In Schrodinger Equation Local $U(1)$ Gauge Invariance, How is Laplacian Simplified?

Question

So I'm trying to derive the conditions necessary for local $U(1)$ gauge invariance in the Schrodinger equation, and I don't understand how Laplacian of the wavefunction is simplified the way it is. So the transformation is:

\begin{equation} \psi \rightarrow \psi'=\psi e^{i\lambda(x,t)} \end{equation}

Which gives the following Laplacian which I understand:

\begin{equation} \nabla^2\psi'=(\nabla^2+2i\nabla\lambda\cdot\nabla+i\nabla^2\lambda - (\nabla\lambda)^2)\psi. \tag{1} \end{equation}

According to every textbook I read, the next step is to simplify it the following way:

\begin{equation} \nabla^2\psi'=(\nabla+i\nabla\lambda)^2\psi \end{equation}

\begin{equation} (\nabla+i\nabla\lambda)^2\psi=(\nabla^2+i\nabla\lambda\cdot\nabla+i\nabla^2\lambda-(\nabla\lambda)^2)\psi. \tag{2} \end{equation}

However, equation (1) and equation (2) are not the same because the 2 present in the second term of equation (1) is not in equation (2). Is there some piece of algebra I am missing? Here is a reference: http://www.niser.ac.in/~sbasak/p303_2010/23.11.pdf (page 2)

Answer 1

Okay I found a little error in my calculation, I basically neglected the product rule in the $\nabla(i(\nabla\lambda)\psi)$ term because I treated the operator as a scalar. So I really should be thinking about simplifying this in terms of operating on the function so I don't make this mistake again:

$$(\nabla+i\nabla\lambda)^2\psi=\nabla^2\psi+\nabla(i(\nabla\lambda)\cdot\psi)+i(\nabla\lambda)\nabla\psi-(\nabla\lambda)^2\psi$$ $$=\nabla^2\psi+i(\nabla^2\lambda)\psi+i(\nabla\lambda)\nabla\psi+i(\nabla\lambda)\nabla\psi-(\nabla\lambda)^2\psi$$ $$=\nabla^2\psi+i(\nabla^2\lambda)\psi+2i(\nabla\lambda)\nabla\psi-(\nabla\lambda)^2\psi$$ Therefore: $$(\nabla+i\nabla\lambda)^2=\nabla^2+i(\nabla^2\lambda)+2i(\nabla\lambda)\nabla-(\nabla\lambda)^2$$