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 $|\langle\psi(0)|\lambda(0)\rangle|^{2}$.


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 $|\langle\psi(\epsilon)|\lambda(\epsilon)\rangle|^{2} = \frac{1}{2}$.


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?

  • $\begingroup$ @AaronStevens yes indeed. Hadamard is unitary. $\endgroup$ – Nimish Mishra Jun 27 '19 at 4:56
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    $\begingroup$ @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 $|\psi(0)\rangle$ $\endgroup$ – Jahan Claes Jun 27 '19 at 18:29
  • $\begingroup$ @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. $\endgroup$ – Nimish Mishra Jun 28 '19 at 8:03
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    $\begingroup$ My point is, $|\lambda\rangle$ is NOT an eigenstate of the Hamiltonian that creates the Hadamard gate. How do you know? Because it evolves! $\endgroup$ – Jahan Claes Jun 28 '19 at 17:09
  • $\begingroup$ @JahanClaes I get it. I guess the matter was I didn't think about the Hamiltonian of the Hadamard before $\endgroup$ – Nimish Mishra Jun 29 '19 at 6:29

The Hadamard gate is generated by a hamiltonian which is not diagonal in the computational basis you're using, so it is not true that you're comparing against an eigenstate of the hamiltonian in use.

In other words, the reason you're getting green on one side and orange on the other is that you're comparing apples and oranges.

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  • $\begingroup$ I get it. Didn't think of that before. Thank you $\endgroup$ – Nimish Mishra Jun 28 '19 at 8:04

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