1
$\begingroup$

I have one probably very silly confusion about a footnote in the paper "2D Kac-Moody symmetry of 4D Yang-Mills theory ". In section (4) the authors consider ${\cal O}_k(E_k,z_k,\bar{z}_k)$ an operator which creates or annihilates a colored hard particle with energy $E_k\neq 0$ crossing the $S^2$ on ${\mathscr{I}}$ at the point $z_k$. In a footnote they say that for scalar particles we would have: $${\cal O}_k(E_k,z_k,\bar{z}_k)=-\frac{4\pi}{E_k}\int_{-\infty}^\infty du e^{iE_k u}\partial_u \lim_{r\to\infty}[r\phi_k(u,r,z_k,\bar{z}_k)]\tag{1}.$$

Now the way I understood this is that ${\cal O}_k(E_k,z_k,\bar{z}_k)$ is just one creation/annihilation operator written in terms of the field data at $\mathscr{I}$. I have tried to take one large $r$ limit of a scalar field and obtain (1).

In that case I have considered the simplest example possible: one massless scalar field $\phi(x)$. Decomposing into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3 p}{(2\pi)^32\omega} (a(p)e^{ipx}+a^\dagger(p)e^{-ipx}),\tag{2}$$

I considered the $r\to \infty$ limit with $(u,z,\bar{z})$ fixed employing the plane wave decomposition into spherical Bessel functions plus the asymptotic behavior of such functions. As a result I have obtained $$\phi(u,r,z,\bar{z})=-\dfrac{i}{8\pi^2 r}\int_0^\infty [a(\omega\hat{x}(z,\bar{z}))e^{-i\omega u}-a^\dagger(\omega\hat{x}(z,\bar{z}))e^{i\omega u}] d\omega+O\left(\frac{1}{r^2}\right)\tag{3}.$$

Now using (1) the result is exactly $a(\omega\hat{x}(z,\bar{z}))$. So it seems to confirm that ${\cal O}$ is really just the familiar creation/annihilation operators, just written in terms of ${\mathscr{I}}$ data.

But if that is the whole point (write the creation/annihilation operators in terms of ${\mathscr{I}}$ data) then why instead of dividing by the energy and taking $\partial_u$ we don't just take $${\cal O}(\omega,z,\bar{z})=4\pi i \int_{-\infty}^\infty e^{i\omega u}\lim_{r\to \infty}(r\phi(u,r,z,\bar{z}))du\tag{4}.$$

I mean (4) does the same job and it seems more natural. So is there any reason to use (1) instead? Why use (1) instead of (4)?

$\endgroup$
2
$\begingroup$

Both do the job, as long as one can freely integrate by parts on ${\mathscr I}$.

At large $|u|$ the fields on ${\mathscr I}$ are finite, i.e. $$ \lim_{u\to\pm\infty}\phi(u,z,{\bar z})= \phi_\pm(z,{\bar z}). $$ You can check that if $C = \phi_+ + \phi_- = 0$, then we can freely integrate by parts (one would have to be careful about the phases $e^{i\omega u}$ but that can be handled by use wave-packets). In this case, both formulae give rise to the same result.

However, $C$ is typically not zero and in fact, plays a crucial role in soft physics. It is the symplectic conjugate of the soft mode $N = \phi_+ - \phi_-$ so you can't have one without the other. Formula (1) is the correct formula when $C \neq 0$.

Aside - If you use start with equation (2) and attempt to calculate $C$ in terms of creation and annihilation modes, you will find that $C = 0$. This is because $C$ cannot be written in terms of the creation annihilation modes. It is a completely separate soft mode that is not captured by mode expansions.

We can think about it like this - the annihilation modes $a(\omega)$ are symplectically paired on the phase space with the creation modes $a(\omega)^\dagger = a(-\omega)$. However, the zero mode $a(0)$ does not have a symplectic conjugate within the creation-annihilation operators. In other words, the phase space is not even-dimensional. To make it even-dimensional, we either project out $a(0)$ (reduce dimension by 1) or we introduce a new mode $C$ (increase dimension by 1) which is the conjugate of $a(0)$. Doing the first implies a trivial soft theorem which is not consistent with charge conservation (related to large gauge transformations discussed in the paper) so we are left with only the second possibility.

The absence of $C$ in the standard discussion of massless quantum field theories is THE reason for infrared divergences. Once we expand the phase space by including $C$ in the phase space, these issues can be resolved and it is possible to define an IR finite $S$-matrix as shown in Infrared divergences in QED revisited.

$\endgroup$
2
  • $\begingroup$ Thanks @Prahar ! So this is related to the result that a function which has a Fourier transform must have boundary values summing to zero, right? So in principle when $C\neq 0$ we wouldn't be able to use (4) and then the point would be to split $\phi(u,z,\bar{z}) = \hat{\phi}(u,z,\bar{z}) + C(z,\bar{z})$ and use the $\partial_u$ in (1) to extract $\hat{\phi}(u,z,\bar{z})$ for which can invert the Fourier transform, am I correct? $\endgroup$ – Gold Jul 13 '20 at 4:39
  • $\begingroup$ @user1620696 that is correct. To be clear, a function which has a Fourier transform that has at most a simple pole at $\omega = 0$ must have boundary values summing to zero. $\endgroup$ – Prahar Mitra Jul 13 '20 at 4:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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