Q1(a). Particle in a box: probability in the centre half

1. Ground state (n=1) of a particle in a box with infinite walls at x=0 and x=L: ψ₁(x)=√(2/L) sin(π x)/(L), normalised so that ∫₀ᴸψ₁² dx=1.
2. The centre half is the middle 50 % of the length: x=L/4 to x=3L/4. By the Born interpretation the probability is P=∫_(L/4)^(3L/4)2/L²(π x)/(L) dx
3. Use ²θ=12(1-2θ): P=1/L∫_(L/4)^(3L/4)(1-cos(2π x)/(L))dx=1/L[x-(L)/(2π)sin(2π x)/(L)]_(L/4)^(3L/4)
4. Limits: sin(3π)/(2)=-1 and sin(π)/(2)=+1, so P=1/L[L/2-(L)/(2π)(-1-1)]=1/2+1π P=12+1π=0.818 (81.8 %)
5. The two outer quarters together hold 1-0.818=0.182, that is 0.091 each. The density |ψ₁|² is largest at the centre (x=L/2) and falls to zero at both walls.
6. A classical particle moves at constant speed and spends equal time everywhere, so it would give exactly 0.50. The quantum result differs most in the ground state. For large n the oscillations of |ψₙ|² crowd together and P→0.50 (correspondence principle).

Application. the box is the simplest model of π electrons in a conjugated chain: the electrons of a polyene sit mostly in the middle of the chain, and E=n²h²/8mL² gives the absorption wavelength of dyes such as the cyanines.