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
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. None of the above
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