The quantum mechanical operator for the momentum of a particle moving in one dimension is given by
A. $$i\hbar \frac{d}{{dx}}$$
B. $$ - i\hbar \frac{d}{{dx}}$$
C. $$i\hbar \frac{\partial }{{\partial t}}$$
D. $$ - \frac{{{h^2}}}{{2m}}\frac{{{d^2}}}{{d{x^2}}}$$
Answer: Option B
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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