A particle of mass m is confined in the potential
\[V\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{2}m{\omega ^2}{x^2},}&{{\text{for }}x < 0} \\
{\infty ,}&{{\text{for }}x \leqslant 0}
\end{array}} \right.\]
Let the wave function of the particle be given by $$\psi \left( x \right) = - \frac{1}{{\sqrt 5 }}{\psi _0} + \frac{2}{{\sqrt 5 }}{\psi _1}$$
where $${\psi _0}$$ and $${\psi _1}$$ are the eigen functions of the ground state and the first excited slate respectively. The expectation value of the energy is
A. $$\frac{{31}}{{10}}\hbar \omega $$
B. $$\frac{{25}}{{10}}\hbar \omega $$
C. $$\frac{{13}}{{10}}\hbar \omega $$
D. $$\frac{{11}}{{10}}\hbar \omega $$
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
Join The Discussion