The Commutator in Group Theory vs QM

[chickenz]

The commutator for group theory is
$$[X,Y]=X^{-1}Y^{-1}XY$$ whereas the quantum commutator is $$[X,Y]=XY-YX$$.

At first glance, the two commutators seem to be totally unrelated because the quantum commutator speaks of two binary operations whereas group theory has one binary operation. However, are they related?

Also the commutator in QM has a simple interpretation: in some weird sense, it's how far two observables are from being simultaneously measurable. Any simple interpretation to the anti-commutator $$[X,Y]_{+}=XY+YX$$?

 

[micromass]

The are related in a way: they share many of the same properties. So do all the commutators of a group G generate a subgroup of G, called the commutator subgroup (or derived subgroup). If you quotient G w.r.t. the commutator subgroup, then your quotient will be abelian. Moreover, G is the smallest subgroup with that property.

The quantum commutator (which I interpret as the commutator of a ring) has the same property. Quotienting w.r.t. the commutator ideal is an abelian ring, and it is the smallest such ideal.

So commutators in a way measure how abelian everything is. And this is the same for every commutator. That's the link between the two notions of commutators. Perhaps there is another one, but that's all I know...

 

[Deveno]

let's put it this way:

suppose G is non-abelian, and you wanted a subgroup H of G, such that G/H is abelian.

so Hxy = Hyx, which means that H = Hyxx^-1y^-1 = Hxyx^-1y^-1 = H[x,y],

so [x,y] has to be in H for all x,y in G.

so the normal closure of the set of all commutators is the subgroup we want.

it turns out the subgroup generated by all the commutators is normal, so its normal closure is itself.

now let's look at rings.

we have a non-commutative ring, R, and we want to define an ideal A, so that R/A is commutative.

so (x + A)(y + A) = (y + A)(x + A) the fact that rings have two operations makes this equation:

xy + A = yx + A, so we must have all elements of the form xy - yx in A, so

the ideal generated by all [x,y] = xy - yx is the ideal we want.

 

[chickenz]

Thanks!

So, now my question is more of a physics one:
Suppose we have a quantum mechanical Hilbert Space H. Is there an interpretation to the set H/[H,H]? (other than the fact that it's the smallest quotient of H that is Abelian - I'm looking for a physics interpretation here)

 

[Deveno]

i'm not much of a physicist, but i think there may be.

one interpretation of AB is: first apply B, then apply A.

if AB - BA = 0, this means that it doesn't matter which order you apply the operators in, so we have "time invariance".

so H/[H,H] ought to be a space of "time-invariant operators". in general, you should be able to make a Hilbert space which is asymmetric with respect to "blank (fill in the appropriate thing here)", symmetric with respect to "blank", by taking a suitable commutator.

to use your example, if commutators measure how far events are from being simultaneously measurable, then mod the commutator, every event should be simultaneously measurable (we've "shrunk down the distance (time?) to 0").

 

[chickenz]

i'm not much of a physicist, but i think there may be.

one interpretation of AB is: first apply B, then apply A.

if AB - BA = 0, this means that it doesn't matter which order you apply the operators in, so we have "time invariance".

so H/[H,H] ought to be a space of "time-invariant operators". in general, you should be able to make a Hilbert space which is asymmetric with respect to "blank (fill in the appropriate thing here)", symmetric with respect to "blank", by taking a suitable commutator.

to use your example, if commutators measure how far events are from being simultaneously measurable, then mod the commutator, every event should be simultaneously measurable (we've "shrunk down the distance (time?) to 0").

Hmm... I'm not sure I buy that, but thanks for trying.

Maybe it's not worthwhile, from a physics standpoint, to think about what [H,H] means.

 

[lpetrich]

The QM definition of commutator is the definition for a Lie Algebra. That's the set of generators for many continuous groups.

I'll show how it works. Let

g1 = I + e*a1*L1
g2 = I + e*a2*L2

where e is a small parameter.

Their commutator:
g1.g2.inv(g1).inv(g2) = I + e²*a1*a2*(L1.L2 - L2.L1) + ...