By Tobias Kienzler


2010-08-24 08:21:23 8 Comments

Noether's theorem states that, for every continuous symmetry of an action, there exists a conserved quantity, e.g. energy conservation for time invariance, charge conservation for $U(1)$. Is there any similar statement for discrete symmetries?

9 comments

@user566 2011-04-12 17:14:33

For continuous global symmetries, Noether theorem gives you a locally conserved charge density (and an associated current), whose integral over all of space is conserved (i.e. time independent).

For global discrete symmetries, you have to distinguish between the cases where the conserved charge is continuous or discrete. For infinite symmetries like lattice translations the conserved quantity is continuous, albeit a periodic one. So in such case momentum is conserved modulo vectors in the reciprocal lattice. The conservation is local just as in the case of continuous symmetries.

In the case of finite group of symmetries the conserved quantity is itself discrete. You then don't have local conservation laws because the conserved quantity cannot vary continuously in space. Nevertheless, for such symmetries you still have a conserved charge which gives constraints (selection rules) on allowed processes. For example, for parity invariant theories you can give each state of a particle a "parity charge" which is simply a sign, and the total charge has to be conserved for any process, otherwise the amplitude for it is zero.

@Keenan Pepper 2011-04-13 20:24:59

Isn't this called Pontryagin duality or something?

@Tobias Kienzler 2011-04-14 09:58:28

@KeenanPepper: Pontryagin duality? I only looked briefly, but it seems to be about generalized Fourier transforms

@Tobias Kienzler 2011-04-18 09:10:20

can you provide references on this?

@Nikos M. 2015-10-01 18:30:05

Actually there are analogies or generalisations of results which reduce to Noether's theorems under usual cases and which do hold for discrete (and not necesarily discretised) symmetries (including CPT-like symmetries)

For example see: Anthony C L Ashton (2008) Conservation Laws and Non-Lie Symmetries for Linear PDEs, Journal of Nonlinear Mathematical Physics, 15:3, 316-332, DOI: 10.2991/ jnmp.2008.15.3.5

Abstract We introduce a method to construct conservation laws for a large class of linear partial differential equations. In contrast to the classical result of Noether, the conserved currents are generated by any symmetry of the operator, including those of the non-Lie type. An explicit example is made of the Dirac equation were we use our construction to find a class of conservation laws associated with a 64 dimensional Lie algebra of discrete symmetries that includes CPT.

The way followed is a succesive relaxation of the conditions of Noether's theorem on continuous (Lie) symmetries, which generalise the result in other cases.

For example (from above), emphasis, additions mine:

The connection between symmetry and conservation laws has been inherent in all of mathematical physics since Emmy Noether published, in 1918, her hugely influential work linking the two. ..[M]any have put forward approaches to study conservation laws, through a variety of different means. In each case, a conservation law is defined as follows.

Definition 1. Let $\Delta[u] = 0$ be a system of equations depending on the independent variables $x = (x_1, \dots , x_n)$, the dependent variables $u = (u_1, \dots , u_m)$ and derivatives thereof. Then a conservation law for $\Delta$ is defined by some $P = P[u]$ such that: $${\operatorname{Div} P \; \Big|}_{\Delta=0} = 0 \tag{1.1}$$

where $[u]$ denotes the coordinates on the $N$-th jet of $u$, with $N$ arbitrary.

Noether’s [original] theorem is applicable in the [special] case where $\Delta[u] = 0$ arises as the Euler-Lagrange equation to an associated variational problem. It is well known that a PDE has a variational formulation if and only if it has self-adjoint Frechet derivative. That is to say: if the system of equations $\Delta[u] = 0$ is such that $D_{\Delta} = {D_{\Delta}}^*$ then the following result is applicable.

Theorem (Noether). For a non-degenerate variational problem with $L[u] = \int_{\Omega} \mathfrak{L} dx$, the correspondence between nontrivial equivalence classes of variational symmetries of $L[u]$ and nontrivial equivalence classes of conservation laws is one-to-one.

[..]Given that [the general set of symmetries] is far larger than those considered in the classical work of Noether, there is potentially an even stronger correspondence between symmetry and conservation laws for PDEs[..]

Definition 2. We say the operator $\Gamma$ is a symmetry of the linear PDE $\Delta[u] \equiv L[u] = 0$ if there exists an operator $\alpha_{\Gamma}$ such that: $$[L, \Gamma] = \alpha_{\Gamma} L$$ where $[\cdot, \cdot]$ denotes the commutator by composition of operators so $L \Gamma = L \circ \Gamma$. We denote the set of all such symmetries by $sym(\Delta)$.

Corollary 1. If $L$ is self-adjoint or skew-adjoint, then each $\Gamma \in sym(L)$ generates a conservation law.

Specificaly, for the Dirac Equation and CPT symmetry the following conservation law is derived (ibid.):

enter image description here

@Qmechanic 2015-10-01 19:47:24

Comment to the answer (v1): Note that the article is talking about symmetries of equations of motion rather than of the action.

@Nikos M. 2015-10-01 21:36:58

@Qmechanic, correct the point is that when the conditions of Noether's thrm are satisfied it reduces to the same results for continuous (Lie) symmetries of the action functional (Lagrange functional). In this sense it is a generalisation with the usual N. thm as special case (sth pointed out in the referenced article). Intuitively it is easy to understand why a symmetry induces sth invariant (a "conservation") even a discrete one, these "generalised theorems" state that

@user41670 2014-03-03 03:28:15

Electric charge conservation is a "discrete" symmetry. Quarks and anti-quarks have discrete fractional electric charges (±1/3, ±2/3) electrons, positrons and protons have integer charges.

@Qmechanic 2014-03-03 09:32:21

Comments to the answer (v1): (i) The action is not invariant under a discrete change of electric charge $Q\to Q+1$. Thus the transformation $Q\to Q+1$ is not a symmetry. (ii) Noether's theorem shows that global gauge symmetry (which is a continuous symmetry) imply that electric charge $Q$ is conserved. (iii) The fact that electric charge $Q$ only takes discrete values is tied to the predicted existence of magnetic monopoles.

@Tobias Kienzler 2014-03-05 10:26:41

I'm afraid you're mixing up symmetry and conserved quantity here

@user41670 2014-03-06 23:02:33

As indicated in one of the answers above, Emmy Noether is the source of the wonderful mathematics that became symmetry, and it all began with conservation of energy and moment, but it got much better, of course. CP symmetry is conservation of charge and parity.

@user41670 2014-03-06 23:05:58

The fractional charges of quarks are one of the few places that QCD is quite specific. It doesn't matter whether the quantization is an elementary charge, or a fractional charge, except in the case of the electron, which, if there were any, would presumably be the entity that gives rise to both magnetic dipoles and monopoles. As far as I'm aware, Maxwell's equations still forbid those types of monopoles, even if Dirac saw a potential loophole.

@Qmechanic 2011-04-12 18:32:27

Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action $S$ implies a local on-shell conservation law. By the words on-shell and off-shell are meant whether Euler-Lagrange equations of motion are satisfied or not.

Now the question asks if continuous can be replace by discrete?

It should immediately be stressed that Noether Theorem is a machine that for each input in form of an appropriate symmetry produces an output in form of a conservation law. To claim that a Noether Theorem is behind, it is not enough to just list a couple of pairs (symmetry, conservation law).

Now, where could a discrete version of Noether's Theorem live? A good bet is in a discrete lattice world, if one uses finite differences instead of differentiation. Let us investigate the situation.

Our intuitive idea is that finite symmetries, e.g., time reversal symmetry, etc, can not be used in a Noether Theorem in a lattice world because they don't work in a continuous world. Instead we pin our hopes to that discrete infinite symmetries that become continuous symmetries when the lattice spacings go to zero, can be used.

Imagine for simplicity a 1D point particle that can only be at discrete positions $q_t\in\mathbb{Z}a$ on a 1D lattice $\mathbb{Z}a$ with lattice spacing $a$, and that time $t\in\mathbb{Z}$ is discrete as well. (This was, e.g., studied in J.C. Baez and J.M. Gilliam, Lett. Math. Phys. 31 (1994) 205; hat tip: Edward.) The velocity is the finite difference

$$v_{t+\frac{1}{2}}:=q_{t+1}-q_t\in\mathbb{Z}a,$$

and is discrete as well. The action $S$ is

$$S[q]=\sum_t L_t$$

with Lagrangian $L_t$ on the form

$$L_t=L_t(q_t,v_{t+\frac{1}{2}}).$$

Define momentum $p_{t+\frac{1}{2}}$ as

$$ p_{t+\frac{1}{2}} := \frac{\partial L_t}{\partial v_{t+\frac{1}{2}}}. $$

Naively, the action $S$ should be extremized wrt. neighboring virtual discrete paths $q:\mathbb{Z} \to\mathbb{Z}a$ to find the equation of motion. However, it does not seem feasible to extract a discrete Euler-Lagrange equation in this way, basically because it is not enough to Taylor expand to the first order in the variation $\Delta q$ when the variation $\Delta q\in\mathbb{Z}a$ is not infinitesimal. At this point, we throw our hands in the air, and declare that the virtual path $q+\Delta q$ (as opposed to the stationary path $q$) does not have to lie in the lattice, but that it is free to take continuous values in $\mathbb{R}$. We can now perform an infinitesimal variation without worrying about higher order contributions,

$$0 =\delta S := S[q+\delta q] - S[q] = \sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_t + p_{t+\frac{1}{2}}\delta v_{t+\frac{1}{2}} \right] $$ $$ =\sum_t \left[\frac{\partial L_t}{\partial q_t} \delta q_{t} + p_{t+\frac{1}{2}}(\delta q_{t+1}- \delta q_t)\right] $$ $$=\sum_t \left[\frac{\partial L_t}{\partial q_t} - p_{t+\frac{1}{2}} + p_{t-\frac{1}{2}}\right]\delta q_t + \sum_t \left[p_{t+\frac{1}{2}}\delta q_{t+1}-p_{t-\frac{1}{2}}\delta q_t \right].$$

Note that the last sum is telescopic. This implies (with suitable boundary conditions) the discrete Euler-Lagrange equation

$$\frac{\partial L_t}{\partial q_t} = p_{t+\frac{1}{2}}-p_{t-\frac{1}{2}}.$$

This is the evolution equation. At this point it is not clear whether a solution for $q:\mathbb{Z}\to\mathbb{R}$ will remain on the lattice $\mathbb{Z}a$ if we specify two initial values on the lattice. We shall from now on restrict our considerations to such systems for consistency.

As an example, one may imagine that $q_t$ is a cyclic variable, i.e., that $L_t$ does not depend on $q_t$. We therefore have a discrete global translation symmetry $\Delta q_t=a$. The Noether current is the momentum $p_{t+\frac{1}{2}}$, and the Noether conservation law is that momentum $p_{t+\frac{1}{2}}$ is conserved. This is certainly a nice observation. But this does not necessarily mean that a Noether Theorem is behind.

Imagine that the enemy has given us a global vertical symmetry $\Delta q_t = Y(q_t)\in\mathbb{Z}a$, where $Y$ is an arbitrary function. (The words vertical and horizontal refer to translation in the $q$ direction and the $t$ direction, respectively. We will for simplicity not discuss symmetries with horizontal components.) The obvious candidate for the bare Noether current is

$$j_t = p_{t-\frac{1}{2}}Y(q_t).$$

But it is unlikely that we would be able to prove that $j_t$ is conserved merely from the symmetry $0=S[q+\Delta q] - S[q]$, which would now unavoidably involve higher order contributions. So while we stop short of declaring a no-go theorem, it certainly does not look promising.

Perhaps, we would be more successful if we only discretize time, and leave the coordinate space continuous? I might return with an update about this in the future.

An example from the continuous world that may be good to keep in mind: Consider a simple gravity pendulum with Lagrangian

$$L(\varphi,\dot{\varphi}) = \frac{m}{2}\ell^2 \dot{\varphi}^2 + mg\ell\cos(\varphi).$$

It has a global discrete periodic symmetry $\varphi\to\varphi+2\pi$, but the (angular) momentum $p_{\varphi}:=\frac{\partial L}{\partial\dot{\varphi}}= m\ell^2\dot{\varphi}$ is not conserved if $g\neq 0$.

@Edward 2011-04-13 10:14:28

This paper may be useful for the discrete action ideas you suggest: arxiv.org/abs/nlin.CG/0611058 A "No-Go" Theorem for the Existence of an Action Principle for Discrete Invertible Dynamical Systems. I haven't read through it yet, but it sounds interesting.

@Vladimir Kalitvianski 2011-04-13 22:09:45

If you solve the simple gravity pendulum problem, you can construct two independent conserved quantities. They can be combined in a quantity known as the total energy in this case.

@Qmechanic 2013-07-26 17:41:20

On my to-read-when-I-get-the-time-list: 1. Hydon & Mansfield arxiv.org/abs/1103.3267. 2. Bartosiewicz & Torres arxiv.org/abs/0709.0400 3. Torres arxiv.org/abs/1106.3597. It seems the papers roughly speaking consider discrete horizontal directions, while keeping the vertical directions continuous; and differentiation in horizontal directions are replaced by differences. Horizontal symmetry transformation are made continuous, which seems to ruin the discrete ideology.

@Qmechanic 2014-11-07 16:16:38

Comment to the answer (v7): Since we consider point mechanics (as opposed to field theory) we may replace Noether current $j_t$ with Noether charge $Q_t$.

@Qmechanic 2014-11-07 22:04:16

Comment to the answer (v7): What seems to be true is that for discrete horizontal space and a continuous vertical space and if we only consider continuous vertical symmetry transformation $q_t=\varepsilon Y_t$, then we have a version of Noether's theorem: The full Noether charge $Q_t= p_{t-\frac{1}{2}}Y(q_t)-f^0_t$ is conserved in time on-shell. This relies on the fact that it is possible to prove a version of the algebraic Poincare lemma for finite differences.

@Vladimir Kalitvianski 2011-04-13 17:12:36

Sobering thoughts:

Conservation laws are not related to any symmetry, to tell the truth. For a mechanical system with N degrees of freedom there always are N conserved quantities. They are complicated combinations of the dynamical variables. Their existence is provided with existence of the problem solutions.

When there is a symmetry, the conserved quantities get just a simpler look.

EDIT: I do not know how they teach you but the conservation laws are not related to Noether theorem. The latter just shows how to construct some of conserved quantities from the problem Lagrangian and the problem solutions. Any combination of conserved quantities is also a conserved quantity. So what Noether gives is not unique at all.

@kakaz 2011-04-13 17:36:43

Vladimir - but this N conserved quantities are just initial values of trajectory of motion ( chosen one, from infinite possibilities) so they are trivial, and completely different for different trajectories. Symmetry transform this trajectories among others, so there are interesting constants of motion, not only trivial ones.

@Vladimir Kalitvianski 2011-04-13 18:39:12

Wrong, they are non trivial combinations of dynamical variables expressed via initial data.

@anna v 2011-04-14 06:31:38

@kakaz how is Vladimir's comment different from the paragraph "methods of identifying constants of motion" in wikipedia? en.wikipedia.org/wiki/Constant_of_motion ? Look at the fourth dot. In Goldstein classical mechanics , second edition,page 594, in the discussion of Noether's theorem, there is the clear statement that fulfilling the theorem is sufficient for a conserved quantity, but it is not necessary.

@anna v 2011-04-14 06:34:58

General comment: there must be something missing in the current generation's education. The past three yeasr I have been following scientific blogs, I find that most difficulties and misunderstandings arise because people cannot understand or see the difference between necessary and sufficient conditions. How is mathematics taught at present bemuses me.

@kakaz 2011-04-14 07:53:15

@anna_v - I am the old fashioned guy - and I have obtained old fashioned education;-) I suppose there is fundamental misunderstanding what kind of systems are disputed here (Hamiltonian or Lagrangian mechanics vs general mechanics etc.). In the former integral of motion means that trajectories lays on certain hypersurfaces which forms differential manifolds - and then Hamiltonian flow defines sufficient structure for forming Noether theorem (such mechanism is called foliation, please take a look here: en.wikipedia.org/wiki/… ).

@kakaz 2011-04-14 07:55:53

Cont. Then we have theory that for Hamiltonian systems if N integrals of motion exists - system is "integrable" So Vladimir statement in a case of Hamiltonian dynamics is wrong. Of course that there exists constants of motion not related to symmetry. But they are not related to structure of phase space and there is no foliation so in certain meaning they are particular, accidental one. And the may be represented ( after mathematical transformation) as initial conditions of well defined system.

@Tobias Kienzler 2011-04-14 08:36:41

I think you read to much in my question - I did not assume that the inversion of Noether's theorem, i.e. "For every conserved quantity there exists a continuous symmetry", was true (though I wonder If all conserved quantities of a system are known, can they be explained by symmetries?)

@anna v 2011-04-14 08:46:04

@Tobias Kienzler I was commenting on the specific answer by Vladimir, and the discussion to this. It is tangent to your question, which was between discrete and continuous systems. I think you got good answers to that, and I learned something from them.

@Tobias Kienzler 2011-04-14 08:52:49

@annav: so did I (learn something). with my previous you I actually meant @Vladimir, but your discussion is very interesting in itself. I have to admit there was rather a shortcoming in group theory in most lectures, it seems to be not everyone's taste :-7

@anna v 2011-04-14 15:40:28

I would like to add that Goldstein whom I referenced above, in the chapter on Noether's theorem, discusses the conservation laws outside the theorem, and connects them with soliton solutions. He also derives a form for discrete systems , where only time remains a parameter.

@Tobias Kienzler 2011-04-15 07:00:56

@anna v: thanks, I'll have a look at it. wouldn't this make an answer of its own?

@Daniel 2011-04-13 01:29:06

As was said before, this depends on what kind of 'discrete' symmetry you have: if you have a bona fide discrete symmetry, as e.g. $\mathbb{Z}_n$, then the answer is in the negative in the context of Nöther's theorem(s) — even though there are conclusions that you can draw, as Moshe R. explained.

However, if you're talking about a discretized symmetry, i.e. a continuous symmetry (global or local) that has been somehow discretized, then you do have an analogue to Nöther's theorem(s) à la Regge calculus. A good talk introducing some of these concepts is Discrete Differential Forms, Gauge Theory, and Regge Calculus (PDF): the bottom line is that you have to find a Finite Difference Scheme that preserves your differential (and/or gauge) structure.

There's a big literature on Finite Difference Schemes for Differential Equations (ordinary and partial).

@user3080 2011-04-13 00:05:44

Maybe,

http://www.technologyreview.com/blog/arxiv/26580/

I am by no means an expert, but I read this a few weeks ago. In that paper they consider a 2d lattice and construct an energy analogue. They show it behaves as energy should, and then conclude that for this energy to be conserved space-time would need to be invariant.

@Lagerbaer 2011-04-12 17:58:18

You mentioned crystal symmetries. Crystals have a discrete translation invariance: It is not invariant under an infinitesimal translation, but invariant under translation by a lattice vector. The result of this is conservation of momentum up to a reciprocal lattice vector.

There is an additional result: Suppose the Hamiltonian itself is time independent, and suppose the symmetry is related to an operator $\hat S$. An example would be the parity operator $\hat P|x\rangle = |-x\rangle$. If this operator is a symmetry, then $[H,P] = 0$. But since the commutator of an operator with the Hamiltonian also gives you the derivative, you have $\dot P = 0$.

@mathphysicist 2010-08-24 20:26:14

No, because discrete symmetries have no infinitesimal form which would give rise to the (characteristic of) conservation law. See also this article for a more detailed discussion.

@Tobias Kienzler 2010-08-26 08:32:02

Unfortunately I can't access that article. But your answer sounds plausible. I still wonder if discrete symmetries offer some other advantage (compared to not having any symmetry at all) besides Bloch waves.

@Lagerbaer 2011-04-12 19:37:15

Who says that conservation laws can arise onyly from infinitesimal forms?

@SRS 2018-02-01 12:54:38

@Lagerbaer But proof of Noether's theorem is based on having an infinitesimal version of the symmetry transformation which does always exist for a continuous symmetries.

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