A spin half particle is in the state $${S_z} = \frac{{{\hbar ^2}}}{2}.$$ The expectation values of $${S_x},\,S_x^2,\,{S_y},\,S_y^2$$ are given by
A. $$0,\,0,\,\frac{{{\hbar ^2}}}{4},\,\frac{{{\hbar ^2}}}{4}$$
B. $$0,\,\frac{{{\hbar ^2}}}{4},\,\frac{{{\hbar ^2}}}{4},\,0$$
C. $$0,\,\frac{{{\hbar ^2}}}{4},\,0,\,\frac{{{\hbar ^2}}}{4}$$
D. $$\frac{{{\hbar ^2}}}{4},\,\frac{{{\hbar ^2}}}{4},\,0,\,0$$
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. e-ax2 (e-ax1 - e-ax2)
A. 0.75
B. 0.50
C. 0.35
D. 0.25
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