Infinite symmetric potential well in one dimension

[M.A.M.Abed]

1. The problem statement.
for Infinite symmetric well -a/2 < x < a/2 in one dimension
show that wave function Ψ = Acos(kx) + Bsin(kx)
is not physically accepted solution although its mathematically accepted

Homework Equations

∫ψ(x)* ψ(x) dx=1

 

[Asim Riaz]

The Attempt at a SolutionThe wave function Ψ = Acos(kx) + Bsin(kx) is mathematically accepted because it satisfies the normalization condition, i.e., ∫ψ(x)* ψ(x) dx=1However, this wave function is not physically accepted solution because it does not satisfy the boundary conditions. The boundary conditions for an infinite symmetric well -a/2 < x < a/2 in one dimension are:ψ(-a/2) = 0 and ψ(a/2) = 0However, the wave function Ψ = Acos(kx) + Bsin(kx) does not satisfy the boundary conditions sinceψ(-a/2) ≠ 0 and ψ(a/2) ≠ 0