For a spin $$\frac{1}{2}$$ particle, the expectation value of sxsysz, where sx, sy and sz are spin operators, is
A. $$\frac{{i{\hbar ^3}}}{8}$$
B. $$ - \frac{{i{\hbar ^3}}}{8}$$
C. $$\frac{{i{\hbar ^3}}}{{16}}$$
D. $$ - \frac{{i{\hbar ^3}}}{{16}}$$
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. e-ax2 (e-ax1 - e-ax2)
A. 0.75
B. 0.50
C. 0.35
D. 0.25
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