From Wikipedia:

The dynamical symmetry group of the $n$-dimensional quantum harmonic oscillator is the special unitary group $SU(n)$. As an example, the number of infinitesimal generators of the corresponding Lie algebras of $SU(2)$ and $SU(3)$ are three and eight respectively. This leads to exactly three and eight independent conserved quantities (other than the Hamiltonian) in these systems. The two dimensional quantum harmonic oscillator has the expected conserved quantities of the Hamiltonian and the angular momentum, but has additional hidden conserved quantities of energy level difference and another form of angular momentum

How can I show that $\mathbf{H} = \hbar \omega \left(\vec{a}^\dagger \vec{a} + \frac{N}{2}\right)$ has dynamical symmetry of $SU(N)$?

Which operator/generator do I need to show to commute with $H$?

  • 1
    $\begingroup$ I would guess they are things like $a_i^\dagger a_i - a_{i+1}^\dagger a_{i+1}$ and, for $i \neq j$, $a_i^\dagger a_j +a_j^\dagger a_i$ or $\sqrt{-1} (a_i^\dagger a_j - a_j^\dagger a_i)$. $\endgroup$ – user2309840 Mar 10 '17 at 2:55
  • $\begingroup$ The answer to several of your recent questions is the Jordan map. Mastery of its properties resolves any and all questions of this ilk. $\endgroup$ – Cosmas Zachos Mar 11 '17 at 2:04

Let $$ \vec b=U\,\vec a\, ,\qquad \vec b^\dagger =\vec a^\dagger U^\dagger $$ then $$ \vec b^\dagger \cdot \vec b= \vec a^\dagger U^\dagger U\,\vec a=\vec a^\dagger \vec a \quad \Leftrightarrow \quad U^\dagger U=\hat 1\, , $$ which defines $U$ as unitary matrix.

Actually, $SU(N)$ is NOT the dynamical symmetry group of the harmonic oscillator. This dynamical symmetry group is $Sp(N,\mathbb{R})$ (also called $Sp(2N,\mathbb{R})$ depending on notations). $U(N)$ (or $SU(N)$) is the simply the symmetry group of the degenerate states of the H.O.

$Sp(N,\mathbb{R})$ is the real symplectic group in $N$ dimensions, with algebra $sp(N,\mathbb{R})$ spanned by $\{a_k^\dagger a_j^\dagger, a_k^\dagger a_j, a_k a_j\}$ with $k,j=1,\ldots, N$. The subset $\{ a_k^\dagger a_j\}$ spans $u(N)$, making $u(N)$ a subalgebra of $sp(N,\mathbb{R})$.

Note that $sp(N,\mathbb{R})$ is the dynamical algebra because, in terms of $x$ and $p$, it is spanned by $x_kx_i$, $p_kp_i$ and $x_kp_i+p_kx_i$. Any observable (such as the kinetic energy) expressed as a polynomial in these basic observables will act within a single $sp(N,\mathbb{R})$ irrep.

Thus wiki is not quite correct. In general the symmetry algebra of $H$ includes all operators that commute with $H$ and close on an algebra, whereas the dynamical algrebra contains $H$ and itssymmetry algebra, but also includes operators that need not commute with $H$ but still close on a (larger) algebra. For 1d h.o. this would include $x^2$, $p^2$ and $xp+px$ which do not all commute with $H$. Clearly $H$ is in there as a linear combination of $x^2$ and $p^2$.

  • $\begingroup$ Could you give a definition of the dynamical symmetry group? I also can't find any references that talk about the $Sp(N,\mathbb{R})$ symmetry of the harmonic oscillator. Can you provide any? $\endgroup$ – level1807 Oct 22 '17 at 14:30
  • 1
    $\begingroup$ @level1807 1. Moshinsky and Quesne. "Linear canonical transformations and their unitary representations." J. Math. Phys 12.8 (1971): 1772-1780 2. Rosensteel and Rowe. "On the algebraic formulation of collective models III. The symplectic shell model of collective motion." Ann. Phys. (N.Y.) 126.2 (1980): 343-3 3. Rowe "Microscopic theory of the nuclear collective model." Rep. Prog. Phys 48.10 (1985): 1419. 4. Chap.3 of Rowe and Wood. Fundamentals of nuclear models: Foundational models. World Scientific 2010. There is also lost of work in the 60’s and 70s by Barut et al. $\endgroup$ – ZeroTheHero Oct 22 '17 at 14:43
  • $\begingroup$ Thank you! How different is the situation in the classical oscillator? Is the dynamical group just $Sp\cap SO=U(N)$? $\endgroup$ – level1807 Oct 22 '17 at 14:50
  • $\begingroup$ I’ve not seen this notion discussed for classical systems but I will ask. $\endgroup$ – ZeroTheHero Oct 22 '17 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.