How to interpret quantum fields?

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?

• Nature doesn’t care how you interpret it’s laws, but in practice you usually either take the classical limit and recover classical field theory, or use the “particle interpretation” where some (combinations of) field operators act as operators that create and destroy particles Commented Dec 19, 2022 at 1:24
• @Prof.Legolasov Does that mean in the particle interpretation an expression of the form $\varphi | 0 \rangle$ means a particle is being created and destroyed in the vacuum? Commented Dec 19, 2022 at 1:27
• @Prof.Legolasov Nature doesn't have an obligation to make sense to us, but we have an obligation to make sense of nature. Commented Dec 19, 2022 at 2:37

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.

• The point you made in the beginning of your post is exactly the error I was making, I had thought that all fields must correspond to observables but now I know this is not the case. Thank you for all your help! Commented Dec 19, 2022 at 4:06
• +1 But I think this takes the viewpoint that particles are fundamental. This viewpoint is not necessary because, even in non-relativistic QM, most operators aren't observables. We could just say that only some functions of field variables are observables, like field energy and momentum. Fields would still be fundamental, and we wouldn't have to manually construct the interaction Hamiltonian relying on simplicity arguments. Everything would directly come from canonical quantisation, like Fermion and boson statistics, conserved quantities, Fock space, etc. Commented Dec 20, 2022 at 10:10
• All these experimental facts, like particle creation and annihilation, come for free in field quantisation, as opposed to manual construction of fields. This points towards fields being fundamental. Fields also generalize well to curved spacetimes. Commented Dec 20, 2022 at 10:14
• @Tom, I know there is this opinion, but I think in the end of the day this is very personal. I have tried starting QFT from the fields first approach and even though I was able to calculate things, I simply couldn't understand what was going on. Then I tried the particles first approach of Weinberg and it worked much better and the "fields first" textbooks became clear. But anyway, I'm a Mathematician and I was only able to feel I understood classical mechanics after I learned differential geometry.
– Gold
Commented Feb 28, 2023 at 16:23
• Moreover, I do understand that Weinberg's textbook might be difficult for someone to go through it alone the first time. If the person is guided, though, I see no issue at all. For example, there is a Physics institute in my country where they offer one "pre-QFT course" which covers Weinebrg's Chapters 2 to 5 plus the canonical quantization chapter. Then they follow with standard QFT. It works pretty well, and people end up being able to follow through the other chapters of Weinberg later on their own if they so wish.
– Gold
Commented Feb 28, 2023 at 16:24

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 $$\varphi(x)$$: $$$$\phi(x)|\varphi(x)\rangle = \varphi(x) |\varphi(x)\rangle$$$$

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 $$$$\langle 0 | \phi(x) \phi(y) | 0 \rangle, \langle 0 | \phi(x) \phi(y) \phi(z) |0 \rangle, \cdots$$$$ 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 $$x\rightarrow y$$ $$$$\lim_{x\rightarrow y} \langle 0 | \phi(x) \phi(y) | 0 \rangle = \infty$$$$ Therefore, one typically "smooths out" the correlation function by integrating the field against a test function.

$$^\star$$ 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.

• Thank you! When you say that the field $\phi(x)$ represents the value of the field at $x$, what is this value exactly? For example, I picture the free field $\phi$ to just be a collection of particles. So if we apply this free field to some state, what exactly are we measuring? What is the corresponding observable? Commented Dec 19, 2022 at 2:24
• @CBBAM The free field does not represent a collection of particles. It is a mathematical tool use to build particle states. When the field acts on the vacuum state, it gives us a particle state that is “around” some point x (but not perfectly a delta function).
– user310742
Commented Dec 19, 2022 at 2:46
• @CBBAM "What the value is" depends on what the field represents; in condensed matter physics, $\phi(x)$ might represent, say, the density of a material. In general, $\phi(x)$ is just some number you can measure associated with the position $x$. The particle interpretation is most clear in the Fock basis (which is different from the field eigenstate basis). Then, the eigenstates represent particles with different momenta, where a particle is an elementary excitation in the field. The particle and field povs are different ways of looking at the same object, rather like wave-particle duality. Commented Dec 19, 2022 at 2:53
• @CBBAM It just depends on the context. The free field is just a toy model that doesn't really have any intrinsic meaning. You can use fields to model all kinds of things, like properties of condensed matter systems, or particles in an accelerator, or density perturbations during inflation. The interpretation will depend on the context. Commented Dec 19, 2022 at 2:59
• @SarahMesser On the left hand side, $\phi(x)$ is an operator, and on the right hand side, $\varphi(x)$ is an ordinary function. I could also have been more explicit by putting a hat on the $\phi$ to make it more clear it is an operator, so then the equation would be $\hat\phi(x)|\varphi(x)\rangle = \varphi(x)|\varphi(x)\rangle$, analogously to $\hat{x}|x\rangle=x|x\rangle$. It's just meant to represent an eigenvalue equation. Commented Dec 19, 2022 at 21:37