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I know that QFT contains quantum mechanics, but the fields do not have a similar probabilistic interpretation as the wavefunction in Quantum mechanics. From my not-so-deep knowledge of QFT, I expext a general state to collapse to an eigenstate of some operator once a measurement of the appropriate quantity is made, but I cannot clearly see this analogy to wavefunction collapse. The wavefunction collapse is an axiom of quantum mechanics as far as I know, but when building up the formalism for QFT, no such thing was mentioned in my course.

So, the question is: what is a way to showcase this analogy to wavefunction collapse in QFT?

I don't need a way to show that wavefunction collapse is contained in QFT, but I want an explanation of how to see the fact that when we make a measurement of some quantity in QFT (say the angular momentum of the bound state of some elementary particles that cannot be described by quantum mechanics), then the quantum state that lives in the corresponding Fock space collapses to an appropriate state.

TheQuantumMan
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    I don't quite understand why you're asking about a wavefunction in this context. You seem to grok that the state just after a measurement is an eigenstate of the measured observable. The wavefunction in QM is just the state (vector) on the position basis. But consider the quantum harmonic oscillator in a superposition of two or more number states. Do we need a wavefunction description of the QHO to understand that the state just after a measurement of E(N) will be an eigenstate of the number operator? – Alfred Centauri Dec 17 '20 at 23:01

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A similar question was asked a couple of days ago: How does wave function collapsing fit into QFT?

I know that QFT contains quantum mechanics, but the fields do not have a similar probabilistic interpretation as the wavefunction in Quantum mechanics.

This is not true. The probability interpretation is the same. What's different is the degrees of freedom that the wavefunction is a function of. For example, in QFT we don't have a photon wavefunction that can be expressed as $\Psi(x,t)$, nor do we have an observable for position.

For example, suppose that we have a state in QFT that is a superposition of the form $(|0\rangle+|e\rangle)/\sqrt{2}$, where 0 is the vacuum and e is some excited state with a different energy. If you then measure the energy of the state, you have a probability 0.5 of measuring 0, and 0.5 for e.

The wavefunction collapse is an axiom of quantum mechanics as far as I know,[...]

Not really. If it's an axiom, it's an optional axiom. You can instead just say that the system becomes entangled with the observer, and the observer is left in a Schrodinger's cat state.

But if you want a description that involves collapse, then in the example above, measuring the energy collapses the state to whatever you measure.

user282979
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    "The probability interpretation is the same". Sure, but fields in QFT are operators. I still don't see where a wavefunction description is present in QFT. Also, on the last paragraph: if we do measure the energy and find a non-zero value, then does the system collapse to an eigenstate of the Hamiltonian that corresponds to that energy (assuming the vacuum has zero energy)? – TheQuantumMan Dec 17 '20 at 19:40
  • Could you perhaps provide an answer on https://physics.stackexchange.com/questions/600703/how-does-wave-function-collapsing-fit-into-qft as well? Would be highly appreciated! – user3397129 Dec 17 '20 at 21:00