A particle of mass m is confined in an infinite potential well \[V\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{0,}&{{\text{if }}0 < x < L} \\
{\infty ,}&{{\text{otherwise}}}
\end{array}} \right.\]
It is subjected to a perturbing potential $${V_P}\left( x \right) = {V_0}\sin \left( {\frac{{2\pi x}}{L}} \right)$$ within the well. Let E(1) and E(2) be the corrections to the ground state energy in the first and second order in V0.

Which of the following is correct?
A. E(1) = 0; E(2) < 0
B. E(1) > 0; E(2) = 0
C. E(1) = 0; E(2) depends on the sign of V0
D. E(1) < 0; E(2) < 0
Answer: Option C
Related Questions on Quantum Mechanics
A. In the ground state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half
B. In the first excited state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half This also holds for states with n = 4, 6, 8, . . . .
C. For an arbitrary state $$\left| \psi \right\rangle ,$$ the probability of finding the particle in the left half of the well is half
D. In the ground state, the particle has a definite momentum
A. (e-ax1 - e-ax2)
B. a(e-ax1 - e-ax2)
C. e-ax2 (e-ax1 - e-ax2)
D. None of the above

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