The probability P(a2)P \left( a_2 \right)P(a2) of getting the value a2a_2a2 is calculated by the standard quantum-mechanical formula
We have ⟨a2∣ψ⟩=i/2\left\langle a_2 | \psi \right\rangle = i / 2⟨a2∣ψ⟩=i/2, ∣⟨a2∣ψ⟩∣2=∣i/2∣2=1/4\left| \left\langle a_2 | \psi \right\rangle \right|^2 = |i / 2|^2 = 1/4∣⟨a2∣ψ⟩∣2=∣i/2∣2=1/4, ⟨ψ∣ψ⟩=∣1∣2+∣i/2∣2=5/4\left\langle \psi | \psi \right\rangle = |1|^2 + |i/2|^2 = 5/4⟨ψ∣ψ⟩=∣1∣2+∣i/2∣2=5/4, so that
Answer: 0.2.
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