where k = 2 m E and k ′ = 2 m ( V − − E ) . This is solved for given A and α by taking the boundary conditions at the both barrier edges, at x = 0 and x = l , where both Ψ Ψ and its derivative must be equal on both sides. For k ′ l ≫ ≫ 1 , this is easily solved by ignoring the time exponential and considering the real part alone (the imaginary part has the same behavior). We get, up to factors depending on the phases which are typically of order 1, and up to factors of the order of k / k ′ = E / ( V − − E ) (assumed not very large, since V is greater than E not marginally):