# Uncertainty, Symmetry, and Commutators

• friend
In summary, the conversation discusses the relationship between the uncertainty principle, commutators between operators, and the symmetry of the action integral. It is questioned how universal this relationship is and if it can be derived from continuous symmetries of the action integral. It is also considered if commutators and uncertainty principles can be developed between other variables, such as electric charge and phase angle. It is suggested that commutation relations can be derived from symmetries, but some responses argue that this may not always be the case. The idea that symmetries should produce uncertainty principles is questioned and it is noted that this intuition may be faulty. Further exploration of these concepts is suggested.
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.

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.

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.

Last edited:
friend said:
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.

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.

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.

Nabeshin said:
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?

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.

friend said:
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?

atyy said:
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.

friend said:
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_(physics) , 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/vorlesungen/hauptseminar/ss2011/1_Vortrag_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.

Last edited:
atyy said:
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.

atyy said:
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/vorlesungen/hauptseminar/ss2011/1_Vortrag_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.

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.

Last edited:
atyy said:
[...]

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.

dextercioby said:
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?

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.

dextercioby said:
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.

dextercioby said:
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?

atyy said:
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

atyy said:
...,and every conserved quantity corresponds to a Noether symmetry (which I'm just learning).
No, only those quantities that generate CONTINUEOUS (Lie) group. Parity can be conserved (i.e., commute with H) but it is not a Noether charge, i.e., there is NO parity current.

Also, does everything carry over from classical to quantum?
Again, No. Not all classical symmetries stay intact in the quantum theory.

@Sam, thanks! What happens to Noether's theorem where the classical theory has a symmetry that leads to a conserved quantity, but the corresponding quantum theory doesn't?

atyy said:
@Sam, thanks! What happens to Noether's theorem where the classical theory has a symmetry that leads to a conserved quantity, but the corresponding quantum theory doesn't?

Could it be a matter of being an "observable", whatever that means? For example there's a conserved charge for an invariance wrt phase. But I don't think that phase is observable. Still I wonder if there is a commutator between charge and phase that might prove useful in calculations.

friend said:
Could it be a matter of being an "observable", whatever that means? For example there's a conserved charge for an invariance wrt phase. But I don't think that phase is observable. Still I wonder if there is a commutator between charge and phase that might prove useful in calculations.

In this article http://arxiv.org/abs/hep-th/9907162 about the chiral anomaly, they give an example in which the classical Noether current is not conserved in the quantum theory. So naively it seems that Noether's theorem fails, but I'm not sure if that is so, and whether there is a generalization.

friend said:
Could it be a matter of being an "observable", whatever that means? For example there's a conserved charge for an invariance wrt phase. But I don't think that phase is observable. Still I wonder if there is a commutator between charge and phase that might prove useful in calculations.

I tried to mention this at the beginning, but the phase is not an operator, it is a number, like $\pi$, like $e^{i\pi}$. It has a vanishing commutator with everything.

jfy4 said:
I tried to mention this at the beginning, but the phase is not an operator, it is a number, like $\pi$, like $e^{i\pi}$. It has a vanishing commutator with everything.

So is it true then, that the uncertainty principle relates only observables that are represented as operators (no scalars) in the non-zero commutator? The question then is how do we know when to use operators in the Lagrangian that we then put into the commutator? Do we first recognize, by other means, that something is an observable and represented as an operator, and then put it into the lagrangian? Or can we recognize observables in the lagrangian, based on first principles such as symmetries, before puting them into the commutator?

Observable quantities are represented in quantum mechanics as hermitian operators. For something to be observable, it must be just that, measurable. For instance, the phase of a state vector, $e^{i\theta}|\psi \rangle$ is not an observable (try and devise an experiment where you can measure it). But $|\psi|^2= \langle \psi |e^{-i\theta}e^{i\theta}|\psi \rangle$ is an observable, but now the phase is gone... Does this help?

jfy4 said:
Observable quantities are represented in quantum mechanics as hermitian operators. For something to be observable, it must be just that, measurable. For instance, the phase of a state vector, $e^{i\theta}|\psi \rangle$ is not an observable (try and devise an experiment where you can measure it). But $|\psi|^2= \langle \psi |e^{-i\theta}e^{i\theta}|\psi \rangle$ is an observable, but now the phase is gone... Does this help?

Yes, I know all that. But what about, say, the color charge of quarks. Is that observable? It seems we will never be able to isolate a quark to measure its properties.

But what I'm really trying to get at is a generalization of what an observable is. Is an observable something we've already measured and then backward incorporate it into the lagrangian? Or are there more fundamental things lurking in the lagrangian that we should be able, at least in principle to make a measurement of?

## 1. What is uncertainty in physics?

Uncertainty in physics refers to the limitations in our ability to know the exact position and momentum of a particle at the same time. This is described by Heisenberg's uncertainty principle, which states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa.

## 2. How does symmetry play a role in physics?

Symmetry is an important concept in physics that describes the balance and patterns found in nature. It helps us understand the underlying laws and principles that govern the behavior of physical systems. Symmetry is often used to simplify complex problems and make predictions about the behavior of physical systems.

## 3. What are commutators in quantum mechanics?

Commutators are mathematical operators used in quantum mechanics to describe the relationship between two physical quantities. They represent the difference between measuring one quantity before or after another. Commutators are important in understanding the uncertainty principle and the behavior of quantum systems.

## 4. How does uncertainty affect our understanding of the physical world?

Uncertainty is a fundamental aspect of quantum mechanics and affects our understanding of the physical world in a profound way. It challenges our classical intuition and forces us to think about the probabilistic nature of reality. Without uncertainty, we would not be able to explain many fundamental phenomena, such as the behavior of subatomic particles.

## 5. Why is symmetry important in the study of particles and forces?

Symmetry plays a crucial role in the study of particles and forces because it allows us to make predictions about the behavior of physical systems. Symmetry principles such as conservation of energy and momentum are fundamental to understanding the laws of nature. Additionally, symmetries in particle interactions can reveal underlying symmetries in the laws of physics, leading to new discoveries and advancements in our understanding of the universe.

• Quantum Physics
Replies
17
Views
1K
• Quantum Physics
Replies
33
Views
2K
• Quantum Physics
Replies
1
Views
981
• Quantum Physics
Replies
2
Views
1K
• Quantum Physics
Replies
8
Views
1K
• Quantum Physics
Replies
5
Views
866
• Quantum Physics
Replies
2
Views
646
• Quantum Physics
Replies
12
Views
945
• Quantum Physics
Replies
18
Views
2K
• Quantum Physics
Replies
25
Views
771