The wave function of a one-dimensional harmonic oscillator is $${\psi _0} = A\exp \left( {\frac{{ - {\alpha ^2}{x^2}}}{2}} \right)$$     for the ground state $${E_0} = \frac{{\hbar \omega }}{2},$$   where $${\alpha ^2} = \frac{{m\omega }}{\hbar }$$   in the presence of a perturbing potential of $${E_0}{\left( {\frac{{\alpha x}}{{10}}} \right)^4},$$   the first order Change in the ground state energy is
$$\left[ {{\text{Given, }}\Gamma \left( {x + 1} \right) = \int_0^\infty {{t^x}\exp \left( { - t} \right)dt} } \right]$$

A. $$\left( {\frac{1}{2}{E_0}} \right){10^{ - 4}}$$

B. $$\left( {3{E_0}} \right){10^{ - 4}}$$

C. $$\left( {\frac{3}{4}{E_0}} \right){10^{ - 4}}$$

D. $$\left( {{E_0}} \right){10^{ - 4}}$$

Answer: Option C