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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 $H$?

0x90
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2 Answers2

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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 $p^2$.

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

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

user398568
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