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

\begin{equation}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 \end{equation}

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?

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    $\begingroup$ 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 $B=0$). $\endgroup$
    – LPZ
    Apr 18, 2023 at 16:09
  • $\begingroup$ @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? $\endgroup$ Apr 18, 2023 at 21:04
  • $\begingroup$ 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. $\endgroup$
    – LPZ
    Apr 18, 2023 at 21:40

2 Answers 2


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 \begin{equation} a_i \to M_{ij} a_j. \end{equation} 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 \begin{equation} a'_i = M_{ij}a_j. \end{equation} As $M_{ij}$ is unitary, the operators $a'_i, a'^\dagger_i$ obey the commutation relations \begin{equation} [a'_i, a'^\dagger_j] = \delta_{ij}, \end{equation} \begin{equation} [a'_i, a'_j] = 0, \end{equation} \begin{equation} [a'^\dagger_i, a'^\dagger_j] = 0. \end{equation} (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: \begin{equation} |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. \end{equation} 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: \begin{equation} 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' \end{equation} On the other hand, \begin{equation} a_i'|k_1\dots k_N\rangle' = \sqrt{k_i}|k_1\dots, k_i - 1,\dots k_N\rangle' \end{equation} 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 \begin{equation} U a_i U^\dagger = a_i' = M_{ij} a_j \end{equation} Obviously, the operator $U$ preserves the Hamiltonian of $N$ oscillators from the question.

  • $\begingroup$ Thank you very much! There is a small typo in the formula for the new Fock basis (the one under the commutators), right? $\endgroup$ Apr 19, 2023 at 20:40
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    $\begingroup$ Yes, I see, thank you! $\endgroup$
    – E. Anikin
    Apr 19, 2023 at 20:46

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.

  • $\begingroup$ Thank you for your answer! I'll try to work out the details as soon as I can $\endgroup$ Apr 19, 2023 at 20:35

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