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I was watching some lectures of the theoretical minimum program, by prof. Leonard Susskind. There he introduces the notion of fields (in the context of QFT) in a very nice and intuitive way.

Suppose one has a basis of states $\{\psi_{1}(x),\psi_{2}(x),...\}$ which I assume here is the solution of the one-particle Schrödinger equation. You can think of these vectors as states $\psi_{i}(x)$ living on $L^{2}(\mathbb{R}^{d})$ each with well-defined energy $E_{i}$. A basis element of the Fock space is: $$|n_{1},n_{2},...\rangle$$ where each $n_{j}$ represents the number of particles of the system in the state $\psi_{j}(x)$. When, say, a creation operator $a_{k}^{\dagger}$ hits such a state, leads to: $$a_{k}^{\dagger}|n_{1},n_{2},...\rangle = |n_{1},n_{2},...,n_{k-1},n_{k}+1,n_{k+1},...\rangle$$ and, according to prof. Susskind, this is to be interpreted as a creation of a particle in the state $\psi_{k}(x)$, that is, with well-defined energy $E_{k}$. In this context, fields operators are given by: $$\Psi(x) = \sum_{i}\psi_{i}(x)a_{i} \quad \mbox{and} \quad \Psi^{\dagger}(x) = \sum_{i}\bar{\psi}(x)a_{i}^{\dagger}$$

This was very intutive, but my experience tells me that this is not standard procedure in QFT. In the above setting one thinks of many (possibly infinitely many) particles whose states are given in terms of Schrödinger equation. In QFT one usually starts with a classical field theory, so one has an action $S = S[\phi]$ for fields and the Hamiltonian given in terms of fields too. The fields satisfy some field equation and this equation is obtained by using Euler-Lagrange equations to minimize $S[\phi]$.

I would like to know whether it is possible to reconcile these two approaches somehow in a way that the second one (using classical fields as starting point) gives an intuition in terms of particles being created by the action of field operators. To my eyes there should be a connection between these two approaches, but I am struggling to understand how to do it because the interpretation of the Hamiltonian in the two pictures seem to be different: the first one is a Hamiltonian in the sense of quantum mechanics and the second one is a Hamiltonian in terms of fields.

PS: I am obviously thinking about non-interacting theories.

Qmechanic
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IamWill
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    This amounts to a change of basis in Hilbert space. You have the "field eigenstate basis" and you have the "particle number or Fock space basis". Have you worked through a QFT book at the level of Chapter 2 of Tong's lecture notes? The relationship between these approaches is covered in great detail there. – Andrew Aug 30 '22 at 01:27
  • Yeah, this is an extremely common question (I think every thoughtful student will have it when first encountering QFT). What you're looking for is the "Schrodinger wavefunctional" approach and you'll find many questions on it here if you search. – knzhou Aug 30 '22 at 03:38
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    The first approach is very artificial in my opinion. Fields yield everything naturally, from conserved quantities to interaction Hamiltonians. The first approach achieves the same things but very artificially by assembling operators into space-dependent fields – Ryder Rude Aug 30 '22 at 08:25

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