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With the product rule, one has to limit the scope of operators (without using $f'$ or $\partial f/\partial x$): $$ \frac\partial{\partial x} u(x) v(x) = \left(\frac\partial{\partial x} u(x)\right) v(x) + u(x) \frac\partial{\partial x} v(x) $$

I have seen this pretty often, and the parenteses should limit the scope of the first partial derivative onto the function $u$. However, in Physics, there is the Schrödinger equation which does quite the opposite: $$ E \psi(x) = \left(\frac{\hbar}{2m} \nabla^2 + V(x) \right) \psi(x) $$

There, you apply the $\nabla^2$ onto $\psi$, although the latter is outside of the parenteses.

So what is the appropriate way to write this product rule, then?

Martin Ueding
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2 Answers2

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It's just a question of clarity - in the Schrödinger equation it's quite clear that's $\nabla^2$ is operating on $\Psi(x)$ and not on $V(x)$.

If it helps, think of a operator as a matrix in some Hilbert space, thus $\nabla^2 + V(x)$ is actually an operator on it's own right since the operator $V(x)$ is actually just a scalar times the identity matrix.

Nathaniel Bubis
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If I understand well your doubt, it's just a matter of definition. Within the Schrödinger framework one uses the convention that for example $$ (\partial_x + \partial_y ) \psi (x,y) = \partial_x \psi (x,y) + \partial_y \psi (x,y). $$ So in the case of the Schrödinger Equation think of $\nabla^2$ as if it were $\partial_x$ and $V(x)$ as $\partial_y$. It has nothing to do with the product rule of differentiation, there you consider following your example two scalar function $ u(x) $ and $v(x)$, but in the Schrödinger Case you consider two operators, i.e., $\nabla^2$ and $V(x)$.

Also and for consistency, one finds usurally another convention which differs from yours, i.e., one writes the product rule as $$ \frac\partial{\partial x}( u(x) v(x) )= v(x)\frac\partial{\partial x} u(x) + u(x) \frac\partial{\partial x} v(x) $$ and it seems necessary to write the first parentheses.

Ognan
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