Answer to Question #118511 in Quantum Mechanics for Moel Tariburu

Question

For the infinite square well potential, calculate the probability that a particle in its second excited state is in each third of the one-dimensional box:
0≪x≪L/3, L/3≪x≪2L/3,2L/3≪x≪L,
Check to see that the sum of the probabilities is one.

Answer 1

Wave function of particle in infinite square well potential is $\psi_n(x) = \sqrt{\frac{2}{L}} \sin \left( \frac{ \pi n}{L} x\right)$.

Hence, for $n=2$:

$p(0 \leq x \leq L/3) = \int_0^{L/3} |\psi_2(x)|^2 dx = \frac{2}{L}\int_0^{L/3} \sin^2 {\frac{2 \pi}{L} x} dx = \frac{1}{3} + \frac{\sqrt{3}}{8 \pi}$,

$p(L/3 \leq x \leq 2L/3) = \int_{L/3}^{2L/3} |\psi_2(x)|^2 dx = \frac{2}{L}\int_{L/3}^{2L/3} \sin^2 {\frac{2 \pi}{L} x} dx = \frac{1}{3} - \frac{\sqrt{3}}{4 \pi}$,

$p(2L/3 \leq x \leq L) = \int_{2L/3}^L |\psi_2(x)|^2 dx = \frac{2}{L}\int_{2 L/3}^{L} \sin^2 {\frac{2 \pi}{L} x} dx = \frac{1}{3} + \frac{\sqrt{3}}{8 \pi}$.

The sum of probabilities is one, because terms with $\sqrt{3}$ cancel out, and one thirds sum to one.