Why is it possible to cancel a function in $\frac{\partial f}{\partial x}$?

Question

I am a freshman engineering student. In my university, in the first semester we studied differentiation and continuity, infinite series and conic sections in mathematics and some thermodynamics in physics. In second term we study integration and Linear algebra in mathematics and Schrödinger equations in physics. I don’t know if this is standard or normal for other countries but all my friends in other universities have to take Schrödinger in second semester. This seems like we are trying to run before learning how to walk.

$\require{cancel}$ I want to ask how we can do this $\frac{ \partial w(x,t)}{ \partial x}= \frac{2i\pi w(x,t) P}{h} $ then $\frac{ \partial \cancel{ w(x,t)}}{ \partial x}= \frac{2i\pi \cancel{ w(x,t)} P}{h} $ in other words $\frac{ \partial}{ \partial x}= \frac{2i\pi P}{h} $

I want to ask, how is it possible to cancel a function this way and what does $\frac{\partial }{\partial x}$ even mean without a function? When I tried to ask my professor, he told me something about linear operators and I didn’t understand a word from him. When I googled this, it was something related to linear algebra. Is it possible to explain this to someone who didn’t study multi variable calculus or linear algebra or ODE or PDE?

Another question is: Is it normal to take Schrödinger equations in the second semester of the first year without proper mathematics?

Answer 1

Seems like eta reduction https://sookocheff.com/post/fp/eta-conversion/. In other words, if $f(x) = g(x)$ for all $x$, then we could just be more concise and write $f = g$. In this case, we have two "operators", namely $(f \mapsto \frac{\partial f}{\partial x})$ and $(f \mapsto \frac{2\pi i f P}{h})$. In the context of the wave equation, since $\frac{\partial \psi}{\partial x} = \frac{2i\pi \psi P}{h}$ always, we could just shorten things and write $\frac{\partial}{\partial x} = \frac{2i\pi P}{h}$, knowing that whenever we see $\frac{\partial \psi}{\partial x} (= \frac{\partial }{\partial x}(\psi))$, it is the same as $\frac{2i\pi \psi P}{h}$ (at least in the QM setting).

From this perspective, the cancellation is just aesthetic. Nonetheless, if you want to think about linear operators, the cancellation says that $\frac{\partial}{\partial x} = \frac{2i\pi P}{h}$ as functions on the space of waves $\psi(x,t)$. Instead of functions taking in points, there are "operators" which take in functions. It does seem a bit odd at first to consider $\frac{\partial}{\partial x}$ as a function taking in inputs. Nonetheless, derivatives are exactly linear maps on functions.

Answer 2

tl;dr: By itself, $ \partial / \partial x $ is a differential operator. (It happens to be a linear one, which is useful for some things, but this fact is not particularly relevant right now, so that's why searching for ‘linear operator’ didn't get you where you needed to go.) It's an operator because it takes a function as input and returns a function as output; and it's a differential operator because it calculates the output function by (at least in part) differentiating the input function.

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In general, we can classify maps (things that take inputs to outputs) into types based on the nature of their inputs and outputs. A function, by default, takes a number as input and returns a number as output (although this term is also used generally for any map whatsoever); a binary function such as $ w $ takes two numbers as input and returns one number as output, and so on. Beyond these, a functional take a function (or binary function etc) as input and returns a number as output, and an operator takes a function as input and returns a function as output.

There's an operator that takes a binary function as input and returns as output the partial derivative of this function with respect to its first input. (Like functions generally, this operator is undefined on some inputs; this one is only defined on differentiable functions.) In a context where the first input number of a binary function is consistently written $ x $, you can call this operator $ \partial / \partial x $ (although some other notation such as $ \mathrm D _ 1 $ is better if you don't want to commit to particular variables for the input numbers). So that's what $ \partial / \partial x $ is, by itself.

Not every branch of mathematics uses these terms ‘functional’ and ‘operator’ in this way. In linear algebra, they're usually used differently: a functional takes a vector as input and returns a scalar (that is a number) as output, while an operator takes a vector as input and returns a vector as output; and both of these are usually assumed to be linear. But since functions form an infinite-dimensional vector space, and since $ \partial / \partial x $ is linear, it actually fits both uses of ‘operator’.

Besides differentiating functions, operators can also multiply their values by numbers or by the outputs of other functions, among other things. In the quantum mechanics of a point particle moving in $ 1 $ dimension, states in the Schrödinger picture are given by binary functions like $ w $, and observables are given by operators like $ \partial / \partial x $. Specifically, the position operator $ \hat X $ is multiplication by $ x $: $$ \bigl ( \hat X w \bigr ) ( x , t ) = x \, w ( x , t ) \text ; $$ and the momentum operator $ \hat P $ is differentiation with respect to $ x $ multiplied by the number $ h / ( 2 \pi \mathrm i ) $: $$ \bigl ( \hat P w \bigr ) ( x , t ) = \frac h { 2 \pi \mathrm i } \frac { \partial w ( x , t ) } { \partial x } \text . $$ For short, we write $ \hat X = x $ and $ \hat P = h / ( 2 \pi \mathrm i ) \, \partial / \partial x $. (You can write the last one even shorter as $ \hat P = - \mathrm i \hbar \partial _ x $, where $ \hbar $ means $ h / ( 2 \pi ) $ and $ \partial _ x $ means $ \partial / \partial x $.) But you have to remember that these are intended to mean operators; even though $ \hat X = x $ looks like $ \hat X $ is a number, it really means the operator that takes a function as input and returns as output a function whose output number is the output number of the input function multiplied by the first input number (which is quite a mouthful, so that's why we use symbols).

Now, when you write $ 2 \mathrm i \pi \, w ( x , t ) P / h $ instead of $ 2 \mathrm i \pi \, \hat P w ( x , t ) / h $, I suspect that your $ P $ is not quite the same as my $ \hat P $ (partly because it has no hat but mostly because it's on the right). If your $ P $ is just a number, then your equation is only true when $ w $ is an eigenstate of momentum with that number as eigenvalue (physically, a state whose momentum is always measured to have the value $ P $). Then the operator equation $ \partial / \partial x = 2 \mathrm i \pi P / h $ only holds on the eigenspace, the vector space consisting of those functions that are such eigenstates. This fact is then motivation for defining the momentum operator $ \hat P $ so that $ \partial / \partial x = 2 \mathrm i \pi \hat P / h $. (But it's also possible that your $ P $ is the same as my $ \hat P $, and you just wrote it on the right for some reason.)