2
$\begingroup$

I am learning about the established formalism used in the literature of IR divergences and dressed states, and I invariably come across an argument of the following form when evaluating a (photon) dressing factor of the form: $$\exp\left(-e \int \frac{d^3 k}{(2\pi)^3 (2\omega)} \frac{p^\mu}{p\cdot k} \left[a_\mu (k) e^{i k \cdot x} + a^\dagger _\mu (k) e^{-ik \cdot x} \right] \right)$$ at large times $t$, which means that the complex exponentials are evaluated at large phase. I understand the intuition that follows of a certain "averaging-out" of the contributions across the $k$ integral, when a form of the Riemann-Lebesgue lemma is invoked.

What I don't understand is the statement that gets made next (all the way back to Faddeev and Kulish themselves) that the exponentials can just be substituted by an IR function $\phi(k)$ such that: $$\phi(k) \to 1 \text{ when } |k| \to 0$$ It is one thing to say that the leading contributions to the integral are realized when $k\to 0$. I am not sure though that the leading behavior for large $t$ is well-represented by a Dirac delta-like function as described by the $\phi$ above. Isn't the leading term supposed to also vanish for large $t$ (by definition of R-L), with the $\phi$ going as $1/t$ instead when $|k|$ goes to zero? Roughly along the following line: $$\int_0 ^\infty f(k) e^{ikt} dk = \left.\frac{f(k)}{it}e^{ikt}\right\vert_{k=0}^\infty + O(t^{-2}) $$

I assume I am missing a key part of the Faddeev-Kulish argument, and I am wondering what that might be.

References:

  1. Initially Eq. 18 in the original Faddeev-Kulish paper,

  2. similarly 2.18 in this later paper.

  3. There seems to be a more formal construction in 4.4 of this reference too.

$\endgroup$
3
  • 1
    $\begingroup$ Ηi, I intend to answer this question, as I have read the original Faddeev-Kulish paper. However, I would like to clarify that the thing you are in a disagreement with is the replacing of the exponential with any function s.t. $\phi(k)\rightarrow1$ as $k\rightarrow0$ and $\phi(k)\rightarrow0$ for $k$ that are not infinitesimally small... $\endgroup$
    – schris38
    Mar 8, 2023 at 17:33
  • $\begingroup$ That is essentially it @schris38, yes. $\endgroup$
    – N.E.
    Mar 8, 2023 at 17:50
  • $\begingroup$ @N.E. I have provided a reply. I hope you find it helpful. If you have any "counter-questions", please do let me know. Also, I would like to ask you some "get-to-know-you" questions if that is okay with you, since I am always keen on meeting people that have similar interests with mine. If you are interested, I think there is some sort of chat here in PSE and we can use it to exchange emails or something like that in hopes of interacting a bit further (suggesting things to read to each other, asking questions etc) $\endgroup$
    – schris38
    Mar 9, 2023 at 10:06

1 Answer 1

2
+25
$\begingroup$

First of all, I would like to discuss some of the things you write I am not sure I am in total agreement with.

  1. The dressing operator, that originates from studying the asymptotic dynamics is not of the form you are writting. Instead, it has a charge density operator and the exponentials have a different exponent, namely $\pm ip\cdot k t/p^0$. The correct form of the dressing operator is the following $$\text{exp}\bigg(e \int\frac{d^3\vec{p}}{(2\pi)^3(2\omega_p)}\rho(\vec{p}) \int\frac{d^3\vec{k}}{(2\pi)^3(2\omega_k)} \frac{p^{\mu}}{p\cdot k} \Big[a^{\dagger}_{\mu}(\vec{k})e^{-ip\cdot k t/p^0}- a_{\mu}(\vec{k})e^{ip\cdot k t/p^0}\Big] \bigg)$$

  2. Replacing the exponentials with a function $\phi(k)$ which is equal to 1 for $|k|\rightarrow0$ and which is equal to zero otherwise, does not mean that the function $\phi$ is the Dirac function. It does not diverge as $|k|\rightarrow0$.

  3. No one performs any integral wrt $k$, as there are functions we do not know how to integrate. So, I am not sure I 100% agree with your argument that the leading term is supposed to also vanish for large $t$. I am not sure, to be honest, whether or not the leading term vanishses when taking the $|t|\rightarrow$ limit strictly, but even if that is the case, we are studying the leading contribution. So, if you like, loosely speaking, you can think of the expansion as the expansion for large $|t|$, but this does not necessarily mean that $|t|\rightarrow\infty$.

Now, let me refresh your memory with repeating some of the key arguments Faddeev and Kulish are providing on how the dressing operator emerges from studying the asymptotic dynamics. If you are familiar with all that you can skip the following 2 paragraphs.

First, we have to acknowledge that the Hamiltonian governing the asymptotic dynamics, $H_{\text{as}}$, will be comprised by the free Hamiltonian and by another potential term, with the latter being dominated by contributions that arise by considering the limit of large $|t|$.

If we denote this potential as $V_{\text{as}}$, then we substitute for the mode expansions of the various spinor and vector fields. Upon expanding those fields, there are two types of contributions in $V_{\text{as}}$: the first type of contribution contains exponentials of the form $$\sim\int\frac{d^3\vec{p}}{(2\pi)^3(2\omega_p)} \int\frac{d^3\vec{k}}{(2\pi)^3(2\omega_k)} \exp\bigg(-i\Big[\sqrt{\vec{p}^2+m^2}+ \sqrt{(\vec{p}-\vec{k})^2+m^2}\pm k^0\Big]t\bigg)$$ and $$\sim\int\frac{d^3\vec{p}}{(2\pi)^3(2\omega_p)} \int\frac{d^3\vec{k}}{(2\pi)^3(2\omega_k)} \exp\bigg(-i\Big[\sqrt{\vec{p}^2+m^2}- \sqrt{(\vec{p}-\vec{k})^2+m^2}\pm k^0\Big]t\bigg)$$

The first type of contributions have an exponent that never vanishes (as the energy of the respective fermions is never zero and we have addition of such energies). So, when $|t|$ is large, one can employ the RL lemma and state that the rapidly oscillating contributions will be "averaged-out" as you say. The other type of contributions, however, has the property that it depends on the difference of two fermionic energies, namely you can show that $\sqrt{\vec{p}^2+m^2}-\sqrt{(\vec{p}-\vec{k})^2+m^2}-k^0\approx \frac{\vec{p}\cdot{\vec{k}}}{p^0}-k^0=\frac{p\cdot k}{p^0}$ (where I distinguish four-vectors with their spatial components by including the vector arrow in the latter).

For a sufficiently small $k^0\omega_k$ (even smaller than the $|t|$ variable (which we consider to be large), the exponential $\exp(\pm i\frac{p\cdot k}{p^0}t)$ reduces to $1$. Hence, the oscillatory behavior of these contributions is being suppressed and there must be some cutoff scale, let's say $\Lambda$, above which the oscillatory behavior is not suppressed and the "averaging-out" indeed takes place. So, we can restrict the upper integration limits for the energy carried by the photons created by $a^{\dagger}_{\mu}(\vec{k})$ and destroyed by $a_{\mu}(\vec{k})$ to be the cutoff scale $\Lambda$.

Alternatively, since the only use of the exponentials are to make the contributions from the photons carrying energy $\omega_k>\Lambda$ to vanish and to make the contributions from the photons carrying energy $\omega_k\le\Lambda$ to count (i.e. by being approximate by 1 at the photon low energy limit), one can exchange the exponents by a function $\phi(k)$, s.t. $$\phi(k)\rightarrow1\ \text{as} |k|\rightarrow0\ (\text{or}\ |k|\le\Lambda)$$ and otherwise being zero.

This is how I understand things. I hope my comments are helpful. If you have any counter-questions, please do let me know.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.