# Uncertainty, Symmetry, and Commutators

by friend
Tags: commutators, symmetry, uncertainty
 P: 982 It seems the uncertainty principle, the commutator between operators, and the symmetry of the action integral are all related. And I wonder how universal this is. For example, the action integral is invariant with respect to time, and this leads to conserved quantity of energy. This means that the energy will remain the same for any time translation. In other words, if we know the energy exactly, then the time variable could be anything and we don't know the time variable with any precision. This sounds like the uncertainty principle between time and energy. Or again, if the action integral is invariant wrt space translations, then the momentum is conserved. And when momentum is conserved, then we know what the momentum is exactly, but the space varaible could be anything. And this sounds like the uncertainty principle between position and momentum. It seems the uncertainty principle can be derived from commutation relations, as shown here. For the commutator not being zero means we cannot measure both observables at the same tiem, which mean if we know one precisely, then we can't know the other precisely. And as I recall, commutation relations can be derived from symmetries, as shown here. So can commutators and uncertainty principles be developed from any continuous symmetry of the action integral? For example, could a commutator and an uncertainty principle be derive between electric charge and phase angle? For electric charge is the conserved quantity of phase invariance of the action integral.
 P: 647 I think the idea is pretty cool, and honestly, I don't know enough to say much on the matter at the moment. I will think a little more about it though, cause it's a cute idea. As for the phase and charge commutator, the conserved quantity for, say, a charged scalar field with a global U(1) invariance is something like $$N = \text{number of +1s} + \text{number of -1s}=\text{total charge}$$ however, the phase angle for a global symmetry is a c-number, it's just a good ole fashion number... so it commutes with everything. Even if you make the symmetry local (which buys you nothing), the phase as a function of the coordinate is still a scalar function, not an operator, and so it commutes also.
P: 2,193
 Quote by friend It seems the uncertainty principle, the commutator between operators, and the symmetry of the action integral are all related. And I wonder how universal this is. For example, the action integral is invariant with respect to time, and this leads to conserved quantity of energy. This means that the energy will remain the same for any time translation. In other words, if we know the energy exactly, then the time variable could be anything and we don't know the time variable with any precision. This sounds like the uncertainty principle between time and energy. Or again, if the action integral is invariant wrt space translations, then the momentum is conserved. And when momentum is conserved, then we know what the momentum is exactly, but the space varaible could be anything. And this sounds like the uncertainty principle between position and momentum.
This isn't right. A classical theory can have a time and position translation invariant action, giving conserved momenta and position which can be simultaneously measured with infinite precision.

 P: 1,020 Uncertainty, Symmetry, and Commutators I have not seen anywhere that commutators are a result of symmetry.Commutators and uncertainty principles are result of First quantization.They are not related to symmetry as far as I know.
 P: 21 Most sources as far as I know use the commutation relations as the most basic principle. All else builds on that. Continuous symmetries can be related through the generators of symmetry, for example, the generator of x-translation is exp^(i p a); 'a' being the distance shifted. This however can be deduced easily from the commutation relation.
P: 982
 Quote by Nabeshin This isn't right. A classical theory can have a time and position translation invariant action, giving conserved momenta and position which can be simultaneously measured with infinite precision.
Yes, even classically time translation gives energy conservation, and that does not prevent measurement of both time and energy with exact precision. So the question is what is the difference between classical and quantum time and energy variables such that classically they can both be measured exactly but quantum mechanically they can not?

I think one answer may be that classically time and energy are treated as real numbers. Then any commutator between them is zero. But in QM they are treated as operators, or at least energy is treated as an operator (the Hamiltonian), and that's where commutators become non-zero. Does this sound right?
 Sci Advisor HW Helper P: 11,958 Energy and time can be measured (also) in a quantum setting with any precision one desires, i.e. with the maximum precision that the measurement apparatus can offer. There's no bound here.
P: 8,802
 Quote by friend And as I recall, commutation relations can be derived from symmetries, as shown here.
There are two definitions on that page. Are you using the ones for groups or rings?
P: 982
 Quote by atyy There are two definitions on that page. Are you using the ones for groups or rings?
Actually, I don't know. I'm asking a question. When do symmetries of the lagrangian produce commutators? When do commutators produce an uncertainty principle? My intuition tells me that symmetries should produce uncertainty principles as I explained in post 1. So I thought it might make for a good question here. And I was hoping to gain some insight into all this.

What I'm beginning to take from the responses so far is that my intuition may be faulty at this point. But I think it is worth considering.
P: 8,802
 Quote by friend Actually, I don't know. I'm asking a question. When do symmetries of the lagrangian produce commutators? When do commutators produce an uncertainty principle? My intuition tells me that symmetries should produce uncertainty principles as I explained in post 1. So I thought it might make for a good question here. And I was hoping to gain some insight into all this. What I'm beginning to take from the responses so far is that my intuition may be faulty at this point. But I think it is worth considering.
I have not heard that the commutator is related to the symmetry of the Lagrangian.

The commutator in QM is the commutator mentioned under rings, not groups. But it is usually groups that are related to symmetry, so that seems a false relation to me.

The other possible false relation is that the commutator is related to a quantity called the "action" in semi-classical Bohr-Sommerfeld quantization. The Lagrangian, to which Noether's theorem about symmetry applies, is also related to a quantity called the "action". However, these are not the same (maybe very indirectly related). In the wikipedia article http://en.wikipedia.org/wiki/Action_%28physics%29 , the action related to the Lagrangian is described under the heading "Action (functional)" while the "action" that is quantized in semiclassical quantization is described under the heading "Action of a generalized coordinate".

Usually the commutator is defined in the Hamiltonian or "canonical" formalism. The commutator is defined between "canonically conjugate" variables or operators. The Hamiltonian is then specified in terms of the canonically conjugate operators. From the Hamiltonian you can get to a Lagrangian and the path intergal formalism. Since the commutator between canonically conjugate variables is defined before specifying the Hamiltonian, it doesn't seem to me that any symmetries are needed to specify the commutator.

Alternatively, if you start from a Lagrangian, you can define canonically conjugate variables, eg. the first 2 lines on p3 of http://itp1.uni-stuttgart.de/lehre/v...ortrag_Bek.pdf. Once you have canonically conjugate variables, the standard procedure is that these become operators and are related via a commutator. However it again doesn't seem that one needs any symmetry of the Lagrangian to define canonically conjugate variables. Perhaps your best bet is that later on p23-24 of the same document, Bek describes semiclassical "torus quantization" via the "action" of an "integrable" system - and "integrability" usually means the system has lots of symmetries. However, in full quantum theory, you can write the commutator even for non-integrable Hamiltonians.
P: 982
 Quote by atyy I have not heard that the commutator is related to the symmetry of the Lagrangian. The commutator in QM is the commutator mentioned under rings, not groups. But it is usually groups that are related to symmetry, so that seems a false relation to me.
Perhaps! I'm not there yet. I'd like to know your source for this statement. As I read about rings here, it tells me that "a ring is an abelian group...". So I fail to see the distinction that you are trying to make.

 Quote by atyy Usually the commutator is defined in the Hamiltonian or "canonical" formalism. The commutator is defined between "canonically conjugate" variables or operators. The Hamiltonian is then specified in terms of the canonically conjugate operators. From the Hamiltonian you can get to a Lagrangian and the path intergal formalism. Since the commutator between canonically conjugate variables is defined before specifying the Hamiltonian, it doesn't seem to me that any symmetries are needed to specify the commutator. Alternatively, if you start from a Lagrangian, you can define canonically conjugate variables, eg. the first 2 lines on p3 of http://itp1.uni-stuttgart.de/lehre/v...ortrag_Bek.pdf. Once you have canonically conjugate variables, the standard procedure is that these become operators and are related via a commutator. However it again doesn't seem that one needs any symmetry of the Lagrangian to define canonically conjugate variables.
So which one is more fundamental, the Lagrangian or the Hamiltonian formulation? I take it that the Newtonian formulation is the starting point, right? Can you get the Hamiltonian directly from Newton's equations of motion? But I have seen where you can get the lagrangian formulation from Newton's equations of motion and visa versa. And I have seen where you can get the Hamiltonian from the Lagrangian. But I've not seen the Hamiltonian directly from the Newtonian formulation, have you? But if all these formulations are essentially equivalent, then it shouldn't matter which formulation you talk about.

It would seem that as soon as one sees commutators one would try to think in terms of generators of a group that represent the symmetry of something. And then when you see that these generators in the commutator are linked to symmetries of the Lagrangian, like space invariance and conserved momentum, or like time invariance and conserved energy, then you would try to find a connection between the generators in the commutators and symmetries in the Lagrangian. I find it strange that this not made obvious in books as to whether this connection is or is not the case.
 Sci Advisor P: 8,802 The commutator that is usually related to a conserved quantity involves the Hamiltonian. If a quantity commutes with the Hamiltonian, then it is conserved. I don't know if the Noether symmetries of the Lagrangian produce all conserved quantities.
HW Helper
P: 11,958
 Quote by atyy [...] I don't know if the Noether symmetries of the Lagrangian produce all conserved quantities.
But of course they do. For any dynamical system the Hamiltonian formalism and the Lagrange formalism are completely equivalent, even for degenerate/constrained systems.
P: 8,802
 Quote by dextercioby But of course they do. For any dynamical system the Hamiltonian formalism and the Lagrange formalism are completely equivalent, even for degenerate/constrained systems.
The part I wasn't sure about wasn't the equivalence of the Hamiltonian and Lagrangian systems. I wasn't sure whether Noether's method gets all conserved quantities. The other thing I'd like to know is: I usually think of integrability of a Hamiltonian system as being associated with a conserved quantity. Is that correct? And if it is, are those conserved quantities associated with Noether symmetries?
 Sci Advisor HW Helper P: 11,958 The integrability of the system of Hamilton's EOM through so-called 'prime integrals' is equivalent to the integrability of the system of Euler-Lagrange's EOM through the same 'prime integrals', since the 2 systems of ODE's are equivalent i.e. one system can be transformed into the other.
P: 982
 Quote by dextercioby The integrability of the system of Hamilton's EOM through so-called 'prime integrals' is equivalent to the integrability of the system of Euler-Lagrange's EOM through the same 'prime integrals', since the 2 systems of ODE's are equivalent i.e. one system can be transformed into the other.
That's interesting. Perhaps the commutators which involve the Hamiltonian can be equated to the commutators involving the lagrangian, although I suspect some derivatives of commutators is probably involved.
P: 8,802
 Quote by dextercioby The integrability of the system of Hamilton's EOM through so-called 'prime integrals' is equivalent to the integrability of the system of Euler-Lagrange's EOM through the same 'prime integrals', since the 2 systems of ODE's are equivalent i.e. one system can be transformed into the other.
Thanks. So let me paraphrase to make sure I understand: Noether's theorem holds in both directions - ie. every Noether symmetry corresponds to a conserved quantity (which I knew), and every conserved quantity corresponds to a Noether symmetry (which I'm just learning).

Also, does everything carry over from classical to quantum?
P: 926
 Quote by atyy The part I wasn't sure about wasn't the equivalence of the Hamiltonian and Lagrangian systems. I wasn't sure whether Noether's method gets all conserved quantities. The other thing I'd like to know is: I usually think of integrability of a Hamiltonian system as being associated with a conserved quantity. Is that correct? And if it is, are those conserved quantities associated with Noether symmetries?
Any body who has taken a course on QFT should be able to show the following
$$[ i \mathcal{ H } ( x ) , Q ] = \partial_{ \mu } J^{ \mu } =0 ,$$
where
$$H = \int d^{ 3 } x \ \mathcal{ H } ( x ) ,$$
is the Hamiltonian,
$$Q = \int d^{ 3 } x \ J^{ 0 } ( x ),$$
is the time-independent Noether charge of a symmetry, and $J^{ \mu }$ is the conserved Noether current of that symmetry. Integration followed by exponentiation yield the invariance of the Hamiltonian under the symmetry group in question:
$$H = U^{ \dagger } H U ,$$
where
$$U = \exp ( - i \alpha Q ) .$$

Sam

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