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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

1 Answer 1


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 $\boldsymbol{x}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.