$${{\bf{\hat A}}}$$ and $${{\bf{\hat B}}}$$ are two quantum mechanical operators. If $$\left[ {{\bf{\hat A}},\,{\bf{\hat B}}} \right]$$ stands for the commutator of $${{\bf{\hat A}}}$$ and $${{\bf{\hat B}}}$$ then $$\left[ {\left[ {{\bf{\hat A}},\,{\bf{\hat B}}} \right],\,\left[ {{\bf{\hat B}},\,{\bf{\hat A}}} \right]} \right]$$ is equal to
A. $${\bf{\hat A\hat B\hat A\hat B}} - {\bf{\hat B\hat A\hat B\hat A}}$$
B. $${\bf{\hat A}}\left( {{\bf{\hat A\hat B}} - {\bf{\hat B\hat A}}} \right) - {\bf{\hat B}}\left( {{\bf{\hat B\hat A}} - {\bf{\hat A\hat B}}} \right)$$
C. zero
D. $${\left( {\left[ {{\bf{\hat A}},\,{\bf{\hat B}}} \right]} \right)^2}$$
Answer: Option C
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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