In the traditional presentation, Quantum Fields are usually presented as operator valued fields defined on spacetime, in the sense that $\varphi : M\to \mathcal{L}(\mathcal{H})$ for some Hilbert space $\mathcal{H}$.

On the other hand, recently I've read that Quantum Fields should be properly defined as operator valued distributions, and I believe that if I understood correctly this has to do with a way to deal with the problem of infinities that is usually associated to QFT. The only difference is that now we have a space of functions $\mathcal{D}(M)$, so that $\varphi : \mathcal{D}(M)\to \mathcal{L}(\mathcal{H})$.

Anyway, independently if we define fields as functions of events $\varphi(x)$ or functionals on functions $\varphi(f)$, the point is that $\varphi$ returns one operator.

Now, in Quantum Mechanics, there are two main kinds of operators which have a very clear meaning. The observables are hermitian operators, and those represent physical quantities by one postulate of the theory.

So whenever we have one observable $A$ is because we know of a physical quantity $A$ and the operator represents the quantity so that its spectrum is the set of allowed values and its eigenkets are the states with definite values of the said quantity. Everything is perfectly clear.

The other kind of operator I speak of are the unitary operators that usually represent transformations acting on the system, like translations, rotations and so forth. The meaning is also clear.

Now, in QFT, provided we have a field $\varphi: \mathcal{D}(M)\to \mathcal{L}(\mathcal{H})$, $\varphi(f)$ is one operator. What is the meaning of said operator?

I've heard it is one observable. But if it is one observable, what physical quantity it represents?

It is not at all clear for me what the operators $\varphi(f)$ represent, nor what is their action. And finally, how all this connects to the traditional Quantum Mechanics?

  • $\begingroup$ Are you familiar with the creation and annihilation (ladder) operators for the quantum harmonic oscillator? $\endgroup$ Oct 11, 2016 at 23:12
  • $\begingroup$ @AlfredCentauri yes I'm familiar with those. Actually, I was reading Merzbacher's book on QM and in the chapter about second quantization there appears something alike. I got the idea of the operator that "creates a particle at $\mathbf{r}$ with a set of quantum numbers", but not all fields are this field right? By the way, I've heard that the second quantization picture only works for free fields. $\endgroup$
    – Gold
    Oct 11, 2016 at 23:26
  • $\begingroup$ Re your final sentence; are you thinking of Haag's theorem? $\endgroup$ Oct 11, 2016 at 23:35
  • $\begingroup$ @AlfredCentauri, actually I didn't know about this theorem. The first time I've actually seems fields being used was in the treatment of identical particles. Studying that I got the impression that fields always acted on the Fock space, with the basis $|n_1,\dots,n_k,\dots\rangle$ of occupation numbers. When I asked here (physics.stackexchange.com/questions/284512/…), in which space a field acts, though, in the question it was said that the Fock space is only the space used for free fields. That's where that sentence came from. $\endgroup$
    – Gold
    Oct 12, 2016 at 16:49

2 Answers 2


What $\phi(x)$ represents depends on what "kind" of field it is: If it is the electric field of electromagnetism, then $\phi(x)$ corresponds to the observable "electric field at $x$". If it is a Dirac fermion field like the electron field or a complex scalar, then $\phi(x)$ is not Hermitian and thus no observable - and I don't know of any other "meaning" for it. If it's a real scalar or a real vector field then it's an observable, but it's not necessarily clear to which quantity we can actually measure it corresponds.

However, there is one very important and very generic meaning to the field operators: Through the LSZ reduction formula, the expectation values of products of field operators are the essential ingredient in the scattering amplitudes.

Lastly, it is difficult to answer the question of what the "action" of a field operator is because the Hilbert spaces of most interacting field theories are unknown in the sense that we don't know how to actually construct them. Most states you will see in a standard quantum field theory course live in the asymptotic free Fock spaces, where the field is the Fourier transform of the creation/annhiliation operators and its action is reasonably clear.

  • 1
    $\begingroup$ This is an excellent answer. One could add however that to the Dirac field there correspond 16 bilinear Hermitian observables $\bar\psi\Gamma\psi$ and at least some of them have simple interpretations: density of particles, total current, spin density... $\endgroup$ Oct 12, 2016 at 6:18

Complementary to ACuriousMind's answer (in particular, the crucial observation that what particle physics experiments measure are scattering amplitudes, hence the experimental relevance of the quantum fields stems primarily from the LSZ formula), and in the spirit of Robin Ekman's comment, you can also think of quantum fields as building bloc operators, from which to construct actual observables on a Fock space.

For a free, massive, complex scalar field $\Psi(x) = \Psi_{\text{part}}(x) + \Psi^{\dagger}_{\text{anti-part}}(x)$ on the Fock space $\mathcal{F} = \mathcal{F}_{\text{part}} ⊗ \mathcal{F}_{\text{anti-part}}$, this works as follow:

One-particle subspace

Let $\mathcal{H}_{\text{1 part}}$ be the subspace of the Fock space $\mathcal{F}_{\text{part}}$ spanned by states containing exactly 1 particle. A pseudo-basis of it can be labeled by impulsions $p$ on the mass-shell, with the normalization (using Weinberg's conventions): $$ \left\langle p \middle| p' \right\rangle = \delta^{(3)}(\vec{p} - \vec{p}^{\prime}) $$

Let $A$ be some Hermitian operator on $\mathcal{H}_{\text{1 part}}$ (ie. a 1-particle observable) and, for simplicity, assume that $A$ has discrete spectrum, so we have an orthonormal basis of $\mathcal{H}_{\text{1 part}}$ made of eigenvectors: $$ \left| \psi_k \right\rangle = \int d^{(3)}\vec{p}\; \tilde{\psi}_k(\vec{p}) \left| p \right\rangle $$ with corresponding eigenvalues $\lambda_k$.

The integral kernel of $A$ is then: $$ \tilde{A}(\vec{p},\vec{p}^{\prime}) := \sqrt{2 E(\vec{p}) 2 E(\vec{p}^{\prime})} \sum_k \tilde{\psi}_k(\vec{p}) \tilde{\psi}^*_k(\vec{p}^{\prime}) $$ or, in position representation (on the $t=0$ time slice): $$ A(\vec{x},\vec{x}^{\prime}) := \int \frac{d^{(3)}\vec{p} \, d^{(3)}\vec{p}^{\prime}}{(2\pi)^3} e^{i(\vec{p}\cdot\vec{x} - \vec{p}^{\prime}\cdot\vec{x}^{\prime})} \tilde{A}(\vec{p},\vec{p}^{\prime}) $$

Creation/annihilation operators

A basis of the Fock space $\mathcal{F}_{\text{part}}$ built on $\mathcal{H}_{\text{1 part}}$ can be given in terms of occupation numbers over the basis $\big( \left| \psi_k \right\rangle \big)_k$: $$ \left| \left( N_k \right)_k \right\rangle := \text{normalized symmetrization of } \left| \psi_1 \right\rangle^{(1)} \otimes \dots \otimes \left| \psi_1 \right\rangle^{(N_1)} \otimes \dots \otimes \left| \psi_K \right\rangle^{(N_K)} $$ Switching from the standard impulsion pseudo-basis into this $A$-adapted basis, one can check that the operators: $$ a_{\text{part},k} := \int d^{(3)}\vec{p}\; \tilde{\psi}^*_k(\vec{p}) a_{\text{part}}(p) \;\&\; a^{\dagger}_{\text{part},k} := \int d^{(3)}\vec{p}\; \tilde{\psi}_k(\vec{p}) a^{\dagger}_{\text{part}}(p) $$ annihilate, resp. create, a particle in the state $\left| \psi_k \right\rangle$.

Second quantization of $A$

The particle part of the quantum complex scalar field is given as: $$ \Psi_{\text{part}}(x) := \int \frac{d^{(3)} \vec{p}}{(2\pi)^{3/2} \sqrt{2E(\vec{p})}} e^{i p \cdot x} a_{\text{part}}(p) $$ so, putting everything together, we get: $$\boxed{ \hat{A} := \int_{t=0} d^{(3)}\vec{x}\, d^{(3)}\vec{x}^{\prime}\; \Psi^{\dagger}_{\text{part}}(\vec{x},0)\, A(\vec{x},\vec{x}^{\prime})\, \Psi_{\text{part}}(\vec{x}^{\prime},0) = \sum_k λ_k \hat{N}_{\text{part},k} }$$ where $\hat{N}_{\text{part},k} := a^{\dagger}_{\text{part},k} a_{\text{part},k}$ measures the number of particles in the state $\left| \psi_k \right\rangle$.

For example, for a region $U$ on the $t=0$ time slice, $$ \int_{t=0, \vec{x} \in U} d^{(3)}\vec{x}\; \Psi^{\dagger}_{\text{part}}(\vec{x},0) \Psi_{\text{part}}(\vec{x},0) $$ measures the number of particles in $U$ (the corresponding 1-particle operator $A$ has spectrum $\{0,1\}$ with the $(λ=1)$-eigenspace being spanned by wave-functions supported on $U$ only).

Assorted remarks

  • Of course, the $t=0$ slice is not distinguished: since Poincaré-invariance is built-in, you can switch to any space-like slice of Minkowski space (and, assuming I got all normalization factors right, this should go smoothly; no guarantee that I did, though...).

  • The "half-field" $\Psi_{\text{part}}(x)$ can be itself be reconstructed from $\Psi(x), \partial_t \Psi(x)$ (this is more easily seen in Fourier-transform). In a similar way, the underlying $\Psi_{\text{part}}(x)$ can be reconstructed from a real scalar field $\Phi(x) = \Psi_{\text{part}}(x) + \Psi^{\dagger}_{\text{part}}(x)$.

  • I am fairly confident that this reasoning can be generalized to other kinds of free fields, by carefully using the appropriate representation intertwinner to convert between spin numbers and field components (even if the field is fermionic, the observables constructed in this way, being quadratic in the field, are bosonic, so should be physically ok).

  • As mentioned by ACuriousMind, for interacting fields, this Fock picture is only valid asymptotically.


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