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The Time Dependent Schrodinger Equation has the form

$$i\hbar\frac{\partial}{\partial{t}}\Psi=-\frac{\hbar^2}{2m}\left(\nabla^2+V\right)\Psi$$

and the Time Independent Schrodinger Equation has the form

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

I know that $V$ is the potential operator and is a function of the position, and $Psi$ is the wavefunction. I also know that $\hbar$ is a constant, and $m$ is the mass of the particle.

What does $\partial$ mean, and what does $\nabla$ mean?

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First of all if you know what $y'(x)$, means then you can think of

$$\frac{\partial{y}}{\partial{x}}$$ as just another way of writing $y'(x)$

and you can use other variables, so you could have

$$\frac{\partial{A}}{\partial{B}}$$

which is the same as $A'(B)$

So if for example

$A=B^2$ then the answer to

$$\frac{\partial{A}}{\partial{B}}$$

is

$$\frac{\partial{A}}{\partial{B}}=\frac{\partial(B^2}{\partial{B}}=\left(B^2\right)'=2B$$

If on the other hand I $\Lambda=\cos(\gamma)$

then the answer to

$$\frac{\partial\Lambda}{\partial\gamma}$$

is

$$\frac{\partial\Lambda}{\partial\gamma}=\frac{\partial\cos(\gamma)}{\partial\gamma}=\left(\cos(\gamma)\right)'=-\sin(\gamma)$$

$$\frac{\partial}{\partial{N}}w=\frac{\partial{w}}{\partial{N}}$$

and if for instance $w=\ln(N)$

then the answer to

$$\frac{\partial}{\partial{N}}w$$

is

$$\frac{\partial}{\partial{N}}w=\frac{\partial{w}}{\partial{N}}=\frac{\partial\ln(N)}{\partial{N}}=\frac{1}{N}$$

Now if you have seen the laplacian in the form $\nabla^2$ then it can help to see the equation

$$\nabla^2\Psi=\left(\frac{\partial\frac{\partial{\Psi}}{\partial{x}}}{\partial{x}}+\frac{\partial\frac{\partial{\Psi}}{\partial{y}}}{\partial{y}}+\frac{\partial\frac{\partial{\Psi}}{\partial{z}}}{\partial{z}}\right)$$

so the Time Dependent Schrodinger Equation is

$$i\hbar\frac{\partial\Psi}{\partial{t}}=-\frac{\hbar^2}{2m}\left(\frac{\partial\frac{\partial{\Psi}}{\partial{x}}}{\partial{x}}+\frac{\partial\frac{\partial{\Psi}}{\partial{y}}}{\partial{y}}+\frac{\partial\frac{\partial{\Psi}}{\partial{z}}}{\partial{z}}\right)+V\Psi$$

while the Time Independent Schrodinger Equation is

$$E\Psi=-\frac{\hbar^2}{2m}\left(\frac{\partial\frac{\partial{\Psi}}{\partial{x}}}{\partial{x}}+\frac{\partial\frac{\partial{\Psi}}{\partial{y}}}{\partial{y}}+\frac{\partial\frac{\partial{\Psi}}{\partial{z}}}{\partial{z}}\right)+V\Psi$$

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    $\begingroup$ The notation $\frac{\partial y }{\partial x }$ is not another way to write $y'(x)$ unless somewhere it's explicitly stated that y solely depends on x, but then one would just use $\frac{d y}{dx}$ instead. All the partial derivative let's you know is that the derivative is with respect to x and all other variables are assumed to be held constant. $\endgroup$
    – Triatticus
    Commented Sep 20, 2021 at 11:57

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