# Domain issues for products of bosonic creation and annihilation operators

Consider the bosonic Fock space $$F$$ and let $$a^\dagger$$ and $$a$$ denote the "usual" creation operators. As far as I know, these are defined on the dense domain $$\mathcal D(N^{1/2}):=\{ \psi \in F| \sum\limits_{n=0}^\infty n\, ||\psi_n||^2 < \infty\}$$, where $$N$$ denotes the number operator, cf. Ref. 1.

Question: What is the domain (such that all expressions are well-defined, they are the adjoint of each others and the CCR hold) of a string of length $$K\in \mathbb N$$ of creation and annihilation operators: $$a^\#(f_1) \, a^\#(f_2)\,\cdots\, a^\#(f_K)\quad , \tag{1}$$

where $$a^\#$$ denotes (in each case) either a creation or annihilation operator. I am mostly interested in the case where $$K$$ is even and the string takes the form

$$a^\dagger(f_1) \,a^\dagger(f_2)\, \cdots \,a^\dagger(f_{K/2})\,a(f_{K/2+1})\,a(f_{K/2+2})\cdots a(f_{K}) \quad . \tag{2}$$

My guess here is that the domain is $$\mathcal D(N^{K/2}):=\{ \psi \in F| \sum\limits_{n=0}^\infty n^{K}\, ||\psi_n||^2< \infty\}$$. If so, how to prove it? Are these domains dense?

Closely related to this is the following:

What is the (largest possible) domain such that for $$\psi\in F$$ it holds that

$$\langle \psi, a^\#(f_1) \,a^\#(f_2)\,\cdots\, a^\#(f_K)\,\psi\rangle_F\tag{3}$$

is well-defined?

Background:

In Ref. 1 it is stated that , although the creation and annihilation operators are unbounded, domain questions are "unproblematic", since one can take the domain of a sufficiently large power of the number operator.

In Ref. 2 a quantity is defined as $$\langle \psi,a^\dagger(f_1)\, a(f_2)\, \psi\rangle_F$$ for $$\psi \in F$$ with $$\langle \psi, N\psi\rangle_F <\infty$$. For me this condition means that we need that $$\psi \in \mathcal D(N)$$ such that $$N\psi \in F$$ and further that $$\psi\in \mathcal D(N^{1/2})$$ (which I think automatically follows). However, following this math stack exchange post, the form domain of an operator (which I think here is meant with the finite particle expectation value) can be larger than the operator domain. So the question remains: What is the (largest possible) domain $$\mathcal D$$ such that the above expression makes sense for $$\psi \in \mathcal D$$.

Maybe in our particular case it follows that the form domain is indeed $$\mathcal D(N^{1/2})$$ - or more generally for strings of length $$K$$ that the form domain is $$\mathcal D(N^{K/2})$$?

References:

1. Benedikter, Niels, Marcello Porta, and Benjamin Schlein. Effective evolution equations from quantum dynamics. Cham: Springer International Publishing, 2016. Chapter 3.

2. Solovej, J. P. Many body quantum mechanics. Lecture Notes 2007. chapter 8.

• If you have two unbounded operators $A:\mathcal{D}(A)\to\mathcal{H}$ and $B:\mathcal{D}(B)\to\mathcal{H}$ in some Hilbert space $\mathcal{H}$, then the product $AB$ is usually understood to be defined on the domain $\mathcal{D}(AB):=\{\psi\in\mathcal{D}(B)\mid B\psi\in\mathcal{D}(A)\}$, if not explicitly stated otherwise. Maybe its the same in this case? Do you know what the image of the operators $a,a^{\dagger}$ under the domain $\mathcal{D}(N^{1/2})$ is? Dec 9, 2022 at 7:50
• @G.Blaickner Thank you for your comment. Yes, you're right and I have tried to work it out, but I am not sure at the moment. This is where my guess came from, actually. But I could not find any reference regarding this. Dec 9, 2022 at 7:52
• @G.Blaickner I've added a (partial) answer. If you have time I'd appreciate if you take a look and let me know what you think. Dec 9, 2022 at 18:21

Let $$\psi \in \mathcal D(N)$$; in particular, it follows that $$\psi \in \mathcal D(N^{1/2})$$. We want to show that these conditions are sufficient such that e.g. $$a(f)\psi \in \mathcal D(N^{1/2})$$. Then the product $$a^\dagger(g) a(f) \psi$$ would be well-defined.

To do so, I think we can use the fact that for all $$\psi \in \mathcal D(N^{1/2})$$ it holds that$$^\ddagger$$

$$||a(f)\psi||_F \leq ||f||_{\mathfrak h}\, ||N^{1/2}\psi||_F \quad ,\tag{1}$$

where $$\mathfrak h$$ denotes the underlying one-particle space and $$f\in \mathfrak h$$. We proceed by noting that for $$M\in \mathbb N$$ we have \begin{align} ||f||^2_\mathfrak h \sum\limits_{n=1}^{M+1} n^2 ||\psi_n||^2_n = ||f||^2_\mathfrak h \sum\limits_{n=1}^{M+1} n \underbrace{||N^{1/2}\psi_n||^2_n}_{=||N^{1/2}\psi_n ||_F^2} &\geq \sum\limits_{n=1}^{M+1}n\underbrace{||a(f)\psi_n||^2_{F}}_{=||(a(f)\psi)_{n-1}||_{n-1}^2} \\ &=\sum\limits_{n=0}^M n||(a(f)\psi)_n||^2_n + \sum\limits_{n=0}^M ||(a(f)\psi)_n||_n^2 \quad , \end{align} where in some intermediate steps we have identified $$\psi_n$$ with a vector in $$F$$ in an obvious way. We thus find

$$||f||^2_\mathfrak h \sum\limits_{n=0}^{M+1} n^2 ||\psi_n||^2_n \geq \sum\limits_{n=0}^M n||(a(f)\psi)_n||^2_n \tag{2} \quad .$$

For $$\psi \in \mathcal D(N)$$, the LHS of $$(2)$$ converges for $$M\to \infty$$ and therefore the RHS converges too, which shows that indeed $$a(f)\psi \in \mathcal D(N^{1/2})$$.

I further think that the domains $$\mathcal D(N^{K/2})$$ for all $$K\in \mathbb N$$ are dense subspaces, since the dense subspace $$F_{\mathrm{fin}}\subset F$$, consisting of all vectors $$\psi=(\psi_n)_n$$ with only finitely many non-zero $$\psi_n$$, is a subspace of these domains, i.e. we have $$F_{\mathrm{fin}} \subset \mathcal D(N^{K/2}) \subset F$$ and the denseness of $$F_{\mathrm{fin}}$$ in turn implies the denseness of $$\mathcal D(N^{K/2})$$.

$$^\ddagger$$ It remains to show that $$(1)$$ holds in the said domain without using some result we want to prove. In Ref. 1. a proof is sketched, which seems plausible to me.

• I hopefully did not to any mistake in the derivation between $(1)$ and $(2)$... If someone spots a mistake, let me know. Dec 11, 2022 at 18:14