What exactly is a quantum field?

As I understand it, a relativistic quantum field is an operator-valued function of spacetime that transforms under some finite dimensional irreducible representation of the Lorentz* group: $$$$U(\Lambda)\Psi^{\alpha}U(\Lambda^{-1}) = D(\Lambda^{-1})^\alpha{}_{\beta}\Psi^{\beta}(\Lambda x)$$$$ Particle states (which I will write as $$|\mathbf{p},\lambda \rangle = a^\dagger (\mathbf{p},\lambda) |0\rangle$$, with $$\mathbf{p}$$ being the 3-momentum and $$\lambda$$ being the helicity), on the other hand, must preserve unitarity due to Quantum Mechanics and therefore should transform under unitary representations of the Poincaré* group

$$$$\Lambda|\mathbf{p},\lambda\rangle = |\mathbf{p}',\lambda'\rangle D^{s}(R(\Lambda,p))^{\lambda'}{}_{\lambda}$$$$

where $$p'^{\mu} = \Lambda^{\mu}{}_{\nu}p^\nu$$ and $$\lambda$$s are the helicities. Note that $$\alpha, \beta$$ are the Lorentz indices. The reason fields are introduced is because we can embed internal particle degrees of freedom transforming under an irreducible unitary representation (and hence infinite-dimensional) in a finite dimensional representation as

$$$$\Psi^{\alpha}(x) = \sum_{\lambda}\int \tilde{d}p[a(\mathbf{p},\lambda)u^{\alpha}(\mathbf{p},\lambda)e^{ipx} + \text{ negative energy term}]$$$$ where $$\tilde{d}p$$ is the Lorentz invariant integration measure. My question is:

How does one make contact with quantum mechanics? The book I am reading (Wu Ki Tung Group Theory in Physics, Chapter 10) says that the wavefunction for a state $$|\phi\rangle$$ is given by $$$$\phi^\alpha(x) = \langle 0 | \Psi^{\alpha}(x) |\phi\rangle.$$$$ So I assume this object describes the particle in the position basis. According the Peskin and Schroeder (chapter 2, eq. 2.42), the field operator creates a particle at position $$x$$ as such. But then what is the wavefunctional of the system, i.e., the object that describes the probability density corresponding to a particular configuration of the fields all throughout spacetime.

• Please do not ask multiple questions per post. I've edited out your second question since it would have been a duplicate of physics.stackexchange.com/q/286078/50583, anyway. Commented Feb 1 at 17:28
• Linked. The "wave function" is something different than what you expect. Commented Feb 1 at 18:03
• Beware, there is no position operator in QFT and hence no position basis. The object that is called wavefunction therefore has a different nature than in non-relativistic QM: it is NOT a probability amplitude for finding a particle around some given point. Rather it is just an object on which we project the quantum field using a suitable inner product (like the KG inner product) to obtain creation and annihilation operators for states. Its usefulness is revealed in the context of the LSZ reduction formula.
– Gold
Commented Feb 1 at 18:29
• Related: What is a quantum field?. Commented Feb 2 at 7:32
• @Gold There is no position basis in QFT, but there is field basis in QFT, right? QFT state is a superposition of field, just like simple particle quantum mechanics state is a superposition of position. Almost no textbook talks about expressing qft state in the field basis when it seems like such a natural thing to do. I know scattering cross sections is probably easier obtained through a and a-dagger basis but working with natural definition\notation\basis in mathematical tend to give good things, at least some good things. Commented Feb 3 at 16:52

Let's address first your title question "What exactly is a quantum field?" The answer is essentially given in the first five chapters of Weinberg's textbook: a quantum field is a convenient mathematical object that we introduce in order to facilitate the construction of relativistic quantum theories which obey the cluster decomposition principle.

The bottom line is that you want to construct an interaction Hamiltonian $$V$$ such that $$V(t)=\int d^3\mathbf{x} {\cal H}(t,\mathbf{x}),\tag{1}$$ in which $${\cal H}(t,\mathbf{x})$$ transforms a scalar under Lorentz transformations. Moreover, this $$V(t)$$ should be built in terms of creation and annihilation operators obeying a certain regularity condition so that cluster decomposition is obeyed.

It turns out that the easiest way to accomplish this is by first building quantum fields and using the quantum fields to build $${\cal H}(t,\mathbf{x})$$. Finally one realizes that it is even possible to bypass the manual construction of $${\cal H}(t,\mathbf{x})$$ from quantum fields if one starts with a classical field theory and quantizes.

Note that this can be compared to something simpler. If you want a Lorentz scalar, the easiest way is to have at your disposal tensors and then construct the possible contractions. For example, given two vectors $$p_\mu$$, $$q_\mu$$ and a tensor $$A_{\mu\nu}$$ you know you can build things like $$p_\mu q^\mu$$, $$p_{\mu}q_\nu A^{\mu\nu}$$, $$A_{\mu\nu}A^{\mu\nu}$$, etc. They are all guaranteed to be Lorentz invariant because each of the building blocks has a definite transformation property that renders these contractions scalars. The same is true for quantum fields and the construction of scalars $${\cal H}(x)$$.

Now, a comment is that in his book Weinberg makes it clear that these are not necessary conditions. They are more like the simplest way of satisfying relativistic symmetry and clustering. For example, (1) is the simplest way to get a Lorentz invariant interaction, but not the unique one. Likewise, the way to obey clustering is never claimed to be the only possible way.

So quantum fields arise simply as convenient mathematical tools that give you the simplest possible way to construct the kind of theories that we would like to consider.

Finally, your comments on a "position basis" deserve an explanation. In QFT there is no position basis. Honestly I fail to grasp why some textbooks insist on this concept which is clearly erroneous. What happens is that while the momentum observable naturally generalizes to the relativistic setting, the position operator does not.

In the absence of the position operators, it is not possible to define a position basis. As a result, it is not possible to construct position space wave functions that give probability amplitudes for the position of particles.

Instead, in QFT wave functions have a different nature. The object $$\psi_f(x)= \langle 0|\phi(x)|f\rangle,$$

where $$|f\rangle$$ is a one-particle state, is a solution to the linearized field equations that has the property that if $$(,)_{\rm KG}$$ is the Klein-Gordon inner product, then $$(\psi_f,\phi)_{\rm KG}$$ will give you either the creation or annihilation operator for the state $$|f\rangle$$. This is very useful in the LSZ reduction formula, that being the reason to introduce these wave functions in the first place.

Note, however, that $$\phi(x)|0\rangle$$ are not to be interpreted as eigenkets $$|x\rangle$$ of some Hermitian operators whose eigenvalues are $$x$$.

There are formulations of QFT in terms of wavefunctionals but they are often rather useless, both for technical reasons and for more pragmatic ones. For a discussion of the technical issues, see this answer of mine. There are also technical issues with the position-space "wavefunction" of a one-particle state in a fully relativistic theory because relativistic position operators don't really exist, but for massive particles and in the non-relativistic limit this description is fine.

As for the pragmatic issues: Ultimately, QFT is still mostly a theory about particles when it comes to what it is supposed to do. There is no classical electron field, no physical experiment asks the question "What is the value of $$\psi^\mu(x)$$?".

QFT, at least in the manner it is usually done in high-energy physics, has as its primary objective to describe the outcome of collider experiments and to tame the particle zoo we first found when we started smashing the tiniest bits of matter into one another with increasingly high energies. So the description in terms of wavefunctionals is not what most people are interested in in practice, really: What people want to know are the scattering cross sections, QFT corrections to anomalous magnetic moments of particles, etc. - all questions about particles, not fields or their wavefunctionals.

Things look a bit different in condensed matter, but there, too, the description in terms of wavefunctionals is usually not the interesting one, and more e.g. the description in terms of expectation values of the fields in some thermal state.

The core construction of hep-th QFT - the thing that makes contact with physics, really - is the idea that the fields possess an asymptotic description in terms of creation and annihilation operators of particle states, and it is the behaviour of these particle states that we are interested in, not the behaviour of the field, which may be viewed just as an efficient encoding of the way these particle states may interact (this view of QFT is particularly close to the way Weinberg's QFT books construct QFT in the first place).

• The Jordan-Lee-Preskill algorithm for simulating scalar QFT on a quantum computer relies on formulating QFT in terms of wavefunctionals. I'm not really disputing your claim that wavefunctions are usually "rather useless." Just providing one example where they are not. arxiv.org/abs/1111.3633 Commented Feb 1 at 18:51

What exactly is a quantum field?

When physicists use the term "field" they usually mean a function of space or a function of space and time.

For example, in classical electrodynamics the electric field is a function of space and time $$\vec E(\vec x, t)$$ and its value at a particular point in space and time ($$\vec x_0$$, $$t_0$$) tells you the force on a particle of charge $$q$$ that is placed at that particular point via: $$\vec F = q\vec E(\vec x_0, t_0)\;.$$

In classical mechanics or classical electrodynamics, the fields of interest are functions of space and time whose values are real or complex numbers (or collections of real or complex numbers, like real vectors in the case of the electric field.)

In quantum field theory, the fields (i.e., the functions of space and time) are functions whose values are quantum operators.

So, for example, in quantum electrodynamics the value of the free quantum electric field operator at a point $$\vec x_0$$ and $$t_0$$ is a triple of operators, not a triple of real numbers: $$\hat{\vec E}(\vec x_0, t_0) = \int \sum_{\rho}\frac{d^3k}{(2\pi)^2}i\omega_{k} \left( \vec \epsilon_{\vec k}^{\rho}\hat{a}_{\vec k}^\rho e^{i\vec k\cdot \vec x_0-i\omega_k t_0} - \vec {\epsilon_{\vec k}^{\rho}}^*\hat{a^\dagger}_{\vec k}^\rho e^{-i\vec k\cdot \vec x_0+i\omega_k t_0}\;, \right)$$ where we can see that $$\hat{\vec E}(\vec x_0, t_0)$$ is a quantum operator since the $$a$$ and $$a^\dagger$$ are quantum operators.

A quantum field does not have to transform under any particular representation of the Lorentz group, but in many cases it is useful to construct such a field. However, quantum field theory can also be applied to non-relativistic theories (and often are in condensed matter theory) and such fields do not usually have any particularly important Lorentz transformation properties.

One certainly makes contact with quantum mechanics via the wave function: $$\langle 0|\Psi|\phi\rangle.$$ Indeed, physical theories must be invariant under the full Poincare group, however, for many special cases, one only needs Lorentz invariance. Since most experiments can be devised in such a way as to get rid of the constant shift of the Poincare transformation, the emphasis on Lorentz invariance is such as it is in most texts. For example in most intro special relativity courses texts only make use of the invariance under boosts apart from the full Lorentz invariance by assuming that the relative motion is along one axis alone which is aligned with the unprimed coordinate axis and has a common origin at t=0. In this special, though useful, case; one does away with the constant shift of the Poincare group and the rotations of the Lorentz to require boosts alone.

"But then what is the wavefunctional of the system"

In simple particle quantum mechanics, the wavefuctional is a function of position $$\psi (x)$$. Its time derivative is given by Schrodinger equation based on spacial derivatives $$\partial / \partial x$$ of the wavefuctional. The particle position and momentum become operators following canonical commutation relation in the canonical quantization process. The quantum state is a superposition of positions.

In QFT, the wavefuctional $$\Phi (\phi)$$, I think, is a function of field. Its time derivative is given by some kind of Schrodinger equation based on its functional derivatives $$\delta / \delta \phi$$. The field strength and field momentum become operators following canonical commutation relation in the canonical quantization process. The quantum state is a superposition of fields.

Now the question is why don't standard textbook talk about qft wavefuctional as a function of field? My guess (and what @ACuriousMind said) is that such teaching doesn't have practical use, since scattering cross sections are easier obtained through quantum field states expressed in basis of $$a$$ and $$a^{\dagger}$$. Quantum field states expressed in the basis of superposition of field is just not nearly as useful. By the way, I disagree with this teaching philosophy. It punishes students who actually want to have a more coherent understanding of qft which might be discovered to be useful in the future.

I'm just a grad student. I am describimg my understanding of the "wavefunctional" of qft. I can be wrong.

• The wave function is not necessarily a function of position as you can see in the example of entangled spins. It is completely irrelevant where those spins are. The same is true for spectroscopy of any kind. "Where" the hydrogen atom is when it emits a photon of energy doesn't matter. The energy is completely sufficient to determine which states are initial and final. Neither do we care about position in any high energy physics experiment. We only care about incoming and outgoing momenta and charges. Commented Feb 1 at 21:01
• @FlatterMann I agree with the observation-motivated argument in favor of certain formalism. However expressing quantum field as a superposition of field, just sounds like such a natural thing to do. In math, when one works with something in a natural way (like work with quantum mechanics in vector notation instead of functions and their partial derivatives), good things tend to happen. Some proofs become simpler. Even if such formalism might over-assume and over-restrict possible theory, I feel like it can still give insight to a general theory from a more restricted but natural looking theory Commented Feb 2 at 1:27
• I agree that using a position-representation is a natural thing to do, it just doesn't have to make for good physics. For one thing quantum mechanics didn't originate from such ideas (neither Planck radiation nor the photoelectric effect are position sensitive). Far worse, the very idea of quanta having position is unphysical. The only time position "enters" from an experimental perspective is when we are introducing position sensitive energy detectors. WE are restricting nature to work differently at x_1 than at x_2. This leads to endless ontological confusion IMHO. Commented Feb 2 at 3:19

I am going to write something absolutely silly now in response to the OP's first point, and you can all laugh about that.

A quantum field is a physical system that interacts with external systems only through the exchange of quanta of energy. In practice it's the kind of system that makes "blip-blip-----blip----blip-blip-blip" in our photomultiplier tubes and other quantum detectors. Quantum field theory is a ridiculously complex theory for a ridiculously simple phenomenology (no offense intended to either theorists or experimentalists here, the devil is in the detail, I am just simplifying to make a point).

The way the theory makes contact with reality is by predicting the relative frequencies of these quanta as a function of energy, momentum, angular momentum and charges (and, yes, occasionally position). The only thing we "care to get" from the theory are the distributions of these quanta (and their correlations) because that is the only information that experiments can return.

The problem with the first question about the wave function is that the wave function is NOT a proper representation of most experiments. Most of our experiments are, at the end of the day, scattering experiments. In atomic physics, nuclear physics and solid state physics these experiments are usually called "spectroscopy", but they are not fundamentally different from what we are doing in high energy physics with accelerators. Only the light sources (accelerators) are a lot larger and more complicated and so are the detectors. Everything else is, up to trivial naming conventions, pretty much the same.

No branch of physics ever measures a quantity called $$\psi$$ or its modulus squared (directly). Those quantities only exist in textbooks. We are never measuring quantum mechanical "states", either. The translation from concrete spectra to abstract states usually takes place (more or less ad-hoc) in experimental physics textbooks and papers and it is poorly documented (if not outright ignored) in many quantum mechanics textbooks for all I can tell.

Unless introductory QM textbooks have changed substantially in the past decades (I admit that I haven't kept up with the releases in the past few decades), they are mostly focused on isolated systems that can be described with a nicely normalized wave function. The downside is that these systems do NOT map easily on (open) spectroscopic measurements that experimentalists are dealing with. Absorption and emission processes are, at most, treated with time dependent perturbation theory. At least in the past we used to introduce Fermi's golden rule hush-hush during a single lecture and that was probably the most formal "translation" between wave functions and in-coming and out-going waves that I ever got to see in undergrad physics personally.

The one system in a typical beginner's class on non-relativistic QM that is useful to develop a deeper understanding of scattering phenomena is the potential barrier that is being used to explain tunneling... and sometimes it seems to be a "system of shame" because it does not have a normalizable wave function. Instead of teaching that scattering is actually "the normal", we pretend that it's an outlier to "rescue" von Neumann's formalism.

I would ask those of you who are actively participating in undergrad education to please, please, please, tell students how to go from normalizable wave functions to spectra formally correctly. If I am not mistaking theorists made that transition between roughly 1927 (a Dirac paper on collision processes) and between the late 1930s and early 1940s (Wheeler and Heisenberg), i.e. during a time during which von Neumann's book was barely dry in its original German lecture/paper form (1932). We should start with that and then specialize on the wave function as a theory of (almost completely) isolated quantum systems. I believe it would remove a lot of confusion about quantum mechanics in general.

End rant.

• The experimentalist's point of contact with the wave function stems from the result that the spectra measured in an experiment do not occur with random frequency, but with an expectation value that is weighted according to the statistics of the wave function. Thus, experimentalists are not only concerned with observables and their eigenvalues, but with quantum states as well. Commented Feb 1 at 17:23
• By the way, I posted the above comment not because I believed that you needed it, but for the benefit of a hypothetical undergrad that might read it and go away thinking that they can forget all about wave functions. It was a good rant, so I will vote it up even though this is probably not the place to be ranting. Commented Feb 1 at 17:39
• @AlbertusMagnus I know where I come in contact with wave functions, it's just not well explained and "formal enough" in undergrad classes. Fermi's golden rule is one "interface" between non-interacting and interacting quantum systems that I am aware of in undergrad education (at least as it was practiced decades ago). There are lots of missing pieces like symmetries which are absolutely vital to atomic physicists and chemists which are barely touched on in QM101. I know that it would take an additional semester to go through that in addition to the standard material. Commented Feb 1 at 17:39
• @AlbertusMagnus That's fair. I am not saying that wave functions aren't important and useful. They are just not the entire story. If anything I have come to believe that they are the smaller part of the story. They happen to be the mathematically far easier to digest part. I think we can agree about that. Commented Feb 1 at 17:41