Suppose we have a (time and space) translationally invariant system with the Fock space for a Hilbert space. Temporal and spatial translation invariance implies that energy and momentum are good quantum numbers. In the free theory, the momenta and energies of one-particle states simply add to get the total momentum and total energy of the system. Why is it that, for an interacting theory, we can (always?) add the momenta of the one-particle states to obtain the total momentum given that we can't do the same for the energy?
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$\begingroup$ Both energies and momenta of asymptotic states in S-matrix elements, and also one-particle states appearing in interaction vertices inside Feynman diagrams are always conserved, so that the total energy and momenta add as one would expect. Note this has nothing to do with whether these are onshell or not. So i’m not sure i understand your question. $\endgroup$– WakabaloolaMar 5, 2018 at 10:09
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$\begingroup$ I realized that I could have probably written the question in a simpler form, so I'm sorry if my intention is unclear. My question is that: given two 1-particle eigenstates of momentum, the combined two-particle state has total momentum equal to the sum of the individual momenta. However the total energy is not equal to the sum of the individual energies. Why is this the case? I can try to edit my question to make this clearer. $\endgroup$– pianyonMar 5, 2018 at 10:57
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$\begingroup$ so your question is about why the energy of bound states is different from the sum of energies of the constituents, right?, as i don’t know of any other situation where this would occur. here the energy difference is the binding energy (when one goes to the rest frame). in a boosted frame the momenta will also differ (by Lorentz covariance). at least this is my understanding, maybe somebody can add more detail. $\endgroup$– WakabaloolaMar 5, 2018 at 17:15
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$\begingroup$ (by the way, time translation invariance is not sufficient for energy-momentum conservation, you also need spatial translation invariance) $\endgroup$– WakabaloolaMar 5, 2018 at 17:19
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2$\begingroup$ The momentum doesn’t naively add at all; for instance the gluon field binding quarks together carries lots of momentum, so the momentum of a proton is not even close to the momenta of the free quarks. $\endgroup$– knzhouMar 5, 2018 at 18:40
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1 Answer
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In quantum mechanics, the linear momentum is represented by a one-body operator: the momentum of the $i$-particle only depends on the variables of this particle, $\hat{\bf p}_i = -i\hbar \nabla_i.$ So, the total momentum of the system is simply defined by the sum of the momentum of each particle. Of course, if the particles are immersed in some field (like an electromagnetic one), it would be necessary to sum the field momentum to the momenta of the particles.
On the other hand, the interaction energy terms correspond to two-body operators, for example, the repulsion between two electrons corresponds to $\hat{V}({\bf \hat{r}}_1, {\bf \hat{r}}_2) = \frac{e^2}{|{\bf \hat{r}}_1-{\bf \hat{r}}_2|},$ and it involves, naturally, the variables of both particles. For these reason, the total energy is not additive, as the interaction terms "belong" to the two-particles, and not to each particle separately. It doesn't matter if the particles form bound states. Other energies are additive (one-body operators), for example, the kinetic energy, the potential energy in an external field.
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$\begingroup$ Thanks! Does this cause any problem in a relativistic theory, where the energies and momenta can mix together? Does momenta remain strictly a one-body operator in that case? My intuition says that it does, but I can't see how that would actually work. $\endgroup$– pianyonMar 19, 2018 at 16:28