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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.

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    $\begingroup$ 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? $\endgroup$ Dec 9, 2022 at 7:50
  • $\begingroup$ @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. $\endgroup$ Dec 9, 2022 at 7:52
  • $\begingroup$ @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. $\endgroup$ Dec 9, 2022 at 18:21

1 Answer 1

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A partial answer:

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.

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

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