The normalized eigen states of a particle in a one-dimensional potential well \[V\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0,}&{{\text{if }}0 \leqslant x \leqslant a} \\ {\infty ,}&{{\text{otherwise}}} \end{array}} \right.\]      are given by $${\psi _n}\left( x \right) = \sqrt {\frac{2}{a}} \sin \left( {\frac{{n\pi x}}{a}} \right)$$      where, n = 1, 2, 3, . . .
The particle is subjected to a perturbation $$\eqalign{ & V'x = {V_0}\cos \left( {\frac{{\pi x}}{a}} \right),\,{\text{for }}0 \leqslant x \leqslant \frac{a}{2} \cr & \,\,\,\,\,\,\,\,\,\,\, = 0,\,\,\,{\text{otherwise}} \cr} $$
The shift in the ground state energy due to the perturbation, in the first order perturbation theory, is

A. $$\frac{{2{V_0}}}{{3\pi }}$$

B. $$\frac{{{V_0}}}{{3\pi }}$$

C. $$ - \frac{{{V_0}}}{{3\pi }}$$

D. $$ - \frac{{2{V_0}}}{{3\pi }}$$

Answer: Option A