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I have always been confused about Leibniz notation. Not the notation itself, but the fact that it treats the differential operators ($d$, $\partial$) as being multipliable. The most famous example would probably be the Schrödinger equation, which if often denoted something like this: $$E\psi=\left(-\frac{h^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right)\psi=-\frac{h^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x)\psi$$ The problem with this is that is makes use of multiplication to expand the expression, $(a+b)c=a*c+b*c$, which would imply that $\partial *f=\partial f$. Personally, I would define the differential operators as functions: $$d(f)=\lim_{h\rightarrow \infty}{\frac{f(x+h)-f(x)}{h}}$$ So the above statement would make no sense at all. First I just accepted it as mathematical laziness, but then I stumbled upon this monstrosity:

Observe that $$\left(v^2\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial t^2}\right)y=0$$ can be factored as (which is what you probably mean by "squaring" in the question) $$\left(v\frac{\partial}{\partial x}+\frac{\partial}{\partial t}\right)\left(v\frac{\partial}{\partial x}-\frac{\partial}{\partial t}\right)y=0$$

What is happening here???

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    $\begingroup$ You can define "multiplication" of linear operators as composition. So $(aS+T)(aS-T) = a^2S^2-aST+aTS-T^2 = a^2S^2 - T^2$ as the mixed partial derivatives are equal when y is nice. $\endgroup$ – RandomWalker Aug 30 '16 at 22:18
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What is happening is that the physicists writing out the equation are using an operator notation, where $\frac{\partial}{\partial x}$ is a shorthand for the operator of taking the partial derivative in the $x$ direction of whatever appears on the right of the operator.

The reason this appears like multiplication is that $\frac{\partial}{\partial x}$ is a linear operator, so that for any objects $\mathcal{O}$ and $\mathcal{P}$ and scalar (real or complex number) $k$ $$ \frac{\partial}{\partial x} (\mathcal{O}+ \mathcal{P}) =\frac{\partial}{\partial x} (\mathcal{O}) + \frac{\partial}{\partial x} (\mathcal{P})\\ \frac{\partial}{\partial x} (k\mathcal{O}) = k\frac{\partial}{\partial x} (k\mathcal{O}) \\ $$ That first property allows you to write things that look like you are using the distributive law of multiplication over addition.

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The operator is not "multiplied with", it is "applied to", with the meaning $$\left(\frac\partial{\partial x}\right)\psi:=\frac{\partial\psi}{\partial x}.$$

The exponent is a convenient notation for the iterated operator,

$$\left(\frac\partial{\partial x}\right)^2\psi:=\left(\frac\partial{\partial x}\right)\left(\frac\partial{\partial x}\right)\psi=\left(\frac\partial{\partial x}\right)\frac{\partial\psi}{\partial x}=\left(\frac\partial{\partial x}\right)\frac{\partial\psi}{\partial x}=\frac{\partial^2\psi}{\partial x^2}.$$

From these definitions, you can observe that the factorization is indeed possible with the natural distributivity extension

$$\left(\frac\partial{\partial x}+\frac\partial{\partial y}\right)\left(\frac\partial{\partial x}-\frac\partial{\partial y}\right)\psi=\left(\frac\partial{\partial x}+\frac\partial{\partial y}\right)\left(\frac{\partial\psi}{\partial x}-\frac{\partial\psi}{\partial y}\right)\\ =\left(\frac\partial{\partial x}\right)\frac{\partial\psi}{\partial x}+\left(\frac\partial{\partial y}\right)\frac{\partial\psi}{\partial x}-\left(\frac\partial{\partial x}\right)\frac{\partial\psi}{\partial y}-\left(\frac\partial{\partial y}\right)\frac{\partial\psi}{\partial y}\\ =\frac{\partial^2\psi}{\partial x^2}-\frac{\partial^2\psi}{\partial y^2}=\left(\left(\frac\partial{\partial x}\right)^2-\left(\frac\partial{\partial y}\right)^2\right)\psi.$$

The same principle is sometimes used in the resolution of ODEs, like

$$y'''-4y''+5y'-2y=f(x)$$ can be written

$$(D^3-4D^2+5D-2)y=(D-2)(D-1)^2y=f(x),$$ where the characteristic polynomial appears.

By setting $z=(D-1)y$ and $w=(D-1)z$, you solve first order equations $$(D-2)w=f(x)$$ then $$(D-1)z=w,$$ $$(D-1)y=z.$$

Symbolically (and taking care of the integration constants), you can even write

$$w=(D-2)^{-1}f(x),$$$$z=(D-1)^{-1}(D-2)^{-1}f(x),$$$$y=(D-1)^{-2}(D-2)^{-1}f(x).$$

You can see a direct link with operational calculus and the Laplace transform.

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$-\frac{h^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)$ is the sum of two operators. $V(x)$ is understood to be the multiplication operator, that is, the operator $V(x)$ applied to $\psi$ is the product of the functions $V(x)\psi(x)$. The sum of operators is defined as $(A+B)\psi = A\psi+B\psi$.

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