A one-dimensional random walker takes steps to left or right with equal probability. The probability that the random walker starting from origin is back to origin after N even number of steps is
A. $$\frac{{N!}}{{\left( {\frac{N}{2}} \right)!\left( {\frac{N}{2}} \right)!}}{\left( {\frac{1}{2}} \right)^N}$$
B. $$\frac{{N!}}{{\left( {\frac{N}{2}} \right)!\left( {\frac{N}{2}} \right)!}}$$
C. $$2N!{\left( {\frac{1}{2}} \right)^{2N}}$$
D. $$N!{\left( {\frac{1}{2}} \right)^N}$$
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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