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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)

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

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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.

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