Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question regarding an equation in Polchinski's "String Theory, Volume 1, An introduction to the bosonic string". The equation is (4.3.27) on p.135.

This section is about the brst-cohomology of the open bosonic string. My question in particular is about the N=1 level and the derivation of the cohomology.

By calculating $Q_B |\psi_1\rangle$ and eliminating all terms containing $c_0$ and $b_n$ where $n \geq 0$ since these annihilate the states, I end up with the following equation:

$$0 = Q_B |\psi_1\rangle = (c_{-1}L^{(m)}_{1} + c_{1}L^{(m)}_{-1})(e_\mu\alpha^\mu_{-1}+\beta b_{-1} + \gamma c_{-1})|0;\boldsymbol{k}\rangle$$

which by further multiplication simplifies to:

$$ 0 = (e_\mu c_{-1}L^{(m)}_{1}\alpha^\mu_{-1}+ \beta c_{-1}L^{(m)}_{1} b_{-1} + e_\mu c_{1}L^{(m)}_{-1} \alpha^\mu_{-1} + \beta c_{1}L^{(m)}_{-1} b_{-1} + \gamma c_{1}L^{(m)}_{-1} c_{-1})|0;\boldsymbol{k}\rangle $$

where $c_{-1}c_{-1}=0$ was used.

In Polchinski eq (4.3.27), this turns out to be:

$$ 0 = \sqrt{2\alpha'}(k^\mu e_\mu c_{-1} + \beta k_\mu \alpha_{-1}^\mu)|0;\boldsymbol{k}\rangle $$

I don't quite see how the Virasoro operators and the ghost operators act on the state to produce this result, other than that it is probably the first and fourth term that gives the result with the others being zero. I've been trying to figure it out, but I haven't been able to, would someone be able to enlighten me?

Thank you!

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.