Let $n_F(\omega) = \large \frac{1}{e^{\beta (\omega)} + 1}$ be the Fermi function.

A fermionic reservoir correlation function is given by:

$$C_{12}(t) = \int_{-\infty}^{+\infty} d\omega~ \tag{5}J_R(\omega) \, n_F(\omega) \, e^{-i\omega t}$$

The Fermi function here is given in terms of Chebyshev polynomials.

The coefficients of the Chebyshev polynomials are given by:

$$c_k = \frac{2}{\pi}\int_{0}^{\pi}f(\cos\theta)\cos(k\theta) \;d\theta$$And the Fermi function itself is (Chebyshev approximated, if you will):

$$n'_F(x) = \sum_{k=0}^{n}\left[ \frac{2}{\pi}\int_{0}^{\pi}\frac{\cos k\theta \; d\theta}{e^{\beta(E_F-\cos\theta)}+1} \right] T_k(x) \tag{6}$$ Where $T_k(x)$ are the Chebyshev polynomials of the first kind.

Now, since the Fermi function here has no poles(as it's given in terms of Chebyshev interpolation polynomials which do not have any poles), the poles in the correlation function are only those of the Spectral Density: $$J_R(\omega) = \sum_{k=1}^{m}\frac{p_k}{4\Omega_k(\omega-\Omega_k)^2+\Gamma_k^2}\tag{7}$$

There's only pole at: $\omega = \Omega_k - i\Gamma_k=\Omega_k^-$, and the residue is $\left.\frac{1}{(\omega - \Omega_k) - i\Gamma_k}\right|_{\omega=\Omega_k^-} = \frac{1}{-2i\Gamma_k}$.

Or the residue of $J_R(\omega)$ at $\omega=\Omega_k^-$:

$$\mathop{\text{Res}}\limits_{\omega=\Omega_k^-} J_R(\omega) = \frac{p_k}{4 \Omega_k(-2i\Gamma_k)}\tag{8}$$

My question is: How can the integral of the correlation function now be solved using the theorem of residues/Jordan's lemma? Is it still possible or another scheme should be employed?

If the Fermi function was given in terms of Matsubara frequency sum, it would have had poles and then it's residues could be calculated. Now it does not, and I can't see how the integral can be solved now. If the Fermi function had poles, we could have said:

Noting that Poles of $J_R(\omega)$: $\Omega_k^-$, and the poles of $n_F(\omega)$: $\nu_{k'}^*$, we could have gotten:

$$C_{12}(t) = (-)(2i\pi) \left \lbrace \sum_{k=1}^m \mathop{\text{Res}}\limits_{\omega=\Omega_k^-}\left[ J_R(\omega) \right] n_F(\Omega_k^-)e^{-i\Omega_k^- t} \\+ \sum_{k'} \mathop{\text{Res}}\limits_{\omega=\nu_{k'}^*} \left[ n_F(\omega) \right] J_R(\nu_{k'}^*)e^{-i\nu_{k'}^* t} \right \rbrace\tag{9}$$

And then the residues could have been calculated.

A bit different Version of the same problem(If you decide to answer, kindly answer this first):

Here's a function:

$$C_{12}(t) = \int_{-\infty}^{+\infty} d\omega~ \tag{1}J_R(\omega)n_F(\omega)e^{-i\omega t}$$

Here's $n'_F(x)\approx n_F(\omega)$:

$$n'_F(x) = \sum_{k=0}^{n}\left[ \frac{2}{\pi}\int_{0}^{\pi}\frac{\cos k\theta \; d\theta}{e^{\beta(E_F-\cos\theta)}+1} \right] T_k(x) \tag{2}$$ Where $T_k(x)$ are the Chebyshev polynomials of the first kind.

Also, $n'_F(x)$ has no poles.

And, $$J_R(\omega) = \sum_{k=1}^{m}\frac{p_k}{4\Omega_k(\omega-\Omega_k)^2+\Gamma_k^2}\tag{3}$$ Where $\Omega$, P and $\Gamma$ are only some numbers.

Are the prerequisites of Jordan's Lemma fulfilled? That is, can equation (1) be written as equation (4) after inserting (2) and (3) in (1) and then applying Jordan's Lemma?

$$C_{12}(t) = (-)(2i\pi) \left \lbrace \sum_{k=1}^m \mathop{\text{Res}}\limits_{\omega=\Omega_k^-}\left[ J_R(\omega) \right] n'_F(\Omega_k^-)e^{-i\Omega_k^- t}\right \rbrace\tag{4}$$

  • $\begingroup$ I'm probably missing something, but why is fewer poles a problem? You still close the contour with a 0-contribution loop and then find the residues of the integrand - whatever they are - inside the closed contour. If there are no residues of $n_\mathrm{F}$ to sum over, then so be it - there are still other residues for the integrand as a whole, and it's not like $n_\mathrm{F}$ had no effect on the answer. $\endgroup$ – user10851 Nov 1 '13 at 7:11
  • 2
    $\begingroup$ That seems like something that could be true. But the thing is that can we still apply theorem of residues/jordan's lemma if there are no poles? If it's possible to, like you say, then I just have to evaulate $n_F$ over the pole of the spectral density and I'm done, right? $\endgroup$ – Hasan Nov 1 '13 at 7:43
  • $\begingroup$ You know what, you're right. It's possible to close the contour and evaluate the residues of $J_x(\omega)$ only. But what happens to $n_F$ then.. just what I wrote above or something else. $\endgroup$ – Hasan Nov 1 '13 at 8:18
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/79521/2451 $\endgroup$ – Qmechanic Nov 1 '13 at 21:13
  • $\begingroup$ I do not see $\omega$ on the r.h.s. of your expansion of the Fermi function (the un-numbered equation after eq. 2) $\endgroup$ – Dave Nov 1 '13 at 21:36

I have just found out that it is fundamentally wrong to approximate the fermi function with chebyshev polynomials. Representation of the Fermi function on the real axis by Chebychev polynomials might be okay, but the representation of the Fermi function in the complex plane and especially close to the poles of the Fermi function is certainly very poor (no pole versus pole). By applying Jordan's lemma to this poor representation we miss the most important ingredients, i.e. the poles, will certainly result in a poor result.

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Where are you getting at with the Chebyshev approximation? Each $T_k$ satisfies

$$ (1-x^2)T_k''(x)-xT_k'(x)+k^2T_k(x) = 0 $$

so from $(2)$

$$ \left[(1-x^2)\left(\frac{d}{dx}\right)^2-x\frac{d}{dx}\right]n'_F(x) \\ = \sum_{k=0}^{n}\left\{\frac{2}{\pi}\int_{0}^{\pi}\frac{\cos (k\theta) d\theta}{e^{\beta(-\cos\theta+E_F)}+1} \left[(1-x^2)\left(\frac{d}{dx}\right)^2-x\frac{d}{dx}\right]T_k(x)\right\} \\ = \sum_{k=0}^{n}\left\{\frac{2}{\pi}\int_{0}^{\pi}\frac{\cos (k\theta) d\theta}{e^{\beta(-\cos\theta+E_F)}+1} \left[-k^2T_k(x)\right]\right\} \tag{*} $$


$$ c_k = \frac{2}{\pi}\int_{0}^{\pi}\cos (k\theta) n_f(-\cos\theta + E_F) d\theta $$

we get by double partial integration

$$ \left(\frac{\partial}{\partial E_F}\right)^2c_k = \frac{2}{\pi}\frac{\partial}{\partial E_F}\int_{0}^{\pi}\cos (k\theta) \dot n_F(-\cos\theta + E_F) d\theta \\ = \frac{2}{\pi}\frac{\partial}{\partial E_F}\left\{-\int_{0}^{\pi}-k\sin (k\theta) n_F(-\cos\theta + E_F) d\theta + [\cos (k\theta) n_F(-\cos\theta + E_F)]_{0}^{\pi}\right\} \\ = \frac{2}{\pi}\left\{\int_{0}^{\pi}k\sin (k\theta) \dot n_F(-\cos\theta + E_F) d\theta + \frac{\partial}{\partial E_F}\left[(-1)^k n_F(-1 + E_F)- n_F(1 + E_F)\right]\right\} \\ = \frac{2}{\pi}\left\{-\int_{0}^{\pi}k^2\cos (k\theta) n_F(-\cos\theta + E_F) d\theta + [k\sin (k\theta) n_F(-\cos\theta + E_F)]_{0}^{\pi} + \left[(-1)^k \dot n_F(-1 + E_F)- \dot n_F(1 + E_F)\right]\right\} \\ = -k^2c_k + \frac{2}{\pi}\left[ - \dot n_F(1 + E_F) + (-1)^k \dot n_F(-1 + E_F) \right] \\ = -k^2c_k - \frac{2\dot\beta}{\pi}\left[ e^{\beta(1 + E_F)}n_F^2(1 + E_F) - (-1)^k e^{\beta(-1 + E_F)}n_F^2(-1 + E_F) \right] $$

(dot denotes the usual derivative). Together with $(*)$ we get

$$ \left[(1-x^2)\left(\frac{d}{dx}\right)^2-x\frac{d}{dx}\right]n'_F(x) = \sum_{k=0}^{n}\left\{\left(\frac{\partial}{\partial E_F}\right)^2c_k + \frac{2\dot\beta}{\pi}\left[\cdots \right]\right\} T_k(x) \\ = \left(\frac{\partial}{\partial E_F}\right)^2n'_F(x) + \frac{2\dot\beta}{\pi}\left[e^{\beta(1 + E_F)}n_F^2(1 + E_F)\sum_{k=0}^{n} T_k(x) - \sum_{k=0}^{n} e^{\beta(-1 + E_F)}n_F^2(-1 + E_F) (-1)^k T_k(x) \right] $$

If you estimate that the 'long' term is negligible, then

$$ \left[(1-x^2)\left(\frac{d}{dx}\right)^2-x\frac{d}{dx}\right]n'_F(x) \approx \left(\frac{\partial}{\partial E_F}\right)^2n'_F(x) $$

I don't know if this helps.

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  • $\begingroup$ What you did might be correct but it's irrelevant. $\endgroup$ – Hasan Nov 6 '13 at 8:25

You need a theorem of the form:

  • given H(z)=F(z)G(z),
  • F(z) is analytic everywhere (this is the Fermi distribution),
  • G(z) has a pole at $z_0$ (this is your correlation function)


  • $Res[ H(z_0) ] = F(z_0) Res[ G(z_0)]$

This follows from the definition of the residue as

$$ \begin{aligned} Res[ H(z_0) ]&= \lim_{z\rightarrow z_0} (z-z_0) H(z) \\ &= \left[ lim_{z \rightarrow z_0} F(z) \right] \left[ lim_{z \rightarrow z_0} (z-z_0)G(z) \right] \\ &= F(z_0) Res[ G(z_0)] \end{aligned} $$

where I can pull the $F$ out of the limit since I know it is analytic everywhere.

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  • $\begingroup$ But the correlation function and the fermi function do not multiply like this.. The correlation function contains an integral of the fermi function multiplied by the spectral density. equation (1). $\endgroup$ – Hasan Nov 1 '13 at 21:25
  • $\begingroup$ I though that your goal was to evaluate the integral in (1) via the theorem of residues; to me it looks like the integrand is $J\times n \times e^{-i \omega t}$ $\endgroup$ – Dave Nov 1 '13 at 21:28
  • $\begingroup$ Yes that's what it is.. but J is the spectral density, and it has a pole. So what I did was, (if you look at (4)), take only the first summation in (4) and evaluate the chebyshev approximated fermi function over the pole of J. Is this what you mean too? $\endgroup$ – Hasan Nov 1 '13 at 21:30
  • $\begingroup$ What I'm saying is that we know that the Fermi function does not have any poles; thus there are no terms in the second summation of Eq. (3). $\endgroup$ – Dave Nov 1 '13 at 21:38
  • $\begingroup$ Yes, that's true. $\endgroup$ – Hasan Nov 1 '13 at 21:41

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