3
$\begingroup$

I am trying to obtain the polchinski's equation 4.3.16 which is following

$ Q_B^2 = \frac{1}{2}\{ Q_B, Q_B \} = -\frac{1}{2}g^{K}_{IJ}g^{M}_{KL}c^Ic^Jc^L b_M =0$

Where $ Q_B = C^I(G_I^m +\frac{1}{2}G_I^g)$ and $C^I$, $b^J$ are anticommuting(ghosts)

and $[G_I, G_J]=ig^K_{IJ} G_K$,
$G_I^g = -ig^K_{IJ} C^Jb_K $ are ghost parts and $G_I^m$ are matter part and they satisfy above commutation relations

What I have done are

$\{ Q_B, Q_B \} = \{ C^I(G_I^m +\frac{1}{2}G_I^g), C^J(G_J^m +\frac{1}{2}G_J^g)\} =\{C^IG_I^m, C^J G_J^m\} +\frac{1}{2} \{C^IG_I^m, C^JG_J^g\} +\frac{1}{2} \{C^I G_I^g, C^JG_J^m\} +\frac{1}{4} \{C^IG_I^g, C^JG_J^g \} = C^IC^J [G_I^m, G_J^m ] +\frac{1}{2} C^IC^J [G_I^m, G_J^g] +\frac{1}{2}C^IC^J[G_I^g,G_J^m]+\frac{1}{4}C^IC^J[G_I^g, G_J^g] =C^IC^J [G_I^m, G_J^m ]+\frac{1}{4}C^IC^J[G_I^g, G_J^g] = C^IC^J ig^K_{IJ}G_K^m+\frac{1}{4}C^IC^J ig^{K}_{IJ}G_K^g =C^IC^J ig^K_{IJ}G_K^m+\frac{1}{4}C^IC^J g_{IJ}^K g^M_{KL}C^Lb_M $

compare with the textbook $\{ Q_B, Q_B \} = -g^{K}_{IJ}g^{M}_{KL}c^Ic^Jc^L b_M$

My calculation is something wrong. How can I fix it?

$\endgroup$
5
  • $\begingroup$ I figure out how to do it! $\endgroup$
    – phy_math
    Aug 27 '14 at 5:15
  • 1
    $\begingroup$ OK. I suppose that you have forgotten that the commutator $[c^I,G^g_J]$ is not zero because the anti-commutator $\{c^i,b_J\}$ is not zero. You have then extra-term which cancel with your original $G_m$ term, and other extra-term which gives you an other $-\frac{1}{4}$ factor equals to your original last term, so you get an overall $-\frac{1}{2}$ factor. $\endgroup$
    – Trimok
    Aug 27 '14 at 8:13
  • $\begingroup$ Hi @phy_math: If you figured it out, your are encouraged to answer your own question. $\endgroup$
    – Qmechanic
    Sep 1 '14 at 13:29
  • $\begingroup$ @Qmechanic, the reply for $Trimok$ is right procedure. Or one can just calculate explicitly. After some algebra, using Jacobi identity in the end, i obtained the desired answer $\endgroup$
    – phy_math
    Sep 6 '14 at 14:14
  • $\begingroup$ @phy_math: If you have solved your own question, you are encouraged to write an answer, so that the post doesn't end as an orphan. $\endgroup$
    – Qmechanic
    Sep 27 '14 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.