Dictionary between interpretations of field operators

Question

For now, let $\hat{\phi}(x)$ be a quantization of a classical, real scalar field $\phi(x)$.

My understanding is that, for fixed $x$, there are three ways to regard the operator $\hat{\phi}(x)$:

1. The value of the classical field $\phi$ at $x$ is an observable, so $\hat{\phi}(x)$ is simply the operator corresponding to this observable.
2. The operator $\hat{\phi}(x)$ is a creation operator, which creates the particle corresponding to $\phi$ at position $x$. (Or, one might say it is the superposition of all ways to create a particle of this species at position $x$; it should be the Fourier transform of various $a(p)$ )
3. Finally, in some treatments, one remains cynical about the role of $\hat{\phi}$, regarding it as a purely formal mathematical object which allows us to build Hamiltonians from certain creation operators.

My question is: How accurate are perspectives 1 and 2 (If I've misrepresented these, feel free to correct me, and then tell me how accurate the correct formulation is), and are these two perspectives compatible with (or even equivalent to?) each other? Why might one want to remain cynical and use perspective 3.

Moreover, how would perspective 2 be formulated when $\phi$ is a vector-valued field?

Answer 1

Assuming that you are interested in the physical interpretation of the hermitean field operator of a free relativistic scalar theory in the Heisenberg picture, its Fourier decomposition (omitting the hat indicating the operator) is given by $$\begin{align} \phi(t, \vec{x})&=\int\limits_{\mathbb{R}^3} \!\frac{d^3 p}{(2\pi)^3 2 \omega(\vec{p})} \left(a(\vec{p})e^{-i \omega(\vec{p})t}e^{i\vec{p} \cdot \vec{x}}+ a^\dagger(\vec{p}) e^{i \omega(\vec{p}) t} e^{-i\vec{p} \cdot \vec{x}} \right), \\[5pt] \omega(\vec{p})&=\sqrt{\vec{p}^2+m^2}, \end{align} \tag{1} \label{1}$$ where the annihilation and creation operators $a(\vec{p})$ and $a^\dagger(\vec{p})$ satisfy the commutation relations $$\left[a(\vec{p}), a^\dagger (\vec{p}^\prime) \right]=(2\pi)^3 2 \omega(\vec{p}) \delta^{(3)}(\vec{p}-\vec{p}^\prime), \quad \left[a(\vec{p}, a(\vec{p}^\prime) \right]=0. \tag{2} \label{2}$$ Usually \eqref{1} is written in the more compact form $$\phi(x)= \int \! d\mu(p) \left (a(p) e^{-ip\cdot x}+ {\rm h.c.} \right) \tag{3} \label{3}$$ with $$x=(t, \vec{x}), \quad p^0=\omega(\vec{p}), \quad p=(p^0, \vec{p}), \quad d\mu(p)=\frac{d^3 p}{(2\pi)^3 2 p^0}. \tag{4} \label{4}$$

1. Your first assertion, interpreting \eqref{3} as an (hermitean) operator associated with an observable field, is essentially correct, apart from the technical complication that $\phi(x)$ is a highly singular object kicking a normalized state vector $|\psi\rangle \in \mathcal{H}$ ($\mathcal{H}$ denotes the Fock space of the theory) out of your Hilbert space, i.e. $\phi(x) | \psi \rangle\notin \mathcal{H}$. This disease can be cured by considering smeared-out operators $$\phi_f= \int \! d^4x \, f(x) \phi(x), \tag{5} \label{5}$$ where $f$ is a a suitable function with compact support on some finite domain of space-time such that $\phi_f|\psi\rangle \in \mathcal{H}$. For this reason, $\phi(x)$ is referred to as an "operator-valued distribution".

2. Your second assertion is wrong. As can be seen from \eqref{3}, $\phi(x)$ is a linear combination of creation and annihilation operators. On top of that, interpreting $$\phi(t, \vec{x}) |0\rangle= \int \! d\mu(p) e^{-ip\cdot x} a^\dagger(p) |0\rangle \tag{6} $$ as a (non-normalized) state of a particle located at the point $\vec{x}$ at the time $t$ is misleading as $$ \langle0 |\phi(t, \vec{x}^\prime) \phi(t, \vec{x})|0\rangle \ne \delta^{(3)}(\vec{x}-\vec{x}^\prime), \tag{7} \label{7}$$ in contrast to the field operator $\Psi(t, \vec{x})$ of the second-quantized nonrelativistic Schrödinger theory, where $\Psi^\dagger(t=0, \vec{x}) |0\rangle$ corresponds indeed to the "state" $|\vec{x}\rangle$, where a single particle is located at the point $\vec{x}$. Note that \eqref{7} reflects the fact that a single particle in a relativistic quantum field theory cannot be localized at a point $\vec{x}$, demonstrating that the notion of "state-vectors" $|\vec{x}\rangle$ makes only sense in the nonrelativistic approximation. However, any element $|\psi\rangle$ of the Hilbert space $\mathcal H$ of the relativistic theory can be written as a linear combination of products of $\phi_f$'s acting on the vacuum, $|\psi \rangle= \sum_n c_{i_1 \ldots i_n} \phi_{f_{i_1}} \ldots \phi_{f_{i_n}}|0\rangle$. According to the Reeh-Schlieder theorem, the supports of the functions $f_{i_k}$ in the previous sum may even be restricted to a common finite (open) region $\Omega$ of space-time, i.e. ${\rm supp} \, f_{i_k} \subset \Omega$ for all $i_k$.

3. It is not clear to me what you mean by "cynical" and which "treatments" you have in mind. Of course, $\phi(x)$ plays an important role as the basic building block in constructing all other operators corresponding to observables of the theory. Because of the aforementioned singular behaviour of the field operator, one encounters additional technical complications defining products of $\phi(x)$ at the same space-time point (requiring the necessity of "renormalization" already in the free quantum field theory). A simple example, found in all text-books, is the energy-momentum operator $P^\mu$ written as a linear combination of normally ordered products of $\phi(x)$, $$ P^\mu = \int \! d^3x \,: \phi(x) i \partial^\mu \phi(x): =\int \! d\mu(p) p^\mu a^\dagger(p) a(p).\tag{8}$$

Edit: In view of some of the comments, let me add a few remarks on "observables" in quantum field theories. Apart from the obvious requirements of hermiticity (more precisely, self-adjointness), being of "bosonic" type (reducing the candidates for observables in fermionic field theories to bilinears of fermion field operators and products thereof) and gauge invariance in local gauge theories, the class of operators qualifying as observables is - to some extent - a matter of taste. In my opinion, excluding the field operator of a hermitean scalar field theory (or its smeared-out cousins) from the list of observable quantities (as suggested by @TobiasFünke in a comment) seems to be too strict.

Let me give you a well known example where measuring $\phi(x)$ (or $\phi_f$) makes perfectly sense. Consider the coherent state $$\begin{align} | c \rangle &= e^{ \int \!d\mu(k) \left(c(k) a^\dagger(k)-c(k)^\ast a(k)\right)} |0\rangle \\[5pt] &= e^{\int \!d\mu(k) |c(k)|^2/2}\, e^{\int \! d\mu(k) c(k) a^\dagger(k)} |0\rangle, \quad \langle c|c\rangle =1,\end{align} \tag{9} \label{9}$$ describing a "classical" state of the scalar field. The expectation value of the field operator in the state $|c\rangle$ is given by $$\langle c |\phi(x) |c\rangle = \int \! d\mu(k) \left(e^{-ik \cdot x} c(k) +e^{i k \cdot x} c^\ast(k) \right), \tag{11} \label{11}$$ being a classical solution of the Klein-Gordon equation. At the same time, \eqref{11} implies $$\begin{align}\langle c |\phi_f |c\rangle &= \int \! d\mu(k) \left(\tilde{f}(k) c(k)+ \tilde{f}^\ast(k) c^\ast(k) \right), \\[5pt] \tilde{f}(k) &= \int \!d^4x \, e^{-ik\cdot x}f(x),\end{align} \tag{12} \label{12}$$ for the expectation value of $\phi_f$.

It is (in principle) straightforward to adapt these ideas to the case of the electromagnetic field, replacing the scalar field by the photon field. The analogue of the state \eqref{9} of the electromagnetic field corresponds to a coherent superposition of energy-momentum eigenstates with all possible numbers of photons, describing a classical electromagnetic wave (e.g. the electromagnetic field produced by a radio transmitter). The "quantum-field-o-meter" is a receiver measuring the electric field $\vec{E}$ averaged over a certain space-time volume (corresponding to its space-time resolution). This observable can be modelled by introducing a suitable normalized weight function $f(x)$ (in the spirit of \eqref{5}) leading to a space-time averaged operator $\vec{E}_f(t, \vec{x})$ of the electric field, taming at the same time (potentially divergent) quantum fluctuations in in expressions like $\langle c | \vec{E}_f(t, \vec{x})^2 |c\rangle-\langle c |\vec{E}_f(t,\vec{x})|c\rangle^2$.

Answer 2

Perhaps there is a simple confusion. As you may notice yourself your points 1. and 2. are not compatible with each other. What you can say is that the quantization of a real scalar field is given by an Hermitian operator. The same holds true for operators in general without the need of having to talk about fields (which, as @Hyperon mentions, are complicated from the math side). Roughly speaking complex conjugation corresponds to taking adjoint at operator level, so

$$ A=\overline{A} $$

at classical level identifies a general potentially complex scalar, as a real scalar (overline denotes complex conjugate). At the quantum level the above becomes

$$ \hat{A}=\hat{A}^\dagger $$

Meaning that the operator is Hermitian.

In a more mathy way you can say that observables in classical worlds are given by the abelian algebra of real functions (i.e. complex functions satisfying $f=\overline{f}$). In the quantum world this becomes a non commutative algebra with an involution ($\dagger$).

Note that the procedure of quantizing a classical problem has other problems. Non commutativity in the quantum side implies that there are several possible quantum Hamiltonians corresponding to a given classical one. For example a term like

$$ P^2 Q $$

can corresponds to $\hat{P}^2 \hat Q$ but also to $\hat P \hat Q \hat P$. Anyway this is another story.