Meaning of "the symmetry group of an $N$-dimensional quantum isotropic oscillator is $U(N)$"

Symmetry of this system has been discussed here but I'm still confused.

Consider a $$N$$-dimensional isotropic harmonic oscillator, with hamiltonian

$$H = \hbar \omega \left(a^\dagger_i a_i + \frac{N}{2} \right).$$

$$a^\dagger_i$$ and $$a_i$$ being the creation and annihilation operators for dimension $$i$$, and using summation convention on repeated indices.

Some textbooks I've read say that the system possesses an $$U(N)$$ symmetry. The given justification is that if $$M$$ is an $$N \times N$$ unitary matrix then the transformation

$$a_i \to a'_i = M_{ij} a_j \quad i=1,\dots,N \tag{1}$$

leaves the hamiltonian invariant, since $$(M_{ij} a_j)^\dagger M_{ik} a_k = a^\dagger_j {M}^*_{ij} M_{ik} a_k = a^\dagger_j \delta_{jk} a_k = a^\dagger_j a_j .$$ However I've been taught that symmetries are realized in Quantum Mechanics through (anti-)unitary operators on a Hilbert space, $$U : \mathcal{H} \to \mathcal{H}$$. What I don't understand is how the previous symmetry trasformation is realized in practice.

I suspect it's something like an operator $$U$$ which acts on the annihilation operator as

$$$$U a_i U^\dagger = M_{ij} a_j \tag{2} \\ \implies U a^\dagger_i U^\dagger = (U a_i U^\dagger)^\dagger = M^*_{ij} a^\dagger_j$$$$

where $$M$$ is a $$N\times N$$ unitary matrix, $$M \in U(N)$$. Which leaves me to a first question: does there exist, for every matrix $$M \in U(N)$$, a unitary operator $$U:\mathcal{H} \to \mathcal{H}$$ which transforms $$a_i$$ like (2)?

In some special cases, e.g. parity which changes sign of $$x_i$$ and $$p_i$$ and thus of $$a_i$$ and $$a^\dagger_i$$ the answer is obviously yes, but I'm not sure it works in the general case.

If it's true, I can understand the original claim, being: to each matrix $$M \in U(N)$$ we can define an "associated" operator $$U$$ in $$\mathcal{H}$$ such that $$U H U^\dagger = H$$ because $$U H U^\dagger = \hbar \omega \left( U a_i^\dagger a_i U^\dagger + \frac{N}{2} \right) = \hbar \omega \left( U a_i^\dagger U^\dagger U a_i U^\dagger + \frac{N}{2} \right) = \hbar \omega \left(a^\dagger_j a_j + \frac{N}{2} \right) = H$$ so there exists a group of unitary transformations homomorphic to $$U(N)$$ which leave $$H$$ invariant, i.e. which represent a symmetry of the system.

Perhaps, instead, the correct $$U$$ is not the one in eqn (2) but another one. Is the existence of a unitary transformation in $$\mathcal{H}$$ that "mimics" the "change of basis" in eqn (1) implied by Wigner's theorem somehow instead?

• The given answer outlines the essential idea behind the Stone-von Neumann theorem (which is not restricted to linear transformations). If you are interested in constructing $U$ from the operators $a_i,a_i^\dagger$, you can check out my answer to a more general question (in your case $B=0$).
– LPZ
Commented Apr 18, 2023 at 16:09
• @LPZ thank you! I'm trying to understand the connection with the theorem. Are you referring to the fact that equation (1) in my question implies $x_i \to x'_i$ and $p_i \to p'_i$ which satisfy $[x'_i, p'_j] = i \hbar \delta_{ij}$? Is it sufficient to conclude that the transformation of the operators is unitary? Commented Apr 18, 2023 at 21:04
• Yes, you can reason directly with the $a_i$. Since $M$ is unitary, $$[a_i,a_j^\dagger]=[a_i',a_j'^\dagger] = \delta_{ij}$$ (in fact it is not only sufficient, it is also necessary) the SvN theorem guarantees the existence of a unitary operator liking the two: $$a_i' = Ua_iU^\dagger$$ and the proof essentially uses the idea of E. Anikin's answer. Note that the linear relation between $a_i'$ and $a_i$ is not needed, only the conservation of the CCR's is.
– LPZ
Commented Apr 18, 2023 at 21:40

I've made a sketch of a proof that for every unitary matrix $$M_{ij}$$, there exists an operator in the Hilbert space $$\mathcal{H}$$ which transforms the annihilation operators as $$$$a_i \to M_{ij} a_j.$$$$ Such an operator $$U$$ can be constructed in the following way. The Hilbert space of $$N$$ oscillators is spanned by the Fock states $$|k_1, \dots k_N\rangle$$ with the operators $$a_i, a^\dagger_i$$ acting on them. Now, let us define a new set of creation/annihilation operators $$a_i, a'^\dagger_i$$ as $$$$a'_i = M_{ij}a_j.$$$$ As $$M_{ij}$$ is unitary, the operators $$a'_i, a'^\dagger_i$$ obey the commutation relations $$$$[a'_i, a'^\dagger_j] = \delta_{ij},$$$$ $$$$[a'_i, a'_j] = 0,$$$$ $$$$[a'^\dagger_i, a'^\dagger_j] = 0.$$$$ (This can be checked by a simple calculation.) Therefore, they can be used to construct a new Fock basis for the system of coupled oscillators: $$$$|k_1\dots k_N\rangle' = \frac{1}{\sqrt{k_1!\dots k_N!}} (a'^\dagger_1)^{k_1}\dots (a'^\dagger_N)^{k_N}|0\dots 0\rangle.$$$$ The unitary $$U$$ which transforms a basis set $$\{|k_1\dots k_N\rangle\}$$ to $$\{|k_1\dots k_N\rangle'\}$$ is the unitary that you asked for. Indeed, let us consider the action of $$Ua_iU^\dagger$$ on an arbitrary Fock state: $$$$U a_i U^\dagger |k_1\dots k_N\rangle' = U a_i |k_1\dots k_N\rangle = U\sqrt{k_i} |k_1, \dots, k_i - 1, \dots , k_N\rangle = \sqrt{k_i}|k_1\dots, k_i - 1,\dots k_N\rangle'$$$$ On the other hand, $$$$a_i'|k_1\dots k_N\rangle' = \sqrt{k_i}|k_1\dots, k_i - 1,\dots k_N\rangle'$$$$ by construction. So, the operator $$U a_i U^\dagger$$ acts on all Fock states the same way as $$a_i'$$, which proves the equality $$$$U a_i U^\dagger = a_i' = M_{ij} a_j$$$$ Obviously, the operator $$U$$ preserves the Hamiltonian of $$N$$ oscillators from the question.
Maybe it's simpler to provide an explicit construction. Consider the following unitary operator: $$U = e^{-i \sum_{ij} a^{\dagger}_i H_{ij} a_j}$$ for a Hermitian matrix $$H_{ij}$$. Using the Campbell-Baker-Hausdorff formula, it shouldn't be too hard to convince yourself that $$U a_i U^{\dagger} = \sum_j[e^{iH}]_{ij} a_j$$ where $$[e^{iH}]_{ij}$$ are the matrix elements of the matrix-exponential $$e^{iH}$$. If your desired matrix $$M$$ is an element of $$SU(N)$$, then it can always be realized as the exponential of a Hermitian matrix. I believe getting the rest of $$U(N)$$ from there should not be too difficult.