Classical Warm-up

Let S_N be the classical permutation group on N symbols. The algebra of continuous functions C(S_N) on S_N is generated by indicator functions:

\displaystyle \mathbf{1}_{j\to i}(\sigma)=\delta_{i,\sigma(j)}.

This algebra is commutative, but we will use some non-commutative algebraic analogues. Let \widetilde{\pi}\in M_p(S_N) be the uniform probability distribution on S_N and by \pi the associated state on C(S_N):

\displaystyle\pi(f)=\sum_{t\in G}f(t)\,\widetilde{\pi}(\{t\})=\frac{1}{|S_N|}\sum_{t\in G} f(t)\qquad (f\in C(S_N))

When choosing a permutation at random what is the probability that it sends 1\to 1? Well, this will equal, in some notation we won’t fully explain here:

\displaystyle \mathbb{P}[\pi(1)=1]=\pi(\mathbf{1}_{1\to 1})=\frac{1}{N}.

Once this has been observed, there is a conditioning of the state \pi\mapsto \pi_{1\to 1}:

\displaystyle \pi_{1\to 1}(f)=\dfrac{\pi(\mathbf{1}_{1\to 1}f\mathbf{1}_{1\to 1} )}{\pi(\mathbf{1}_{1\to 1})}=N \pi(\mathbf{1}_{1\to 1}f\mathbf{1}_{1\to 1} )) \qquad (f\in C(S_N)).

The state was given by \pi but is now given by \pi_{1\to 1}. Now, after having observed this random permutation mapping 1\to 1, we can ask it another question.

Consider a subset S\subseteq \{3,\dots,N\} and define:

\displaystyle\mathbf{1}_{3\to S}:=\sum_{s\in S}\mathbf{1}_{3\to s}.

This is an observable, which basically asks of a permutation, do you map three into S? One for yes, zero for no. So let us ask this of \pi_{1\to 1}:

What is the probability that a random permutation maps three into S after having been observed mapping one to one?

We find, with a fairly elementary calculation that the answer to this question is:

\displaystyle \mathbb{P}[\pi_{1\to 1}(3)\in S]=\frac{|S|}{N-1}.

If \pi_{1\to 1}(3)\in S, we get further conditioning, to let us say \nu:

\displaystyle \nu(f)=\frac{N(N-1)}{|S|}\pi(\mathbf{1}_{1\to 1}\mathbf{1}_{3\to S}f\mathbf{1}_{3\to S}\mathbf{1}_{1\to 1})\qquad (f\in C(S_N))

What is the probability that after seeing \pi_{1\to 1}(3)\in S after \pi(1)=1 we see \nu(2)=2? We find it is:

\displaystyle \mathbb{P}[\nu(2)=2]=\frac{1}{N-2}.

Therefore, where \succ means after, and the state-conditioning implicit

\displaystyle \mathbb{P}[(\pi(2)=2)\succ (\pi(3)\in S)\succ(\pi(1)=1)]

\displaystyle =\frac{1}{N}\cdot \frac{|S|}{N-1}\cdot\frac{1}{N-2}=\frac{|S|}{N(N-1)(N-2)}

Now let us ask a ridiculous question. What was the probability that \nu(1)=2? But \nu is just a conditioning of \pi_{1\to 1}, which mapped one to one with probability one:

\displaystyle \mathbb{P}[\pi_{1\to 1}(1)=1]=\dfrac{\pi(\mathbf{1}_{1\to 1}\mathbf{1}_{1\to 1}\mathbf{1}_{1\to 1})}{\pi(\mathbf{1}_{1\to 1})}=1.

Of course the answer is zero. Can we actually see it in the above framework: well, yes, because the algebra of functions is commutative. You end up with evaluating a state at

\displaystyle \mathbf{1}_{1\to 1}\mathbf{1}_{3\to S}\mathbf{1}_{1\to 2}\mathbf{1}_{3\to S}\mathbf{1}_{1\to 1},

but commutativity sees \mathbf{1}_{1\to 1}\mathbf{1}_{1\to 2}… and no permutation maps one to one and one to two (you’d need a relation to do that), and so \mathbf{1}_{1\to 1}\mathbf{1}_{1\to 2}=0, so is the length five monomial above, and so is the probability we spoke about.

We don’t have to do state conditioning and multiplication to calculate the probability that a random permutation maps one to one, then maps three into S, then maps two to two though. Where |f|^2=f^*f:

\displaystyle \mathbb{P}[(\pi(2)=2)\succ (\pi(3)\in S)\succ(\pi(1)=1)]

=\pi(|\mathbf{1}_{2\to 2}\mathbf{1}_{3\to S}\mathbf{1}_{1\to 1}|^2)=\dfrac{|S|}{N(N-1)(N-2)}.

Let us remark that this probability is increasing in |S|\in\{0,1,\dots,N-2\}: the less we specify about the event, in this case represented by the projection \mathbf{1}_{3\to S}, the more general it is, the larger the probability.

Let’s Go Quantum

In the case of quantum permutations, so talking S_N^+ the ridiculous question of asking what is the probability that a quantum permutation maps one to two after it had previously mapped one to one… is no longer zero. Where q(N)=N(N-1)(N^2-4N+2), the probability of the analogue of a “random” quantum permutation, the Haar state h doing this is:

\displaystyle \mathbb{P}[(h(1)=2)\succ(h(3)=3)\succ (h(1=1))]=\frac{N-3}{q(N)}

Anyway, the quantum versions of the \mathbf{1}_{j\to i} are entries u_{ij}\in M_N(C(S_N^+)) of a universal magic unitary. We say:

\mathbb{P}[(h(2)=2)\succ (h(3)\in S)\succ (h(1)=1)]:=h(|u_{22}u_{S,3}u_{11}|^2),

where u_{S,3}=\sum_{s\in S}u_{s3}. We find this probability is:

\displaystyle \mathbb{P}[(h(2)=2)\succ (h(3)\in S)\succ (h(1)=1)]=\frac{|S|(N^2-5N+5+|S|)}{(N-2)q(N)}.

This is also increasing in |S|. Again, the less specificity about the event, the greater the probability.

This quantity is related to the classical probability above, and there are asymptotic similarities. Writing |S|=N-2-|S^c|, where S^c is the complement of S in \{3,4,\dots,N\}:

\displaystyle \mathbb{P}[(h(2)=2)\succ (h(3)\in S)\succ (h(1)=1)]

\displaystyle=\mathbb{P}[(\pi(2)=2)\succ (\pi(3)\in S)\succ(\pi(1)=1)]\cdot\left[1-\frac{|S^c|+1}{q(N)}\right]

I guess this also says, the larger |S|, the more similar the classical and quantum probabilities.

A twist

What if instead of looking at h(2)=2 after h(3)\in S after h(1)=1 but instead we asked about h(1)=2? This should be non-zero, but what is the dependence on |S|?

From our classical intuition, we would probably just expect, well, the same really. The larger the value of |S|, the less we are saying about the event. The larger the probability. But in fact this is not the case. The probability is largest when |S| is close to (N-2)/2. We will write down the probability, and then explain, mathematically, why the symmetry with respect to |S|\leftrightarrow N-2-|S|, with respect to S\leftrightarrow S^c:

\displaystyle \mathbb{P}[(h(1)=2)\succ (h(3)\in S)\succ (h(1)=1)]=\dfrac{|S|(N-2-|S|)}{q(N)}.

Proof: Consider u_{21}u_{11}=0. Insert between them 1_{C(S_N^+)}=\sum_{k=1}^Nu_{k3}:

\displaystyle u_{21}\sum_{k=1}^Nu_{k3} u_{11}=\sum_{k=1}u_{21}u_{k3}u_{11}=0.

Like in the classical case, by definition here, u_{21}u_{23}=0=u_{13}u_{11}=0. So split into two sums:

\displaystyle \sum_{s\in S}u_{21}u_{i3}u_{11}+\sum_{j\in S^c}u_{21}u_{j3}u_{11}=0.

Now, classically, this is just a sum of zero terms equal to zero, but not in the quantum world where we have these u_{21}u_{j3}u_{11}\neq 0. They are not positive though. We can now split:

\displaystyle \sum_{s\in S}u_{21}u_{i3}u_{11}=-\sum_{j\in S^c}u_{21}u_{j3}u_{11}.

Now take the |\cdot|^2 of both sides to show that:

\displaystyle |u_{21}u_{S,3}u_{11}|^2=|u_{21}u_{S^c,3}u_{11}|^2

This yields the strange probability above. I think any intuition I have for this would be very much post-hoc. I think I will just let it hang there as something weird, cool and mysterious about quantum permutations…