## Quantisation

I had been of the understanding that a quantisation looks as follows. There is some process or property $P(X)$ of a space $X$ which we want to examine. Depending on the type of space $X$, from a suitable algebra of complex functions on $X$ $F(X)$, we can recover and examine many of the properties of $P(X)$ by instead looking at $F(X)$: we essentially have the identification $X\leftrightarrow F(X)$. When the process/ property $P(X)$ is about a space then it is said to be classical or commutative because for any $x\in X$ and $f,\,g\in F(X)$ we have that $fg=gf$ because $fg(x)=f(x)g(x)=gf(x)$ as the $f(x),\,g(x)$ lie in the commutative algebra $\mathbb{C}$.

Now from $X$ we know about $F(X)$ and vice versa. Now a suitably chosen $F(X)$ is just a commutative C*-algebra so what about a non-commutative C*-algebra $F$ — can we examine it’s “underlying space” in the same way?

So essentially, this means that I thought you quantised objects, such as Markov chains, by replacing a commutative C*-algebra $F(X)$ with a not-necessarily commutative one. (This roughly follows my interpretation as per this)

## Quantising a Markov Chain

So where is the C*-algebra in a Markov Chain? Well let $\xi$ be a Markov chain on a set $X=\{x_1,\dots,x_N\}$ with initial distribution $\nu$ and transition probabilities $\textbf{P}[\xi_{k+1}=x_j|\xi_{k}=x_i]=p(x_i,x_j)=p_{ij}$.

The probability distribution of this walk, after $n$ steps, is given by $\nu P^n$. However the probability measures, $M_p(X)$, on $X$, which lie in the dual of $F(X)$ ( $M_p(X)\subset \mathbb{R}^{N}$ is equipped with the 1-norm while $F(X)$ is equipped with the supremum norm). In fact we can go further, the probability measures comprise the states of $A$ as for any $\theta\in M_p(X)\subset F(X)^*$, $\|\theta\|_1=1$ and $\theta\geq0$.

Actually in this case ( $X$ is a finite set) the positivity of the functional $\theta$ has three equivalent definitions (the first is the same as saying $\theta(x)\geq 0$ for all $x\in X$):

1. The dual basis to $(\mathbf{1}_{\{x_1\}},\cdots,\mathbf{1}_{\{x_N\}})$, is the basis of delta measures $(\delta^{x_1},\cdots,\delta^{x_N})$. In this basis, the coefficients of $\theta$ are all positive.
2. With respect to the involution $(a_1,\cdots,a_N)^*=(a_1^*,\cdots,a_N^*)$ $\theta$ is a positive element in the *-algebra $F(X)^*$ — where multiplication is defined pointwise.
3. $\theta(f)\subset \mathbb{R}^+$ for all positive functions $f\in F(X)^+$.

Usually when we talk about functionals on C*-algebras being positive we are talking about definition 3. i.e. a linear map $\varphi:A\rightarrow B$ between C*-algebras is said to be positive if $\varphi(A^+)\subset B^+$. I’m not sure if $\varphi$ is then necessarily positive in the C*-algebra $L(A,B)$, the linear maps from $A$ to $B$ (note that $L(A,B)$ may not necessarily be a C*-algebra — I think)?

The states of a C*-algebra, $A$, are given by: $S(A)=\{\varphi\in A^*:\|\varphi\|=1,\,\varphi\geq 0\}$.

It is an easy exercise to show that stochastic operators are characterised as being $M_p(X)$-stable. In this picture — which looks at the deterministic evolution of $\{\nu P^k:k=0,1,\dots,N\}$ — rather than the random variable picture of the $\xi_k:(Y,\mu)\rightarrow X$ (see p.11 here for this construction), we thus have an initial distribution $\theta\in S(A)$, a stochastic operator: $P:F(X)^*\rightarrow F(X)^*$ $\theta\mapsto \theta P$

which is $S(F(X))$-stable, and we look at the set of distributions: $\{P^k(\theta):k=1,\dots,n\}$,

to tell us everything we want to know about the Markov chain. For example, if $P^k(\theta)$ is convergent then we know the walk converges and a fixed point of the stochastic operator is a stationary distribution.

The only thing left to do is to put some conditions on a stochastic operator being ‘ $S(A)$‘-stable — how about the stochastic operator being positive and isometric?

The C*-algebra quantisation is then as follows. Let $A$ be a C*-algebra with dual $A^*$. Choose an element $\psi\in S(A)$ and a positive linear isometry $T:A^*\rightarrow A^*$ — which is automatically $S(A)$-stable. The distribution of the quantum Markov chain generated by $\psi$ and $T$ after $k$ steps would then be given by $T^k(\psi)$.

In my interpretation, this is what a quantisation of a Markov chain looks like.

## The Coalgebra Picture

This whole piece is written because I don’t understand coalgebras and duality in category theory — and I was trying to do it a different way. At times I can see that the coalgebra is encoding a lot (see some remarks here and here), but really I haven’t a clue what’s going on!

Wikipedia says turn all the arrows around — it also seems to say turn Cartesian products into tensor products (??). The best I can do out of this is I can see that, roughly (with $k+1$ elements in the cartesian product), the classical Markov chain random variables look like: $\xi_k:X\times X\times X\times\cdots\times X\rightarrow X$,

while, roughly, their quantum counterparts (with $k+1$ elements in the tensor product) $j_k:A\rightarrow A\otimes A\otimes\cdots\otimes A$.

(the $\{j_k\}$ are constructed in Franz & Gohm).

## Problems

My construction is leaning towards the fact that the evolution of the $T^k(\psi)$ can tell us if the Markov chain is stationary or if the Markov chain has limiting behaviour (well, in a quantum way). However, I’m looking at the “co” approach of Franz & Gohm, and I think they get to this picture another way. In other words they do their quantisation ‘down there’ first and then ‘lift’ — I have ‘lifted’ first and then quantised ‘up there’. Very precise I know! This (terrible attempt at a) diagram should explain what I mean, the right arrows are quantisations; mine on top — their’s on the bottom: $\begin{array}{ccc}(M_p(X))P^k &\overset{q}{\rightarrow} & T^k (S(A))\\ \uparrow & & \uparrow\\ \{\xi_k\} & \overset{q}{\rightarrow} &\{j_k\}\end{array}$

Suppose that $G$ is a group — then it is kind of clear to me that we need the “co” picture to encode the group laws ‘up’ in $F(G)$ — so if we’re going to encode group laws in a not-necessarily commutative algebra $A$ (quantum group), we will need the “co” picture. In otherwords my approach can’t deal with quantum random walks on groups.