# Bound states in QFT

I was studying this article on consequences of invariance under charge conjugation symmetry(in particular, in positronium decay) where in section IV it says

The state function of any state of positronium can be expanded in terms of free-particle state functions,$$\Psi_\text{positronium}=(\sum_{\vec p,s_1,s_2}c(\vec p,s_1,s_2)a_{s_{1}}^*(\vec p)b_{s_{2}}^*(-\vec p) +\sum_r...a_{r_{1}}^*a_{r_{2}}^*b_{r_{3}}^*b_{r_{4}}^*+...)\Psi_\text{vac}$$ where the second term represents the effect of virtual pair production.

I want to know about the nature of the terms that have been left out and make sure that the claim is actually true. It is confusing to me because, in the introductory lectures, we have been told that the (free) bound states of combinations of some or all possible particles should be taken as a separate sector of the Fock space in the sense, the creation and annihilation operators for these states(positronium atom creation/annihilation operator in this case) are independent of the creation and annihilation operators for the constituent particles. And I guess this is the spirit behind Weinberg's remark(in his book of Quantum Theory of Fields) in section 3.1:

Also, any relevant bound states in the spectrum of $$H$$ should be introduced into $$H_0$$ as if they were elementary particles**

**Alternatively, in non-relativistic problems we can include the binding potential in $$H_0$$. In the application of this method to rearrangement collisions, where some bound states appear in the initial state but not the final state, or vice-versa, one must use a different split of $$H$$ into $$H_0$$ and $$V$$ in the initial and final states.

Another related query is, are positronium decay process and pair annihilation process the same?

Any comment/clarification is welcome!

Walter Greiner in his text on field quantisation tries to justify this exact same point in example 10.2:

It is very difficult to find an exact description of the bound states of the $$e^+e^-$$ system since this would amount to solving the relativistic two-body problem...

According to the general principle of Lorentz invariance the state vector of positronium $$|Ps\rangle$$ can be classified by the eigenvalues of the operators $$P^\mu$$, $$\mathbf J^2$$ and $$J^z$$. In addition we have the operators of space inversion P and charge conjugation C, which commute with the set of kinematic operators. The corresponding eigenvalues are $$P|Ps\rangle = \pi_P |Ps\rangle, C|Ps\rangle = \pi_C |Ps\rangle\tag{1}$$ where $$\pi_P = ±1$$ denotes space parity and $$\pi_C = ±1$$ denotes charge parity. Now we make the following ansatz for the state vector of positronium in the center-of-mass system $$(P = 0)$$: $$|Ps\rangle=\int d^3\vec p\sum_{s,s'}R(\vec p,s,s')b^\dagger_{\vec p,s}d^\dagger_{-\vec p,s'}|0\rangle\tag{2}$$ where $$R(\vec p,s,s')$$ is the wave function in momentum space. Here $$s$$ and $$s'$$ are the projection of the electron and positron spins onto the z axis. The state (2) contains one electron-positron pair with a combined momentum of zero and is an approximation to the true bound state since the particle number in QED is not a conserved quantiy. In general higher multi-pair configurations with total charge zero, such as $$b^\dagger b^\dagger d^\dagger d^\dagger|0\rangle$$ etc., can contribute to the state vector. In positronium, however, such admixtures are very small, owing to the essentially nonrelativistic nature of this system. In any case, for the purpose of classifying the bound states it is sufficient to use the ansatz (2); any complicated higher-order admixtures would have the same symmetry properties.

For now, I'll take this as an explanation which invalidates my previous understanding

the creation and annihilation operators for these states(positronium atom creation/annihilation operator in this case) are independent of the creation and annihilation operators for the constituent particles.

Since now the creation operator of positronium atom, for the positron and for the electron don't commute pairwise.