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I'm reading a paper and don't understand some of the calculations. We are given an integral equation with asymptotic boundary conditions

$\rho_+(u)=\frac{1}{2\pi} \int\limits_{|v|>\mu}^{}\mathrm{d}v\,\frac{2\hbar \rho_+(v)}{(u-v)^2+\hbar^2}$

$\rho_+(u)=\ln(|u|)-\frac{1}{2} \ln\left(\Xi\right)+O(u-1),\,u\rightarrow\infty$

$\rho_+$ is a density of zeros an $\ln(\Xi)$ can be seen as a density of particles. But the exact meaning is not important for further calculations. The first integral ranges over $\mathbb{R}$ exceot the interval $[-\mu,\mu]$, where we have $\rho_+=0$.

Now a new function $\rho_-$ eis defined as $\rho_-(u)=\begin{cases} 0 &\mbox{if } |u|\geq\mu \\ -\frac{1}{\pi \hbar}[\rho_+(u)-\frac{1}{\pi}\int\limits_{|v|>\mu}^{}\mathrm{d}v\,\frac{\hbar \rho_+(v)}{(u-v)^2+\hbar^2} & \mbox{if } |u|<\mu \end{cases}$

with which we can rewrite the first integral equation as

$\pi\hbar\rho_- +(1-K_+)\rho_+=0$

with $K_+\rho(u)=\int\limits_{-\infty}^{\infty} k_+(u-v)\rho(v) \mathrm{d} v$ and $k_+(u)=\frac{1}{\pi} \frac{\hbar}{u^2+\hbar^2}$.

So far everything is clear. Obviously a constant function is an eigenfunction fo $K_+$ with eigenvalue 1 and so $(1-K_+)$ ist invertible. It is the said, that an operator $K_-$ can be definied on functions vanishing at $\infty$ by on functions vanishing at $\infty$

$1-K_-=(1-K_+)^{-1}$

This is still quite clear although I don't really understand, why the functions must vanish at $\infty$.

Then it ist said, that due to the degeneracy of the operator $K_+$, $K_-$ ist just defined up to a constant. What ist meant by this? Is it just due to the fact that every constant function is an eigenfunction?

Now without any calculations it is said, that the kernel $k_-(u)$ is given by

$k_-(u)=\frac{1}{\pi\hbar} \Psi(1+i\frac{u}{\hbar})+\Psi(1-i\frac{u}{\hbar})+\Delta$

where $\Psi(u)=\Gamma\,'(u)/\Gamma(u)$ is the digamma-function.

It is only said, that this can easily obtaind by fourier transform. I don't have really much knowledge about integral-equations and do not see how this result can be found. Could please someone show me?

I have some other questions concerning the following calculations, but maybe I stop here until this problem is solved.

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Which paper are you reading? –  Qmechanic Nov 2 '13 at 12:38
1  
I'm reading a paper by Sklyanin about the quantization of the toda lattice. I don't know if I'm allowd to attach the pdf, I'll do it anyway. You can download it here: dropbox.com/s/tju18r0jvb80zdk/Skly85_Toda.pdf The part I'm struggling with starts at page 221. –  Wessi990 Nov 2 '13 at 12:55
    
I tried to use $(1-K_-)(1-K_+)\rho=\rho$. This gives one integral which contains only $k_-$, one with onliy $k_+$ and one double integral. I don't see, how using fouriertransform and inverse fouriertransform would lead to isolating $k_-$. I knwo how to use fouriertransform to solve partial differential equations, but I have no idea how it could be used here. –  Wessi990 Nov 2 '13 at 16:43
    
Note that $\hbar\equiv\eta$. –  Qmechanic Nov 5 '13 at 15:49

1 Answer 1

up vote 1 down vote accepted

We start with the Lorentzian distribution

$$ \tag{4.11} k_{+}(u)~:=~ \frac{1}{\pi} \frac{\eta}{u^2+\eta^2}; $$

with Fourier transform

$$ \tag{4.11'} \tilde{k}_{+}(x) ~:=~\int_{\mathbb{R}}\!du ~e^{-ixu} k_{+}(u)~=~e^{-\eta|x|}; $$

with integral operator

$$ \tag{4.10} (K_{\pm}\rho)(u)~:=~\int_{\mathbb{R}}\!dv~ k_{\pm}(u-v)\rho(v) ~=~(k_{\pm}\ast\rho)(u). $$

The Fourier transformed operator is a multiplication operator

$$ \tag{4.10'} (K_{\pm}\rho)^{\sim}(x)~:=~\tilde{k}_{\pm}(x) \tilde{\rho}(x). $$

Repeated application of the integral operator leads to repeated convolutions

$$ \tag{A} (K_{\pm}^n\rho)(u) ~=~(\underbrace{k_{\pm}\ast\ldots \ast k_{\pm}}_{n\text{ factors}}\ast\rho)(u), \qquad (K_{\pm}^n\rho)^{\sim}(x)~:=~\tilde{k}_{\pm}^n(x) \tilde{\rho}(x). $$

Note that for a constant function $\rho \propto 1$ is an eigenfunction with eigenvalue 1 for the integral operator $K_{+}$:

$$\tag{B} (K_{+} 1)(u) ~=~\int_{\mathbb{R}}\!dv~ k_{+}(u-v)~=~1. $$ Since we want the operator $1-K_{+}$ to be invertible, we will from now on only consider integrable functions $\rho:\mathbb{R}\to \mathbb{C}$, such that $\int_{\mathbb{R}}\!du~\rho(u)=0$. In particular, we will exclude constant functions $\rho \propto 1$. This means that the kernels $k_{\pm}(u)\to k_{\pm}(u)+\Delta_{\pm}$ are only defined modulo additive constants $\Delta_{\pm}$.

Furthermore,

$$\tag{C} 1-K_{-} ~=~ \frac{1}{1-K_{+}}~=~\sum_{n=0}^{\infty}K_{+}^n ,$$

and therefore, naively,

$$\tag{D} \tilde{k}_{-}(x)~=~\left(1- \frac{1}{\tilde{k}_{+}(x)}\right)^{-1} ~=~\frac{1}{1-e^{\eta|x|}} . $$

However the expression (D) is not integrable at $x=0$. The solution is to regularize $k_{-}(u)$ with an infinite additive constant $\Delta_{-}=\int_{\mathbb{R}}\!\frac{dx}{2\pi}~\frac{e^{-\eta|x|}}{\eta|x|}=\infty$, so that

$$ k_{-}(u)~=~ \int_{\mathbb{R}}\!\frac{dx}{2\pi} \left[\frac{e^{-\eta|x|}}{\eta|x|}+ \frac{e^{ixu}}{1-e^{\eta|x|}}\right] ~=~\int_{0}^{\infty}\!\frac{dx}{\pi} \left[\frac{e^{-\eta x}}{\eta x} +\frac{\cos(xu)}{1-e^{\eta x}} \right]$$

$$\tag{E} ~\stackrel{t=\eta x}{=}~\sum_{\pm}\int_{0}^{\infty}\!\frac{dt}{2\pi\eta} \left[\frac{e^{-t}}{t}+\frac{\exp\left(\pm\frac{ iut}{\eta}\right) }{1-e^{t}}\right]~=~\frac{1}{2\pi\eta}\sum_{\pm}\psi(1\pm\frac{iu}{\eta}), $$

cf. the Digamma function and Abramowitz & Stegun eq. (6.3.21). Note that the final formula (E) differs from Sklyanin's formula (4.12) by a factor 2 in the overall normalization.

References:

  1. E.K. Sklyanin, The quantum Toda chain, Lecture Notes in Physics, 226 (1985) 196.
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Perfect, I talked to one of my professors about that problem and he also gave the adivice, that I just need to use the convulition theorem and an integral representation of the digamma function. Nevertheless I didn't really succeed, thanks a lot for this answer! –  Wessi990 Nov 5 '13 at 15:20
    
Hello, I have another problem with the paper. Right after the calculation the asymptotic conditions for $\rho_-$ are derived. I'don't see, where they come from. I thought they would be trivial, but now when I look closer at them, that's no longer the case and I'm quite unsure how Sklyanin found them. Could you maybe help me again? –  Wessi990 Dec 5 '13 at 0:22

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