A one-dimensional harmonic oscillator carrying a charge -q is placed in a uniform electric field $$\overrightarrow {\bf{E}} $$ along the positive X-axis. The corresponding Hamiltonian operator is
A. $$\frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}}}{{d{x^2}}} + \frac{1}{2}k{x^2} + qEx$$
B. $$\frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}}}{{d{x^2}}} + \frac{1}{2}k{x^2} - qEx$$
C. $$ - \frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}}}{{d{x^2}}} + \frac{1}{2}k{x^2} + qEx$$
D. $$ - \frac{{{\hbar ^2}}}{{2m}}\frac{{{d^2}}}{{d{x^2}}} + \frac{1}{2}k{x^2} - qEx$$
Answer: Option D
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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