In quantum field theory a soft particle is a low energy particle. If I'm not mistaken one introduces a energy treshold $\Lambda$ and calls soft any particle with energy $\omega < \Lambda$.

If $a^\dagger(\mathbf{p},\sigma)$ is the creation operator, we say that $a^\dagger(\mathbf{p},\sigma)$ creates a soft particle if $$\omega = \sqrt{|\mathbf{p}|^2+m^2}<\Lambda.$$

For massless particles, $\omega_p = |\mathbf{p}|$. In that case, if $\hat{\mathbf{p}}$ is the unit vector in the momentum direction $\mathbf{p} = \omega \hat{\mathbf{p}}$ and we can write $a^\dagger(\omega\hat{\mathbf{p}},\sigma)$.

In this paper, Andrew Strominger considers operators of the type

$$\mathcal{O}=\lim_{\omega \to 0^+} a^\dagger(\omega \hat{\mathbf{p}},\sigma).$$

My doubt is how to understand this limit (what topology on the space of operators it is defined with respect to) and the states obtained by acting with $\mathcal{O}$ on the vacuum.

I mean suppose we take the vacuum $|0\rangle$ and operate with this:

$$|\Phi\rangle=\lim_{\omega \to 0^+} a^\dagger(\omega \hat{\mathbf{p}},\sigma)|0\rangle.$$

What this state will be? Although usually soft means low energy, which we can make precise with an energy treshold, what Strominger seems to call soft is energy zero.

But there are no states with zero energy in the single particle state space.

I've considered the momentum representation so that the state space is based on $L^2(H_m^+,d\Omega_m;\mathbb{C}^{2s+1})$ where $H_m^+$ is the set of momenta with $p^2=m^2$ and $p^0 >0$ and $d\Omega_m$ is the Lorentz invariant measure.

So states are square integrable functions $\Phi : H_m^+\to \mathbb{C}^{2s+1}$. In that setting, what would be for massless particles

$$\Phi_\sigma(p)=\langle p,\sigma | \Phi\rangle = \langle p,\sigma | \lim_{\omega\to 0^+}a^\dagger(\omega\hat{\mathbf{p}},\sigma)|0\rangle$$

and how do we make sense of this limit?

  • 1
    $\begingroup$ I can't help with your primary question about the nature of the state $| \Phi \rangle$, but it seems to me like Andy is coming at the problem from another angle. Checking out equation (2.8.21), the "standard soft theorem," it seems like what he calls the "soft limit" is not really to be thought of as the scattering of zero-energy states, but the scattering of states in the limit as their energy becomes small. Whether or not $| \Phi \rangle$ appears in your Hilbert space doesn't then seem relevant to the calculations Andy performs. $\endgroup$ – Diffycue Nov 11 '18 at 23:08

Your Answer

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

Browse other questions tagged or ask your own question.