# Unitary and linear operator in quantum mechanics

1. Dec 6, 2015

### spaghetti3451

Given a transformation $U$ such that $|\psi'>=U|\psi>$, the invariance $<\psi'|\psi'>=<\psi|\psi>$ of the scalar product under the transformation $U$ means that $U$ is either linear and unitary, or antilinear and antiunitary.

How do I prove this?

$<\psi'|\psi'>$
$= <U\psi|U\psi>$

For $U$ unitary and linear, we have
$<U\psi|U\psi>$
$<U(\alpha\psi_{a}+\beta\psi_{\beta})|U(\alpha\psi_{a}+\beta\psi_{\beta})>$
$<(\alpha U\psi_{a}+\beta U\psi_{\beta})|(\alpha U\psi_{a}+\beta U\psi_{\beta})>$
$=(<\psi_{a}|(U^{\dagger})\alpha^{*}+<\psi_{b}|(U^{\dagger})\beta^{*})(\alpha U|\psi_{a}>+\beta U|\psi_{\beta}>)$
$=|\alpha|^{2}(<\psi_{a}|(U^{\dagger})(U)|\psi_{a}>+|\beta|^{2}(<\psi_{b}|(U^{\dagger})(U)|\beta>$
$=|\alpha|^{2}+|\beta|^{2}$

Is this how the proof should go for $U$ linear and unitary?

Last edited: Dec 6, 2015
2. Dec 6, 2015

### dextercioby

Yes. That's right.

3. Dec 6, 2015

### A. Neumaier

But you don't need to go into a basis: $\langle U\phi|U\psi\rangle =\langle \phi|U^*U\psi\rangle=\langle \phi|\psi\rangle$ is enough.

4. Dec 7, 2015

### spaghetti3451

I wanted to show the property of linearity.