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In first quantization, a state of system is represented by wavefunction (w.f.) $\phi(x)$ (a representation of a state $|\phi\rangle$ in Hilbert space). The way I understand it is that $|\phi(x)|^2$ gives probability of finding a particle at position $x$. So, $|\phi\rangle$ is a column matrix (written in some basis). Understandable to me!

In second quantization, the many-body state of system is represented by field operators. According to Wikipedia, field operators are given in terms of creation and annihilation operators $$\Psi = \sum_\nu \psi_\nu \hat{a}_\nu \quad ; \quad \Psi^\dagger = \sum_\nu \psi_\nu^* \hat{a}_\nu^\dagger$$ where $\psi$ is ordinary first quantization w.f. and $\hat{a} (\hat{a}^\dagger$) is annihilation (creation) operator.

I don't understand that how does the field operators represent a state? How can I intuitively think about it? How to relate a field operator representation with physical system? What is physical meaning for a field operator?

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    $\begingroup$ Where did you see the claim that the state of a many body system is represented by field operators? $\endgroup$
    – J. Murray
    Jan 16, 2020 at 22:36
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    $\begingroup$ An operator isn't the same thing as a state. Just think in terms of undergraduate quantum mechanics: does $\hat{x}$ represent the state $|\psi \rangle$? $\endgroup$
    – knzhou
    Jan 16, 2020 at 22:40
  • $\begingroup$ @J.Murray so there is no concept of "state of system" in field theory? $\endgroup$ Jan 16, 2020 at 23:06
  • $\begingroup$ @knzhou actually that is what confuse me. An "operator" is not same as a "state". From undergrad quantum mechanics: a system is represented by a "state"; and an operator is applied on that state to get information from the system. So, where is that concept of "state" in second quantization? $\endgroup$ Jan 16, 2020 at 23:09
  • $\begingroup$ I mean, the state is still there, in basically the exact same way. Why don't you think there is a state? $\endgroup$
    – knzhou
    Jan 16, 2020 at 23:19

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Summarising what is being said in the comments and adding some info: the many-body state of a system can be represented by field operators acting on some state, but it would be incorrect to say that field operators represent a many-body state. If the annihilation and creation operators diagonalize the hamiltonian, then it is convenient to use the basis of Fock states, written in terms of $\hat{a}_\nu$, $\hat{a}^\dagger_\nu$ acting on the ground state $|0\rangle$. The strength of the second quantization formalism is here. It allows to treat many-body states in a much simpler way.

Finally, the field operators do have a physical meaning as it is not hard to show that the field operators destroy/create a particle at point $\boldsymbol{x}$.

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