# Position probability in infinite wells

An electron is on the third energy level of an infinite square well between $0 \leq x\leq L$. What is the probability that the particle is situated in a region centered at (a) x=L/4, (b) x=L/2, and (c) x=2L/3 if the region width is $0.004 L$. (d) When L=10 nm please find the particle energy.

For the infinite square well that starts at x=0 and has the width L, the wave functions corresponding to energy levels E1,2,… are

$\psi_n (x,t)=\sqrt{2/L}*\sin(k_n x)*e^{-i\omega t}$

Since each energy level accommodates a standing wave (on E1 there is just half of a complete wavelength- or equivalent just a $\pi$ shift of phase) we write kn as

$k_n=n\pi/L$

Since k is a wave vector it can be written also as

$k_n=2\pi/\lambda_n =2\pi/(h/p_n )=\sqrt{2mE_n }/\hbar$ and thus

$E_n=(\hbar^2 k_n^2)/2m=(n^2 \pi^2 \hbar^2)/(2mL^2 )$

For L =10 nm and n=3 we have $E_3=5.428*10^{-21} J=0.034 eV$

The probability of finding the particle in space between L1 and L2 is just:

$P=\int_(L_1)^{L_2} |\psi_3 (x,t) |^2*d x ≈ |\psi(L_1,t) |^2*(L_2-L_1)$

For $L1 =x =L/4$ we have

$P=2/L*[\sin(3π/L*L/4)]^2=2/L*(\sqrt{2}/2)^2=1/L$

For $L1 =x=L/2$ we have

$P=2/L*[\sin(3π/L*L/2)]^2=2/L*1^2=2/L$

For $L1 =x =2L/3$ we have

$P=2/L*[\sin(3π/L*2L/3)]^2=2/L*0^2=0$