Proof that QFT ladder operators produce a complete set of Eigenvectors

Question

Consider a free scalar field theory in Quantum Field Theory, which has a Lagrangian given by: $$ \mathcal{L} = \frac{1}{2}(\partial_{\mu}\phi)^2 - \frac{1}{2}m^2\phi^2. $$

The typical textbook then proceeds to write the Fourier transform of $\phi$: $$ \phi(x) = \int \frac{d^3\mathbf{p}}{(2\pi)^{3/2}} \frac{1}{\sqrt{2E_{\mathbf{p}}}} \left( a(\mathbf{p}) e^{ip \cdot x} + a^\dagger(\mathbf{p}) e^{-ip \cdot x} \right). $$

Imposing the commutation relations gives the following commutation relation for the ladder operators: $$ [a(\mathbf{p}), a^\dagger(\mathbf{p}')] = (2\pi)^3 \delta^{(3)}(\mathbf{p} - \mathbf{p}'). $$

Finally, with the observation that the (normal ordered) Hamiltonian can be written in terms of the ladder operators: $$ H = \int \frac{d^3\mathbf{p}}{(2\pi)^{3/2}} E_{\mathbf{p}} \left( a^\dagger(\mathbf{p}) a(\mathbf{p}) \right). $$

Textbooks typically then state that the all energy eigenvectors are given by applying an arbitrary amount of ladder operators to the vacuum. For example, kets are created by applying creation operators for arbitrary $n$, such as: $$ |\mathbf{p}_1, \mathbf{p}_2, \dots, \mathbf{p}_n \rangle = a^\dagger(\mathbf{p}_1) a^\dagger(\mathbf{p}_2) \dots a^\dagger(\mathbf{p}_n) |0\rangle. $$ which obviously is an eigenvector of the Hamiltonian.

My question is: How to prove that all these eigenvectors, that can be created by applying arbitrary numbers and arbitrary combinations of raising operators to the vacuum, form a complete basis, that spans the whole space of the theory?

Edit for further clarification: To phrase things another way, the space of the theory is, by-constructed, a Fock space defined as the sum of the n-particle Hilbert spaces, for all $n \ge 0$, but how do we prove that such a space is the most general space we can possibly create for a free scalar field theory in QFT?

Answer 1

One can show that the usual Fock space constructed from the one-particle space of states $a^\dagger(f)\lvert 0\rangle$ where $a^\dagger(f) = \int a^\dagger(\vec p)f(\vec p)\mathrm{d}\Omega_p$ ($\mathrm{d}\Omega_p$ is shorthand for the usual Lorentz-invariant measure) creates states with momentum wavefunction $f(\vec p)$ is an irreducible representation of the CCR algebra, and hence the action of neither $H$ nor the field $\phi$ or any other observable built out of the c/a operators can lead outside of it (otherwise it would not be an irrep). In this sense it is "the full space" - if you start with any state within it, i.e. either the vacuum or any state containing at most finitely many particle, you will remain within it for all eternity, it is hence the appropriate setting for doing physics. If you just tried to "extend" this space, you'd just extend it be regions you can never visit. The closedness under the action of the free $H$ is not difficult to see, since it is the number operator and hence leaves each n-particle space invariant on its own.

Note that in contrast to the question we are careful here to specify the states being created by $a(f)$, not the distribution $a(p)$. This just mirrors the usual subtlety that the "eigenvectors" of operators with continuous spectrum like momentum cannot lie inside the Hilbert space. In fact, one may define first $a(f)$ by its creation/annihilation action on the space of momentum wavefunctions, then infer the existence of the distribution $a(p)$ after the fact.

What remains is the question whether there is an appropriate different setting that is inequivalent to this, i.e. is there a second irrep of the CCR that justifiably is that of a free field in some sense yet is not unitarily equivalent to the standard Fock representation above/from the question.

The answer for the free field is no, this is the Jost-Schroer theorem (cf. thm. 4-15 in Streater and Wightman): Any field $\phi$ that has as its two-point correlation function the usual propagator of the free field shares all its n-point functions with the free field, and is hence unitarily equivalent to the standard presentation of the free field (i.e. the standard Fock construction from above/the question) by the Wightman reconstruction theorem.

Answer 2

First there is the annoying little issue of what constitutes a valid state in the Hilbert space. They are required to have a finite norm ($L^2$ norm). Therefore, the eigenvectors of the Hamiltonian produced by applying creation operators $\hat{a}^{\dagger}(\mathbf{p})$ with various $\mathbf{p}$'s do not belong to the Hilbert space. The question whether these eigenvectors provide a complete basis for the Hilbert space it therefore ill-posed.

There are two ways around this problem. One does not need to use the eigenvectors of the Hamiltonian to define a basis for the Hilbert space. Instead, one can define creation operators in terms of normalizable parameter functions $$ \hat{a}_F = \int \hat{a}^{\dagger}(\mathbf{p}) F(\mathbf{p}) \text{d}\mathbf{p} , $$ where $F$ is a normalized complex function. One can now define a complete basis of such parameter functions and then produce Fock states in the usual way with these creation operators.

The other way is to treat the situation with a rigged Hilbert space, where you define your states in terms of spectra $$ |\psi\rangle = \int \psi (\mathbf{p}_1,\mathbf{p}_2,...,\mathbf{p}_n) \hat{a}^{\dagger}(\mathbf{p}_1) \hat{a}^{\dagger}(\mathbf{p}_2) ... \hat{a}^{\dagger}(\mathbf{p}_n) |\text{vac}\rangle\ \text{d}\mathbf{p}_1\ \text{d}\mathbf{p}_2 ...\text{d}\mathbf{p}_n , $$ where the spectra $\psi (\mathbf{p}_1,\mathbf{p}_2,...,\mathbf{p}_n)$ are normalized.

The question of the completeness of these bases now depends on what one considers to be the complete set of all states of a free scalar field theory. If one can provide a formal definition of such a space then one can investigate the completeness of the basis with the aid of such a definition. Therefore, the space of all such states is often defined by construction in terms of the basis. The argument is that all the states are those with a finite number of quanta (particles) having arbitrary spatiotemporal degrees of freedom. It is not too difficult to address the spatiotemporal degrees of freedom, because that is just given by the functions of the $\mathbf{p}$ which we can define in terms of complete sets of functions (perhaps easier in the first approach). To deal with the particle number degrees of freedom, one needs to build them up progressively from 1 to $n<\infty$ as tensor products of Fock states (as defined above) or ladder operators. So, in the end we have a complete set, based on what we assumed to be the set of all states.