Definition of unitary operators Let $\phi, \psi \in \mathcal{H}$ be some element from a hilbert space $\mathcal{H}$ and $U$ a linear operator $U: \mathcal{H} \rightarrow \mathcal{H}$. Does
$$ \forall \phi: \| U \phi \|^2 = \| \phi \|^2 = (\phi, \phi) $$
imply that $U$ is unitary? I suspect not, and I think I have to require the more general relation:
$$ \forall \phi, \psi: (U \phi , U \psi) = (\phi, \psi)$$
However, I cant see why this is more general then the condition above. In other words, is there a $U$ that is not unitary and fulfills the first equation but not the second?
 A: Both above conditions are equivalent and define an isometry. A unitary operator satisfies 
$$ U^*U = UU^* = I,$$
while the isometry satisfies only $V^*V=I$. 
So for a unitary operator apart from the condition which you wrote we also have it for its adjoint, that is, 
$$ \left<U^*x, U^*y\right> = \left<x, y\right>.$$
Example of a map which is an isometry, but not unitary: 
Let $V \colon \ell^2 \rightarrow \ell^2$ be given by $V(x_n)=(y_n)$, where $y_1=0$ and $y_{n+1}=x_n$ for every $n \in \mathbb{N}$. Then $V$ is an isometry, that is 
$$ \left<Vx, Vy\right>=\left<x,y\right>,$$
but it fails to be a unitary operator (it is not onto). 
Proof that both conditions which you wrote are equivalent:
One direction is obvious. 
Let us assume that the following holds
$$ \|Vx\| = \|x\|$$
we would like to show that for any $x$, $y \in \mathcal{H}$ we have also
$$ \left<Vx, Vy\right> = \left<x, y\right>.$$
Let $x$, $y \in \mathcal{H}$ and let $r \in \mathbb{C}$, then
$$ LHS=\| x+ry\|^2 = \|V(x+ry)\|^2 = \| Vx + rVy\|^2=RHS.$$
Now note that 
$$LHS = \|x\|^2 + 2 \mathrm{Re} \overline{r}\left<x,y\right> + \|y\|^2$$
and
$$RHS = \|Vx\|^2 + 2 \mathrm{Re} \overline{r}\left<Vx,Vy\right> + \|Vy\|^2=
\|x\|^2 + 2 \mathrm{Re} \overline{r}\left<Vx,Vy\right> + \|y\|^2.
$$
Therefore we obtain that
$$\mathrm{Re} \overline{r}\left<Vx,Vy\right>=\mathrm{Re} \overline{r}\left<x,y\right>.$$
We end the proof by first taking $r=1$ and then taking $r=i$, note that if your Hilbert space was over $\mathbb{R}$, then you would just take $r=1$.
