Symmetries in quantum mechanics I have been studying symmetries in quantum mechanics and I have come across two types:


*

*Given some transformation on a Hilbert space of states $\mathcal{H}$, an operator $U: \mathcal{H} \rightarrow \mathcal{H}$ is a symmetry operator if 
$$ U^\dagger H U = H$$
where $H$ is the Hamiltonian of the system.

*A symmetry transformation is either a unitary or anti-unitary operator on $\mathcal{H}$ (Wigner's Theorem).


Are these two definitions of symmetry equivalent?
 A: There is some confusion between symmetries and dynamical symmetries. A Wigner symmetry is a bijective map $S$ associating a pure state, i.e. a normalized vector of the Hilbert space $\cal H$ up to phases, namely  a set  $[\psi]= \{e^{ia} \psi | a \in \mathbb C\}$, to a normalized vector up to phases $[\psi']$ (another pure state) preserving the probability transition: $$|\langle \psi| \phi \rangle |^2 = |\langle \psi'| \phi' \rangle |^2\:. \tag{1}$$
Wigner theorem proves that for every such $S$ there is an operator $U_S : {\cal H}\to {\cal H}$ which is unitary or antiunitary depending on $S$ and it is determined up to phases by $S$ itself, such that $$[U_S\psi] = S[\psi]\:.$$
It is evident that every unitary or antiunitary operator induces a map satisfying (1). This is the reason why one says that symmetries are all of unitary or antiunitary operators over ${\cal H}$.
Since $\langle U\psi| A U\psi\rangle = \langle \psi| (U^*A U)\psi\rangle$, acting on pure states with symmetries $\psi \to U\psi 
$ is equivalent to acting on observables $A \to U^*AU$. Hence, one can define a "dual" action on symmetries on observables:
$$A \to U^*AU$$
All that makes sense if no superselection rules occur so that every unit vector represent a pure state (up to phases), otherwise a different notion of symmetry due to Kadison is more appropriate (equivalent to Wigner's one in the standard case).
A dynamical symmetry is a (unitary) symmetry $U$ with the following property with respect to the temporal evolutor of the system $V_t = e^{-itH}$: if I modify the initial state $\psi$ with $U$, the evolution $\psi_t = V_t\psi$ of $\psi$ changes with the same rule
$V_t U\psi  = U\psi_t$ (this is by no means obvious for arbitrary symmetries and evolutors, in general $U\psi_t$ is not the evolution of any initial state). In other words $$V_tU = UV_t \quad \mbox{for every $t \in \mathbb R$}\tag{2}\:.$$ 
(2) is equivalent to 
$$U^*V_tU = V_t \quad \mbox{for every $t \in \mathbb R$}\tag{2'}\:.$$ 
Taking the (strong) derivatives of both sides at $t=0$ we get
$$U^*HU=H\:.$$
It is possible to prove that this identity is equivalent to (2') (there are subtleties with domains).
A: In the first definition you have not made $U$ preserve the $\mid \langle \psi \mid \phi \rangle \mid$. As an example, suppose $H=0$ (a trivial system), then you would get all operators $U$ as symmetries even if they aren't even unitary or anti-unitary. Those shouldn't have been called symmetries. You want both conditions not either.
The implications go the other way
Symmetry => Unitary or Anti-unitary
Symmetry => Preserves Hamiltonian under $U^\dagger H U$
You need both conditions to say something might be called a symmetry.
