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Nonrelativistic QM can be applied to bound states like a hydrogen atom. QFT is used for free particles (whatever one means by particles) that shortly interact with each other and are free again after the interaction. Can QFT also be used for bound states?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/26960/2451 and links therein. $\endgroup$ – Qmechanic May 11 '16 at 8:21
  • $\begingroup$ Of course. What makes you think that it can't? $\endgroup$ – CuriousOne May 11 '16 at 8:22
  • $\begingroup$ Because of the V in the Schrödinger equation. How can arrive at that V from QFT? And because QFT only applies to non-bound states, and not bound states like the hydrogen atom. Has the derivation really been made? Are there sources that show how? $\endgroup$ – descheleschilder May 11 '16 at 8:42
  • $\begingroup$ How can you derive the V (assuming it´s an electric potential) in the Schrödinger equation, when the wavefunctions for photons already contain V? Isn´t that circular? $\endgroup$ – descheleschilder May 11 '16 at 11:07
  • $\begingroup$ There is no fundamental difference between quantum mechanics and quantum field theory. Quantum field theory is just the quantum mechanics of fields. When you consider the specific problem of fermions (or their composites) at low energies, then the dynamics of single-particle states of QFT are well approximated by quantised point particles a la undergraduate QM. The $V(r)$ in Schroedinger's equation is an approximation to certain kinds of field states (usually bosonic fields in coherent states). In quantum optics it is fairly well established when this is a good approximation or not. $\endgroup$ – Mark Mitchison May 11 '16 at 11:10