# Why is particle number conserved in non-relativistic limit of QED?

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?

The small-coupling expansion treats QED as a series of corrections to a different model, namely "free" QED without the interaction term. In free QED, the operators $$c,c^\dagger,d,d^\dagger$$ are indeed the creation and annihilation operators for individual particles, but that is no longer true in the original model with the interaction term. As a result, the operator $$N_{e^-}$$ shown in the OP is not really the number operator for electrons. It is in free QED, but not in the presence of the interaction term. To see this, notice that the vacuum state in QED (with the interaction term) is not annihilated by $$c_{ks}$$, so with $$N_{e^-}$$ defined as in the OP, the operator $$N_{e^-}$$ has a non-zero expectation value in the vacuum state, even though the vacuum state is empty by definition.
In this approach, we appeal to the model's symmetries and other general intuition to come up with a list of fields that the non-relativistic approximation should involve, like one two-component spinor field for the electron, one for the positron, and an EM gauge field. Then we write down the most general Lagrangian that we can write down using those fields, using non-relativistic kinetic terms and discarding terms that we know will be negligible at sufficiently low energies. (This can be inferred by considering the units of each term's coefficient.) Then, if $$n$$ denotes the number of independent coefficients in the remaining Lagrangian, we calculate $$n$$ different physical quantities and compare them to the same $$n$$ physical quantities calculated using the original model (relativistic QED in this case) in order to fix the values of the $$n$$ coefficients. The result is a Lagrangian that qualifies as a non-relativistic approximation to QED. This technique (and the result) is called NRQED.