We remain with E k e i k ⋅ ⋅ x u k ( x ) = − − ℏ ℏ 2 2 m ∇ ∇ ⋅ ⋅ ( i k e i k ⋅ ⋅ x u k ( x ) + e i k ⋅ ⋅ x ∇ ∇ u k ( x ) ) + U ( x ) e i k ⋅ ⋅ x u k ( x ) E k e i k ⋅ ⋅ x u k ( x ) = − − ℏ ℏ 2 2 m ( i k ⋅ ⋅ ( i k e i k ⋅ ⋅ x u k ( x ) + e i k ⋅ ⋅ x ∇ ∇ u k ( x ) ) + i k ⋅ ⋅ e i k ⋅ ⋅ x ∇ ∇ u k ( x ) + e i k ⋅ ⋅ x ∇ ∇ 2 u k ( x ) ) + U ( x ) e i k ⋅ ⋅ x u k ( x ) E k e i k ⋅ ⋅ x u k ( x ) = ℏ ℏ 2 2 m ( k 2 e i k ⋅ ⋅ x u k ( x ) − − 2 i k ⋅ ⋅ e i k ⋅ ⋅ x ∇ ∇ u k ( x ) − − e i k ⋅ ⋅ x ∇ ∇ 2 u k ( x ) ) + U ( x ) e i k ⋅ ⋅ x u k ( x ) E k u k ( x ) = ℏ ℏ 2 2 m ( − − i ∇ ∇ + k ) 2 u k ( x ) + U ( x ) u k ( x )