# Are superposition and time-evolution of a quantum system unrelated?

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?

• @AaronStevens yes indeed. Hadamard is unitary. – Nimish Mishra Jun 27 '19 at 4:56
• @NimishMishra The state $|\lambda\rangle$ you're using in your second section isn't "non-evolvable". In fact, you know exactly how it evolves--it starts out the same as $|\psi(0)\rangle$, so it must evolve exactly like $|\psi(0)\rangle$ – Jahan Claes Jun 27 '19 at 18:29
• @JahanClaes By being non-evolvable, I meant since it is an eigenstate, it remains an eigenstate throughout the evolution. So there is no change in the state. – Nimish Mishra Jun 28 '19 at 8:03
• My point is, $|\lambda\rangle$ is NOT an eigenstate of the Hamiltonian that creates the Hadamard gate. How do you know? Because it evolves! – Jahan Claes Jun 28 '19 at 17:09
• @JahanClaes I get it. I guess the matter was I didn't think about the Hamiltonian of the Hadamard before – Nimish Mishra Jun 29 '19 at 6:29