Consider a single particle (a single qubit if you will) in some arbitrary state $|\psi\rangle$ and an eigenvector $|\lambda\rangle$ corresponding to the eigenvalue $\lambda.$ Consider the time evolution of this system in some infinitesimal time $\epsilon$ to be given by a unitary operator U: $|\psi(\epsilon)\rangle = U|\psi(0\rangle)$.
Time-evolution preserving the inner product:
Consider the following statements holding that time evolution preserves inner product $\langle\psi|\lambda\rangle$. I think $\lambda$ is non-evolvable, or $\lambda(\epsilon) = \lambda(0)$, or $U$ does nothing on it. Then the following are true:
$\langle\psi(\epsilon)| = \langle\psi(0)|U^{\dagger}$.
$\implies$ $\langle\psi(\epsilon)|\lambda(\epsilon)\rangle = \langle\psi(0)|U^{\dagger}U|\lambda(0)\rangle = \langle\psi(0)|\lambda(0)\rangle$.
So when you measure $|\psi(\epsilon)\rangle$, you get $|\lambda\rangle$ with probability $|\langle\psi(\epsilon)|\lambda(\epsilon)\rangle|^{2}$ which is equal to $|\langle\psi(0)|\lambda(0)\rangle|^{2}$.
Superposition
If you start with $|\psi(0)\rangle = |0\rangle$ and apply Hadamard operation to it, you get $|\psi(\epsilon)\rangle = \frac{|0\rangle + |1\rangle}{2^{1/2}}$. If you consider $|\lambda(0)\rangle = |\lambda(\epsilon)\rangle = |0\rangle$, then $|\langle\psi(0)|\lambda(0)\rangle|^{2} = 1$ and $|\langle\psi(\epsilon)|\lambda(\epsilon)\rangle|^{2} = \frac{1}{2}$.
Question
Have I done something wrong or is there some problem in my understanding of the time evolution of a quantum system? Is Hadamard-ing a state not considered in the class of operations that qualify as time evolution of a quantum system? In short, why are these probabilities different?
