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
This Question Belongs to Engineering Physics >> 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. e-ax2 (e-ax1 - e-ax2)
A. 0.75
B. 0.50
C. 0.35
D. 0.25
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