After a number of years out of grad school and in industry I'm trying to brush up on my fundamentals. Going through Schwartz's "Quantum Field Theory and the Standard Model", I'm having trouble understanding the calculation in (4.18) -- (4.22) of a particular matrix element that arises as part of a term in the the OFTP expansion of the "electron electron" scattering transfer matrix, though ignoring spin and charge and treating the electron and photon as real scalars. Changing notation slightly, this matrix element is
$$ V^{(R)} = \frac{e}{2} \int d^3 x \langle \psi^3 \phi^\gamma | \Psi(x) \Phi(x) \Psi(x) | \psi^1 \rangle = e (2\pi)^3 \delta^3(p_1-p_3-p_\gamma)\tag{4.18+21} $$ where $\phi^\gamma$ represents the state with a photon with three-momentum $\gamma$; $\psi^i$ represents a state with an electron with three-momentum $p_i$; and $\Psi$ and $\Phi$ are the real scalar operators for these fields.
However, trying to make this calculation myself, I end up with an extra nonsense, divergent term $$ e (2\pi)^3 \omega_1 \delta^3(p_1-p_3) \delta^3(p_\gamma) \int \frac{d^3 q}{2 \omega_q} $$ coming ultimately from a leftover term proportional to $ \langle 0 | \Psi(x)^2 | 0\rangle$ that I get after applying [creation/annihilation operator, field operator] commutation relations.
First, since the $\Psi$ and $\Phi$ operators commute with each other, the integrand in the matrix element breaks down into the product of elements in the two fields Fock spaces: $$ \langle \psi^3 \phi^\gamma | \Psi(x) \Phi(x) \Psi(x) | \psi^1 \rangle = \langle \psi^3 | \Psi(x)^2| \psi^1 \rangle \langle \phi^\gamma | \Phi(x)| 0 \rangle = e^{-ip_\gamma x}\langle \psi^3 | \Psi(x)^2| \psi^1 \rangle $$ Then, using $|\psi^i\rangle = \sqrt{2\omega_{p_i}} a_{p_i}^{\dagger} |0\rangle$ and $[\Psi,a_{p_i}^{\dagger}] = e^{ip_i x}/\sqrt{2\omega_{p_i}}$, I get $$ \langle \psi^3 | \Psi(x)^2| \psi^1 \rangle = 2e^{ip_1x}\langle \psi^3 |\Psi(x) | 0 \rangle + 2\sqrt{\omega_1\omega_3}\langle 0 | a_{p_3} a_{p_1}^{\dagger} \Psi(x)^2 | 0 \rangle $$ The first term becomes $2e^{i(p_1-p_3)x}$ and so to the right answer the second term above should become zero. However, since $[a_{p_3},a_{p_1}^\dagger] = (2\pi)^3 \delta^3(p_1-p_3)$, I expand it to $$ 2\sqrt{\omega_1\omega_3} (2\pi)^3 \delta^3(p_1-p_3)\langle 0 | \Psi(x)^2 | 0 \rangle + 2\sqrt{\omega_1\omega_3}\langle 0 | a_{p_1}^{\dagger} a_{p_3} \Psi(x)^2 | 0 \rangle $$ Since $a_{p_1} | 0 \rangle = 0$ the latter subterm is zero, leaving the term proportional to $\langle 0 | \Psi(x)^2 | 0 \rangle$. But inserting a complete set of states, and since $\Psi^\dagger = \Psi$, $$ \langle 0 | \Psi(x)^2 | 0 \rangle = \frac{1}{(2\pi)^3} \int \frac{d^3 q}{2 \omega_q} \langle 0 | \Psi(x) | q \rangle \langle q | \Psi(x) | 0 \rangle = \frac{1}{(2\pi)^3} \int \frac{d^3 q}{2 \omega_q} $$ which diverges. Inserting this back into the integral over $x$, while I should get $e(2\pi)^3 \delta^3(p_3-p_1-p_2)$, I instead get that plus the divergent term $$ e (2\pi)^3 \delta^3(p_3-p_1) \delta^3(p_2) \omega_1 \int \frac{d^3 q}{2 \omega_q} $$ Maybe interpreting this pictorially could give me some clues to where this is going wrong. Diagramatically this would be a term with a 0 momentum photon and constant momentum electron with no interaction vertex. If I'm winding up with a term with no vertex, perhaps I went wrong in the first place in factoring the matrix element into two pieces? Also, although the above calculation is for OFPT, this brings to mind old lectures about $S$-matrix scattering amplitude calculations and how disconnected / bubble diagrams factor out -- perhaps I am ignoring other terms that will cancel this one?
