# Explanation of notation change between 1D and 3D T.I.S.E.?

I am currently reviewing this webpage, near the end it shows how the one-dimensional time independent Schrondinger equation can easily be extended to three dimensions.

My question is specifically with the change in notation used between the 1D case

$$-\frac{\hbar^2}{2m} \frac{d^2 \Psi(x)}{dx^2} + V(x) \Psi(x) = E\Psi(x)$$

and the 3D case

$$-\frac{\hbar^2}{2m} \nabla^2\Psi(\mathbf{r}) + V(\mathbf{r}) \Psi(\mathbf{r}) = E\Psi(\mathbf{r})$$

Specifically, in mathematical "plain English" how is the change from $$\frac{d^2 \Psi}{dx^2}$$ to $$\nabla^2\Psi$$ read and interpreted?

From how I understand it it is read as "the rate of change, of the rate of change" which is interpreted in a single dimension. Versus, "The gradient" (which to mean is essientially the same but on a manifold instead of a line) which is interpreted in 3 dimensions.

I also do want to point out that I am aware that $$\mathbf{r}$$ is a vector of the $$i,j,k$$ basis in Euclidean space. (But I am not sure if I stated that exactly correctly)

• Note the $\nabla^2$ is the Laplace operator and not the gradient operator – Hal Hollis Oct 12 '18 at 18:19

Let's just stick with Cartesian coordiantes for now. In 1 dimension, the function $$\Psi$$ is a function of only one variable $$x$$: $$\Psi(x)$$, in 3 dimensions it is now a function of 3 variables: $$\Psi(x,y,z)$$. What is $$\nabla^2$$ you ask, well it is the divergence of the gradient (not simply a gradient): $$\nabla^2\Psi(x,y,z)=\vec{\nabla}\cdot\vec{\nabla}\Psi(x,y,z)$$. It's name is the "Lapacian". The gradient of a function is a vector which points in the direction of steepest ascent: $$\vec{\nabla}\Psi(x,y,z)=\hat{x}\frac{\partial\Psi}{\partial x}+\hat{y}\frac{\partial\Psi}{\partial y}+\hat{z}\frac{\partial\Psi}{\partial z}.$$ The divergence of a vector field (say, as an example, $$\vec{v}=v_x\hat{x}+v_y\hat{y}+v_z\hat{z}$$) is a scalar field giving the quantity of that vector field's "source" at each point: $$\vec{\nabla}\cdot\vec{v}=\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}$$ Finally, when you take the divergence of the gradient, you get the Laplacian:$$\vec{\nabla}\cdot\vec{\nabla}\Psi=\nabla^2\Psi=\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}.$$ Hopefully now you can see the direct analogy between the 3-D case and the 1-D case. All you're doing is adding the 3 different second (partial) derivatives together. Perhaps it's easiest to see it from the perspective that the 1-D case is simply a special circumstance of the 3-D case where there aren't $$y,z$$ directions (this is indeed the logical way to see things).

That derivation you linked to does not really explain the connection between momentum and energy and operators and it helps if you go back to classical mechanics where you get :

$$H=T+V=\frac{p^2}{2m}+V$$

In multiple dimensions $$p^2\to\mathbf{p\cdot p}$$ using vector momentum and the dot product instead of squaring.

When the TISE is built it uses an operator for the analogue of momentum, we get an operator equation :

$$H\psi=E\psi$$

and in one dimension $$p=-i\hbar \frac d{dx}$$ and because that's an operator $$p^2$$ results in a second order differential. That is, we apply the $$p$$ operator twice.

But in multiple dimensions we use $$\mathbf{p}=-i\hbar\nabla$$ instead. $$\nabla$$ is a vector operator, but applying it twice results in the scalar Laplacian operator.