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I divided the time interval $[t_0=:t_i,t_f:=t_N]$ into $N$ steps $[t_{k-1},t_{k}],\, k=1,\dots, N$ and applied the resolution of unity for coherent states \begin{equation} \mathbb{I}=\int_\mathbb{C}\frac{dzd\bar{z}}{2\pi i}\exp\left\lbrace-z\bar{z}\right\rbrace\lvert z\rangle\langle z\rvert \end{equation} at each step; this yields the following \begin{multline} \langle{z_f}\lvert\left( \exp\lbrace-iH\epsilon\right\rbrace)^N\rvert z_i\rangle=\dots=\lim_{N\to\infty}\int\prod_{j=1}^{N-1}\frac{dz_jd\bar{z_j}}{2\pi i}\\ \exp\left\lbrace\sum_{k=0}^{N-1}\bar{z}_{k+1}z_k-\sum_{k=1}^{N-1}z_k\bar{z_k}-i\epsilon\sum_{k=0}^{N-1}H(\bar{z}_{k+1},z_k)\right\rbrace,\quad(*) \end{multline} where $\epsilon:=\frac{t_f-t_i}{N}$, and with boundary conditions $z_0=z_i,\,\bar{z}_N=\bar{z}_f$. Then I need to put $(*)$ in this form \begin{align} \int\mathcal{D}(z,\bar{z})&\exp\bigg\lbrace\frac{\bar{z}_fz_f+\bar{z}_iz_i}{2}+\frac{1}{2}\sum_{k=0}^{N-1}\big[z_k(\bar{z}_{k+1}-\bar{z}_k)-\bar{z}_k(z_{k+1}-z_k)+\\ &-i\epsilon H(\bar{z}_{k+1},z_k)\big]\bigg\rbrace \end{align} but I can't understand how to transform the argument of the exponential.

Any ideas?

Thanks

Source: Itzykson - Zuber, page 438 with line 2 corrected as shown in the errata (you can find it here: http://www.lpthe.jussieu.fr/~zuber/corrize.pdf).

edit: @ ACuriousMind Then I don't understand how to rewrite it: is the following right? \begin{equation} \frac{1}{2}\sum_{k=0}^{N-1}\big[z_k(\bar{z}_{k+1}-\bar{z}_k)-\bar{z}_k(z_{k+1}-z_k)\big]=\frac{1}{2}\sum_{k=0}^{N-1}\big[z_k\bar{z}_{k+1}-\bar{z}_k z_{k+1}\big]=\sum_{k=0}^{N-1}\big[z_k\bar{z}_{k+1}\big] \end{equation} if this were right, I wouldn't know how to cope with this \begin{equation} -\sum_{k=1}^{N-1}z_k\bar{z}_k \end{equation}

If it is so straightforward to you, could you please rewrite it step-by-step?

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  • $\begingroup$ It's just a rewriting of the sums, there's no "transformation" required. $\endgroup$
    – ACuriousMind
    Jan 15, 2016 at 18:26
  • $\begingroup$ Related: physics.stackexchange.com/q/210138/2451 $\endgroup$
    – Qmechanic
    Jan 15, 2016 at 20:00
  • $\begingroup$ @Qmechanic My doubt is not about the path integral itself... it's not obvious to me how to rewrite it in the symmetric form proposed by Itzykson - Zuber... Brown's form of coherent state path integral, which you cite, is slightly different, thus, it doesn't answer my question... $\endgroup$
    – Red Lex
    Jan 15, 2016 at 20:09
  • $\begingroup$ Hint: The sought-for manipulation of the action is a discrete version of changing $\lambda=0$ to $\lambda=\frac{1}{2}$ in eq. (7) in my Phys.SE answer here. $\endgroup$
    – Qmechanic
    Jan 15, 2016 at 20:41
  • $\begingroup$ You're right... but there is no derivation of that formula in your post. Can you provide a reference? $\endgroup$
    – Red Lex
    Jan 15, 2016 at 20:49

1 Answer 1

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At last I think I managed to rearrange the terms in the sums... (which wasn't a monumental task after all ;-) ) \begin{equation} \sum_{k=0}^{N-1}\bar{z}_{k+1}z_k-\sum_{k=1}^{N-1}z_k\bar{z}_k=(*) \end{equation} In the notation here:

  • $\lambda=0$: \begin{equation} (*)=\sum_{k=0}^{N-1}\bar{z}_{k+1}z_k-\sum_{k=0}^{N-1}z_{k+1}\bar{z}_{k+1}+z_N\bar{z}_N=-\sum_{k=0}^{N-1}\bar{z}_{k+1}(z_{k+1}-z_k)+z_f\bar{z}_f,\,(1) \end{equation}
  • $\lambda=1$: \begin{equation} (*)=\sum_{k=0}^{N-1}\bar{z}_{k+1}z_k-\sum_{k=0}^{N-1}z_k\bar{z}_k+z_0\bar{z}_0=\sum_{k=0}^{N-1}z_k(\bar{z}_{k+1}-\bar{z}_k)+z_i\bar{z}_i,\,(2) \end{equation} Taking $(2)+(1)$ we obtain $2(*)$, and finally
  • $\lambda=1/2$ \begin{equation} \sum_{k=0}^{N-1}\bar{z}_{k+1}z_k-\sum_{k=1}^{N-1}z_k\bar{z}_k=\frac{1}{2}\sum_{k=0}^{N-1}\big[z_k(\bar{z}_{k+1}-\bar{z}_k)-\bar{z}_{k+1}(z_{k+1}-z_k)\big]+\frac{z_i\bar{z}_i+z_f\bar{z}_f}{2} \end{equation} but there was a $+1$ missing in one of the $z$'s subscripts, if I'm right.
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