Given a Hilbert space, let us denote by $\mathcal{T}(\mathcal{H})$ the full Fock space $$ \mathcal{T}(\mathcal{H}) = \mathbb{C} \Omega \oplus \bigoplus_{n \geq 1} \mathcal{H}^{\otimes n}. $$ For $\xi \in \mathcal{H}$ we the so-called creation operator, $$ l(\xi): \mathcal{T}(\mathcal{H}) \rightarrow \mathcal{T}(\mathcal{H}): \begin{cases} \eta \mapsto \xi \otimes \eta \\ \Omega \mapsto \xi \end{cases}.$$ In any notes I can find about free probability theory, the operator $$ s(\xi) = \frac{l(\xi)+l(\xi)^\ast}{2},$$ which is the real part of $l(\xi)$, plays a somewhat central role. I know that the von Neumann algebra it generates is isomorphic to $L^\infty(\mathbb{R})$ and that free products of these von Neumann algebras thus give a model for the von Neumann algebra of the free groups $L(\mathbb{F}_n)$. I also know that the distribution of $s(\xi)$ w.r.t. to the vacuum state $x \mapsto \langle x \Omega, \Omega \rangle$ is the semi-circular law. Although I undererstand these facts, I don't have the intuition and I don't know what Voiculescu motivated to look at the operators $l(\xi)$ and $s(\xi)$ above. I'd like to have an answer containing the reason why we study these.

  • $\begingroup$ What is $\mathbb C \Omega$? $\endgroup$
    – lisyarus
    May 28, 2019 at 16:52
  • $\begingroup$ The subspace generated by the vacuum vector which is of norm $1$. $\Omega$ is just a symbol and it is a canonical unit vector in $\mathcal{T}(\mathcal{H})$. $\endgroup$
    – abcdef
    May 28, 2019 at 17:07
  • $\begingroup$ Ok. Guess this is sometimes denoted $\mathcal H^{\otimes 0}$. $\endgroup$
    – lisyarus
    May 28, 2019 at 17:35
  • $\begingroup$ One of the motivations is to proof that the $R$ transform, i.e., the cumulant generating formula is linear with respect addition of free operators. $\endgroup$
    – Phicar
    May 28, 2019 at 18:10
  • $\begingroup$ Indeed, once you have these operators all constructions are relatively natural. However I'm asking about the other direction. How did one come up with these operators? What is so natural about them that they are destined to play a central role in the theory? $\endgroup$
    – abcdef
    May 28, 2019 at 18:16

1 Answer 1


I do not know precisely what motivated Voiculescu, but I can give some reasons why one might consider this setup.

One way to think of the operators $l$ and $l^*$, is as a variation on the shift operators $R$ and $L$ on $\ell^2(\mathbb{N})$, where for a sequence $x = (x_1,x_2,\dots) \in \ell^2(\mathbb{N})$, $Rx = (0,x_1,x_2,\dots)$ and $Lx = (x_2,x_3,\dots)$. The operators $R$ and $L$ are canonical (anti-)examples in function analysis, and hence interesting in their own right (for instance $R$ is an injective linear map that is not surjective, an impossibility in a finite dimensional vector space). Given $\mathcal{H}$ and $\xi \in \mathcal{H}$, $\ell(\xi)$ is then the operator that 'shifts by $\xi$' on $\mathcal{T}(\mathcal{H})$, etcetera.

As pointed out in the comments, Fock spaces originate in physics, where they model many-body quantum systems. In this context, $\mathcal{H}$ is intepreted as the Hilbert space of single-particle quantum states and the vacuum state $\Omega$ represents an empty system. The operator $\ell(\xi)$ then has the interpretation of creating a particle in the state $\xi$, and $\ell^*(\xi)$ removes (or annihilates) a particle in this state. However, physicists impose relations on creation/annihilation operators coming from different vectors of $\mathcal{H}$ (modelling 'particle exchange'). Let $e_1,\dots,e_n$ be an ONB for $\mathcal{H}$ (I will assume that $\mathcal{H}$ is finite dimensional for simplicity), and write $\ell(e_i) = \ell_i$ for brevity. Then the two most common such relations are the 'canonical commutation relations' $\ell_i^* \ell_j - \ell_j \ell_i^* = \delta_{ij}$ and the 'canonical anti-commutation relations' $\ell_i^* \ell_j + \ell_j \ell_i^* = \delta_{ij}$ (these model 'Bosons' and 'Fermions' respectively). You can check that in our case, the operators satisfy instead $\ell_i^* \ell_j = \delta_{ij}$, which is the 'average' of the previous two relations.

Now for the operators $s_i = s(e_i)$.

Mathematically, the relation just derived implies that the $\ell_i$ are not normal: $\ell_i^*\ell_i = \mathrm{id}$, but $\ell_i \ell_i^* \neq \mathrm{id}$ since $\ell_i \ell_i^* \Omega = 0$. Hence, we do not have a priori access to the powerful tools of spectral theory. Additionally, the vacuum state is not faithful on the von Neumann algebra generated by the $\ell_i$, again since $\ell_i \ell_i^* \Omega = 0$, but one can check that $\ell_i \ell_i^* \neq 0$. Considering the self-adjoint part of $\ell_i$ will certainly fix the first problem, and one can check that it also fixes the second (the vacuum state even becomes a faithful trace when considered on the von Neumann algebra generated by the $s_i$, and $\Omega$ itself a cyclic and separating vector). By for instance solving the associated moment problem, one can then show that the measure on $\sigma(s_i)$ given by sandwiching the resolution of the identity induced by $s_i$ with the vacuum $\Omega$ is precisely the semi-circle law (as you pointed out). When considering joint moments of the $s_1,\dots,s_n$, one will notice that they can always be expressed in terms of moments of order 2 (Wick's/Isserlis' theorem), which is also true for an $n$-tuple of 'classical' random variables with a joint Gaussian distribution (hence the name free Gaussian functor). This observation allows for a simple proof that $s$ operators associated to orthogonal vectors in $\mathcal{H}$ are free.

Physically, in the cases of the canonical (anti-)commutation relations, it is true that the $s_i$ satisfy the same relations as the $\ell_i$. For the canonical anti-commutation relations, the $s_i$ are the famous Majorana Fermions. While it does not work for our case, it might still be motivation to consider the $s_i$.


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