# What conserved quantity causes degeneracy of spectrum in rational polygonal billiards?

Apart from Laplace-Runge-Lenz vector conservation in Coulombic and something similar in harmonic and other central potentials, something leads to existence of periodic trajectories in such systems as rational polygonal billiards (although not all trajectories are periodic)$^\dagger$.

I believe in quantum case and in the case of vibrating membrane this results in degeneracy such that e.g. for a square membrane non-standing-wave eigenmodes are possible, like these:

This degeneracy isn't lifted even in the presence of a particular form of linearly-changing potential, see e.g. this question.

So I suppose there must be some quantity, which is conserved in such billiards. Its quantum operator, as I understand, should be the generator of the wave motions presented above (like momentum operator is the generator of translations, angular momentum operator is the generator of rotations, etc.).

So what is this conserved quantity which leads to periodic trajectories?

$^\dagger$ In fact, after some thinking I'm not sure that periodic trajectories have anything to do with this. They are possible even in irrational-side-ratio rectangular billiard, it's just that their projections of velocities also have irrational ratio to make rationally-related periods of oscillation in different directions.

• May we have an Hamiltonian/explicit system description? I may be able to help you on that one. – G. Bergeron Dec 11 '16 at 11:33
• @G.Bergeron the membrane vibration animations were generated from solutions of Helmholtz equation in a square: $u_{nm}(x,y)=\sin(nx)\sin(my)$. Helmholtz equation is what Schrödinger's equation in 2D square with constant potential represents. I.e. the Hamiltonian is merely kinetic energy operator with zero Dirichlet conditions on boundaries of a square. The animations represent real part of time-dependent solutions $(u_{1,2}(x,y)+iu_{2,1}(x,y))\exp(i\omega t)$ and $(u_{3,1}(x,y)+iu_{1,3}(x,y))\exp(i\omega t)$. – Ruslan Dec 11 '16 at 11:54
• So it's simply 2D free particle in the box? There's obvious discrete simmetry, I dub it $\hat{R}$ in the form of $\pi/2$ rotations. In QM it corresponds to exchange $n\leftrightarrow m$ in $u_{n,m}$ It has either one eigenvalue $\hat{R}u_{m,m}=u_{m,m}$ and then there's no degeneracy, or two eigenvalues with eigenstates $\hat{R}\Big(u_{m,n}+u_{n,m}\Big)=+\Big(u_{m,n}+u_{n,m}\Big)$ and $\hat{R}\Big(u_{m,n}-u_{n,m}\Big)=-\Big(u_{m,n}-u_{n,m}\Big)$. It can't give all degeneracy though because of Pythagorean triples $n^2+0^2=a^2+b^2$. Don't really know what is interpretation for that symmetry. – OON Dec 12 '16 at 10:54
• @OON we can break rotational symmetry by adding $(x-y)Q$ potential, and there'll still be degeneracy — see this unanswered question. So the symmetry is trickier than it appears. – Ruslan Dec 12 '16 at 11:07
• @Ruslan Interesting. However note that initial problem has symmetry under both exchange (coming from aforementioned rotations) $x\leftrightarrow y$ AND reflections $x\mapsto -x$ and $y\mapsto -y$. Your new term breaks exchange and reflections individually but not combined $x\mapsto -y$, $y\mapsto -x$. Also concerning degeneracy it's not only Pythagorean triples but also e.g. $7^2+4^2=8^2+1^2=65$, sometimes multiplying degeneracy in two or even three. I wonder what your extra term does with them. – OON Dec 12 '16 at 11:32

The case of square membrane is simple. Whenever the total Hamiltonian $\hat H$ is a sum of independent Hamiltonians for each direction, i.e.

$$\hat H=\hat H_x+\hat H_y,$$

both energies of the subsystems are conserved: $E_x$ and $E_y$. In the case of identical Hamiltonians for $x$ and $y$ we also have $x\leftrightarrow y$ exchange symmetry, i.e. if we consider the operator $\hat S$ such that

$$(\hat Sf)(x,y)=f(y,x),$$

then total Hamiltonian $\hat H$ commutes with it. But Hamiltonians $\hat H_x$ and $\hat H_y$ don't. For example, for $\hat H_x=-\frac1m\partial_x^2(\cdot)+U(x)(\cdot)$ we have

$$(\hat H_x\hat Sf)(x,y)=-\frac1m f^{(0,2)}(y,x)+U(x)f(y,x),$$ $$(\hat S\hat H_xf)(x,y)=-\frac1m f^{(2,0)}(y,x)+U(y)f(y,x),$$

$$\left(\left(\hat H_x\hat S-\hat S\hat H_x\right)f\right)(x,y)=\frac1m\left(f^{(2,0)}(y,x)-f^{(0,2)}(y,x)\right)+(U(x)-U(y))f(y,x).$$

So we have the following pairs of non-commuting conserved quantities:

• energy $E_x$ and $(x-y)$ parity $S$,

• energy $E_y$ and $(x-y)$ parity $S$.

Since all they individually commute with total Hamiltonian, there must be degeneracy in the spectrum. These conserved quantities exist not only for simple particle in a 2D box, but also for the case of $(x-y)Q$ potential, which separates into $Qx$ and $Qy$ parts (after coordinate change $y\to-y$), as well as for any other potential of the form $U(x)+U(y)$.

I guess that for general rectangular membrane one would have to somehow generalize $\hat S$ to explain the degeneracy in the case of rational sides ratio.