The wave function of a particle in a one-dimensional potential at time t = 0 is $$\psi \left( {x,\,t = 0} \right) = \frac{1}{{\sqrt {15} }}\left[ {2{\psi _0}\left( x \right) - {\psi _1}\left( x \right)} \right]$$ where, $${\psi _0}\left( x \right)$$ and $${\psi _1}\left( x \right)$$ are the ground arid the first excited states of the particle with corresponding energies E0 and E1. The wave function of the particle at a time t is
A. $$\frac{1}{{\sqrt 5 }}{e^{ - i\left( {{E_0}{E_1}} \right)t/2h}}\left[ {2{\psi _0}\left( x \right) - {\psi _1}\left( x \right)} \right]$$
B. $$\frac{1}{{\sqrt 5 }}{e^{ - i{E_0}t/h}}\left[ {2{\psi _0}\left( x \right) - {\psi _1}\left( x \right)} \right]$$
C. $$\frac{1}{{\sqrt 5 }}{e^{ - i{E_1}t/h}}\left[ {2{\psi _0}\left( x \right) - {\psi _1}\left( x \right)} \right]$$
D. $$\frac{1}{{\sqrt {15} }}\left[ {2{\psi _0}\left( x \right){e^{ - i{E_0}t/h}} - {\psi _1}\left( x \right){e^{ - i{E_1}t/h}}} \right]$$
Answer: Option D
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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