# How can a QFT field act on particle states in Fock space?

Recently I asked a question that was considered a duplicate. However I felt that the related question didn't answer my doubts. After a bit of pondering I have realized the core of my discomfort with the concept of QFT fields, so I am asking a new question, different from the previous one, hoping that this time it won't be considered a duplicate, and hoping not to attract the moderators' wrath.

At a very basic and conceptual level: after second quantization the fields are operators, acting on the space of state vectors (Fock space), right? Some fields are observables, some aren't, some act as ladder operators for things like charge, etc. But they are all operators that do things in Fock space, operating on state vectors.

Vectors in Fock space on the other hand, as we were saying, are state vectors: they represent the state of the physical system. One vector in Fock space could for example represent a state of two bosons, with some charge, momentum, etc. They represent quantum particles.

But the fields also represent the quantum particles! All the idea behind QFT is that we can interpret the excitations of the quantized fields as particles!

This melts my brain a little because we have fields, that represents particles1, that are also operators acting on state vectors, that also represent particles! It appears to me as a bit circular, like a representation of particles acting on another representation of the same thing.

In QM we didn't have this problem: operators weren't fields, they were just operators, acting on state vectors in Hilbert space, vectors that represented the state of the system; the wave function in the equations was just a way to express state vectors in position basis, no problem whatsoever.

So how can I make sense of the mess that we get after second quantization?

[1]: Their excitations represent particles.

• You're being too literal about the phrase "particles are excitations of the field", see e. g. this answer of mine Mar 4 at 16:51
• @ACuriousMind I see your point, but, humbly, I don't think it is so simple: we use a formalism where the fields, so the operators, evolve in time (like Heisenberg representation, it is Heisenberg representation right?), so in a way the state of the system as it evolves trough time is tracked by the fields; the fields are stealing the job that the state vectors (QM) in Hilbert space had in Schrödinger representation. I don't know.. maybe I simply have to little familiarity with Heisenberg representation to make it not melt my brain. Mar 4 at 17:20
• @Noumeno What? You can work in the Heisenberg picture in (non-relativistic) QM, too. In the Heisenberg picture, the observables evolve in time, and the observables are build out of the fields. Quantum Theory (including relativistic QFT) makes predictions via expectation values. The Heisenberg and Schrödinger as well as Dirac picture are just a mathematical way to shuffle around the time-evolution. All pictures are equivalent by construction. All in all, I agree with the comment above: You are too literal about the words. Mar 4 at 17:46

You had encountered exactly the same situation already in elementary quantum mechanics, but maybe you did not recognize the analogies. Consider a single harmonic oscillator with angular frequency $$\omega$$. In the Heisenberg picture, the elongation operator $$\phi(t)$$ can be expressed in terms of the ladder operators $$a, a^\dagger$$ (satisfying the commutation relation $$[a, a^\dagger]=\mathbf{1}$$) by $$\phi(t)=\frac{1}{\sqrt{2 \omega}}\left(e^{-i \omega t}a+ e^{i\omega t}a^\dagger\right) \tag{1} \label{1}.$$ Together with $$\pi(t):= \dot{\phi}(t) = \frac{-i \omega}{\sqrt{2 \omega}} \left( e^{-i \omega t} a - e^{i\omega t}a^\dagger \right), \tag{2} \label{2}$$ the operator $$\phi(t)$$ satisfies the canonical equal-time commutation relation $$\left[\phi(t), \pi(t) \right] = i \mathbf{1} \tag{3} \label{3}$$ The associated Hamilton operator (after the subtraction of the zero-point energy) is given by $$H= \frac{1}{2}\left( \dot{\phi}(t)^2+\omega^2 \phi(t)^2-\omega\right)= \omega \, a^\dagger a. \tag{4} \label{4}$$ The ground state $$|0\rangle$$ of $$H$$ is chararacterized by $$a |0\rangle=0, \quad \langle 0 |0\rangle=1 \tag{5} \label{5}$$ and the other energy eigenstates $$|n \rangle$$ (excited states) are obtained by $$|n\rangle = \frac{(a^\dagger)^n}{\sqrt{n!}} |0\rangle, \quad H |n\rangle = n \omega |n\rangle.\tag{6} \label{6}$$ If you wish, you may introduce eigenstates $$|\varphi\rangle$$ of the elongation operator $$\phi(0)$$ (taken at the time $$t=0$$), defined by $$\phi(0) |\varphi\rangle = \varphi |\varphi \rangle , \quad \varphi \in \mathbb{R}, \quad\langle \varphi| \varphi^\prime \rangle =\delta(\varphi-\varphi^\prime), \quad \int\limits_{-\infty}^\infty \! \!d\varphi | \varphi \rangle \langle \varphi |=\mathbf{1}. \tag{7} \label{7}$$ In this basis, $$\phi(0)$$ acts as a multiplication operator and $$\pi(0)$$ as a differential operator on "wave functions" $$\psi(\varphi):=\langle \varphi |\psi\rangle$$, $$\langle\varphi | \phi(0) |\psi \rangle=\varphi \langle \varphi |\psi\rangle=\varphi \psi(\varphi), \quad \langle \varphi |\pi(0) |\psi \rangle=-i \frac{d}{d \varphi} \langle \varphi |\psi \rangle= -i \psi^\prime(\varphi). \tag{8} \label{8}$$ At this point, you should have realized that the quantum harmonic oscillator is nothing else than a (free) quantum field theory in $$0+1$$ space-time dimensions.

The generalization to a (free) quantum field theory in $$3+1$$ space-time dimension is straightforward with the replacements $$\phi(t) \to \phi(t, \vec{x})$$, $$a \to a(\vec{k})$$, et cetera.

• Well $\phi(0)$ is just proportional to $X$ and $\pi(0)$ to $P$, no? So $|\varphi\rangle$ is just a position ket. I like this answer, but you could explicitly point out that all of this is just the position and moment operator (using the Heisenberg picture) in disguise. Mar 4 at 19:03
• Also, let me emphasize that there are two ways to look at the harmonic oscillator: a) as in introductory quantum mechanics, where you consider a single-particle. The $n$-basis then gives the excited states (as you also said) and b) as a Fock space over a single bosonic mode, i.e. you can consider this system as a system of infinite/undefined number of particles, each of which can be only in one single-particle state (mode); and the $n$-states are just states with $n$ particles occupying this mode. Viewed in this second interpretation, your example is even nicer in this context. Mar 4 at 19:19
• @TobiasFünke I did not write $X$ on purpose to avoid any confusion between the operator and the space coordinates $\vec{x}$ in $d=1+3$. I also wanted to make clear that the operator $\phi$ can be associated with any generalized coordinate (not necessarily the position of a particle). For a similar reason, I did not write $P$ avoiding a possible confusion between the momentum conjugate to $\phi$ and the field momentum operator $\vec{P}=\int d^3x \pi \vec{\nabla} \phi$. Mar 4 at 20:55

The field operators do not represent particles, they are operators that, if acting on a state (a vector in fock space which represents a state with or without particles), can change the state of the system by changing the vector into a new one with a different number of particles. That is, we could say that the field operators creates or anhilate particles (different ones do different jobs).

Usually when you want to describe a state with particles, you use that property of the field operators, so you start with the vacuum state and act on it with as many field operators as you need in order to create your initial state with the number of particles that you want. Also, these field operators might change the number of particles during the time evolution of the system, as they also appear in the Hamiltonian.

Also, the fields, as any operator, can have an expectation value (which are observables only for real fields though). So you can calculate this value for a field operator, for any state in fock space. So you might have a state with a non zero value for the electric field in some region of space but no particles there (think of electrostatics), or calculate the number of particles in some region using the particle density operator, also a function of the field operators, but they are not the same thing.

Last, it might be useful for you to think about QFT by using the QFT of condensed matter systems. There the fields are a limit approximation to the behavior of a huge number of particles, it is just non relativistic quantum mechanics applied to a system of many particles. The field represents the overall background solid, only described as a field of excitations, which are the vibration modes of the solid as a whole. The effective description describes many things you find in QFT: particle like excitations (phonons), antiparticles and thus creation and annihilation of these phonon particles, and a few other things. The bad part of the analogy is that a fundamental field in QFT is likely not the result of excitations of some background invisible condensed matter system, for a few technical resons that are beyond the scope of this answer. But it has helped me in developing intuitions and understand the math better when QFT seems confusing.

The two sentences :

Excitations of the quantum fields are particles

and

The quantum fields acts on the multi-particles states in the Fock space

both encapsulate the same idea because excitations of a field are states of this field.

For a free (real scalar) field, this is the equation : $$\phi(x) = \int \frac{\text d^3 \vec p}{(2 \pi)^{3}\sqrt{2E(\vec p)}} (a_{\vec p}^\dagger e^{-ip\cdot x} + a_{\vec p} e^{ip\cdot x})\tag 1$$ and the Hamiltonian : $$H_0 = \frac{1}{2}\int\text d\vec x : \dot\phi^2 + \nabla \phi^2 + m^2\phi^2: {}=\int\frac{\text d^3\vec p}{(2\pi)^3} E(\vec p)a^\dagger_{\vec p}a_{\vec p}$$ You can read this equation both ways :

• starting from the field, you take its Fourier transform, decomposing it into its eigenmodes, and see that the Hamiltonian is just an infinite collection of harmonic oscillators, one for each value of the spatial momentum. The same construction as in ordinary quantum mechanics produces a family of creation/annihilation operators. They create/annihilate excitations of the field in the corresponding mode. These states span the Fock space of a massive scalar boson.
• starting from the Fock space, it defines an operator(-valued distribution). This operator is a covariant local scalar field, which acts on the multi-particle states we started with. Just as before, those states are the same as the excitations of the field.

One important consequence of $$(1)$$ which will stay true in an interacting theory is : $$\langle \vec p|\phi(x)|0\rangle = \frac{Z}{\sqrt{2E(\vec p)}}e^{-ip\cdot x}$$ (The normalization constant $$Z$$ was equal to $$1$$ for the free theory.)This means that the field operator, acting on the vacuum state, will create one particle states.

NB :

• If the fact that the quantum field is a distribution which depends on space troubles you, the rigorous way to do it is to smear it with a test function : take $$f(x)$$ a complex-valued smooth rapidly decreasing function and consider $$\int f(x)\phi(x) \text dx$$ where $$\phi(x)$$ is defined as before. It can be shown that this rigorously defines an operator on the Fock space.
• What about the wave function ? Any state in the Fock space can be written as a superposition of $$n$$-particle states : $$|\psi\rangle = \sum_{n=0}^{+\infty} \int \text d\vec p_1\ldots \text d\vec p_n \psi_n(\vec p_1,\ldots,\vec p_n)|\vec p_1,\ldots,\vec p_n\rangle$$ so that (up to normalization) $$\langle \vec p_1,\ldots,\vec p_n|\psi\rangle = \psi_n(\vec p_1,\ldots,\vec p_n)$$ is the $$n$$-particle wavefunction of the $$n$$-particle part of $$|\psi\rangle$$. There is also a way to write $$|\psi\rangle$$ as a superposition of states which have a prescribed configuration of the field $$\phi(x)$$, schematically $$|\psi\rangle = \int \mathcal D\varphi \Psi[\varphi]|\varphi\rangle$$ where $$|\varphi\rangle$$ is the state stuch that $$\phi(\vec x,t=0) |\varphi\rangle = \varphi(\vec x)|\varphi\rangle$$ for all $$\vec x$$ and $$\Psi[\varphi]$$ is the so-called wave-functional.

In both quantum mechanics (QM) and quantum field theory (QFT) Heisenberg picture observables give us information about the evolution of measurable quantities. Observables don't directly represent the results of a measurement. Rather the eigenvalues of the observables represent possible measurement results and an observable can be used with the state to get the expectation value of the physical quantity described by the observable. The Heisenberg picture state represents records of the values of observables in the past. The Schrodinger picture munges those records with the evolution of physical quantities and it can and does lead to confusion about what is happening in experiments even in QM:

https://arxiv.org/abs/quant-ph/9906007

https://arxiv.org/abs/1109.6223