# Dynamical symmetry groups

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1. Sep 9, 2015

### phoenix95

Hi,
I am learning classical mechanics right now, Particularly Noether's theorem. What I understood was that those kinds of transformations under which the the Hamiltonian framework remains unchanged, were the key to finding constants of motion.
But here are my Questions:
1. What is dynamical symmetry?
2. What is a symmetry group?
3. What are the matrices group representing these things? please explain. (Its just that I am confused between the group SO(n) and SU(n), which are said to represent symmetry group and dynamical symmetry respectively)
4. Is there a separate transformation called dynamical symmetry transformation?
It is possible that what I am saying is not making sense... Can anybody guide me, please?
I would really appreciate any help....

2. Sep 9, 2015

3. Sep 9, 2015

### Andy Resnick

4. Sep 10, 2015

### phoenix95

Let me say what I have understood by far...
Symmetry is some special property of the dynamical system which renders it invariant to the action of some transformations...,
symmetry group is is the set of all transformations under which the system remains invariant...
Am I right?
I supposed to study relativistic classical mechanics later, so I don't think I would be needing these
https://en.wikipedia.org/wiki/Lorentz_group
https://en.wikipedia.org/wiki/Poincaré_group right now....

I know about scaling transformations, but I don't see how the system is invariant to scaling.... I mean if I scale velocity/momentum in phase space, then the Lagrangian or Hamiltonian should change right? Since the kinetic energy is changed..?? Or am I wrong?

OK, but what are dynamical symmetry transformations??
How are they represented by the group SU(3)?

And no disrespect, but could tell me something rather than showing some wikipages...?? Because sometimes they get too technical that I don't understand them at all...

5. Sep 10, 2015

### DEvens

No disrespect back, but you need to ask specific questions. Shotgun collections of questions indicating blurred understanding isn't really going to get you a sharply targeted response. You claimed to be studying Noether's theorem. The degree of math required to understand that is pretty much in line with the wiki pages I linked to.

6. Sep 12, 2015

### samalkhaiat

Okay, I will assume you know the followings: Infinitesimal Point and Canonical transformations, Poisson brackets, The Variation (Hamilton’s) principle, and a bit of Group theory. If you don’t then don’t bother reading my post.

1) Noether Theorem in the Lagrangian Formalism

In general, symmetry group is the set of all transformations (on the dynamical variables) which leave the action integral unchanged (i.e. invariant).
Now recall the following: If $F(q,t)$ is some function such that $$\int_{t_{1}}^{t_{2}} dt \frac{dF}{dt} = 0 ,$$ then any two Lagrangians, $L$ and $\bar{L}$, related by $$\bar{L}(q,\dot{q}) = L(q,\dot{q}) + \frac{d}{dt}F(q,t) ,$$ are said to be dynamically equivalent, i.e. they lead to the same equation of motion. Therefore, if under some infinitesimal point transformation $$q \to q + \delta q ,$$ the Lagrangian changes (transforms) according to, $$\delta L(q,\dot{q}) = \frac{d}{dt} G(\delta q , t) , \ \ \ \ \ (1)$$ then the action integral remains unchanged and the point transformation is, therefore, a symmetry transformation.
Another expression for the variation $\delta L$ can be obtained as follow $$\delta L = \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \frac{d}{dt} \delta q .$$ If we differentiate the second term by parts and introduce the canonical momentum, $p = \partial L / \partial \dot{q}$, we obtain $$\delta L = \left( \frac{\partial L}{\partial q} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})\right) \delta q + \frac{d}{dt} (p \ \delta q) . \ \ \ (2)$$ Subtracting (1) from (2) leads to the following (Noether) identity $$\left( \frac{\partial L}{\partial q} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}})\right) \delta q + \frac{dC}{dt} = 0 , \ \ \ (3)$$ where
$$C = p \delta q - G(\delta q ,t) . \ \ \ \ (4)$$ Thus, on actual trajectories (i.e. when the dynamical variables $q$’s satisfy the Euler-Lagrange equations), the Noether identity (which follows from the symmetry of the action integral) tells us that $C$ is constant of motion. For each parameter of symmetry group there exits one constant of motion. Indeed, if we factor out the parameters, $\delta q = \epsilon^{a} \delta_{a} q$, ($a = 1,2, \cdots , r$) from Eq(4), we find $$C_{a} = p \cdot \delta_{a}q - G_{a}(q,t) . \ \ \ \ (5)$$

2) Noether Theorem in the Hamiltonian Formalism

Using the Legendre transformation $$H(q,p) = p \cdot \dot{q} - L(q,\dot{q}) , \ \ \ \ (6)$$ we can carry out the variations induced by the infinitesimal canonical transformations $$q \to q + \epsilon^{a}\delta_{a} q , \ \ p \to p + \epsilon^{a}\delta_{a} p ,$$ on both sides of Eq(6):
$$\frac{\partial H}{\partial q} \cdot \delta_{a} q + \frac{\partial H}{\partial p} \cdot \delta_{a} p = \dot{q} \cdot \delta_{a} p - \dot{p} \cdot \delta_{a} q + \frac{d}{dt}(p \cdot \delta_{a} q) - \delta_{a} L . \ \ (7)$$ If the transformations are symmetry transformations, then the variation of action integral vanishes. Thus, we can substitute Eq(1) in Eq(7) and find the Hamiltonian version of Noether identity $$\left( \dot{q} - \frac{\partial H}{\partial p} \right) \cdot \delta_{a} p - \left( \dot{p} + \frac{\partial H}{\partial q} \right) \cdot \delta_{a} q + \frac{dC_{a}}{dt} = 0 .$$ Again, when the (Hamilton) equations of motion are satisfied, the $C_{a}$’s, as given by Eq(5), are the corresponding constants of motion. In this case, i.e., when the canonical pair $(q,p)$ satisfies Hamilton’s equations, we have $$\frac{dC_{a}}{dt} = \frac{\partial C_{a}}{\partial t} + \{ C_{a} , H \} = 0 .$$

3) Canonical Transformations and Integrability

As we said before, in the Hamiltonian formalism, the allowed transformations are the so-called canonical transformations on phase space variables $\mathbb{X} = (q,p)$. In one degree of freedom (for simplicity), the transformations $(q,p) \to (Q,P) \equiv \mathbb{Y}$ are called canonical if and only if they leave the fundamental Poisson brackets unchanged, i.e.,
$$\{q,p\} = 1, \ \{q,q\} = \{p,p\} = 0 ,$$ if and only if $$\{Q , P \}_{(q,p)} = 1, \ \{Q,Q\}_{(q,p)}=\{P,P\}_{(q,p)}=0 .$$
If we denotethe $(2\times 2)$ Jacobian matrix of the transformation by $\partial \mathbb{Y}/ \partial \mathbb{X}$, then (with a bit of work) we can show that the above conditions on the transformations become equivalent to $$(\frac{\partial \mathbb{Y}}{\partial \mathbb{X}})^{T} \ J \ (\frac{\partial \mathbb{Y}}{\partial \mathbb{X}}) = J , \ \ \ \ \mbox{Det}(\frac{\partial \mathbb{Y}}{\partial \mathbb{X}}) = +1 ,$$ where $$J = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} .$$
The set of all unimodular $(2n \times 2n)$ real matrices (i.e. with unit determinant) satisfying $M^{T} J M = J$ forms a matrix Lie group called the symplectic group $Sp (2n , \mathbb{R})$. Thus, our canonical transformations, $(q,p) \to (Q(q,p),P(q,p))$ on the even dimensional phase space $\mathbb{R}^{2}$ belong to the symplectic group $Sp (2 , \mathbb{R})$. Notice that $J$ itself is a symplectic matrix (prove it). This means that transformations with Jacobian matrix equal to $J$ are canonical transformations. Do you know what kind of transformations are these?

A Hamiltonian system is integrable if there exist a closed set $\{C_{i}\}$ of constants of motion (other than the Hamiltonian) such that: A) they Poisson-commute with the Hamiltonian $$\{ C_{i} , H \} = 0, \ \forall i ,$$ and B) they form a closed Poisson-Lie algebra $$\{ C_{i} , C_{j} \} = f_{ij}^{k} C_{k} .$$ The so-called dynamical (some times called hidden) symmetry group is the Lie group generated by the set $\{C_{i}\}$. In some Hamiltonian systems, the dynamical group is a smaller than the simplectic group $Sp (2n , \mathbb{R})$ of canonical transformations. In fact, it is a subgroup of both $Sp (2n , \mathbb{R})$ and the (geometrical) symmetry group of the Hamiltonian. This means (as we will see soon) that i) Not ALL canonical transformations are symmetry transformations, and ii) Not ALL (geometrical) symmetries of the Hamiltonian are canonical. That is to say: if $\mathcal{G}$ is the (geometrical) symmetry group of the Hamiltonian (i.e. it leaves the Hamiltonian invariant $\mathcal{G} H = H$), then the dynamical (also called accidental) symmetry is a group isomorphic to the intersection $\mathcal{G} \cap Sp(2n , \mathbb{R})$.

4) Geometrical and Dynamical Symmetries

Consider the equation $$E = x^{2} + p^{2} .$$
We want to know the symmetry group of $E$, i.e., what kind of transformations keep $E$ invariant? Clearly $E = \mbox{constant}$ is a circle (the 1-dimensional sphere $S^{1}$). This circle is symmetric under all rotations in the $(xp)$ plane. The set of all such rotations forms a Lie group called the Special Orthogonal group in 2-dimension, or $SO(2)$. So, we made an obvious geometric “discovery”: $E$ is invariant under $SO(2)$.
Let us ask similar question about $$H = P_{1}^{2} + P_{2}^{2} + P_{3}^{2} .$$ Here, $H = \mbox{constant}$ is the 2-dimensional sphere $S^{2}$. This sphere stays the same when you rotate it in the planes $P_{1}P_{2}$, $P_{2}P_{3}$ and $P_{3}P_{1}$. The set of all rotations in 3 dimensions form the Special Orthogonal group $SO(3)$. And we made another great geometric discovery: $SO(3)$ is the symmetry group of $H$.
Going to higher dimensions is easy and it may look boring to you. But believe me it is not. As we will see soon, things get really interesting and strange once we hit the number 4 and investigate the symmetries.
As before, we consider the equation $$2H = q_{1}^{2} + p_{1}^{2} + q_{2}^{2} + p_{2}^{2} . \ \ \ (8)$$ Notice that $H$ is the Hamiltonian of 2D isotropic oscillator. The face space of this system is the 4-dimensional Euclidean space $\mathbb{R}^{4}$ in which $(q_{1},p_{1},q_{2},p_{2})$ represents a local coordinate system. Thus, as a Hamiltonian system, the allowed transformations must be canonical, i.e. belong to the simplectic group of our system which is $Sp(4, \mathbb{R})$. However, our Hamiltonian is not invariant under ALL $Sp(4, \mathbb{R})$ transformations. On the other hand, the Hamiltonian is invariant under its geometric symmetry group $SO(4)$: the hyper-surface $H= \mbox{constant}$ is the 3-sphere $S^{3}$ which is invariant under the 4-dimensional rotation group $SO(4)$. But, not ALL $SO(4)$ transformations are canonical. Thus, the real dynamical symmetry group must be isomorphic to the intersection $$SO(4) \cap Sp(4, \mathbb{R})$$. In fact, we know from group theory that any transformation in $\mathbb{R}^{2n}$ which is both orthogonal and simplectic must be unitary. In our case, the dynamical symmetry group is the group of Special Unitary matrices in 2 dimensions: $$SO(4) \cap Sp(4, \mathbb{R}) \cong SU(2) . \ \ \ (9)$$

If you are not shocked by this result, then you should pay attention to the following: 1) The group $SU(2)$, more or less, is the same as the group of rotations in three dimensions, $SO(3)$, the symmetry group of the sphere $S^{2}$! The physical motion of the 2D oscillator is confined in the configuration space of the oscillator which is the plane $q_{1}q_{2}$. So, it is surprising to see a $3D$ rotational symmetry in a system whose physical space is only $2D$.
2) Things get more confusing in the Lagrangian formalism where the symmetry group is just the $2D$ rotational group $SO(2)$ which corresponds to the physical rotations in the physical configuration space and generated by only one constant of motion, the orbital angular momentum, $$L = q_{1}p_{2} - q_{2}p_{1} .$$ But $SU(2) = SO(3)$ has three generators and, therefore, three constants of motion. So, what are the extra two generators? How can we interpret them physically? And, can we visualize them? Of course on PF, I can only give partial answers to those questions.
Digging up the dynamical symmetry group, which is normally hidden in the Hamiltonian, is a very difficult and highly non-trivial task. In fact, it has a long history and still is an active research in both classical and quantum theories of fields and particles. More difficult is the proof that the action of $SU(2)$ on phase space $\mathbb{R}^{4}$ is indeed given by transformations that leave invariant both the Hamiltonian and the Poisson brackets. So, I am not going to give you a rigorus and step by step proof of Eq(9). And I will certainly not going to derive the explicit form of the canonical transformations.
All we need is to construct a set of 3 constants of motion (quadratic in the coordinates), which commute with the Hamiltonian and generate the Lie algebra of the group $SU(2)$. There are many ways to construct that set, but I will just throw them at you using a bit of physics and a bit of mathematics. If you are interested in the details, I will be happy to help you.

Okay, we only have the Hamiltonian, so let us start by rewriting it in the form
$$H = \frac{1}{2}\left( p_{1}^{2} + q_{1}^{2} \right) + \frac{1}{2}\left( p_{2}^{2} + q_{2}^{2} \right) .$$ Clearly, this is the sum of energies of the two coordinates. So, let us try our luck with the energy difference of the two coordinates:
$$J_{1} = \frac{1}{2}\left( p_{1}^{2} + q_{1}^{2} \right) - \frac{1}{2}\left( p_{2}^{2} + q_{2}^{2} \right) .$$ Also, it is easy to show that $$\{ H , J_{1} \} = 0 ,$$ which is a good sign. Next, we must not forget the good old orbital angular momentum of the system $$J_{3} = q_{1} p_{2} - q_{2} p_{1} ,$$ which also Poisson commutes with the Hamiltonian $$\{ H , J_{3} \} = 0 .$$ The last one $J_{2}$ is a tough one. Recall that the energy of a harmonic oscillator depends only on the sum of the squares of the semi-axes of its orbital ellipse. Indeed, our last “generator” does this and a bit more, $$J_{2} = p_{1}p_{2} + q_{1}q_{2} .$$
In fact, it induces a small change in the eccentricity of the orbital ellipse while preserving the orientation of the semi-axes, and preserving the sum of the squares of their lengths.
Finally, I will leave it to you to show that $J_{i}, \ i = 1,2,3$ generate the Poisson-Lie algebra of $SU(2)$ $$\{ \frac{J_{i}}{2} , \frac{J_{j}}{2} \} = \epsilon_{ijk} \frac{J_{k}}{2} ,$$ and $$\{ H , J_{i} \} = 0 , \ \ \forall i .$$ The algebra means that $J_{i}$ is an abstract “angular momentum”. You can also show that the square of the Hamiltonian is the invariant Casmir of the group $$H^{2} = J^{2}_{1} + J^{2}_{2} + J^{2}_{3} .$$ This allows us to interpret $(H , J_{i})$ as a null “4-vector”.

Good Luck

Sam

7. Sep 12, 2015

### samalkhaiat

To be fair, these links do not address any of the particular questions raised in the OP.

8. Sep 12, 2015

### samalkhaiat

This is even worst! Do you even know that the Renormalization group is not a group, let alone symmetry group. So, what exactly is your reason for posting that link?

9. Sep 13, 2015

### phoenix95

@samalkhaiat Thanks a lot sir. That explains a lot more than I could imagine. Thank you for time. I'm really sorry for my post wasn't informative enough . All My questions are cleared now.

10. Sep 13, 2015

### Andy Resnick

Why so angry? The renormalization group is involved with dynamical scaling laws in condensed matter- critical points, for example.