How to interpret quantum fields?

Question

As an analogy of what I am looking for, suppose $f(x,t)$ represents a classical field. Then we may interpret this as saying at position $x$ and time $t$ the field takes on a value $f(x,t)$.

In quantum field theory the fields are now operator valued distributions. Suppose $\varphi$ is a quantum field, thus it must be of the form $\varphi(f)$ where $f$ is some suitably nice test function. What is the physical interpretation here analogous to the classical field case? Is the test function $f$ supposed to represent the state of the system (as it would in quantum mechanics, i.e. $f \in \mathcal{H}$ where $\mathcal{H}$ is some Hilbert space)?

To word things different, what exactly does it mean to apply the resulting operator valued distribution $\varphi$, for example, $\varphi | 0 \rangle$? Physically what does this tell us?

Answer 1

There is one usual confusion about quantum fields which is, at least in my perspective, perhaps caused by one very familiar example which we all have met before studying QFT. This example is that of the electromagnetic field. We are all used from electrodynamics to talk about electric and magnetic fields $\mathbf{E}$ and $\mathbf{B}$ as measurable quantities. As such, we often think that in QFT, the quantum fields should represent observables.

While this is true in some particular cases (the electromagnetic field one for example), it is not true in general. For example, a spinor field, like the electron field $\Psi(x)$, is not even Hermitian, so it does not qualify as an observable as per the postulates of QM.

All that said, in my opinion the best way to interpret quantum fields is that quantum fields are building blocks for observables, introduced as mathematical constructs which facilitate us to build relativistic interactions obeying cluster decomposition. That is quite a bit, so I'll elaborate.

Please, bear in mind that what follows is a very rough summary of a logic that Weinberg constructs through four long chapters in his book. I encourage you to study Chapters 2 - 5 of Weinberg's "The Quantum Theory of Fields" to get the full picture.

Now, the argument is roughly this. We start with relativistic particles. We want to build interactions for these particles which satisfy two properties: (1) they are compatible with the relativistic symmetry encoded in the Poincaré group and (2) they satisfy cluster decomposition, which means that experiments conducted far away should have nothing to do with each other.

The most simple way to satisfy cluster decomposition is to build our interaction potential $V$ as a kind of polynomial in the creation and annihilation operator where the coefficients obey one kind of regularity condition. Likewise, the most simple way to satisfy relativistic invariance is to build $V$ as

$$V(t)=\int d^3 \mathbf{x} {\cal H}(t,\mathbf{x})$$

where ${\cal H}(t,\mathbf{x})$ is a scalar which commutes with itself at spacelike separations.

Now, one must be able to combine the two things. But it is a bit hard because while the creation and annihilation operators $a(p,\sigma)$ and $a^\dagger(p,\sigma)$ transform very simply under translations, they have a complicated transformation law under Lorentz transformation. So the most simple way around this is to repackage these operators into quantum fields, which are demanded to have simple transformation laws under Lorentz transformations. Once that is done, constructing such ${\cal H}(t,\mathbf{x})$ becomes extremely simple.

Weinberg then introduces the formalism of canonical quantization in which you start from a classical Lagrangian density ${\cal L}$ and derive from it the interaction Hamiltonian. I do believe this is the best moment to introduce it, since now we know what is a quantum field and why it is reasonable to use them. Canonical quantization bypasses the need to construct such interaction Hamiltonians by hand, and so what one gets is a framework for constructing relativistic theories. The advantage of Weinberg's point of view is that at this point we already know the correct relation between fields and particles and know which particles can be described by each field. In particular I find it very beautiful that as Weinberg shows, if you try to encode spin one massless particles into a vector field $A_\mu(x)$ you can't really get an object transforming as a vector field. Rather, $A_\mu(x)$ transforms as a vector field up to a gauge transformation, and then the requirement of gauge invariance appears naturally.

So in the end, I would say that today I view this point of view exposed in Weinberg's textbook as the most natural way to interpret a quantum field. They are objects introduced for convenience in order to construct relativistic theories obeying cluster decomposition. It does happen that some of them are observables (like the electromagnetic field $F_{\mu\nu}$), but it is not a general rule (as we see in the electron field $\Psi(x)$), and with this point of view this is not an issue at all.

Answer 2

Perhaps the most direct (but not only) interpretation is to say that $\phi(x)$ represents a local observable.$^\star$ In other words, $\phi(x)$ represents the value of the field at $x$. You can (in principle) perform a measurement to learn the value of the field at $x$.

Just like in normal quantum mechanics, in a general state $|\Psi\rangle$, you cannot predict precisely what the outcome will be of measuring $\phi(x)$. You can only make such predictions in special states, field eigenbasis states. For example, there is an eigenstate $|\varphi(x)\rangle$ where the field will take on the value $\varphi(x)$: \begin{equation} \phi(x)|\varphi(x)\rangle = \varphi(x) |\varphi(x)\rangle \end{equation}

However, in other states, like the ground state (also called the vacuum state) $|0\rangle$, the field does not take on a definite value. There is a superposition of field values, represented by the Schrodinger wavefunctional (which is the generalization of the Schrodinger wavefunction of quantum mechanics, to quantum field theory).

Frequently we are interested in the correlation functions of the field, in some state (usually the vacuum). This is a way to capture the probability distribution over different field configurations. From these correlation functions, we can extract other observables we care about (such as scattering amplitudes in particle physics). Some examples of correlation functions are \begin{equation} \langle 0 | \phi(x) \phi(y) | 0 \rangle, \langle 0 | \phi(x) \phi(y) \phi(z) |0 \rangle, \cdots \end{equation} Note that because of operator ordering ambiguities, in practical applications it is important to specify how the operators are ordered when defining and computing correlation fucntions.

The reason that the field is an operator valued distribution, and not simply an operator valued function, is because it is quite a singular object. For example, the two point function is divergent in the limit $x\rightarrow y$ \begin{equation} \lim_{x\rightarrow y} \langle 0 | \phi(x) \phi(y) | 0 \rangle = \infty \end{equation} Therefore, one typically "smooths out" the correlation function by integrating the field against a test function.

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$^\star$ As Gold mentioned in their answer, this is a simplification. Because of the freedom to do field redefinitions, and in gauge theories the ability to do gauge transformations, the value of the field itself is not a physically invariant quantity. For the field value itself to be meaningful, you have to couple the field to a probe that will measure the field's value.