I am trying to recover ordinary quantum mechanics from QED. One main feature of quantum mechanics is the conservation of particle and antiparticle number separately, i.e. $[N_{e^-},H] = [N_{e^+},H] = 0$. I would like to show this.
The QED Hamiltonian is given by $H = \int d^3 x H_0 + H_{int}$ where $H_{int} = e \bar\psi A^\mu\gamma_\mu \psi$. The free Hamiltonian commutes with the number operators.
The fields are given by $$\psi = \sum_s\int\frac{d^3k}{\sqrt{2\pi}^3\sqrt{2E}} \left(e^{ikx} u_s(k) c_{ks} + e^{-ikx} v_s(k) d^\dagger_{ks}\right) = \psi^+ + \psi^-$$ $$A^\mu = \sum_s\int\frac{d^3k}{\sqrt{2\pi}^3\sqrt{2E}} \left(e^{ikx} \varepsilon_s^\mu a_{ks} + e^{-ikx} \varepsilon_s^{*\mu} a^\dagger_{ks}\right)$$
I already calculated the commutator of the number operator $N_{e^-} = \int d^3k \sum_s c^\dagger_{ks} c_{ks}$ with the fields: $$[N_{e^-}, \psi] = -\psi^+$$ $$[N_{e^-}, \bar{\psi}] = \bar{\psi}^+$$
So the commutator with the interaction Hamiltonian is $$[N_{e^-},\bar{\psi}A^\mu\gamma_\mu\psi] = -\bar{\psi}A^\mu\gamma_\mu\psi^+ + \bar{\psi}^+A^\mu\gamma_\mu\psi = \bar{\psi}^+A^\mu\gamma_\mu\psi^- - \bar{\psi}^-A^\mu\gamma_\mu\psi^+$$
In order for the particle number to be conserved (in the non relativistic case) we need that the commutator vanishes for $\vec{p} \to 0$.
If I insert the field operators and integrate over $d^3x$ I find (spin indices dropped)
$$\bar{\psi}^+A^\mu\gamma_\mu\psi^- = \int \frac{d^3p d^3k}{\sqrt{2\pi}^3\sqrt{2E_p}\sqrt{2E_k}\sqrt{2|p+k|}} \bar{u}(p)\left(\varepsilon \gamma a_{p+k} + \varepsilon^* \gamma a^\dagger_{-p-k}\right)v(k) c^\dagger_p d^\dagger_k$$
For $p,k \to 0$ the denominator become $E_k,E_p \to m$ and $|p-k| \to 0$ while in the nominator $\bar{u} \gamma \varepsilon v = O(1) \cdot m$. So since the photon energy drops to $0$ the commutator becomes infinity.
Therefore there should be no particle number conservation in the non relativistic limit. How can one justify the particle number conservation?
