5
$\begingroup$

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 $H$?

Improve this question
$\endgroup$
2
  • 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
    Commented Mar 10, 2017 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
    Commented Mar 11, 2017 at 2:04

2 Answers 2

Reset to default
10
$\begingroup$

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 $p^2$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$
    – Alex Bogatskiy
    Commented Oct 22, 2017 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
    Commented Oct 22, 2017 at 14:43
  • $\begingroup$ Thank you! How different is the situation in the classical oscillator? Is the dynamical group just $Sp\cap SO=U(N)$? $\endgroup$
    – Alex Bogatskiy
    Commented Oct 22, 2017 at 14:50
  • $\begingroup$ I’ve not seen this notion discussed for classical systems but I will ask. $\endgroup$
    – ZeroTheHero
    Commented Oct 22, 2017 at 15:05
  • 1
    $\begingroup$ I came across this wonderful answer while thinking about the question I asked here. It's pretty cool that both Coulomb and HO in any dimension have residual symmetries. Won't be surprised if Coulomb in 2D ($\log(r)$) has it as well. $\endgroup$
    – mavzolej
    Commented Feb 2, 2022 at 21:21
-2
$\begingroup$

It is easy to show that the following operator commutes with the hamiltionian

$$A_{ij} = \frac{1}{2} \hbar \omega (a_i a_j^+ + a_i^+ a_j)$$

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ ... showing what exactly? How are your $A_{ij}$ related to $U(N)$ or $SU(N)$? $H$ also commutes with the angular momenta, but that doesn't make the symmetry group $SO(N)$. $\endgroup$
    – ZeroTheHero
    Commented Mar 28 at 14:44
  • 1
    $\begingroup$ @ZeroTheHero if $H$ commutes with angular momenta, then one of the symmetry groups is in fact $SO(N)$, it just so happens that the Hilbert space is a reducible unitary representation of $\mathrm{SO}(N)$ where every element is mapped to the identity $\endgroup$
    – John
    Commented Apr 1 at 15:47
  • $\begingroup$ @John certainly not to the identity! The defining 3-dimensional irrep of SU(3) - sometimes called the quark rep, also labelled by $(1,0)$ in the Dynkin notation - contains the $L=1$ irreps of SO(3) and elements are certainly not mapped to the identity. Finding a set of operators - in the case above there are 6 - which do not even close on an algebra does not show anything. $\endgroup$
    – ZeroTheHero
    Commented Apr 1 at 16:39
  • $\begingroup$ Agreed that $SO(3)$ is a subgroup of $U(3)$ but "the symmetry group" is commonly defined as the largest group leaving the Hilbert space invariant. $\endgroup$
    – ZeroTheHero
    Commented Apr 1 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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