# What does $(\nabla \psi)^2$ mean?

This is a question about notation. I know that for a function $\psi(x,t)$ $$\nabla\psi=\left(\frac{\partial\psi}{\partial x},\frac{\partial\psi}{\partial t}\right)$$ (the gradient of $\psi$) and $$\nabla^2 \psi=\frac{\partial^2\psi}{\partial x^2}+\frac{\partial\psi^2}{\partial t^2}$$ (the laplacian of $\psi$).

By what does $(\nabla \psi)^2$ mean?

Notes

• If $(\nabla \psi)^2$ is another way of representing the laplacian than why didn't they just write $\nabla^2 \psi$ or $\Delta \psi$?
• $K$ is a scalar constant.
• If context is needed see below

which is from "Novel Methods in Soft Matter Simulations" chapter 29 pg 21

• It is a short-hand notation for the (square of the) norm of $\nabla\psi$. Physicists, right? – AccidentalFourierTransform Jan 31 '18 at 2:43
• Then why not use $||\nabla \psi||$? – AzJ Jan 31 '18 at 2:44
• beats me.${}{}$ – AccidentalFourierTransform Jan 31 '18 at 2:46
• I've seen it used in math texts also but yeah it is definitely common in physics as it just means $\vec{v}^2 = \vec{v} \cdot \vec{v}$ – Triatticus Jan 31 '18 at 2:55
• So wouldn't it just mean that $(\nabla \psi)^2$=$\nabla^2 \psi$? – AzJ Jan 31 '18 at 2:58

$\nabla \psi \cdot \nabla \psi=(u_x \hat i+u_y \hat j) \cdot(u_x\hat i+u_y\hat j)=(u_x)^2+(u_y)^2$.