Intuitively, quantizing a field implies that there exists a probability for each state of the field denoted as $$\Omega[\psi]$$ So the vector $|\Omega\rangle$ must be a superposition of all possible field states weight with an amount relating to probability $$|\Omega\rangle=\Omega[\psi_0]|\psi_0\rangle + \Omega[\psi_0+\delta\psi]|\psi_0+\delta\psi\rangle+\Omega[\psi_0-\delta\psi]|\psi_0-\delta\psi\rangle+...$$ $$|\Omega\rangle=\Omega[\psi_0]|\psi_0\rangle + \sum_{n=1}^{\infty} \left[ \Omega[\psi_0+n\delta\psi]|\psi_0+n\delta\psi\rangle + \Omega[\psi_0-n\delta\psi]|\psi_0-n\delta\psi\rangle\right]$$ Or written with integral notation $$|\Omega\rangle=\int d\phi \, \Omega[\psi_0 + \phi] |\psi_0 + \phi\rangle$$ Is this the correct formula for the field state and is there any other way of writing the field state in terms of any other basis like energy, particles, and other properties.

  • $\begingroup$ The answer to this sort of question will appear in the first couple chapters (if not pages, historical notes aside) of any quantum field theory textbook. You're more likely to get a detailed (and correct) picture of what's going on from any of those sources than here. If you would like references, the standard is Peskin and Schroeder, but there are many many others with varying levels of detail on any given topic. $\endgroup$ Commented Nov 12, 2020 at 23:44
  • 1
    $\begingroup$ @RichardMyers I've got "Quantum Field Theory: A Modern Introduction" by Michio Kaku, and It doesn't give enough satisfying answers to this question. I'm asking on here because I'm self-taught and books are quite expensive. $\endgroup$ Commented Nov 12, 2020 at 23:55
  • 2
    $\begingroup$ This is called the Schrodinger functional representation, but it's rarely used for computation. A standard introductory QFT book is all about the particle basis for the free 'in' and 'out' fields. $\endgroup$
    – knzhou
    Commented Nov 13, 2020 at 1:18
  • $\begingroup$ The sum in your third equation doesn't really make sense since there's no way to reach all field configurations by moving in fixed steps $\delta \psi$ otherwise this seems accurate if not computationally useful. $\endgroup$
    – jacob1729
    Commented Nov 13, 2020 at 1:22
  • $\begingroup$ @jacob1729 What would be the correct way to write it? Also, how would you write it in a different basis such as the number of particles? $\endgroup$ Commented Nov 13, 2020 at 20:01

1 Answer 1


Particles should be identified with states in the Hilbert space that transform in unitary irreducible representations of the Poincaré group. Single- and multi-particle states are eigenstates of the momentum operator $\hat P^\mu$, with $\hat P^\mu \vert X \rangle = p^\mu \vert X \rangle$ for a set of real numbers $p^\mu$ with $p^0 \gt 0$ and $p^2 \ge 0$, which transform in the 4-vector representation of the Lorentz group. Single-particle states $\vert X \rangle$ transform under unitary irreducible representations of the Lorentz group as well, as $\vert X \rangle \to exp (i \theta_{\mu \nu} S^{\mu \nu}) \vert X \rangle$ where $\theta_{\mu \nu}$ are the boost and rotation angles and $S^{\mu \nu}$ are the generators of the Lorentz group in the representation of that particle. The vacuum $\Omega$ is assumed to be Lorentz invariant and to have zero momentum: $\hat P^\mu \vert \Omega \rangle = 0$.

The completeness relation is
$1 = \sum_X \int d \Pi_X \vert X \rangle \langle X \vert$
where the sum is over all of the single- and multi-particle states $\vert X \rangle$ and
$d \Pi_X = \prod_{j \in X} \frac{d^3 p_j}{(2 \pi)^3} \frac{1}{2 E_j}$

Quantum fields are operators and are defined as integrals over creation and annihilation operators, respectively $a_p^\dagger$ and $a_p$, for each momentum
$\phi (x) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} [a_p (t) e^{-i p x} + a_p^\dagger (t) e^{i p x}]$
$x = (\vec x, t)$
$signature = (+, -, -, -)$
$a_p^\dagger \vert \Omega \rangle = \frac{1}{\sqrt{2 \omega_p}} \vert p \rangle$
and $\vert p \rangle$ is a state with a single particle of momentum $p$.

You have $\phi (x) \vert \Omega \rangle = \vert x \rangle$, that is $\phi (x)$ creates a particle at position $x$.

Note: In QFT the symbol $\Omega$ is reserved for the vacuum state in an interacting theory.


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.