In classical mechanics, the phase space of a single particle is given by the position $x=x(t)$ and its momentum $p=\dot{x}(t)$. More generally, a particle is described by generalized coordinates $q$ and $p = \dot{q}$. In quantum mechanics, this two objects become operators $\hat{x}$ and $\hat{p}$ by means of canonical quantization. Now, observables in QM are represented by operators (as the position and momentum mentioned before) because these are objects that we measure, so operators are closely connected to the act of measuring some physical quantity of the system, even if theoretically one can only calculate mean values.

In QFT, classical fields become operators, or quantum fields. Take a scalar field $\phi(x,t)$ for example. We can interpret $x$ as an index label instead of a coordinate, say $\phi(x,t) = \phi_{x}(t)$. The idea of QFT is to turn each field $\phi_{x}(t)$ into an operator.

Question: Is the interpretation in the second case analogous to the one in the first case (QM)? In other words, $\phi_{x}(t)$ become operators because we are thinking of each $\phi_{x}(t)$ as an observable, i.e. something we can measure at every fixed $t$?


2 Answers 2


That is completely correct. For a real scalar field $\phi$, you can think of the "states" in quantum field theory as being "wave functionals." At a fixed time $t$, they take in a field configuration $\phi(\vec{x})$ and output a complex number. In other words, if $\phi(\vec x)$ is a classical field configuration, $$ \phi : \mathbb{R}^3 \to \mathbb{R}, $$ that is $$ \phi \in C^\infty(\mathbb{R}^3) $$ then the quantum state is a wave functional $\Psi[\phi]$ $$ \Psi : C^\infty(\mathbb{R}^3) \to \mathbb{C}. $$ Then, $|\Psi[\phi]|^2$ has the intepretation of being the probability density of the field "having the values $\phi$", much like how in regular quantum mechanics, $|\psi(q)|^2$ is the probability density that the particle is located at position $q$. The operators $\hat \phi(\vec x)$ and $\hat \pi(\vec x)$ now have the definitions $$ \big( \hat \phi(\vec x) \Psi \big) [\phi] = \phi(\vec x) \Psi[\phi] $$ and $$ \big( \hat \pi(\vec x) \Psi \big) [\phi] = \frac{\hbar}{i} \frac{\delta}{\delta \phi(\vec x)} \Psi[\phi] $$ where $\frac{\delta}{\delta \phi(\vec x)}$ is the functional derivative of the functional $\Psi$ with respect to the variable $\phi(\vec x)$. This is just like the definition of $\hat q$ and $\hat p$ in quantum mechanics.

If we come up with some Hamiltonian $\hat H$, then this wave functional of field configurations on a time slice will evolve in time via the Schrodinger equation.

  • 3
    $\begingroup$ That was such an amazing answer! Thank you so much! $\endgroup$
    – IamWill
    Oct 6, 2020 at 0:16

The issue with this way of viewing things is that while a quantum field can be an observable, as it happens with the electric and magnetic fields, it needs not be an observable, as it happens with the electron field. Perhaps the best way to view a quantum field is rather as a building block for the interaction hamiltonian (and possibly other observables), which is convenient because of its simple transformation properties under Lorentz transformations.

This point of view is heavily emphasized in Weinberg's "The Quantum Theory of Fields", Volume 1. In the first five chapters Weinberg tackles the problem of how to write down interaction hamiltonians which give rise to a Lorentz covariant ${\cal S}$-matrix which further obey the cluster decomposition principle, which roughly speaking, states that experiments conducted far apart must give uncorrelated results. I shall outline the main ideas, but I highly encourage you to study these chapters.

These two conditions imposes constraints on the possible interaction hamiltonians $V$ and in that case, as is always a good idea, Weinberg follows a "criterium of simplicity". In other words: find the simplest way to construct such $V$ obeying the constraints imposed.

It turns out that:

  1. The Lorentz covariance property of the ${\cal S}$-matrix elements can be most simply implemented by writing down $V$ as the integral of a density. In the interaction picture $$V(t) = \int{\cal H}(t,\mathbf{x})d^3\mathbf{x},\quad U_0(\Lambda,a) {\cal H}(x)U_0^{-1}(\Lambda,a)={\cal H}(\Lambda x+a),\quad [\mathcal{H}(x),{\cal H}(x')]=0,\quad (x-x')^2>0$$

  2. The clustering property of the ${\cal S}$-matrix can be most simply implemented by writing down $V$ as a polynomial in the creation and annihilation operators of the asymptotic multiparticle states, with all creation operators to the left of all annihilation operators and with a coefficient containing a total momentum conservation Dirac delta distribution times some sufficiently regular function of the momenta.

Quantum fields then appear as the simplest solution to combine points (1) and (2). To implement (2) you write down ${\cal H}(x)$ as a polynomial in creation and annihilation operators with the regularity conditions obeyed and with $x$-dependent coefficients. Now you need to implement (1). But as it stands it will be quite painful to implement (1) because the creation and annihilation operators do not transform simply under conjugation by Poincaré transformations.

Weinberg then proposes to "repackage" the creation and annihilation operators into objects which transform simply under the Poincaré group. More precisely, instead of working with the bare $a(\mathbf{p},\sigma,n)$ and $a^\dagger(\mathbf{p},\sigma,n)$ you define objects $$\phi_{\ell}^+(x)=\sum_{\sigma n}\int d^3\mathbf{p} u_\ell(\mathbf{p},\sigma,n;x)a(\mathbf{p},\sigma,n),\quad \phi_{\ell}^-(x)=\sum_{\sigma n}\int d^3\mathbf{p} v_\ell(\mathbf{p},\sigma,n;x)a^\dagger(\mathbf{p},\sigma,n),$$

where we demand that these objects transform simply under the Poincaré group: $$U_0(\Lambda,a)\phi_\ell^\pm(x)U_0^{-1}(\Lambda,a)=\sum_{\bar \ell}D_{\bar \ell \ell}(\Lambda^{-1})\phi_{\bar \ell}(\Lambda x+a),$$

where $D_{\ell \bar{\ell}}(\Lambda)$ are finite-dimensional representations of the Lorentz group.

To give one very basic intuition: what is the easiest way to construct scalars? Obviously by contracting tensors, like $p_\mu p^\mu$, $J_{\mu\nu} J^{\mu\nu}$, etc. Take a time to convince yourself that this is the basic idea here. The tensors like $p^\mu$ and $J^{\mu\nu}$ transform in a simple way, as will transform the $\phi_\ell$, and we know how to couple then to form scalars, like the $\phi_\ell$.

This is an outline to show you that quantum fields appear as the building blocks for your interaction hamiltonian (and possibly other observables). Therefore while they can be seem as observable, teaching yourself from the start that they are observables can be misleading.


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