Question

A particle in the ground state is located in a unidimesional square potential well of length `l` with absolutely impenetrable walls `(0 lt x lt l)`. Find the probability of the particle staying within a region `(1)/(3)l le x le (2)/(3)l`.

Answer 1

We look for the solution of Schrodinger eqn. with
`-(ħ^(2))/(2m)(d^(2)Psi)/(dx^(2))=E Psi,0lexlel`….(1)
The boundary condition of impenetrable walls means
`Psi(x)=0` for `x=0` and `x=l`
(as `Psi(x)=0` for `xlt0` and `xgtl`,)
Then solution of (1) is
`Psi(x)=A "sin"sqrt(2mE)/(ħ)x+B "cos"sqrt(2mE)/(ħ)x`
Then `Psi(0)=0implies B=0`
`Psi(l)=0implies A "sin"sqrt(mE)/(ħ)l-0`
`A~~ 0` so
`sqrt(2mE)/(ħ)l=n pii`
Hence `E_(n)=(n^(2)pi^(2)ħ^(2))/(2ml^(2)),n= 1,2,3`....
Thus the ground state wave fucnction is
`Psi(x)=A "sin"(pi x)/(l)`
We evaluate `A` by nomalization
`1=A^(2)int_(0)^(i)"sin"^(2)(pi x)/(l)dx=A^(2)(l)/(pi)int_(0)^(x)sin^(2) theta d theta=A^(2)(l)/(pi).(pi)/(2)`
Thus `A+(sqrt(2)/(l))`
Finally, the probability `P` for the particle to lie in `(l)/(2) le (2l)/(3)` is
`P=P((l)/(3)lexle(2l)/(3))=(2)/(l)int_(l/3)"sin"^(2)(pix)/(l)dx`
`=(2)/(pi)int_(pi//3) sin^(2) theta d theta=(1)/(pi)int_(pi//3)^(2pi//3)(1- cos 2 theta)d theta`
`(1)/(pi)(theta-(1)/(2) sin 2 theta)^(2pi//3)=(1)/(pi)((2pi)/(3)-(pi)/(3)-(1)/(2)"sin"(4pi)/(3)+(1)/(2)`sin`(2pi)/(3))`
`=(1)/(pi)((pi)/(3)+(1)/(2)(sqrt(3))/(2)+(1)/(2)(sqrt(3))/(2))=(1)/(3)+(sqrt(3))/(2pi)= 0.609`