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In Folland's book Quantum Field Theory, he says

...we start out with classical field equations and a relativistically invariant Lagrangian from which they are derived, then replace the classical field variables by quantum fields, which are Fourier integrals of creation and annihilation operators that satisfy suitable commutation or anticommutation relations.

I must have missed it while reading the text, but how do we know that all quantum fields

  1. Involve only creation and annihilation operators and not any other kind of operator.
  2. Are Fourier integrals of creation and annihilation operators. Where does this come from? (Why do we take the Fourier transform?)
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    $\begingroup$ How do we know anything in physics? We test our models against the phenomena. If the model didn't match reality you've probably never have heard of it. $\endgroup$
    – John Doty
    Dec 23, 2022 at 12:58

3 Answers 3

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That the Fourier coefficients of the field are creation and annihilation operators is a consequence of the commutation (or anticommutation for fermion fields) relations imposed on the quantum field. For a free scalar field with Lagrange density $$\mathcal{L}=\frac{1}{2}(\partial^{\mu}\phi)(\partial_{\mu}\phi)-\frac{m^{2}}{2}\phi^{2},$$ the quantized degrees of freedom are the values of the field at each spatial point, $\phi(\mathbf{x})$. The continuum version of the canonical quantization involves imposing an equal-time commutator between each $\phi(\mathbf{x})$ and its canonical conjugate $$\pi(\mathbf{x})=\frac{\delta\mathcal{L}}{\delta\dot{\phi}(\mathbf{x})}=\dot{\phi}(\mathbf{x}),$$ which takes the form $$[\phi(\mathbf{x}),\pi(\mathbf{y})]=i\delta^{3}(\mathbf{x}-\mathbf{y}).$$ This is the continuum version of the usual relation $[x_{i},p_{j}]=i\hbar\delta_{ij}$ (setting, of course, $\hbar=c=1$). The dynamical variable $\phi(\mathbf{x})$ at each point has been promoted to an operator, making $\phi$ itself an operator-valued field.

When you expand the field in Fourier modes, since $\phi$ is now an operator-valued field, the Fourier coefficients are also operators. (We expect the Fourier expansion to be useful, because we know that for free field, energy and momentum will commute, and so the energy eigenstates can be taken to be states of fixed momentum.) Since $\phi$, prior to quantization, was a real field, the quantized field must be Hermitian, so that it only takes on real eigenvalues. The general Fourier expansion of a Hermitian operator can be written as $$\phi(\mathbf{x})=\int\frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2}(p^{2}+m^{2})^{1/4}}\left(a_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}}+a^{\dagger}_{\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}\right),$$ where, at this stage, the $a_{\mathbf{p}}$ are some unknown operators. However, we do know some important things about this operator. The field operator obeys the Klein-Gordon equation, which tells us that the similar expansion of $\pi$ (which is also an operator-valued field) must be $$\pi(\mathbf{x})=\int\frac{d^{3}p}{(2\pi)^{3}}\frac{(p^{2}+m^{2})^{1/4}}{i\sqrt{2}}\left(a_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}}-a^{\dagger}_{\mathbf{p}}e^{-i\mathbf{p}\cdot\mathbf{x}}\right).$$ The other thing we know is the equal-time commutation relation for $\phi$ and $\pi$, and it turns out that in order to satisfy $[\phi(\mathbf{x}),\pi(\mathbf{y})]=i\delta^{3}(\mathbf{x}-\mathbf{y})$, the Fourier mode coefficient operators $a_{\mathbf{p}}$ have to satisfy $$[a_{\mathbf{p}},a^{\dagger}_{\mathbf{q}}]=(2\pi)^{3}\delta^{3}(\mathbf{p}-\mathbf{q})\\ [a_{\mathbf{p}},a_{\mathbf{q}}]=0\\ [a^{\dagger}_{\mathbf{p}},a^{\dagger}_{\mathbf{q}}]=0.$$ And these are precisely the (continuum normalized) commutation relations for creation and annihilation operators. (If you have never worked this through before, it is a very useful exercise.)

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    $\begingroup$ This only holds for free fields though. So the answer to the question in OP is 'not all wisdom $\endgroup$
    – Jojo
    Dec 23, 2022 at 16:59
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    $\begingroup$ Which quantities are time dependent in the Fourier expansion of $\phi(\mathbf x)$? Are all the boldface vectors 3-vectors? I could probably find this in literature easily but it might be relevant for other people as well. $\endgroup$ Dec 24, 2022 at 13:35
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I would like to add to Buzz excellent answer that the development of a quantum field operator in annihilation and creation operators as mentioned in Buzz's answer is only valid for free quantum fields, i.e. fields that do not couple to another quantum field. If a quantum field couple to another field, so that the field equation ("wave equation") is non-homogeneous or non-linear the development is no longer true. In almost all of such cases the development of the field operator is even unknown.

In this context it is important to know that most quantum fields couple to other fields and free quantum fields only exist as toy models in textbooks. The problem how to deal with interacting quantum fields requires more advanced tools as for instance perturbation theory (where in zeroth order the mentioned Fourier development can be used) or non-perturbative QFT.

EDIT

Inspired by the comment of user196574 I would like to add that in case of "most well-known" interacting QFT's inspite of the statement in the first paragraph one can actually still construct an operator $a^\dagger(\mathbf{k})$ (I am following closely and using the formula tagging of the book of M. Srednicki):

$$a^\dagger (\mathbf{k}) = -i \int d^3x e^{ikx} \stackrel{\leftrightarrow}{\partial_0} \varphi(x) \tag{5.2} $$

where $f\stackrel{\leftrightarrow}{\partial_\mu}g = f(\partial_\mu g) - (\partial_\mu f)g$. This construction here is done for an interacting scalar field theory, for other well-known fields as spin-1/2 and spin-1 theories similar expressions apply. However, there might be interacting theories (in particular with strong coupling) where even this construction makes no sense at all (at least at low energy). An example could be QCD.

Although the corresponding operator $a(\mathbf{k})$ still "annihilates the vacuum":

$$a(\mathbf{k})|0\rangle =0 \tag{5.3}$$

$a^\dagger(\mathbf{k})$ no longer only generates an one-particle state. As an a priori unknown vector $a^\dagger(\mathbf{k})|0 \rangle$ in Fock space it can be developed in the following way:

$$a^\dagger(\mathbf{k})|0 \rangle = z|0 \rangle + Z|\mathbf{k}\rangle + \sum_{n,p} b_{n,p} |p,n\rangle $$

where $|p,n\rangle $ denotes any multiparticle state with total three-momentum $\mathbf{p}$ and $n$ is short for all other labels needed to specify the state (Srednicki p.40).

The coefficients $z$, $Z$ and $b_{n,p}$ are difficult to determine, actually $Z$ can be computed with methods of perturbation and in particular renormalisation theory.

To recap: annihilation and creation operators are widely used in particular for free, i.e. non-interacting QFTs, but for interacting QFTs they loose most of their sense, so one should not call them anymore annihilation and creation operators. But they can be still used in many QFTs, but with the corresponding care. It is important to note that the construction (5.2) is actually based on the Fourier development of the field operators (see for instance (3.19),(3.20) and (3.21) of M. Srednicki's book).

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    $\begingroup$ Suppose I have an interacting field, like $\phi^4$ theory. Are the canonical commutation relations no longer satisfied? It seems to me that at worst they are satisfied up to a proportionality constant from renormalization $Z$ factors. Thus I can still write the field as a linear combination of creation and annihilation operators following the same steps in Buzz's answer. The only thing that changes with a non-linear field equation is that the eigenstates of the Hamiltonian are no longer number eigenstates; if I act a couple creation operators on the vacuum, I don't get an eigenstate of $H$. $\endgroup$
    – user196574
    Dec 23, 2022 at 20:00
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    $\begingroup$ @user1965674 After Christmas break I added a section in order to answer to your point. $\endgroup$ Dec 28, 2022 at 13:36
  • $\begingroup$ Thanks, that's nice to see. I realize I have a few wires crossed in regards to this, probably related to the construction of the Hilbert space in interacting theories. On the right side of $a^\dagger(\mathbf{k})|0 \rangle = z|0 \rangle + Z|\mathbf{k}\rangle + \sum_{n,p} b_{n,p} |p,n\rangle$ are eigenstates of the Hamiltonian $H$; we know the eigenstates of $H$ can be characterized as such because of commutation with translation operators. But I think you still have $a^\dagger(\mathbf{k})|0 \rangle' = |\mathbf{k}\rangle'$ where the $'$ refers to the non-interacting, non-physical states. $\endgroup$
    – user196574
    Dec 28, 2022 at 18:03
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They are really expansions in eigenmodes of field equations (such as, e g , Maxwell equations for electromagnetic fields). In vacuum these are plane waves, but when the quantum field theory is developed in a cavity (a typical situation in quantum optics), they are cavity modes, whereas the field theory for non-relativistic electrons is often developed in terms of the eigenstates of the Schrödinger equation in whatever is the confining potential. Creation/annihilation operators describe creation/annihilation of particles in respective quantum states, with the commutation relations determined by usual statistics for identical particles.

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