In the real-time finite temperature formalisms (Schwinger-Keldysh or Thermo-field), the free propagators are often defined with terms like: $$ \mathrm{Dirac\ Delta}\ \times \ \mathrm{Thermal\ Distribution}(|\mathrm{frequency}|) $$

For example, in Thermo-field theory the free propagators for a real scalar field are: $$ - i \Delta_{11}(p;m) = \frac{-i}{-p_0^2 + |\mathbf{p}|^2 + m^2 - i \epsilon} + \frac{2 \pi \delta(-p_0^2 + |\mathbf{p}|^2 + m^2 )}{e^{\beta |p_0|} - 1} \\ - i \Delta_{12}(p;m) \ = \ - i \Delta_{21}(p;m) = \pi \mathrm{csch}\left( \tfrac{\beta|p_0|}{2} \right) \delta(-p_0^2 + |\mathbf{p}|^2+m^2) \\ - i \Delta_{22}(p;m) = \frac{i}{-p_0^2 + |\mathbf{p}|^2 + m^2 + i \epsilon} + \frac{2 \pi \delta(-p_0^2 + |\mathbf{p}|^2 + m^2)}{e^{\beta |p_0|} - 1} $$

So for example the term with the $\frac{\delta(p^2 + m^2 )}{e^{\beta |p_0|} - 1}$ is concerning me.

The reason I am confused is that I have been reading about generalized functions/distributions and a basic fact about these objects is that you cannot multiply two distributions by one another (ie. multiplying two distributions does not yield a well-defined distribution).

The $\delta$ is obviously a distribution and since we've got an absolute bar on $|p_0|$ in the $\frac{1}{e^{\beta |p_0|} - 1}$ I take it that this is a distribution as well?.

Am I misunderstanding the meaning of $\frac{1}{e^{\beta |p_0|} - 1}$? How are the above propagators meaningful in the sense of distributions?


2 Answers 2


Two remarks:

  1. Essentially every function can be regarded as a distribution, although the converse is not true. In this sense, you can indeed regard $(e^{\beta |p_0|} - 1)^{-1}$ as a distributions. Distributions of this kind are known as regular distributions. Non-regular distributions are known as singular distributions.

  2. It is not true that you cannot multiply distributions. For example, the multiplication of regular distributions is trivial – you just multiply their associated functions, which is a perfectly well-defined operation.

    A somewhat less trivial statement concerns singular distributions. You can multiply those with regular distributions just fine (unless the regular distribution is too wild, in which case you have to insist that the singularities of the former do not coincide with the singularities of the latter). This is precisely the case with your propagators, and I invite to think whether the singularities of the objects you are working with coincide or not. You should be able to convince yourself that this is not the case, and so the multiplications are all well-defined.

    Finally, a much less trivial statement concerns the product of two singular distributions. If their singular supports are disjoint, their multiplication is perfectly well-defined and trivial to implement. If their singular supports do overlap, then things get more interesting. In some cases you can multiply singular distributions with non-disjoint singular supports, and sometimes you cannot. The details depend on how fast the distributions decay in Fourier space; to formalise these facts Hörmander introduced the so-called wave front set. We will not discuss this here.


Concerning OP's explicit example $$ \frac{\delta(p^2 + m^2 )}{e^{\beta |p_0|} - 1} , \qquad\beta\neq 0. \tag{D}$$

  1. The massive case $m\neq 0$. Then the singularity of $\frac{1}{e^{\beta |p_0|} - 1}$ does not overlap with the support of $\delta(p^2 + m^2 )$, so the product distribution is mathematically well-defined.

  2. The massless case $m=0$. Then the distribution (D) [and already $\delta(p^2)$ itself!] are mathematically ill-defined per se. Heuristically (D) can be rewritten as $$ \frac{1}{e^{\beta |p_0|} - 1}\frac{1}{2|p_0|} \sum_{\pm}\delta(p_0 \pm |{\bf p}| ),\tag{D'}$$ which has a double pole at $p_0=0$. Thus (D) only makes sense for test functions that have a corresponding double zero at $p_0=0$.

  • $\begingroup$ Thanks for your answer. In the case $\beta \neq 0$ and $m=0$ it would seem that $\delta(p^2)$ has a singularity at $p_0 = \mathbf{p}$ and $\frac{1}{e^{\beta|p_0|} - 1}$ has a singularity at $p_0 = 0$. These coincident when you send $\mathbf{p} \to \mathbf{0}$. Does this mean massless theories are ill-defined? $\endgroup$ Commented Jun 8, 2018 at 11:18
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Jun 8, 2018 at 12:09
  • $\begingroup$ Could you expand on your comment $\delta(p^2)$ being ill defined itself? Does this means that it must always be understood as $\frac{\delta(p_0-|\mathbf{p}|)+\delta(p_0+|\mathbf{p}|)}{2|p_0|}$ and so must come along with test functions that have a single zero at $p_0 =0$ (I take this to mean that the test function $\phi$ must look like $\sim p_0 + \mathscr{O}(p_0^2)$ near $p_0 =0$)? $\endgroup$ Commented Jun 8, 2018 at 12:19
  • 1
    $\begingroup$ $\uparrow $ Right. $\endgroup$
    – Qmechanic
    Commented Jun 8, 2018 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.