For a spin s particle, in the eigen basis of $$\overrightarrow {{{\bf{s}}^2}} ,$$ sz the expectation value $$\left\langle {sm\left| {\overrightarrow {{\bf{s}}_x^2} } \right|sm} \right\rangle $$ is
A. $$\frac{{\hbar \left\{ {s\left( {s + 1} \right) - {m^2}} \right\}}}{2}$$
B. $$\hbar \left\{ {s\left( {s + 1} \right) - 2{m^2}} \right\}$$
C. $${\hbar \left\{ {s\left( {s + 1} \right) - {m^2}} \right\}}$$
D. $${\hbar ^2}{m^2}$$
Answer: Option A
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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