The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$ where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$ are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$ is the angular momentum. The Hamiltonian does not commute with
A. $$\overrightarrow {\bf{L}} + \overrightarrow {\bf{S}} $$
B. $$\overrightarrow {{{\bf{S}}^2}} $$
C. $${L_z}$$
D. $$\overrightarrow {{{\bf{L}}^2}} $$
Answer: Option C
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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