Three operators X, Y and Z satisfy the commutation relations, $$\left[ {X,\,Y} \right] = i\hbar Z,\,\left[ {Y,\,Z} \right] = i\hbar X$$ and $$\left[ {Z,\,X} \right] = i\hbar Y.$$ The set of all possible eigen values of the operator Z, in units of $$\hbar $$ is
A. $$\left[ {0,\, \pm 1,\, \pm 2,\, \pm 3,\,...} \right]$$
B. $$\left\{ {\frac{1}{2},\,1,\,\frac{3}{2},\,2,\,\frac{5}{2},\,...} \right\}$$
C. $$\left\{ {0 \pm \frac{1}{2},\, \pm 1,\, \pm \frac{3}{2},\, \pm 2,\, \pm \frac{5}{2},\,...} \right\}$$
D. $$\left\{ { - \frac{1}{2},\, + \frac{1}{2}} \right\}$$
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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