# QM based on commutators

1. Nov 2, 2011

### ralqs

A footnote to Griffiths reads "In a deep sense all of the mysteries of quantum mechanics can be traced to the fact that position and momentum do not commute. Indeed, some authors take the canonical commutation relation as an axiom of the theory, and use it to derive [itex]p = (\hbar / i)d/dx[\itex]". Could anyone recommend a text that does this?

2. Nov 2, 2011

### homeomorphic

Well, I like Baez's lecture notes on quantization and categorification, but it's possible may contain too much pure math stuff for you. Very interesting, though, and you may skip the parts that you aren't interested in.

http://math.ucr.edu/home/baez/qg-fall2003/

What's very cool is that he traces the non-commutativity down to the combinatorial fact that there's just one less way to take a ball out of a box and put it back in than there is to put one in and take it out. Non-commutativity turns out to be a "decategorification" of this fact.

3. Nov 3, 2011

### Fredrik

Staff Emeritus
This sounds like the algebraic approach to QM. Instead of the usual assumptions about Hilbert spaces, algebraic QM assumes that the observables are members of an abstract C*-algebra, and uses a theorem about representations of C*-algebras to recover the usual Hilbert space stuff. I don't know all the details, but I know that it requires that you know a lot of functional analysis (and to even begin to study functional analysis, you need to know a lot of topology).

I have only read very small parts of these books yet, so I'm not entirely sure they cover this, but the first ones you should check if you really want to see what this stuff looks like are "Introduction to the mathematical structure of quantum mechanics" by F. Strocchi, and "Mathematical theory of quantum fields" by H. Araki.

I'm going to say a few things about this and I'm not sure if all of it is correct. x and p are unbounded operators on a Hilbert spaces, so they do not generate a C*-algebra. I think the C*-algebra in this case is the one generated by their exponentials exp(iax) and e(ibp) where a and b are real numbers. There is a theorem that guarantees that those exponentials make sense, and that they are unitary operators, and therefore bounded. We should therefore be able to define the smallest C*-algebra that contains all these exponentials. To use the algebraic approach to recover Schrödinger's theory, you would start with an abstractly defined C*-algebra that's isomorphic to that particular C*-algebra. There's one theorem that guarantees the existence of a representation (an algebra isomorphism that also preserves the * operation, into the set of bounded linear operators on a Hilbert space) of the kind that allows us to recover the quantum theory of a single spin-0 particle in Galilean spacetime, and another theorem that says that all representations of this particular C*-algebra are equivalent.

4. Nov 3, 2011

### dextercioby

Fredrik is quoting the work by Segal and Mackey. Related to the quote from Griffiths,the author mentions the famous result by Stone and von Neumann improved by Dixmier and re-phrased by Mackey in terms of the algebraic approach using C* algebras. The commutation relations can be axiomatized and the rest derived (I can't think of a rigorous text who develops the theory of QM starting from the Stone-von Neumann theorem), or even deeper, the commutators be derived from the symmetry principle (the representation theory of a central extension of the Galileo group). This is done by Ballentine in his textbook quoting work and results by T.F. Jordan in the '60s and '70s.

5. Nov 3, 2011

### kith

I think most textbooks on QM postulate the canonical commutation relations and derive the position representation of the momentum operator. See Cohen-Tannoudji for example.

Another interesting option is to derive the canonical commutation relations from the properties of translation. This is done in Sakurai for example.

However, I wouldn't say that postulating some commutation relations for observables in the usual Hilbert space framework bases QM on commutators. I don't know much about this C*-stuff, but there seem to be two fundamentally different approaches to QM: starting with a Hilbert space and defining observables as operators on this space or starting with an algebra of observables and recovering the Hilbert space by knowing how these observables can be represented. This bases QM on commutators, because the defining operation for the algebra is the commutator. (Please correct me, if I'm wrong!)

Last edited: Nov 3, 2011
6. Nov 3, 2011

### homeomorphic

You don't need to know any topology to START learning functional. Kreysig's book doesn't have topology prerequisites, and it only has to introduce a few concepts from topology. Topology becomes more important with operator algebras because they have different topologies on them that become important.

The idea of using C* algebras is a little weird because they are Banach algebras, which means they have a norm on them, satisfying some condition, just like operators norms do. As soon as you throw in the commutation relations, you can work out that p acts by differentiation on polynomials in x. From that and the properties of Banach algebras, if you assume that x and p are in a Banach algebra, you can show that the natural numbers are bounded (consider [p, x^n]). That's a contradiction. So, a Banach algebra can't tolerate the commutation relations. So, evidently the solution is to exponentiate. But then, you don't have x and p anymore, just their exponentials. So, there's something tricky going on there.

My small mind would prefer to think of x and p as just being in an algebra, rather than a C* algebra because that can tolerate unbounded operators. Essentially, you have recovered Schrodinger's theory to some extent, once you establish that p acts by differention (times - i hbar) on polynomials in x. For the rest, the Weierstrass approximation theorem says any continuous function can be uniformly approximated by polynomials (at least on an interval) and continuous functions are dense in L^2 (the Lebesgue square integrable functions, which is the relevant Hilbert space for QM). Seems to me, it should be something more along those lines, but I guess there are deeper ideas than that.

Commutators aren't really defining operations, I would say. They are just operations that are used to express the defining relations of the algebra. You can always take commutators (which comes from the more basic operations of multiplication and addition), but the point is what they are equal to.

7. Nov 3, 2011

### homeomorphic

Actually, I think the C* algebra should be a subalgebra of the one that contains p and x. I guess that's probably all it is. The completion (whatever that is) of the one generated by p and x does contain the exponentials.

8. Nov 15, 2011

### Monocles

Dealing with unbounded operators in C*-algebras is tricky at first but there are a few tools to do it with.

There is the aforementioned exponentiation of the unbounded operator to get a unitary operator. Another thing that is done is if the spectrum of the unbounded operator does not have a cluster point at zero we can look at its set of resolvents instead, which will be a bounded. We can also consider a sequence of operators which have the same spectrum as the unbounded operator but only in an increasing sequence of compact intervals and then in some sense we can think of this sequence as converging to the unbounded operator.

I believe there are other methods but I cannot remember them right now and I am no expert. It all seems very messy (at first) but I think that these methods deal with unbounded operators appropriately (though I would be interested in hearing if there are situations where this is not the case).

9. Dec 1, 2011

### A. Neumaier

This second approach is carried out in detail in http://lanl.arxiv.org/abs/0810.1019

10. Dec 1, 2011

### Bill_K

You can't always deduce the Hilbert space from the operators. The correct description of antiparticles came about only when it was recognized that we had the wrong Hilbert space. The negative energy states were removed and replaced by positive energy antiparticle states. Operators and commutation relations remained the same, but the interpretation (creation vs annihilation) was reversed.

11. Dec 1, 2011

### A. Neumaier

This is correct. The same algebra typically allows many different unitary representations.
Restricting to irreducible representations makes things unique for the standard CCR in finite dimensions
(defining a finite-dimensional Heisenberg algebra), but not in most other cases.

The algebra fully describes only the possible observables. Defining the corresponding space of states
is equivalent to picking the right representation. This is studied in algebraic QFT, It is far from trivial in quantum field theory, where finding the correct representation is equivalent to renormalizing the field theory.
This is not correct. When one interchanges commutation and annihilation operators, the sign of the commutator changes.