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?

  • $\begingroup$ I'm guessing one possibility is that the "complete" set of states I added is really just the complete set of one-particle states, but I still believe that should suffice in this case. In either event I think my calculation that $\langle 0 | \mid \Psi^2 \mid 0 \rangle$ must be wrong since this is a typical Lagrangian term. $\endgroup$
    – jwimberley
    Oct 14, 2023 at 23:23
  • $\begingroup$ I'm going to read over physics.stackexchange.com/questions/434163/…, which has broader but includes a pretty similar question about these steps $\endgroup$
    – jwimberley
    Oct 15, 2023 at 19:33
  • $\begingroup$ Is this perhaps an artifact of the matrix element being one for an interacting theory but calculated for the free-theory vacuum, and not the interacting vacuum? In which case the text is sweeping in under the rug for the time being, this section just being a short one about OFPT and the full S-matrix / LS Z /time ordering and interaction vacuum bits coming later? $\endgroup$
    – jwimberley
    Oct 23, 2023 at 15:51

1 Answer 1


The term you get corresponds to a ‘nothing happens’ process where there’s only one particle propagating, this by definition is not a scattering event thus should be discarded. See this post S-matrix element In the context of LSZ reduction.

The interacting vacuum is not an issue since you’re doing tree level perturbative calculations.


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