## Quantisation

I had been of the understanding that a quantisation looks as follows. There is some process or property of a space which we want to examine. Depending on the type of space , from a suitable algebra of complex functions on , , we can recover and examine many of the properties of by instead looking at : we essentially have the identification . When the process/ property is about a space then it is said to be *classical* or *commutative* because for any and we have that because as the lie in the commutative algebra .

Now from we know about and vice versa. Now a suitably chosen is just a commutative C*-algebra so what about a non-commutative C*-algebra — 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 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 be a Markov chain on a set with initial distribution and transition probabilities .

The probability distribution of this walk, after steps, is given by . However the probability measures, , on , which lie in the dual of ( is equipped with the 1-norm while is equipped with the supremum norm). In fact we can go further, the probability measures comprise the *states *of as for any ,

and .

Actually in this case ( is a finite set) the positivity of the functional has *three* equivalent definitions (the first is the same as saying for all ):

- The dual basis to , is the basis of delta measures . In this basis, the coefficients of are all positive.
- With respect to the involution , is a positive element in the *-algebra — where multiplication is defined pointwise.
- for all positive functions .

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

The *states *of a C*-algebra, , are given by:

.

It is an easy exercise to show that stochastic operators are characterised as being -stable. In this picture — which looks at the deterministic evolution of — rather than the random variable picture of the (see p.11 here for this construction), we thus have an initial distribution , a stochastic operator:

,

which is -stable, and we look at the set of distributions:

,

to tell us everything we want to know about the Markov chain. For example, if 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 ‘‘-stable — how about the stochastic operator being positive and isometric?

The C*-algebra quantisation is then as follows. Let be a C*-algebra with dual . Choose an element and a positive linear isometry — which is automatically -stable. The distribution of the quantum Markov chain generated by and after steps would then be given by .

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 elements in the cartesian product), the classical Markov chain random variables look like:

,

while, roughly, their quantum counterparts (with elements in the tensor product)

.

(the are constructed in Franz & Gohm).

## Problems

My construction is leaning towards the fact that the evolution of the 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:

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

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