1
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ @ValterMoretti Nuking a mosquito? :P $\endgroup$
    – Danu
    Commented Oct 25, 2015 at 13:14
  • $\begingroup$ Sorry, but this doesn't really help me :/ $\endgroup$
    – Darius
    Commented Oct 25, 2015 at 13:18

1 Answer 1

3
$\begingroup$

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 $U_\Gamma U_{\Gamma^{-1}}= U_{\Gamma \circ \Gamma^{-1}}= I$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.