# Equivalence of symmetry and commuting unitary operator

Throughout my readings on particle physics, I've encountered a problem with quantum mechanics. Indeed, say we have a transformation given by a unitary operator : $U \rightarrow \psi' = U\psi$ for which the hamiltonian is left invariant, $H' = H$. Is there a way to show that this implies that $[H,U] = 0$, without knowing a priori that $H' = UHU^{\dagger}$ ?

This question comes from the fact that I get $H' = UHU^{\dagger}$ from the Schrodinger equation, although I would think that these symmetry properties might be more general.

• I think it should be $H' = U H U^\dagger$. – Noiralef Dec 21 '16 at 22:55
• – Phoenix87 Dec 21 '16 at 23:26

1. Before you can prove $[H,U]=0$ you certainly have to define what $H'$ is, otherwise the statement $H' = H$ does not have any meaning.
2. You do not need the Schrödinger equation though in order to understand why $H' = U H U^\dagger$: Say we have a transformation $U$ acting on vectors $\psi$ as $\psi' = U\psi$ and we have the operator $H$ acting on $\psi$. Now we want to write down an operator $H'$ that has the same effect on the transformed $\psi'$ that $H$ has on $\psi$. If you think about this, it should be clear that $$(H \psi)' = H' \psi' \qquad (\ast)$$ is the condition we are looking for, i.e. $U(H\psi) = H'(U\psi)$. If this is to hold for all $\psi$, we necessarily get $$H' = U H U^\dagger \;.$$
Note: In more technical terms, we have a transformation $U$ acting on the Hilbert space $\mathcal H$ in some representation. This always induces a unique representation of $U$ on the operator space of endomorphisms $\mathrm{End}(\mathcal H) = \mathcal H^\ast \otimes \mathcal H$ by some general rules so that (*) holds. If $U$ acts in the fundamental representation $\psi' = U\psi$ on $\mathcal H$ then the induced representation $H' = UHU^\dagger$ is called the adjoint representation.