### Convolution including Bose function yields an integral over a Bose function and a Fermi function - is it correct?

51

views

1

Micha had a nice problem concerning self energies in a slave-boson theory. There he had some nasty convolutions of Bose functions with spectral functions. When he asked me how to approach this problem e.g. numerically (problematic due to the divergence of the Bose function) we came up with the Idea of just a coordinate transformation. Although we had not much hope it would work, we pulled it through out of curiousity. I appended the results to this calculation and I now want to know: Is this a known phenomenon? Can we learn something from this transformation? Is it even allowed? Because I see some possible problems with diverging scales. Please note that the final integrals appear to be no convolution integrals anymore.

EDIT: there were sign mistakes in eq. 8, 10 and 11 - the final result is still correct, but now the intermediate solutions should be as well ;)

EDIT: there were sign mistakes in eq. 8, 10 and 11 - the final result is still correct, but now the intermediate solutions should be as well ;)

File attached: Integrals_over_Bose_function_convolutions_v1.1.pdf (46.85 KB)

### 2 Answers

2

Upon a first reading and little bit of thought, I am not sure if we gain anything from this transformation, in theory or practice (numerics).

The major problem is that the transformation \( \epsilon \mapsto \epsilon' \) with \( \epsilon' = \frac{1}{\beta} \ln( 1 - e^{- \beta \epsilon}) \), has a

A secondary problem related to the final conclusions, Eqs. 14—15ff, is that due to the intricate nature of the transformation, it is hard to come up with a physical picture based on it: e.g., “that a convolution of a Bose function with some other function can be split up into an integral of the Bose function with another function over positive energies and a Fermi function with some other function over the whole energy range.” Is \(\epsilon'\) really an “energy”? Hard to say.

Note that there is always very straight-forward relations between the Bose and Fermi distributions; e.g.,

\[

n_B(\varepsilon) = \frac{n_F(\varepsilon)}{2 n_F(\varepsilon) - 1} ~.

\]

So they can be always written in terms of each other. Does it resolve a problem? It depends.

The major problem is that the transformation \( \epsilon \mapsto \epsilon' \) with \( \epsilon' = \frac{1}{\beta} \ln( 1 - e^{- \beta \epsilon}) \), has a

**logarithmic singularity**at \( \epsilon = 0 \). In particular, it*cannot*serve as a*regularization*scheme for integrations on the Bose-Einstein distribution. This singularity in the Bose-Einstein is really deep — perhaps, divine ;) I think one cannot remove it, because essentially, the very singularity leads to the Bose-Einstein condensation.A secondary problem related to the final conclusions, Eqs. 14—15ff, is that due to the intricate nature of the transformation, it is hard to come up with a physical picture based on it: e.g., “that a convolution of a Bose function with some other function can be split up into an integral of the Bose function with another function over positive energies and a Fermi function with some other function over the whole energy range.” Is \(\epsilon'\) really an “energy”? Hard to say.

Note that there is always very straight-forward relations between the Bose and Fermi distributions; e.g.,

\[

n_B(\varepsilon) = \frac{n_F(\varepsilon)}{2 n_F(\varepsilon) - 1} ~.

\]

So they can be always written in terms of each other. Does it resolve a problem? It depends.

1

Regarding the first part of your comment, “the other integral (containing the Fermi-Dirac distribution function) [is not] integrating over a divergence”, I'd say, it is quite common that one disentangles an originally divergent integral to a non-divergent and a divergent part. Yet, how one achieves this is extremely important and determines the applicability of the method.

written
26 days ago by
AlQuemist

0

A closely related topic is the Cauchy principal value of a Hilbert transform. Hilbert transform is one of the most divine transformation we have (see this), and we do it everyday in many-body calculations without naming it.

One of the stable methods to perform a

I believe one can find a similar method for the Bose-Einstein kernels; yet I have not tried it myself.

One of the stable methods to perform a

*finite*Hilbert transform is the using the Cauchy principal value as described here: hilbert_cauchy.pdf (70.58 KB).I believe one can find a similar method for the Bose-Einstein kernels; yet I have not tried it myself.

Please login to add an answer/comment or follow this question.

^{st}I agree, that the transformation has a logarithmic singularity and does not regularize the Bose-Einstein distribution function. This can very easily be seen in the final equations, where we have one integral yielding basically the same problem that we tried to solve in the first place. I find it however interesting, that the other integral (containing the Fermi-Dirac distribution function) does not show this particular problem of integrating over a divergence.

2

^{nd}I also agree that the new coordinates are possibly not "physical" energies, although they are energies by dimension analysis. The idea is, that the original divergence (like \(\frac{1}{\epsilon}\) close to zero, so logarithmic in the integral) should have been converted to a logarithmic divergence in the scale.