# Representation of Symmetry group

Suppose the SE $\boldsymbol{H}\psi=E\psi$ describes a closed system and $G$ is a symmetry group of the system. Then any transformation in $G$ leaves the form of the SE invariant. It seems plausible to state that the unitary operators corresponding to various transformations in $G$ in a representation of $G$ on the full Hilbert Space of physical states of the system would commute with the Hamiltonian. I am trying to show this more rigorously, but can't seem to fix the math involved. Any indication as to how we could show this would be very helpful.

Thanks.

Let $g\in G$ (more strictly let $g$ be an element of a unitary representation of $G$ on our Hilbert space $\mathbb{H}$) then $g$ leaves the SE invariant:

$${\mathbf H} g\psi = Eg\psi$$

Since $E$ is just a number we can commute $E$ and $g$ freely so:

$${\mathbf H} g\psi = gE\psi$$

Left multiplying the original symmetry equation by $g$ gives:

$$g{\mathbf H}\psi = gE\psi$$

Hence:

$$g{\mathbf H}\psi = {\mathbf H}g\psi$$

Now in a finite dimensional Hilbert space the eigenvectors of ${\mathbf H}$ are a basis for the space so, by linearity, we are justified in saying that $\mathbf H$ and $g$ commute on the whole space. In infinite dimensional spaces we can (often) do something similar but it will, in general, depend on the properties of ${\mathrm H}$. In the case where ${\mathrm H}$ is defined everywhere on ${\mathbb H}$ then it is bounded and the result still follows.

Unfortunately most Hamiltonians of interest in quantum mechanics aren't bounded so we have to get a bit more technical. The most general setting I am confident the result follows in is where $\mathrm H$ has a compact inverse. I think we are still fine as long as the spectrum of $\mathrm H$ is non-empty but I could very well be wrong.