A noncommutative de Finetti theorem: Invariance under quantum permutations is equivalent to freeness with amalgamation

Introduction

The de Finetti theorem states that an infinite family of random variables whose distribution is invariant under finite permutations (such a family is called exchangeable) is independent and identically distributed with respect to the conditional expectation onto the tail algebra of the random variables. Since the implication in the other direction is fairly elementary one has the equivalence between exchangeability and conditional independence. See, e.g., [Kal] for an exposition on the classical de Finetti theorem.

In a noncommutative context classical random variables are replaced by, typically noncommuting, operators on Hilbert spaces. The expectation with respect to a probability measure is then replaced by a state on the algebra generated by these operators. The notion of exchangeability makes of course also sense in such a context, as invariance of mixed moments under permutations of the random variables, and one can ask what exchangeability implies in such a more general context. It turns out that in the noncommutative world there are actually many quite different possibilities for exchangeable random variables. It was shown in [Koe1] that they all possess some kind of factorization property; but, as one sees from the variety of examples, one cannot expect that exchangeability implies some fixed kind of independence. Indeed, both independence and freeness provide basic examples for exchangeable random variables. (See also [Leh, Koe2] for more on this.)

However, if one moves into the noncommutative realm, one should also take into account that invariance under permutations is a commutative concept and should be replaced by its noncommutative analogue. To provide such noncommutative analogues of actions of groups was one of the motivations for the creation of the theory of quantum groups, which has been developed very extensively within the last 20 years or so. In particular, Wang introduced in [Wan] the noncommutative analogue of the permutation group $S_{n}$, namely the quantum permutation group $A_{s}(n)$. So if one considers noncommuting random variables, it is natural to replace the requirement of invariance under permutations by the stronger requirement of invariance under quantum permutations. Classical (commuting) independent random variables do not satisfy this stronger form of exchangeability any more and, as we will show in our main theorem, this noncommutative version of exchangeability singles out again a very special situation - namely freeness with amalgamation. In the same way as classical exchangeability is equivalent to conditional independence, quantum exchangeability is equivalent to freeness with amalgamation.

Thus our noncommutative de Finetti theorem is another instance of the general philosophy that freeness plays in the noncommutative world the same role as independence plays in the commutative world. Note that freeness is not a hidden assumption in our de Finetti theorem, but it is a consequence of replacing the commutative permutation group by its noncommutative counterpart.

Here is the statement of our noncommutative de Finetti theorem. All relevant notions will be defined in Sections 2 and 4.

Our paper is organized as follows. In the next section we collect the preliminaries. On one side, we present the definition of the quantum permutation group and the notion of invariance under quantum permutations. On the other side, we recall the basic definitions and relevant results about free independence with amalgamation. In Section 3, we will prove the “easy” implication of our de Finetti theorem, namely that freeness with amalgamation implies invariance under quantum permutations. This is actually not as elementary as in the classical case (where it follows directly from the fact that independence is a rule for expressing mixed moments in terms of moments of the single random variables) and we will have to use some of the basic theory of freeness for this proof. In Section 4, we will define the tail algebra of our sequence of random variables, and show some basic properties of the corresponding conditional expectation. Section 5 will finally give the proof of the other implication of our de Finetti theorem, Theorem 1.1. The paper closes with an example which shows that, as in the classical case, one needs infinitely many random variables in our de Finetti theorem: quantum exchangeability of finitely many random variables does not necessarily imply freeness with amalgamation.

Preliminaries

Here we recall the basic notions of non-commutative probability spaces and distributions of random variables; this is just to have a convenient language for our main statements.

1) A noncommutative probability space $(\mathcal{A},\varphi)$ consists of a unital algebra $\mathcal{A}$ and a unital linear functional $\varphi$. 2) A $W^{*}$-probability space $(\mathcal{A},\varphi)$ is a von Neumann algebra $\mathcal{A}$ together with a faithful normal state $\varphi$ on $\mathcal{A}$.

Note that for a $W^{*}$-probability space we do not require that our state $\varphi$ is a trace.

2. Quantum Permutation Group

Wang introduced in [Wan] the following noncommutative version of the permutation group $S_{n}$.

The quantum permutation group $A_{s}(n)$ is defined as the universal unital $C^{*}$-algebra generated by elements $u_{ij}$ ($i,j=1,\dots,n$) such that we have

each $u_{ij}$ is an orthogonal projection: $u_{ij}^{*}=u_{ij}=u_{ij}^{2}$ for all $i,j=1,\dots,n$

the elements in each row and column of $u=(u_{ij})_{i,j=1}^{n}$ form a partition of unity, i.e., are orthogonal and sum up to 1: for each $i=1,\dots,n$ and $k\not=l$ we have

Note that the above requirements imply in particular that the matrix $u=(u_{ij})_{i,j=1}^{n}$ is orthogonal, i.e., for each $i,j=1,\dots,n$ we have

$A_{s}(n)$ is a compact quantum group in the sense of Woronowicz [Wor]. That this is the right noncommutative version of the permutation group can be seen from the fact that adding commutativity of the $u_{ij}$ to the above definition yields the group algebra of the permutation group and that, by a theorem of Wang [Wan], $A_{s}(n)$ is the biggest Hopf algebra coacting on a space of $n$ points. For more information on $A_{s}(n)$, see [BC, BBC].

where $q_{1}$ and $q_{2}$ are arbitrary projections. If we take them non-commuting, then the $C^{*}$-algebra generated by $q_{1}$ and $q_{2}$, which is a quotient of $A_{s}(4)$, is infinite dimensional.

To say it in other words, invariance under quantum permutations asks for the validity of (1) for any matrix $u=(u_{i,j})_{i,j=1}^{k}$ whose entries are bounded operators on some Hilbert space and satisfy the defining relations of $A_{s}(k)$ from Definition 2.3. Note that we do not apply a state on the elements from $A_{s}(k)$ to get equality in (1), but ask for an algebraic identity in $A_{s}(k)$.

For a permutation $\sigma\in S_{k}$ the permutation matrix $(e_{ij})_{i,j=1}^{k}$ with $e_{ij}=\delta_{\sigma(i)j}$ provides an example of such a $u$, in this case (1) gives

3. Freeness with Amalgamation

Here we collect the basic definitions and needed facts about freeness. For general introductions on free probability theory, see [VDN, NS, HP]. In the classical de Finetti theorem we do not get ordinary independence of the random variables, but have to condition this over the tail algebra. In the same spirit, in our noncommutative de Finetti theorem, we cannot hope for ordinary freeness with respect to the state $\varphi$, but must expect that we have to condition this with respect to the tail algebra of the random variables. Voiculescu introduced such a conditional version of freeness (called operator-valued freeness or freeness with amalgamation) from the very beginning and developed its basic theory in [Voi]. In [Spe] this concept was treated from the combinatorial point of view and it was shown that the theory of free cumulants extends to the operator-valued frame. As our proof of the “easy” direction of theorem (1.1) relies on free cumulants, we will below recall the relevant facts about operator-valued free cumulants.

Let us first give the definition of an operator-valued probability space and freeness. This will be done in a general, algebraic context, as one implication of our de Finetti theorem does only require such general structure.

Recall that a conditional expectation $E:\mathcal{A}\to\mathcal{B}$ (for unital algebras $\mathcal{B}\subset\mathcal{A}$) is a linear map which satisfies $E[b]=b$ for all $b\in\mathcal{B}$ and the bimodule property

1) An operator-valued probability space $(\mathcal{A},E:\mathcal{A}\to\mathcal{B})$ consists of a unital algebra $\mathcal{A}$, a unital subalgebra $\mathcal{B}\subset\mathcal{A}$ and a conditional expectation $E:\mathcal{A}\to\mathcal{B}$. Elements in $\mathcal{A}$ are called (operator-valued) random variables.

2) For a unital algebra $\mathcal{B}$ we denote by $\mathcal{B}\langle X\rangle$ the $\mathcal{B}$-valued polynomials in the formal variable $X$; these are linear combinations of elements of the form $b_{0}Xb_{1}X\cdots b_{n-1}Xb_{n}$ for all $n=0,1,2,\dots$ and all $b_{0},\dots,b_{n}\in\mathcal{B}$. (For $n=0$, this is just $b_{0}$.) Elements from $\mathcal{B}$ do not commute with $X$ (with the exception of $1\cdot X=X=X\cdot 1$). For $p\in\mathcal{B}\langle X\rangle$ and $a\in\mathcal{A}$ (for some algebra $\mathcal{A}$ which contains $\mathcal{B}$ as a subalgebra) we denote by $p(a)\in\mathcal{A}$ the element which one gets by replacing the variable $X$ by $a$.

4. Operator-valued free cumulants

The combinatorial theory of operator-valued freeness [Spe] relies on the notions of non-crossing partitions and free cumulants. We will now recall these notions.

1) A partition $\pi$ of a set $S$ is a decomposition $\pi=\{V_{1},\dots,V_{r}\}$ of $S$ into disjoint, non-empty subsets $V_{i}$. The elements $V_{i}$ are called the blocks of $\pi$. We denote the partitions of $S$ by $\mathcal{P}(S)$. In the case $S=\{1,\dots,n\}$, we just write $\mathcal{P}(n)$.

2) For $\pi,\sigma\in\mathcal{P}(n)$ we say that $\pi\leq\sigma$ if each block of $\pi$ is contained in a block of $\sigma$.

2) Let $S$ be an ordered set. A partition $\pi\in\mathcal{P}(S)$ is called non-crossing if there do not exist two different blocks $V,W$ of $\pi$ such that we have $s_{1}<t_{1}<s_{2}<t_{2}$ and $s_{1},s_{2}\in V$ and $t_{1},t_{2}\in W$. The set of non-crossing partitions of $S$ is denoted by $NC(S)$, or just $NC(n)$ in the case of $S=\{1,\dots,n\}$.

If one draws partitions by connecting elements belonging to the same block by half-circles below the numbers $1,\dots,n$, then the partition is non-crossing if and only if one does not get crossings between different blocks in such a drawing. Another characterization of a non-crossing partition is the following recursive description: $\pi\in\mathcal{P}(S)$ is non-crossing if at least one of the blocks of $\pi$, say $V$, is an interval (i.e., consists of consecutive numbers) and if $\pi\backslash V$ is a non-crossing partition of $S\backslash V$.

Let $(\mathcal{A},E:\mathcal{A}\to\mathcal{B})$ be an operator-valued probability space.

Otherwise, let $V=(i+1,\dots,i+r)$ be an interval of $\pi$. Then, for $a_{1},\dots,a_{n}\in\mathcal{A}$,

As illustration of this definition consider

Note that in the moment-cumulant formula (2) the right hand side is of the form $\kappa_{n}^{E}(a_{1},\dots,a_{n})$ plus products of lower order terms; thus this can indeed recursively be solved for the $\kappa_{n}^{E}$. There is a quite a lot one can say about the structure of the formulas for the $\kappa_{n}^{E}$, but we will not need this here and refer for more information on this to [NS, Spe]. Here are as examples just the first three cumulants:

The main result which we will use about free cumulants is that they characterize freeness via the property “vanishing of mixed cumulants”.

Let $(\mathcal{A},E:\mathcal{A}\to\mathcal{B})$ be an operator-valued probability space and consider, for some index set $I$, random variables $(a_{i})_{i\in I}$. Then the following are equivalent:

The random variables $(a_{i})_{i\in I}$ are free with respect to $E$.

We have the vanishing of mixed operator-valued free cumulants: For all $n\geq 2$, all $i(1),\dots,i(n)\in I$, and all $b_{1},\dots,b_{n-1}\in\mathcal{B}$ we have

whenever there are $1\leq k,l\leq n$ such that $i(k)\not=i(l)$.

If we transfer this characterization from the $\kappa_{n}^{E}$ to $\kappa_{\pi}^{E}$ then freeness of the $a_{i}$ implies that $\kappa_{\pi}^{E}[a_{i(1)},\dots,a_{i(n)}]$ can only be non-zero when all the $i$-indices belonging to the same block are equal. It will be convenient to have a notation at hand which encodes that information.

With this notation we have: if $(a_{i})_{i\in I}$ are free with respect to $E$, then $\kappa_{\pi}^{E}[a_{i(1)},\dots,a_{i(n)}]$ can only be non-zero for $\ker{\bf i}\geq\pi$. Note that $\ker{\bf i}$ is in general a possibly crossing partition.

Operator-valued free random variables are invariant under Quantum Permutations

We will now first prove the “easy” direction of our de Finetti theorem, namely that random variables which are free with respect to a conditional expectation $E$ are invariant under quantum permutations with respect to any $\varphi$ which is compatible with $E$. In contrast to the other direction this can be done in a purely algebraic frame, thus we will treat this implication in the context of an arbitrary non-commutative probability space. Note also that this implication does actually not require that our sequence is infinite. This will only be crucial for the other implication.

Fix $n,k$ and ${\bf i}=(i(1),\dots,i(n))$ with $1\leq i(1),\dots,i(n)\leq k$. We have

Now we note that because of the vanishing of mixed cumulants for free variables the term $\kappa_{\pi}^{E}[x_{j(1)},\dots,x_{j(n)}]$ is only non-vanishing if $\ker{\bf j}\geq\pi$, where ${\bf j}=(j(1),\dots,j(n))$. Furthermore, by the identical distribution with respect to $E$ of our random variables, for any ${\bf j}$ with $\ker{\bf j}\geq\pi$ the term $\kappa_{\pi}^{E}[x_{j(1)},\dots,x_{j(n)}]$ has the same value, which we denote by $\kappa_{\pi}^{E}$. Thus we can continue the above calculation as follows:

The sum over $j(1),\dots,j(n)$ with $\ker{\bf j}\geq\pi$ means that we sum for each block of $\pi$ independently over one $j$-variable. Since $\pi$ is non-crossing at least one of its blocks is an interval, i.e., of the form $\{p,p+1,p+2,\dots,p+s\}$ for some $1\leq p\leq p+s\leq k$. Then we have $j(p)=j(p+1)=\cdots=j(p+s)$, the sum over this variable is independent of the other sums; and it only involves

Because of the orthogonality of different elements in the same row of $u=(u_{ij})_{i,j=1}^{k}$, the term $u_{i(p)j}u_{i(p+1)j}\cdots u_{i(p+s)j}$ is zero for any $j$ unless $i(p)=i(p+1)=\dots=i(p+s)$. In the latter case, $u_{i(p)j}u_{i(p+1)j}\cdots u_{i(p+s)j}=u_{i(p)j}$ and the sum over $j$ just gives $1$. In this way we are left with the same problem as before but with the positions $p,p+1,\dots,p+s$ removed. For $\pi$ we have just removed one of its interval blocks. Since $\pi$ is non-crossing, we can now find another interval block in the new partition and repeat the above argument. In this way we can do all the summations over the blocks of $\pi$ in an inductive way. In each step the $i$-indices must agree on the considered block of $\pi$ to get a non-vanishing contribution. If they do then the summation over the $j$-index for this block gives 1. So we get in the end that

Thus, by recalling that $\kappa_{\pi}^{E}$ is equal to $\kappa_{\pi}^{E}[x_{i(1)},\dots,x_{i(n)}]$ for any $i$ with $\ker{\bf i}\geq\pi$, we have

Properties of the conditional expectation onto the tail algebra

In order to make the step from quantum exchangeability to freeness with amalgamation we need some more analytic structure.

where $vN(x_{k}\mid k\geq n)\subset\mathcal{A}$ is the von Neumann algebra generated by all $x_{k}$ with $k\geq n$.

It is easily seen that the linear map $Q$ enjoys

This shows that $Q(a)=a$ for all $a\in\mathcal{A}_{\text{tail}}$. Thus $Q$ is the conditional expectation of $\mathcal{A}$ onto $\mathcal{A}_{\text{tail}}$ with respect to $\varphi$, which we denote from now on by $E$. ∎

Our main goal will be to show that quantum exchangeability implies freeness with respect to this ($\varphi_{\infty}$-preserving) conditional expectation $E:\mathcal{A}_{\infty}\to\mathcal{A}_{\text{tail}}$. Note that, in the non-tracial case, we do not define $E$ on $\mathcal{A}$, but only on $\mathcal{A}_{\infty}\subset\mathcal{A}$. This is no problem, however, since all our statements on distribution and freeness with respect to E involve only elements from $\mathcal{A}_{\infty}$. So, in the present section and in Section 5, the conditional expectation $E$ will always be understood as introduced in Proposition 4.2. If the reader prefers, she may throughout assume that $\mathcal{A}$ is generated by the considered sequence of random variables, i.e., that $\mathcal{A}=\mathcal{A}_{\infty}$ and $\varphi=\varphi_{\infty}$.

Let us first check that quantum exchangeability with respect to $\varphi$ extends to the same property with respect to $E$.

More generally, for any $p_{1},\dots,p_{n}\in\mathcal{A}_{\text{tail}}\langle X\rangle$ we have

Fix $n$, $k$, $i(1),\dots,i(n)$. Because of $\varphi|_{\mathcal{A}_{\infty}}=\varphi_{\infty}=\varphi_{\infty}\circ E$, (3) will follow if we can show that

Equation (4) follows from (3) by multilinearity and by checking that we have compatibility of our formulas under multiplying two $x_{i}$ together and under inserting a factor $b\in\mathcal{A}_{\text{tail}}$. But this is clear from the relations of the $u_{ij}$; the first compatibility follows from

and the second one from $\sum_{j=1}^{k}u_{ij}=1$. ∎

It is clear that the same arguments work also for the case of exchangeability. Since we will use this version in the proof of Proposition 4.5, let us state it here explicitly for later use.

More generally, for any $p_{1},\dots,p_{n}\in\mathcal{A}_{\text{tail}}\langle X\rangle$ we have

In the next section we will show how the quantum exchangeability of $E$ will imply freeness with respect to $E$. For this we will need as an important ingredient the following factorization property of $E$. This is actually a consequence of the classical exchangeability property with respect to $\varphi$ and was shown in [Koe1] for more general situations. To establish this desired factorization property it is crucial to work with an infinite sequence of random variables (see also Remark 5.2). In order to make the present paper self-contained we provide the proof for this factorization in our case. For more details and generalizations one should see [Koe1]. A related finite version of that result was also considered in Lemma 2.6 of [AL].

whenever $i(l)$ is different from all the other $i(r)$ ($r\not=l$).

Note that exchangeability with respect to $\varphi$ does not imply a factorization property for $\varphi$, but only for the conditional expectation $E$. This is of course responsible for the fact that we get freeness with respect to $E$ and not with respect to $\varphi$ in our noncommutative de Finetti theorem.

The exchangeability of our sequence implies then that we can define on $L^{2}(\mathcal{A},\varphi)$ an isometric shift $\alpha$ given by

Restricted to $\mathcal{A}\subset L^{2}(\mathcal{A},\varphi)$, this shift maps $\mathcal{A}$ into itself and acts there as an endomorphism. Let us denote the fixed point algebra of this shift by

for all $a\in\mathcal{A}$ and hence $b=\alpha(b)$. Since this is true for any $b\in\mathcal{A}_{\text{tail}}$ we actually have that $\mathcal{A}_{\alpha}=\mathcal{A}_{\text{tail}}$. Thus Proposition 4.2 entails that the $\varphi$-preserving conditional expectation $E_{\alpha}$ from $\mathcal{A}$ onto $\mathcal{A}_{\alpha}$ exists and equals the conditional expectation $E$ onto the tail algebra.

We recall next that the mean ergodic theorem of von Neumann implies that we have for all all $a\in\mathcal{A}$

in the strong operator topology (see, e.g., [Koe1]).

Now let us consider the situation as in our proposition. By Proposition 4.4 we have exchangeability of $E$ according to (6); this means that we have in our situation

for any $i>N:=\max\{i(1),\dots,i(n)\}.$ But then we also have

converges by the mean ergodic theorem to $E[p_{l}(x_{N})]=E[p_{l}(x_{i(l)})]$ and thus we get

Note again for the convergence argument that multiplication on norm bounded sets is continuous in the strong operator topology. ∎

Invariance under quantum permutations implies freeness over the tail algebra

We will now provide the proof of the implication ‘$\ref{item:deFinetti-a}\implies\ref{item:deFinetti-b}$’ of Theorem 1.1. Throughout this section $E$ will denote the conditional expectation as introduced in Proposition 4.2.

Let us first address the identical distribution of the $x_{i}$ with respect to E. For this we actually need only the classical exchangeability.

This is just a special case of Proposition 4.4. ∎

Now we will address the freeness property. For this one needs, as in the classical case, an infinite sequence of random variables. One should, however, note that the only way in which this infinity enters is via the factorization property of Proposition 4.5 (which relied in the end on the mean ergodic theorem). If this factorization property is assumed then it is feasible that a more elaborated version of the following arguments is also applicable to finite sequences of random variables.

We will prove this (for fixed $n$ and $p_{1},\dots,p_{n}$) by induction over the number of blocks of $\ker{\bf i}$, starting from the biggest number and going down in steps of one. To get started consider the biggest number of blocks, which is $n$. Then all $i(1),\dots,i(n)$ are different and, by an iterated application of the factorization property, Proposition 4.5, we have

Now assume, for some $r$, we have proved that $E[p_{1}(x_{i(1)})\cdots p_{n}(x_{i(n)})]=0$ whenever $i(1)\not=i(2)\not=\dots\not=i(n)$ and $\ker{\bf i}$ has at least $r+1$ blocks. We want to show the same for the case that $\ker{\bf i}$ has $r$ blocks.

Let us first observe that ${\bf j}$’s where two neighboring indices are the same do not contribute; this follows from the fact that $u_{i(s)j(s)}u_{i(s+1)j(s+1)}=0$ if $j(s)=j(s+1)$ because $i(s)\not=i(s+1)$. Thus we only have to sum over ${\bf j}=(j(1),\dots,j(n)$ in the above sum for which $j(1)\not=j(2)\not=\dots\not=j(n)$. But for those our induction hypothesis applies and thus we see that in the summation over $\pi\in\mathcal{P}(n)$ we can restrict to $\pi$ which have at most $r$ blocks. Since $E[p_{1}(x_{i(1)})\cdots p_{n}(x_{i(n)})]$ is invariant under permutations, we can fix specific (different) $i$-values for the $r$ blocks of $\ker{\bf i}$; let us take $1,3,5,\dots,2r-1$ for them.

Let us now choose $k=2r$ and a specific $u=(u_{ij})_{i,j=1}^{2r}$, namely

where $q_{1},\dots,q_{r}$ are arbitrary projections. With this choice of $u$ we have that for a non-vanishing $u_{ij}$ the $j$-value determines the $i$-value (since we only have the odd numbers as possible $i$-values), i.e., we have $\ker{\bf j}\leq\ker{\bf i}$; thus in the sum (8) we can restrict to $\pi\leq\ker{\bf i}$. But since we also restricted to $\pi$ with at most $r$ blocks, we are just left with the one possibility $\pi=\ker{\bf i}$, i.e., with the above $u$ we can continue (8) as follows:

In the last step we have used the fact that, because of the identical distribution of the $x_{i}$’s with respect to $E$, the term $E[p_{1}(x_{j(1)})\cdots p_{n}(x_{j(n)})]$ depends only on $\ker{\bf j}$.

is different from 1, then this implies that $E[p_{1}(x_{i(1)})\cdots p_{n}(x_{i(n)})]$ has to vanish, and we are done.

Note that if $\ker{\bf i}$ is non-crossing then the sum (10) is actually equal to 1 for any $u$ satisfying the relations of the quantum permutation group. However, if $\ker{\bf i}$ is non-crossing then the condition $i(1)\not=i(2)\not=\dots\not=i(n)$ implies that $\ker{\bf i}$ must have at least one singleton, i.e., one i-index appears only once and then the factorization property (7) gives right away that $E[p_{1}(x_{i(1)})\cdots p_{n}(x_{i(n)})]=0$. Thus we can restrict to crossing $\ker{\bf i}$ when considering (10).

Note also that if all the $u_{ij}$ in (10) commute, then we will actually get 1 (independent of whether $\ker{\bf i}$ is crossing or non-crossing); this shows that invariance under usual permutations is (clearly) not strong enough to imply freeness. We have to invoke some real quantum permutations, i.e., we should choose the $q_{1},\dots,q_{r}$ in (9) as non-commuting. However, it suffices to take just two of them as non-commuting. Since $\ker{\bf i}$ is crossing we can choose two blocks which have a crossing. For those two blocks we choose some non-commuting projections $p$ and $q$, whereas for all the other blocks we choose their projections as 1. Then the sum (10) reduces to

for some $s\geq 2$. It is clear that this is not equal to 1 for generic projections $p$ and $q$. Actually it is fairly easy to see that these expressions are equal to 1 if and only if $p$ and $q$ commute.

This finishes the proof of the implication ‘$\ref{item:deFinetti-a}\implies\ref{item:deFinetti-b}$’ in Theorem 1.1.

As already mentioned before, our noncommutative de Finetti theorem is not true for a finite number of random variables. To infer freeness from quantum exchangeability one needs, as in the classical case, infinitely many variables. To prove that claim we will in the following present an example, which can be considered as an analogue to classical urn models without replacement.

Since $E[u_{11}]$ is selfadjoint this implies that $E[u_{11}]=0$, and thus also

However, $u_{11}=u_{11}u_{11}^{*}$ and $\psi$ is faithful, thus $u_{11}=0$, which is a contradiction.

Acknowledgement

We thank Franz Lehner for some helpful comments on an earlier version of the manuscript.

References