# Rigorous definition of the commutator $[a(k_1), a^\ast(k_2)]$ of creation and annihilation operators in boson quantum field models

In their lecture notes "Boson Quantum Field Models" (in "Mathematics of Contemporary Physics", R.Streater (ed.)), Glimm and Jaffe define an annihilation operator $a(k), k \in \mathbb{R}$ on a certain domain $\mathcal{D}$ in Fock space by $$(a(k)\theta)_n(k_1, \ldots, k_n) = (n+1)^{1/2}\theta_{n+1}(k, k_1, \ldots, k_n) .$$ (This is the equation between equations (4.4) and (4.5) in the paper.) They then make the point that the adjoint, $a^\ast(k)$, is not a well-defined operator on Fock space, although it is a quadratic form. However, in equation (4.9) on the next page, they state that $$[a(k_1), a^\ast(k_2)] = \delta(k_1 - k_2) ,$$ and claim that this commutator can be verified directly''. I have no trouble verifying the commutator directly at the level of formal manipulation, but I am having trouble understanding how to make the calculation rigorous, or even how to make rigorous sense of the term $a(k_1) a^\ast(k_2)$ in the commutator, if $a^\ast(k_2)$ is only a quadratic form. Is there a standard way to represent $a^\ast(k_2)$ as an operator on some larger space in order to make sense of the commutator? Failing that, is there some other rigorous interpretation of the commutator?

(Glimm and Jaffe do not give an explicit formula for $a^\ast(k)$, but it is easy to derive, and can be found in many other references: $$(a^\ast(k)\theta)_n(k_1, \ldots, k_n) = n^{-1/2} \sum_{l=1}^n \delta(k - k_l) \theta_{n-1} (k_1, \ldots, k_{l-1}, \hat{k_l}, k_{l+1}, \ldots k_n) ,$$ where $\hat{k_l}$ indicates that $k_l$ is omitted. I have not yet found a reference that explains a rigorous interpretation of the commutator, however.)

• I thought the normal order, $a^*(k_1)a^*(k_2)\cdots a(q_1)a(q_2)\cdots$, so all creation operators to the left of the annihilation operators, was a well-defined operator in Fock space, and the commutator is just a prescription of normal ordering. – Carlo Beenakker May 24 '17 at 21:54
• Perhaps the right way to go about this is to define $a$ a little more gently? It's a really an operator-valued distribution, so it's a little cruel to evaluate it on points. – user1504 May 25 '17 at 0:34
• It's not well-defined on the Fock space, because it is unbounded. Actually, you cannot write down an everywhere defined operator which is unbounded without using choice. But it is a well-defined operator on Fock states with finite particle number. The same way the commutator holds on this domain. You definitely want to smear $a$ with a Schwartz function to make things rigorous. But it's a standard exercise. – Marcel Bischoff May 25 '17 at 3:25
• Thank you for the replies. It is certainly true that one gets better-behaved operators by smearing with Schwartz functions. My issue is that Glimm and Jaffe define the smeared fields in terms of the pointwise fields, and then claim that the commutation relations for the smeared fields follow from those for the pointwise fields. I assume there is some subspace of distributions on which the necessary calculations become rigorous, but I am having trouble finding that space. – Kevin McLeod May 26 '17 at 2:04

The annihilation and creation operators can be properly defined only as maps from the so-called one particle (or phase) space $(V,\sigma)$ (a real vector space $V$ with a non-degenerate antisymmetric bilinear form $\sigma$) to the closed operators in some (regular) representation of the canonical (anti)commutation relations C*-algebra.
If $(V,\sigma)=(FH,\Im \langle \cdot,\cdot\rangle)$, where $F$ is the forgetful functor and $(H,\langle\cdot,\cdot\rangle)$ is a complex pre-Hilbert space with completion $\mathscr{H}$, then the creation and annihilation operators can be easily explicitly written in the Fock representation $(\Gamma_{s/a}(\mathscr{H}),\pi_F,\Omega_F)$. Let me write their form in the case of $H=\mathscr{H}=L^2(\mathbb{R}^d)$ and symmetric Fock space (they can be defined analogously in the general case): $\forall f\in L^2(\mathbb{R}^d)$, $\forall \Psi\in C_0^{\infty}(d\Gamma(1))$ \begin{align} (a(f)\Psi)_n(x_1,\dotsc,x_n)&= \sqrt{n+1}\int_{\mathbb{R}^d}\bar{f}(x)\Psi_{n+1}(x,x_1,\dotsc,x_n)dx\\ (a^*(f)\Psi)_n(x_1,\dotsc,x_n)&=\frac{1}{\sqrt{n}}\sum_{j=1}^n f(x_j)\Psi_{n-1}(x_1,\dotsc,\hat{x}_j,\dotsc,x_n)\; . \end{align} Here $C_0^{\infty}(d\Gamma(1))$ is the subspace of the Fock space of finite particle vectors. The operators defined above are unbounded, densely defined and closable, and their closures are one adjoint of the other. They satisfy, in addition the canonical commutation relations (when applied on the finite particle vectors subspace) $$[a(f),a^*(g)]=\int_{\mathbb{R}^d}\bar{f}(x)g(x)dx\; .$$
The above definitions are all rigorous, and in fact it is possible to define creation and annihilation operators even for non-Fock representations of the canonical commutation relations, and if $(V,\sigma)$ is not originated by a pre-Hilbert space. However even if $V=F\mathscr{H}$, in any representation that is not normal with respect to the Fock representation, the creation and annihilation operators fail to have a common dense domain of definition for all $f\in\mathscr{H}$, and such a common domain can be defined only when $f$ ranges in finite dimensional subspaces of $\mathscr{H}$.
After this somewhat lengthy preamble, let me discuss the so-called creation and annihilation operator-valued distributions that you denote by $a^{\#}(k)$. I will denote them by $a^{\#}(x)$, following the notation above. These objects are very convenient to make formal manipulations, and are frequently used, especially by physicists. One has however to keep in mind that these objects are only formal, and it is not possible to define them in a completely rigorous fashion neither as operator-valued distributions (even if they are so-called). Of course many of the manipulations that physicists do with them can be rigorously defined. For example, $$\int_{\mathbb{R}^d} a^*(x) P(x,i\partial_x) a(x) dx\; ,$$ with $P(x,i\partial_x)$ a self-adjoint pseudodifferential operator in $L^2(\mathbb{R}^d)$ is rigorously defined by means of the Wick quantization - as the Wick quantization of the (homogeneous) polynomial $p(\cdot)=\langle \cdot , P(x,i\partial_x)\cdot\rangle_2$ - or equivalently applying the second quantization map to $P(x,i\partial_x)$: $$\int_{\mathbb{R}^d} a^*(x) P(x,i\partial_x) a(x) dx= p(\cdot)^{\mathrm{Wick}}=d\Gamma(P(x,i\partial_x)).$$
I have the book right in front of me and from a quick glance at the relevant section it seems that GJ chose a somewhat non-standard presentation. I think the most commonly used approach is the one explained by yuggib or suggested by user1504 which involves smearing and understanding everything in some suitable "sense of distributions". However, as you rightly noticed, GJ are really talking about statements which are pointwise in the momentum variables. Your (formal) formula (I know that sounds bad) for $a^*(k)$ which contains a delta function immediately shows that there is a problem with the interpretation as an unbounded operator on $\mathscr{F}$. GJ are well aware of that and mention on page 96 that the domain of this "operator" would have to be $\{0\}$. Then, they (adroitly!) pirouette this by saying that they do not define $a^*(k)$, for fixed $k$, as an operator but rather as a bilinear form $\mathscr{D}\times\mathscr{D}\rightarrow\mathbb{C}$. For $\theta_1,\theta_2\in \mathscr{D}$, the definition of this bilinear form $$\langle \theta_1, a^*(k)\theta_2\rangle$$ is: compute à la physicist using the "formal formula" and then declare the outcome (which has no delta function) to be the definition. Similarly, $a^*(k)a(k')$ is not the composition of two linear maps (with suitable domains) but a bilinear form defined by the same recipe.