# Commutator in QM vs. Lie brackets in DG

1. Oct 8, 2008

### mhazelm

so, is the commutator relation between two observables just a Lie bracket?

And if so, I have two questions:

I know from differential geometry that the Lie bracket of two vector fields gives me a third vector field. So, what do we mean when we say that [x,p] = i*hbar? In fact, is there at all a way to visualize (for better lack of word) these vector fields (if that is indeed what they are)? And do we always get a vector field back out (i.e. my example i*hbar).

And also, I can think of a vector field as being the infinitesimal generator for some flow. If my Lie bracket vanishes (i.e. my vector fields commute), this implies that I can "flow along" the two corresponding flows in either direction (I can flow along flow 1 and then flow 2, or vice versa, and get the same thing either way). What does this mean physically? What is is saying about the universe?

There's just one thing bothering me about all my reasoning: the operators, as I've understood them, represent linear transformations in a vector space. So, how is it that we can think of them as vector fields as well? Or can we take the Lie brackets of things besides vector fields? Is this kind of how we can represent any linear transformation as a matrix, then consider that matrix as an element of a matrix vector space?

2. Oct 8, 2008

### waht

x, and p are operators that operate on a vector to give another vector. It could be written

$$[x,p]\psi = i \hbar \psi$$

3. Oct 8, 2008

### mhazelm

This makes me feel better about referring to [x,p] as a vector field. Any thoughts on how to physically think of things?

4. Oct 8, 2008

### meopemuk

I don't think your suggested analogy between quantum operators and vector fields serves any useful purpose. Yes, you can define the Lie bracket for both of them, but that's where the similarity ends, in my opinion.

5. Oct 8, 2008

### waht

Well, a Lie bracket comes form lie algebra, which is basically a vector space endowed with an additional operation you can do with the elements of the vector space. The extra rule is the lie bracket.

But a vector space can be more general than just a vector you can visualize with a arrow. Alot of different abstract objects satisfy the axioms of a vector space and do not look like vectors at all. This where the intuition might start to fade.

In quantum mechanics, the commutator relationship means that if it isn't zero, the two operators will yield a different set of eigenvalues when acted on a vector. This will eventually lead to an uncertainly principle. But if the Lie bracket is zero, then the operators are sort of dependent, and share same eigenvalues.

6. Oct 8, 2008

### atyy

This means that measuring position, momentum then position could give you a different result from just measuring position. The commutator is the mathematical statement of the Heisenberg uncertainty principle.

7. Oct 9, 2008

### koolmodee

Simply check if they fulfill

a) skew-symmetry: [x,y] = -[y,x].
b) bilinearity: [x,ay] = a[x,y],
[x,y+z] = [x,y] + [x,z]. (a is any number)
c) Jacobi identity: [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0.

8. Oct 9, 2008

### koolmodee

Write [X,P]=i h I, I is the unit operator.

At each point at your Lie group, a manifold, you got a vector space...read about the connection of Lie algebras and Lie groups.

9. Oct 9, 2008

### strangerep

Since no one has said much in this thread about the relationship
between Lie brackets on classical phase space and commutators
of quantum operators, here's a few words...

In modern treatments of (advanced) classical mechanics,
scalar-valued functions over a phase space manifold are the
observables, and the phase space points are the (classical)
states. In simple examples, the Hamiltonian defines a vector
field (flow) for a given observable via a Poisson bracket.
To describe systems with a more general symmetry group G
with Lie algebra "g", one employs a "moment map" such that
the observables become scalar-valued functions over g* (the
dual of the Lie algebra g). This construction induces
something called a "Lie-Poisson" bracket between any two
such functions. (The ordinary Poisson bracket just
corresponds to the special case of the Heisenberg algebra.)

One attempts to quantize such a system by establishing a map
of such functions to operators on a Hilbert space.
Unfortunately, there's a well-known no-go theorem of
Groenwald-van Hove that it is in principle impossible to
consistently quantize every classical observable on the
phase space $R^{2n}$ in a physically meaningful
way. Similar no-go theorems are known for other phase spaces
math-ph/9809011, or try Google Scholar for others.) For
example, it comes as something of a surprise that one cannot
consistently quantize all the observables on the phase space
of a massive spinning particle.

Faced with such obstructions, one either retreats to
quantizing only a subset of the phase space functions (under
the heading of "geometric quantization"), or else one tries
to construct an algebra of operators without a Hilbert
space (under the heading of "deformation quantization"). One
approach to the latter is to generalize the classical
Poisson bracket to a Moyal bracket (see, e.g.,
http://en.wikipedia.org/wiki/Moyal_bracket).

Summary: the relationship between classical vector fields
on phase space (ie exterior derivatives of function observables
thereon), and quantum operators is quite far-reaching, but
it gets rather difficult rather quickly. :-)

10. Oct 13, 2008

### quantumfireball

The Algebra of The Poisson brackets studied in classical physics is the same as that of the commutator.
ie[a,b]=ab-ba generates the same algebra as the poisson bracket of two classical observables a(p,q),b(p,q)

11. Oct 13, 2008

### meopemuk

As far as I know, the Poisson bracket {..} is just an approximation to the quantum commutator [..]. They are related by the following formula

$$[a,b] = i \hbar \{a,b\} + \ldots$$

where ... contains smaller terms proportional to $$\hbar^2, \hbar^3, \ldots$$. This means that Lie algebras generated by [..] and {..} are not exactly the same.

12. Oct 13, 2008

### quantumfireball

Sir,
But is it not a fallacy to equate operators on one side ,to real valued functions(classical observables) on the other???
Plz Give me the link to the derivation in case i have got the entire concept wrong
By the way im absolutely sure that both PB and commutators define a Lie algebra.
They both satisfy jacobi identity.

Thank you

Last edited: Oct 13, 2008
13. Oct 13, 2008

### meopemuk

Yes, I was a bit sloppy in my notation. The left-hand side [a,b] is an antihermitian operator, while {a,b} is a real function of x and p. In order to match both sides of the equation you need to promote x and p on the right hand side to Hermitian operators. The result of this promotion is not well-defined. It depends on the order in which x and p are present in {a,b}. For classical Poisson bracket this order is completely irrelevant, but for quantum operators it is important. To make the result of this promotion unique we'll need to agree on a definite (but arbitrary) order of x and p factors in quantum operators. For example, we may agree that in the canonical form all x factors are on the left and all p factors are on the right. By changing this convention you'll get different terms proportional to $$\hbar^2, \hbar^3, \ldots$$ on the right hand side of the formula.

You can get more details in subsection 5.3.4 of http://www.arxiv.org/abs/phys/0504062 [Broken]

Yes, both PB and commutators define Lie algebras, but structure constants of these two algebras are not necessarily the same.

Last edited by a moderator: May 3, 2017
14. Oct 13, 2008

### quantumfireball

the above link does not exist.
Anyways thanks for the explanation

15. Oct 14, 2008

### samalkhaiat

16. Oct 14, 2008

### meopemuk

Last edited: Oct 14, 2008
17. Oct 14, 2008

### meopemuk

18. Oct 14, 2008

### strangerep

I think Sam was talking about representations of Lie algebras, whereas
you're now talking about commutators in the larger enveloping algebra.

In fact, the above is an illustration of the Groenwald-van Hove
theorem I mentioned in my post of 8-Oct in this thread. I.e., there is
an obstruction that prevents consistent quantization of every classical
observable on this phase space if we try to rely only on a naive
correspondence between phase space Poisson bracket and Hilbert space
commutator. That's where the Moyal bracket comes in -- it includes
the important higher order terms in $\hbar$.

19. Oct 14, 2008

### meopemuk

I guess you are right. In some simple cases (like generators of rotations $$\mathbf{L}$$) there is one-to-one mapping between classical and quantum algebras of observables. However, in more complex cases (e.g., polynomials in x and p) there is no such mapping. In QM algebra xp and px are two different operators, while in the classical case they are the same. So, the idea of "quantization" (constructing QM algebra from the classical one) seems to be ill-defined from the beginning.

20. Oct 14, 2008

### Haelfix

Yup its ill defined up to ordering ambiguities. Its just one of those things you have to live with, as there are many unitarily inequivalent quantizations that have the same classical limit, at least in the general case (polynomial in x-p and higher)

You can state the problem many different ways. The mathematicians prefer the Groenewald-Van Hove statement (inability to create a unique lie algebra homomorphism)

21. Oct 14, 2008

### ice109

can someone fill me in? what does an operator algebra allow us to do?

22. Oct 15, 2008

### samalkhaiat

Any Lie algebra is uniquely determined by a set of real numbers, known as the structure constants. These constants DO NOT depend on the specific type of the Lie bracket used to express the algebra. So, it is wrong to think that the structure constants of Poincare’ algebra, SU(2)-algebra or any other Lie algebra CHANGE when we change the Lie brackets from commutators to Poisson brackets. This is mathematics and therefore true regardless of the (physical) fact that

$$\{a , b \} = \lim_{\hbar \rightarrow 0} \frac{i}{\hbar}\left[ \hat{a} , \hat{b} \right]$$

[if you are interested in the proof see]

As for your example, frankly I do not know what do you mean by “structure constants”. However, if you use the Lie bracket identity

$$[ab , cd] = ac [b , d] + a [ b , c] d + c [a , d] b + [a , c] db$$

(which holds for Poisson brackets as well as commutators)

you find

$$[xp , p^{2}x] = p^{2}[x , p] x + p [x , p] px - p^{2} [x , p] x$$

Thus

$$[xp , p^{2}x] = p [x , p] px = i p^{2}x$$

Or, we can be a bit smarter and write

$$[xp , p^{2}x] = [ xp - px , p^{2}x] + [ px , p^{2}x] = [px , p(px)]$$

Thus

$$[xp , p^{2}x] = p [px , px] + [px , p] px = [px , p] px$$

or

$$[xp , p^{2}x] = p[x , p] px + [p , p]xpx = ip^{2}x$$

So, your “structure constant” is equal to 1 in both cases ( [ , ] and { , } )! Notice that;
1) No particular representation for x and p has been used in the above derivations.
2) Nothing in the above says that

$$x^{n}p^{m} \rightarrow \hat{x}^{n}\hat{p}^{m}$$

can be achieved without ambiguity.

Regards

sam

23. Oct 15, 2008

### meopemuk

In your example you chose to write products of quantum operators as $$p^nx^m$$ and obtained agreement with Poisson brackets. In my example I chose to write the products as $$x^mp^n$$ and got a disagreement. There is no preferred way to write products of operators, so there is an inherent ambiguity in structure constants calculated with commutators. However, there is no ambiguity in structure constants calculated with Poisson brackets. That's the difference I wanted to point out. I hope we can agree on that.

24. Oct 17, 2008

### samalkhaiat

25. Oct 17, 2008