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In single-particle quantum mechanics, the probability of finding the particle in all space is conserved due to the hermiticity of the Hamiltonians (and remains equal to unity for all times, if normalized).

But in relativistic quantum field theory, particle numbers are not conserved. For example, in QED, an initial state consisting of an electron-positron pair can annihilate into two photons in the final state. There is no trace of the initial electron and the positron in the final state. Similarly, in $\beta$-decay, an initial neutron is converted into a proton, an electron and an anti-electron neutrino in the final state. There is no neutron after the decay takes place though the Fermi theory Hamiltonian is Hermitian.

So it seems that hermitian Hamiltonian in QFT is not responsible for the conservation of the probability. I seem to have a conceptual glitch here which I would like to be clarified. What is the role of hermitian Hamiltonians in relativistic QFT in the time development of states? There must be some constraining feature of hermitian QFT Hamiltonian. Sorry if the question sounds dumb.

Some more thoughts on this for clarification In single-particle quantum mechanics, the norm of a quantum state is the probability of finding the particle in all of space. Is there a similar probability interpretation of the norm of a state in QFT? Since QFT hamiltonians are hermitian, the norm of the state remains preserved under time development but the state itself can change. This confuses me. To take the example given above the norm of the initial state neutron does not change under time development but the state itself can into other states. How do we interpret this in QFT?

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2 Answers 2

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In Q field theory we are quantizing the fields not the particles. In some (but not all) field theories there are excitations that look like particles, but their conservation or lack of has nothing to to do with the conservation of probability. The latter states that given a starting state then the sum of the probabilities of all possible outcomes must be unity. There is no mention of particles in that statement.

In general the Fock-space, many-particle picture is rather misleading, and only approprite for free or weakly coupled systems.

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  • $\begingroup$ > "In general the Fock-space, many-particle picture is rather misleading, and only approprite for free or weakly coupled systems." What is the less misleading alternative? $\endgroup$ May 16 at 16:43
  • $\begingroup$ @Jan Lalinsky: I'd say that focussing on the correlation functions is the best thing. They exist even when there are no particle states (as in conformal field theory, say) and when there are particles you can extract their poperties from the correlators by looking for poles. $\endgroup$
    – mike stone
    May 16 at 17:06
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Things happen in Fock space in QFT, but I don't remember seeing any result looking like total probability equals to 1 over Fock space (does such a sum even converge?)

However, the closest result I know is the unitarity of the S-matrix, although its scope is a bit narrower. It's an operator than relates the initial and final states of a scattering process, meaning that particles are free long before and long after the process. The fact that the S-matrix has to be unitary is related to probability conservation.

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