$$ we derive that $$X^\mu(\sigma, \tau) = X^\mu_R(\tau - \sigma) + X^\mu_L(\tau + \sigma)$$ with $$X^\mu_R(\tau - \sigma) = \frac12 x^\mu + l^2 p^\mu (\tau - \sigma) + i l \sum_{n \ne 0} \frac1n \alpha_n^\mu e^{-i n (\tau - \sigma)},\\ X^\mu_L(\tau + \sigma) = \frac12 x^\mu + l^2 p^\mu (\tau + \sigma) + i l \sum_{n \ne 0} \frac1n \tilde \alpha_n^\mu e^{-i n (\tau + \sigma)}.$$

Using the definition of the Poisson bracket in Hamiltonian field theory I can derive that $$\{X^\mu(\sigma, \tau), \Pi_\nu(\sigma', \tau)\} = \delta^\mu_\nu \delta(\sigma - \sigma'),$$ where the conjugate momentum field $$\Pi_\mu = \frac{\partial \mathcal L}{\partial \dot X^\mu}$$. Indeed, we substitute \begin{align} X^\mu(\sigma) &= \int X^\mu(\sigma'') \delta(\sigma'' - \sigma) d\sigma'',\\ \Pi_\nu(\sigma') &= \int \Pi_\nu(\sigma'') \delta(\sigma'' - \sigma') d\sigma'', \end{align} into $$\{X^\mu(\sigma), \Pi_\nu(\sigma')\} = \int \left( \frac{\delta X^\mu(\sigma)}{\delta X^\lambda} \frac{\delta \Pi_\nu(\sigma')}{\delta \Pi_\lambda} - \frac{\delta \Pi_\nu(\sigma')}{\delta X^\lambda} \frac{\delta X^\mu(\sigma)}{\delta \Pi_\lambda} \right) d\sigma$$ and get the result. Then in many books it is claimed that one can derive from it the Poisson brackets of oscillators $$\{\alpha^\mu_m, \alpha^\nu_n\} = \{\tilde \alpha^\mu_m, \tilde \alpha^\nu_n\} = -i m \delta_{m+n,0} \eta^{\mu \nu}, \qquad \{\alpha^\mu_n, \tilde \alpha^\nu_m\} = 0, \qquad \{x^\mu, p^\nu\} = \eta^{\mu \nu}.$$ (E.g. equation (3.3.17) in Introduction to Superstring Theory by Elias Kiritsis, or equation (2.51) in "String Theory and M-theory" by Becker, Becker, Schwartz.)

And here I am completely lost. First of all just the definitions. We think about $$x^\mu, p^\mu, \alpha^\mu_n, \tilde \alpha^\mu_n$$ as implicit functionals of $$X^\mu, \Pi_\mu$$, right? But what to do next? It seems that it's reasonable to expand $$\begin{multline} \{X^\mu(\sigma,\tau), \dot X^\nu\} = \left\{ x^\mu + 2 l^2 p^\mu \tau + il \sum_{n \ne 0} \frac1n \alpha^\mu_n e^{-in(\tau-\sigma)} + il \sum_{n \ne 0} \frac1n \tilde \alpha^\mu_n e^{-in(\tau+\sigma)},\\ 2 l^2 p^\nu + l \sum_{n \ne 0} \alpha^\mu_n e^{-in(\tau-\sigma')} + l \sum_{n \ne 0} \tilde \alpha^\mu_n e^{-in(\tau+\sigma')} \right\} \end{multline}$$ by linearity to get infinite sum of individual Poisson brackets. But I have no idea how to extract the values of that brackets.

• In order that other people can easily find this answer via internet search, can you include the title/section/equation number of the book you are reading? Commented May 20, 2021 at 17:43
• @user1379857, yeah, nice idea. Added that info. Thank you for your answer. I'll read it thoroughly a bit later. Commented May 20, 2021 at 18:09
• @vanger I know this is an old question now, but I've been doing the same thing recently. I've been stuck on the previous step, so I was wondering if you would be able to share your workings or a source to show how to calculate $\{X^{\mu}(\sigma), \Pi_{\nu}(\sigma')\}$ , because I'm not sure how this is being done. Thanks Commented Jan 6, 2022 at 21:50

OK, I get it. One can present oscillators as functionals of fields and momenta as \begin{align} x^\mu &= \frac{1}{2 \pi} \int_0^{2 \pi} \left( X^\mu - \frac{\tau}{T} \Pi^\mu \right) d \sigma,\\ p^\mu &= \int_0^{2 \pi} \Pi^\mu d \sigma,\\ \alpha_n &= \frac{e^{i n \tau}}{4 \pi l} \int_0^{2 \pi} \left( \frac1T \Pi^\mu - i n X^\mu \right) e^{-i n \sigma} d \sigma,\\ \tilde \alpha_n &= \frac{e^{i n \tau}}{4 \pi l} \int_0^{2 \pi} \left( \frac1T \Pi^\mu - i n X^\mu \right) e^{ i n \sigma} d \sigma. \end{align} And then one can compute Poisson brackets by definition. I checked that it gives the correct result.
I imagine the authors mean the following. If one has $$\sigma \sim \sigma + 2 \pi$$, then one can write the Dirac delta as $$\delta( \sigma ) = \frac{1}{2 \pi} \sum_{n = - \infty}^\infty e^{i n \sigma}.$$ One can then write out the equal time Poisson bracket roughly as $$\{ X(\sigma), \dot X(\sigma')\} = \delta(\sigma - \sigma')$$ You can write the LHS in terms of all the pairs of Poisson brackets of all the oscillator modes $$\alpha$$, and you can write the RHS in terms of the Fourier decomposition of the Dirac delta I wrote above. Both terms will then be expressly written as Fourier series, i.e. as coefficients multiplied by $$e^{i n (\sigma - \sigma')}$$. You then use the fact that if two functions are equal their Fourier series are equal. (For instance, you could take the inverse Fourier transform of both sides, but that isn't necessary in practice.) Because the Fourier coefficient on the RHS is always $$1$$ for every $$n$$, this should work out to the exact same commutation relations you got by other means.