Treating a bipartite Hamiltonian relativistically

For a relativistic bipartite system, you usually do not start with a simple Hamiltonian like$$H = H_A \otimes I_B + I_A \otimes H_B + H_{\text{int}}$$H=HA​⊗IB​+IA​⊗HB​+Hint​

unless you are in a special approximation. The relativistic treatment changes the framework.

The main issue is this:$$\boxed{ \text{Relativity treats space and time together, but ordinary Hamiltonians single out time.} }$$Relativity treats space and time together, but ordinary Hamiltonians single out time.​

So the more natural object is the action or the field theory Lagrangian, not a nonrelativistic two-particle Hamiltonian.

For two relativistic particles/fields, one writes something like$$S = \int d^4x \, \mathcal L_A + \int d^4x \, \mathcal L_B + \int d^4x \, \mathcal L_{\text{int}}$$S=∫d4xLA​+∫d4xLB​+∫d4xLint​

or, schematically,$$\mathcal L = \mathcal L_A + \mathcal L_B + \mathcal L_{\text{int}}$$L=LA​+LB​+Lint​

For two scalar fields,$$\mathcal L_A = \frac12 \partial_\mu \phi_A \partial^\mu \phi_A – \frac12 m_A^2\phi_A^2$$LA​=21​∂μ​ϕA​∂μϕA​−21​mA2​ϕA2​ $$\mathcal L_B = \frac12 \partial_\mu \phi_B \partial^\mu \phi_B – \frac12 m_B^2\phi_B^2$$LB​=21​∂μ​ϕB​∂μϕB​−21​mB2​ϕB2​

and an interaction might be$$\mathcal L_{\text{int}} = -g \phi_A^2 \phi_B^2$$Lint​=−gϕA2​ϕB2​

Then the relativistic dynamics comes from$$\delta S = 0$$δS=0

rather than from a simple two-body Schrödinger Hamiltonian.

For quantum theory, the state lives in a tensor product of field Hilbert spaces:$$\mathcal H = \mathcal H_A \otimes \mathcal H_B$$H=HA​⊗HB​

but the particles are excitations of fields, not little objects with fixed wavefunctions.

A bipartite relativistic state might look like$$|\Psi\rangle = \sum_{p,q} C(p,q) \,a_A^\dagger(p) a_B^\dagger(q) |0\rangle$$∣Ψ⟩=p,q∑​C(p,q)aA†​(p)aB†​(q)∣0⟩

where $a_A^\dagger(p)$aA†​(p) creates particle $A$A with four-momentum $p$p, and $a_B^\dagger(q)$aB†​(q) creates particle $B$B with four-momentum $q$q.

The relativistic energy is$$E_{\mathbf p} = \sqrt{\mathbf p^2 + m^2}$$Ep​=p2+m2​

so the free Hamiltonian is roughly$$H_0 = \int \frac{d^3p}{(2\pi)^3} E_{\mathbf p} \,a^\dagger_{\mathbf p}a_{\mathbf p}$$H0​=∫(2π)3d3p​Ep​ap†​ap​

For two species,$$H_0 = \int \frac{d^3p}{(2\pi)^3} E^A_{\mathbf p} a_A^\dagger(\mathbf p)a_A(\mathbf p) + \int \frac{d^3q}{(2\pi)^3} E^B_{\mathbf q} a_B^\dagger(\mathbf q)a_B(\mathbf q)$$H0​=∫(2π)3d3p​EpA​aA†​(p)aA​(p)+∫(2π)3d3q​EqB​aB†​(q)aB​(q)

Then add an interaction Hamiltonian:$$H = H_A + H_B + H_{\text{int}}$$H=HA​+HB​+Hint​

But now $H_{\text{int}}$Hint​ comes from a Lorentz-invariant field interaction.

So the relativistic version of the bipartite Hamiltonian idea is:$$\boxed{ H_A \otimes I + I \otimes H_B + H_{\text{int}} }$$HA​⊗I+I⊗HB​+Hint​​

but interpreted in QFT, where $H_A$HA​, $H_B$HB​, and $H_{\text{int}}$Hint​ act on field Fock spaces.

The clean hierarchy is:$$\boxed{ \text{Nonrelativistic QM: two-particle Hamiltonian} }$$Nonrelativistic QM: two-particle Hamiltonian​ $$\boxed{ \text{Relativistic QM: constrained and delicate} }$$Relativistic QM: constrained and delicate​ $$\boxed{ \text{Relativistic QFT: fields, Fock space, Lorentz-invariant action} }$$Relativistic QFT: fields, Fock space, Lorentz-invariant action​

So the safest answer is:$$\boxed{ \text{Treat the bipartite system as two interacting quantum fields.} }$$Treat the bipartite system as two interacting quantum fields.​

Not as two particles with an ordinary potential $V(\mathbf x_A-\mathbf x_B)$V(xA​−xB​), because instantaneous potentials generally conflict with relativity.