# CPT invariance and Soft Theorems

I am reading the paper IR Dynamics and Entanglement Entropy, written by Toumbas and Tomaras and I have a question on using the CPT invariance of the QED $$S$$-matrix elements in order to derive the scattering amplitude describing a soft photon absorption, given the amplitude describing a soft photon emission.

The soft photon emission amplitude is given as $$\lim_{|\vec{q}|\to0}(2\omega_q)^{1/2} \langle\beta|a_r(\vec{q})S|\alpha\rangle= \bigg(\sum_{i\in\beta}\frac{e_ip_i\cdot\epsilon_r^*(\vec{q})}{p_i\cdot q}- \sum_{i\in\alpha}\frac{e_ip_i\cdot\epsilon_r^*(\vec{q})}{p_i\cdot q} \bigg)\langle\beta|S|\alpha\rangle$$ where $$\beta$$ denotes the final state, $$\alpha$$ the initial one, $$p_i$$, $$e_i$$ are the momenta and the charges of the incoming/outgoing matter particles and $$q$$ is the momentum of the emitted photon. The authors state that by using CPT invariance and the amplitude above, one can write down the amplitude describing soft photon absorption, $$\lim_{|\vec{q}|\to0}(2\omega_q)^{1/2} \langle\beta|Sa_r^{\dagger}(\vec{q})|\alpha\rangle=- \bigg(\sum_{i\in\beta}\frac{e_ip_i\cdot\epsilon_r(\vec{q})}{p_i\cdot q}- \sum_{i\in\alpha}\frac{e_ip_i\cdot\epsilon_r(\vec{q})}{p_i\cdot q} \bigg)\langle\beta|S|\alpha\rangle.$$

My question is: how is CPT invariance used in obtaining the second amplitude from the first?

• It should be crossing symmetry, not CPT. Commented Jul 18, 2023 at 10:36
• Hi @Prahar and thanks for commenting again. May I ask why should it be crossing symmetry instead of CPT? And if so, how does this symmetry imply the soft theorem describing the photon absorption from the one describing emission? Commented Jul 18, 2023 at 10:39
• CPT symmetry should act on all the particles in the S-matrix. Crossing symmetry on the other hand acts on just one particle (in this case, the photon) and takes a single $in$ state to an $out$ state or vice versa. The extra minus sign is the CPT phase of the photon. Commented Jul 18, 2023 at 10:42
• Should subleading order in the soft momentum terms obey the crossing symmetry as well? Commented Jul 18, 2023 at 10:54
• I'll add an answer to discuss this. Commented Jul 18, 2023 at 10:57

The result actually holds due to crossing symmetry, not CPT. Note that while CPT invariance follows from the CPT theorem, there is no such established theorem regarding crossing symmetry. So we really aren't sure (at the level of a theorem) if the photon operator $$O_a(p)$$ satisfies crossing symmetry. However, what we really need for your case is whether the soft photon operator satisfies crossing symmetry.
In [1907.02808], we work in $$D=d+2$$ dimensions and parameterize a massless momenta by $$p^\mu = \frac{\omega}{2} ( 1 + x^2 , 2x^a , 1 - x^2 ) , \qquad x^a \in {\mathbb R}^d~ (a=1,\cdots,d).$$ where $$x^2 = x^a x^a$$. In $$d+2$$ dimensions, the photon has $$d$$ polarizations. Let $$O^{\pm}_a(\omega,x)$$ be the $$out$$ (+) and $$in$$ (-) annihilation operator for a photon with polarization vector $$\varepsilon^\mu_a(x) = ( x_a , \delta^b_a , - x_a ).$$ In the soft limit, the operators admit the expansion $$O^{\pm}_a(\omega,x) = \frac{1}{\omega} N_a^\pm(x) + B_a^\pm(x) + {\cal O}(\omega) .$$ The operator $$N_a^\pm$$ inserts a leading soft-photon and $$B_a$$ inserts a subleading soft-photon. In equation (2.21) of [1907.02808], it is argued that $$\tag{1} N_a^{\pm\dagger}(x) = N_a^{\pm}(x) , \qquad B_a^{\pm\dagger}(x) = - B_a^{\pm}(x) .$$ Note that this property implies that $$N_a$$ and $$B_a$$ do NOT admit an interpretation of creation or annihilation operators. There is NO particle interpretation for the leading order subleading soft photons.
Let us now see what this implies as far crossing symmetry for the $$S$$-matrix is concerned. Note that the operator which we insert in a scattering amplitude (from the LSZ theorem) is $$\tag{2} Out:~ O^+_a(\omega,x) - O^-_a(\omega,x) , \qquad In:~ O^{-\dagger}_a(\omega,x) - O^{+\dagger}_a(\omega,x)$$ Taking the leading soft limit of the $$out$$ and $$in$$ operators, we find $$Leading~Out:~ N^+_a(x) - N^-_a(x) , \qquad Leading~In:~ N^{-\dagger}_a(x) - N^{+\dagger}_a(x)$$ Then, using (1), we find that $$\boxed{ \text{Leading out soft-photon} = - \text{Leading in soft-photon} }$$ Taking a subleading limit of (2), we find $$Subleading~Out:~ B^+_a(x) - B^-_a(x) , \qquad Subleading~In:~ B^{-\dagger}_a(x) - B^{+\dagger}_a(x)$$ Then, using (1), we find that $$\boxed{ \text{Subleading out soft-photon} = + \text{Subleading in soft-photon} }$$
• Hi @Prahar and thanks for the reply. Despite being a research-level answe, I think I can keep up with some of it. Ok, so I realize you have some photon operators (i.e. Fourier coefficients in the Photon field mode expansion promoted to operators upon quantizing), which should, in principle, add/remove a photon from the in/out states. Those admit an expansion in the soft energy. Why do you suggest that the coefficients of that expansion (namely $N_a,\ B_a$) do not admit an interpretation as creation/annihilation operators, however? Commented Jul 19, 2023 at 7:35
• Also, from your answer I am led to conclude that $\langle\beta |a_r(k)S|\alpha\rangle=-\langle\beta |Sa_r(k)^{\dagger}|\alpha\rangle$ in the leading case, whereas in the subleading one we have $\langle\beta |a_r(k)S|\alpha\rangle=+\langle\beta |Sa_r(k)^{\dagger}|\alpha\rangle$, correct? And recall that $\alpha$ is the initial state of the scattering and $\beta$ the final one, and $S$ is the S-matrix and $a_r(k)$ a photon annihilation operator of polarization $r$ and momentum $k$... Commented Jul 19, 2023 at 7:37