# Show that translation and rotation operator are unitary

I have a problem understanding how to show that operators are unitary if they are not in the "normal" matrix form.

The translation operator is defined as $$(T_v \psi)(x) = \psi(x-v)$$ and the rotation operator is defined as $$(R_{\alpha} \psi)(x) = \psi({R^{-1}}_{\alpha}(x))$$ where $R_{\alpha}$ is a rotation matrix. How can I show, that they are unitary?

• @ValterMoretti Nuking a mosquito? :P – Danu Oct 25 '15 at 13:14
• Sorry, but this doesn't really help me :/ – Darius Oct 25 '15 at 13:18

If $\Gamma : \mathbb R^n \to \mathbb R^n$ is an isometry, define $$(U_\Gamma\psi)(x) := \psi(\Gamma^{-1}x)\quad \forall \psi \in L^2(\mathbb R^n, dx)\:.$$ With this definition you have, using the fact that the Lebesgue measure is $\Gamma$-invariant, $d\Gamma x = dx$ (which is the same as $dx = d\Gamma^{-1}x$) $$\langle U_\Gamma \psi | U_\Gamma \phi \rangle = \int_{\mathbb R^n} \overline{\psi(\Gamma^{-1}x)} \phi(\Gamma^{-1}x) dx = \int_{\mathbb R^n} \overline{\psi(\Gamma^{-1}x)} \phi(\Gamma^{-1}x) d\Gamma^{-1} x = \int_{\mathbb R^n} \overline{\psi(x)} \phi(x) dx = \langle \psi|\phi \rangle\:.$$ This result simultaneously proves that $U_\Gamma (L^2(\mathbb R^n, dx)) \subset L^2(\mathbb R^n, dx)$ just taking $\psi=\phi \in L^2(\mathbb R^n, dx)$, and that $U_\Gamma$ preserves the scalar product of $L^2(\mathbb R^n, dx)$. To conclude that $U_\Gamma$ is unitary, it is enough to establish that it is surjective. Surjectivity is an immediate consequence of $U_\Gamma U_{\Gamma^{-1}}= U_{\Gamma \circ \Gamma^{-1}}= I$.