Unitary and linear operator in quantum mechanics

[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?

 

[dextercioby]

Yes. That's right.

 

[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.

 

[spaghetti3451]

I wanted to show the property of linearity.