Arjun wrote:

4. Will we be only considering time independent observables here?

Yes, more precisely only those without ‘explicit time dependence’. Clearly the values of most interesting observables change with time because the state changes with time. For example, in quantum mechanics the expected value changes with time even though does not.

5. You wrote: “It turns out that 1-parameter unitary groups are either continuous in a certain way, or so pathological that you can’t even prove they exist without the axiom of choice!”. Can you please expand on this ?, in plainer English – I’m interested.

Plainer than what?

You can see this most simply as follows. Suppose you want a function

obeying this equation

for all There are some obvious solutions:

for any real number . All these solutions are continuous. But if you use the axiom of choice, it’s easy to prove there are infinitely many *other* solutions, that are *not* continuous.

However, it is impossible to write down a formula for any of these other solutions! And if the axiom of choice is false, they might not really exist!

For more details, read this:

• Hamel basis and additive functions.

See Theorem 5.

This immediately has consequences for one-parameter unitary groups. Now suppose you want a function

that obeys these three equations:

and

for all and also

(The third equation follows from the second, but I include it just so you can more easily see that can be seen as a 1-parameter unitary group of 1×1 matrices.)

Which functions obey all three equations? The obvious solutions are

where All these functions are continuous. But the axiom of choice implies there are infinitely many *other* solutions, that are not continuous. Again, it’s impossible to write down a formula for any of them.

So, if you use the axiom of choice, you can prove there are lots of 1-parameter unitary groups that are not continuous. However, you can’t write down a formula for any of them, they’re not even measurable (see Theorem 7 in that paper), and if you use some other axioms you can prove they don’t exist. So it’s best to ignore them.

6. If a strongly continuous 1-parameter group is defined by this additional property: , how is a continuous 1-parameter group defined?

There are various different topologies on the set of operators on a Hilbert space or Banach space, so there are different kinds of continuity for 1-parameter groups. The only important ones are strong continuity and uniform continuity, also known as norm continuity. A **uniformly continuous 1-parameter group** has this property:

where the norm here is the operator norm.

Ironically, norm continuity is stronger than strong continuity.

If you want to learn about this, read Reed and Simon’s book *Functional Analysis*, which is full of the analysis that mathematical physicists need to know.

7. Why is finite dimensional for a finite set ?

Because the vector space of complex-valued functions on an -element set is -dimensional and is just the vector space of complex functions on when is finite.

]]>4. Will we be only considering time independent observables here?

5. You wrote: ” It turns out that 1-parameter unitary groups are either continuous in a certain way, or so pathological that you can’t even prove they exist without the axiom of choice!”. Can you please expand on this ?, in plainer English- I’m interested.

6. If a strongly continuous 1-parameter group is defined by this additional property: , how is a continuous 1-parameter group defined?

7. Why is finite dimensional for a finite set X?

8. How do we extend the results for finite sets to countably infinite sets, as is true for all the examples in the previous posts?

]]>Arjun wrote:

1. What is the difference between an operator and a function?

Given sets and , **function** assigns to each element of the set a unique element of the set

When we talk about a function on a set and don’t specify the set we usually mean or These are called ‘real-valued functions’ and ‘complex-valued functions’ if we want to be more clear.

An **operator** is a linear function from a vector space to a vector space.

By the way, definitions of standard math terms can be looked up on Wikipedia (see the links). It’s better if you ask me questions about my work, instead of questions about this standard stuff.

2. […] Now is the vector space with those s as elements for which If this sum is 1, the corresponding is called a

stochastic state. Is all this correct?

It’s almost all correct. One mistake is that since we want to be a vector space, it consists of all functions with

We need the absolute value here!

But stochastic states are elements of with and

just as you said.

Another possible mistake is that you said:

or (where is a matrix with the s as columns, and is a column vector with the s as elements).

I don’t understand this, because I don’t know how you’re making elements of a set into columns of a matrix. More importantly, I never use this way of thinking in any of these blog articles, so it’s probably best if you don’t think this way when trying to understand what I’m saying.

Like shouldn’t the observable be instead of being a function on ?

In quantum mechanics, an observable is an operator

but not just any operator: it needs to be self-adjoint! There are good and well-known reasons for this.

Now we’re talking about stochastic mechanics, which is a subject I’m just inventing. It makes sense to guess that an observable should be an operator

but probably not just any operator!

What kind of operator can be an observable in stochastic mechanics? My guess, made for very good reasons, is that it’s one of this form:

for some measurable function

In Part 11 Brendan and I explained this in the special case where was a *finite set*. In this case any operator

can be described by a matrix

where And the special operators called observables correspond to *diagonal* matrices. So for an observable we have

for some function Then we have

as desired.

But in Part 11 we called the function instead of and we write instead of for this function’s values. This may be confusing, but it’s efficient—it’s annoying to have two names and for two ways of thinking about the same thing! We explained this as follows:

In stochastic mechanics an

observableis simply a function assigning a number to each state .However, in quantum mechanics we often think of observables as matrices, so it’s nice to do that here, too. It’s easy: we just create a matrix whose diagonal entries are the values of the function And just to confuse you, we’ll also call this matrix So:

Whenever I say “just to confuse you”, it’s a joke: I’m actually warning that some notation might be confusing, so pay careful attention!

You can see a more detailed explanation in our paper.

]]>1. What is the difference between an operator and a function ?

2. As I have understood, is the set of states. Then is the set of ordered pairs , which in the case of finite X, can be summarized as or (where is a matrix with the s as columns, and is a column vector with the s as elements). Now is the vector space with those s as elements for which

If this sum is 1, the corresponding is called a stochastic state. Is all this correct?

3. Like , shouldn't the observable be instead of being a function on ? Even if then has to be defined as instead of the usual definition of function composition. Instead for an , if , we can say that is diagonal.

]]>**Puzzle 1.** Suppose is a finite set. Show that every isometry is invertible, and its inverse is again an isometry.

**Answer.** Remember that being an isometry means that it preserves the inner product:

and thus it preserves the norm

given by It follows that if , then so is one-to-one. Since is a linear operator from a *finite-dimensional* vector space to itself, must therefore also be onto. Thus is invertible, and because preserves the inner product, so does its inverse: given we have

since we can write and then the above equation says

Puzzle 2.Suppose is a finite set. Which stochastic operators have stochastic inverses?

**Answer.** Graham wrote:

OK, a sketch proof: A stochastic operator must map the unit simplex into itself. If its inverse is also stochastic operator this must be a bijection. A bijective linear map must take a vertex to a vertex. So must be a permutation matrix.

Expanding that a bit, suppose the set has points. Then the set of stochastic states

is a **simplex**. It’s an equilateral triangle when , a regular tetrahedron when , and so on.

In general, has corners, which are the functions that equal 1 at one point of and zero elsewhere. Mathematically speaking, is a **convex set**, and its corners are its **extreme points**: the points that can’t be written as **convex combinations** of other points of in a nontrivial way.

Any stochastic operator must map into itself, so if has an inverse that’s also a stochastic operator, it must give a bijection . Any linear transformation acting as a bijection between convex sets must map extreme points to extreme points (this is easy to check), so must map corners to corners in a bijective way. This implies that it comes from a permutation of the points in

In other words, any stochastic matrix with an inverse that’s also stochastic is a **permutation matrix**: a square matrix with every entry 0 except for a single 1 in each row and each column. So, Uwe Stroinski‘s intuition was right!

**Puzzle 3.** If is a finite set, show that any operator on that’s both diagonal and infinitesimal stochastic must be zero.

**Answer.** We are thinking of operators on as matrices with respect to the obvious basis of functions that equal 1 at one point and 0 elsewhere. If is an infinitesimal stochastic matrix, the sum of the entries in each column is zero. If it’s diagonal, there’s at most one nonzero entry in each column. So, we must have .

**Puzzle 4.** Show any Markov semigroup is a contraction semigroup.

**Answer.** We need to show

for all and . Here the norm is the norm, so more explicitly we need to show

We can split into its positive and negative parts:

where

Since is stochastic we have

and

so

]]>Uwe wrote:

Puzzle 2 seems to be permutation matrices however my incompetency to prove the nontrivial direction (in finite time) indicates that this might be false :-(

Actually Graham Jones pointed out a mistake in Puzzle 2! I had written:

In quantum mechanics we are mainly interested in invertible isometries, which are called

unitaryoperators. There are lots of these. There are, however, very few invertible stochastic operators:

Puzzle 1.Suppose is a finite set. Show that every isometry is invertible.

Puzzle 2.Suppose is a finite set. What are the invertible stochastic operators ?

In fact there are lots of invertible stochastic operators whose inverses are not stochastic! I didn’t mean to include these in the discussion. So, I’ve corrected my puzzles as follows:

In quantum mechanics we are mainly interested in invertible isometries, which are called

unitaryoperators. There are lots of these, and their inverses are always isometries. There are, however, very few stochastic operators whose inverses are stochastic:

Puzzle 1.Suppose is a finite set. Show that every isometry is invertible, and its inverse is again an isometry.

Puzzle 2.Suppose is a finite set. Which stochastic operators have stochastic inverses?

I won’t give away the answers to these reformulated puzzles yet, but here are two proofs that there are lots of invertible stochastic operators.

First, every stochastic operator that’s ‘close to the identity’ in this sense:

(where the norm is the operator norm) will be invertible, simply because *every* operator obeying this inequality is invertible! After all, if this inequality holds, we have a convergent geometric series:

Second, suppose is a finite set and is infinitesimal stochastic operator on . Then is bounded, so the stochastic operator

will always have an inverse, namely

But for sufficiently small, this inverse will only be stochastic if is infinitesimal stochastic, and that’s only true if .

In something more like plain English: when you’ve got a finite set of states, you can formally run any Markov process backwards in time, but a lot of those ‘backwards-in-time’ operators will involve negative probabilities for the system to hop from one state to another!

]]>Thanks for the comments, Martin.

Since probability distributions on a measure space live in , it seems natural to think of Markov semigroups as consisting operators on . And indeed this is how various papers seem to define them. For example:

• Ryszard Rudnicki, Katarzyna Pichór and Marta Tyran-Kamínska, Markov semigroups and their applications.

Defining them on seems weird – that’s really how they did it when you were a kid?

Whoa, currently I’m not even sure about what the dual of is (or vice versa).

For a -finite measure space the dual of is , but not vice versa. If you have trouble remembering this—and don’t feel bad, I do too!—just remember that a Banach limit is a strange sort of continuous linear functional that lets you define a kind of ‘limit’ for any sequence in . This shows that the dual of contains elements that aren’t in .

I say that Banach limits are ‘strange’ because you can’t actually construct them: you need to use the axiom of choice, or at least some weaker but still nonconstructive principle, to get your hands on them! In fact, I have a vague memory that *no* elements of the dual of can be ‘explicitly given’, other than elements of . Does anyone remember?

But I’m digressing. I’ll look at the paper you pointed me to. Thanks!

]]>((Oops, this stuff has induced/coincided a major flashback in my poor brain. (And its math demon had been sleeping almost all summertime, until a few hours ago.) I’m still clearing up confusion: On first reading there was still the picture in my mind that I got indoctrinated in stochastic analysis, long ago: Markovian being positivity preserving and contractive on ))

It looks you are “only” interested in Markov semigroups which are the dual of a positive contraction semigroup on . So, said Theorem 2.2.1 possibly is it? But the “dissipative conditions” look suspicious. (Plus, I don’t recall ever having known about the paper.)

((Need postpone further flashback and reading by a day or two.))

]]>* Charles J. K. Batty, Derek W. Robinson, Positive one-parameter semigroups on ordered banach spaces, Acta Appl. Math 2 (1984), 221-296

(And Scotty has kindly provided me with a virtual tunnel to read it.)

Theorem 2.2.1 on p. 261 there is **not what you want** (being simply about positivity preserving semigroups). But it’s a start: Methinks **the** start is to study positivity preserving semigroups first – perhaps not necessarily the paper.

I’ll read more of the paper later.

Plus, I’ll check that yellowish wrinkled piece of penciled paper promising a generalized Kato-Simon-Shigekawa criterion for semigroup domination. Alas it is in and symmetric – but there’s a simple formula for the generators which could be generalized to other duality. It’s all about the distributional Laplacian of the norm. (Whoa, currently I’m not even sure about what the dual of is (or vice versa). That more than a dozen years back.)

Which reminds me of the rig in the stochastics vs. quantum picture: Perhaps the point is the ring’s “involution”. E.g. the absolute value in the reals (a nonlinear involution) vs. complex involution. The absolute value involution leads to positivity preserving semigroups. Another involution, cutting off outside the unit interval leads to Markovian semigroups. (Perhaps cf. Reed-Simon, where I guess it’s split up that way.)

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