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???

• 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. – RandomWalker Aug 30 '16 at 22:18

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.

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.

$-\frac{h^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)$ 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 $(A+B)\psi = A\psi+B\psi$.