Starting with the standard particle in a box Hamiltonian as the zero ${ }^{\text {th }}$ order Hamiltonian and the potential of $\mathrm{V}_{1}$ from 'a/2' to 'a' as a perturbation, the first order energy state is
(a)$\mathrm{V}_{1}$
(b)$\mathrm{V}_{1} / 4$
(c)$-\mathrm{V}_{1}$
(d)$\mathrm{V}_{1} / 2$
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
E⁽¹⁾ = V₁ × (probability of being in a/2 < x < a) = V₁/2, since every box eigenfunction has equal probability in the two halves.