Why can we treat differential operators as if they behave like algebraic quantities?

Question

In college, I've come across many instances where we multiply a derivative by a function, and the result somehow becomes the derivative of the function i.e $\frac{d}{dx}\times f=\frac{df}{dx}$— as if we're multiplying "operators" with functions in a purely algebraic way. This has always puzzled me, because it seems to work, but I don't fully understand why it works.

Here are a few examples to illustrate my confusion:

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1. Curl of a vector field:

$$ \text{curl } \mathbf{F} = \nabla \times (M, N, P) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \times (M, N, P) $$

When evaluating the determinant in the cross product definition, it seems like we’re "multiplying" the components of a vector of operators (whatever that means) with the components of a vector field. But that feels mysterious. Why is it okay to do something like:

$$ \left( \frac{\partial}{\partial x} \right) \cdot M $$

and call the result just $\frac{\partial M}{\partial x}$? What allows us to manipulate the gradient operator $\nabla$ as though it were a vector?

If this is valid, why can't I do something like:

$$\operatorname{ pie} (M,N,P) = (!, \sqrt{\ }, \wedge 2) \times (M, N, P) $$

where “!” is the factorial operator, and $\wedge 2$ is the square operator, just for the sake of argument? it doesn't make less or more sense that the defintion of curl in my opinion.

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2. Divergence of a vector field:

$$ \text{div } \mathbf{F} = \nabla \cdot (M, N, P) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right) \cdot (M, N, P) $$

Same idea — we’re dotting a vector of operators with a vector of functions. What justifies treating the differential operator as if it were a vector component that can be dotted and crossed in this way?

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3. Operator methods in ODEs:

In solving linear differential equations like:

$$ a y'' + b y' + c y = f $$

we often introduce the differential operator $ D = \frac{d}{dx} $ and rewrite the equation as:

$$ (aD^2 + bD + c)y = f $$

Then we manipulate this expression algebraically, even dividing by the operator polynomial, e.g.,

$$ y = \frac{1}{aD^2 + bD + c} f $$

In some cases, we go even further and treat operator expressions like $\frac{1}{1 - D}$ (I also have no idea how to define $1/(1-D)$) as a geometric series:

$$ \frac{1}{1 - D} = 1 + D + D^2 + \dots $$

and apply this to a function $f$, obtaining:

$$ f + Df + D^2f + \dots $$

How does this make sense? Why are we allowed to expand differential operators like power series and apply them this way?

More about my doubts about this method here.

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4. Schrödinger equation and time evolution:

In quantum mechanics, I remember (though somewhat vaguely) that we would manipulate expressions like:

$$ i\hbar \frac{\partial}{\partial t} \psi = H \psi $$

and treat $\frac{\partial}{\partial t}$ almost as an algebraic quantity. At one point, I recall solving for $4\psi(t)$ by exponentiating the Hamiltonian operator and applying it to $\psi(0)$, like:

$$ \psi(t) = e^{-iHt/\hbar} \psi(0) $$

This again treats differential operators as algebraic objects, able to be exponentiated, composed, and manipulated.

We also did this :- $\frac{ \partial w(x,t)}{ \partial x} =\frac{ \partial }{ \partial x}\times \cancel{ w(x,t)} =\frac{2i\pi \cancel{ w(x,t)} P}{h} $ To get $\frac{ \partial}{ \partial x}= \frac{2i\pi P}{h} $

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My Question:

All of these cases seem to rely on treating differential operators like algebraic objects — manipulating them in ways that resemble ordinary algebra (adding, multiplying, factoring, even inverting or expanding them in power series). Why is this allowed? What mathematical framework makes this rigorous?

I've asked professors and classmates, but usually get vague or unsatisfying answers like "it just works" or "it's a notation thing.". I feel there must be a more rigorous explanation but I don't know where to look.

Is there a general theory or justification that explains why and when it’s valid to treat differential operators this way?

Answer 1

The short answer is ‘yes’ there is a general theory of these things.

The story starts when you start thinking of collections of functions that all satisfy a certain property (let’s say: smooth real-valued functions on the real line) as infinite dimensional vector spaces. Then one thinks of differential operators as a linear maps between such spaces. Often the space of all linear maps between two spaces is itself a vector space and so one can indeed start to manipulate differential operators as if they are ‘objects’ in their own right eg add them together. And often but not always the space of linear maps is in fact an algebra: you can multiply together the elements, which naturally corresponds to composing the linear maps they represent.

Some keywords to read more: Unbounded operator, Banach space, operator algebra, spectral theory more generally etc.

Answer 2

It comes down to the fact that the derivative is a linear operator. That is, it distributes over addition:

$$ D (f + g) = (D f) + (D g) $$

and commutes with scalar multiplication:

$$ D (a f) = a (D f) $$

Thus it can be notationally treated as a kind of multiplication.

Answer 3

I will try to show how differential operators can be treated like algebraic variables, which simplifies solving linear differential equations. This explanation is intuitive and without detailed proofs.

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Definitions and Notation

Define the differential operator $D$ as:

$$ D^n f = \frac{d^n f}{dx^n} $$

which simplifies notation and allows algebraic manipulation of differential equations.

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Operator Polynomial

Let

$$ \phi(D) = a_n D^n + a_{n-1} D^{n-1} + \cdots + a_1 D + a_0 $$

where the coefficients $a_k$ is constant in all theorems except theorem 1 can be a function.

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Theorems (Without Proof)

Theorem 1:
Any linear differential equation

$$ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = f(x) $$

can be expressed as:

$$ \phi(D) y = f(x) $$

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Theorem 2:
If $a, b$ are constants, then

$$ (a D^m)(b D^n) y = ab\, D^{m+n} y $$

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Theorem 3:
If $\phi(D) = \phi_1(D) \phi_2(D)$ then $$ y = Y_1 + Y_2 $$

where $Y_1$ and $Y_2$ are the general solutions of $\phi_1(D) y = 0$ and $\phi_2(D) y = 0$ respectively.

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Results for Particular Solutions

Result 1:
If $ \phi(D) y = e^{a x} $ and $\phi(a) \neq 0$, then

$$ y = \frac{e^{a x}}{\phi(a)} $$

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Result 2:
If $\phi(a) = 0$ and

$$ \phi(D) = (D - a)^m \phi_1(D), \quad \phi_1(a) \neq 0 $$

then

$$ y = \frac{x^m}{m!} \frac{e^{a x}}{\phi_1(a)} $$

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These two results show the benefit of using the definition:

$$ y = \frac{1}{\phi(D)} e^{ax} $$

to find particular solutions. Using Results 1 and 2 along with this definition, we have:

• For $\phi(a) \neq 0$:
  $$ \frac{1}{\phi(D)} e^{ax} = \frac{e^{ax}}{\phi(a)} $$

• For $\phi(a) = 0$ and $\phi(D)=(D-a)^m \phi_1(D)$ with $\phi_1(a) \neq 0$:
  $$ \frac{1}{\phi(D)} e^{ax} = \frac{x^m}{m!} \frac{e^{ax}}{\phi_1(a)} $$

Result 3:
If $ \phi(D) = \phi_1(D) \phi_2(D) $ and $$ Y_1 = \frac{1}{\phi_1(D)} f(x) $$

then

$$ \frac{1}{\phi(D)} f(x) = \frac{1}{\phi_2(D)} Y_1 $$

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Why does $\frac{1}{1 - D} = 1 + D + D^2 + \cdots$ work?

Consider the equation $ (1 - D) y = f(x) $ This is equivalent to

$$ y = (1 + D + D^2 + \cdots) f(x) $$

The right-hand side means

$$ y = f(x) + f'(x) + f''(x) + \cdots $$

Putting this into the left-hand side:

\begin{aligned} y - y' &= \big(f + f' + f'' + \cdots \big) - \big(f' + f'' + f''' + \cdots \big) \end{aligned} therefore we get $y-y'=f(x)$

which shows the equality holds, so the operator series is valid.

for more generalization lets see theorem 5

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Theorem 5:
If

$$ y = \frac{1}{1 - \phi(D)} f(x) $$

then

$$ y = (1 + \phi(D)+ \phi^2(D) + \cdots) f(x) $$

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Answer 4

This is a comment not an answer.

You can remove a lot of mystery by simply noting that if we had two linear operators $E_1, E_2$ given by:

$$ E_1 = a_0*I + a_1 \frac{d}{dx} + a_2 \frac{d^2}{dx^2} + ... $$ $$ E_2 = b_0*I + b_1 \frac{d}{dx} + b_2 \frac{d^2}{dx^2} + ... $$

Then we can compute

$$E_1 [E_2] = a_0 b_0 + (a_0 b_1 + b_0 a_1) \frac{d}{dx} + ...$$

But now independent of this suppose we had two analytic functions

$$ f(z) = a_0 + a_1z + a_2z^2 + ... $$ $$ g(z) = b_0 + b_1z + b_2z^2 + ... $$

Then $$f(z)\cdot g(z) = a_0 b_0 + (a_0 b_1 + b_0 a_1)\cdot z + ... $$

So, without deeply knowing WHY, we can still SEE CLEARLY, that composing the operators $E_1, E_2$ gives the same coefficients as multiplying the analytic functions $f,g$. From here all the results you want arise intuitively (such as integration being division, factoring operators, the Euler-Maclaurin formula as a jacked cousin of $\frac{1}{1-e^{x}}$ etc...)

The WHY is given by @user76284 's answer. And it's good to explore this.

Yet simultaneously you want to learn how to manipulate these even without understanding too deeply why. See back in the late 1700s and 1800s folks were wildly manipulating these expressions without consciously being aware of such formal considerations, because they noticed the specific pattern I mentioned above. And so getting used to this similar to how a young child learns to add numbers (not understanding what base-N arithmetic IS, just knowing, that you can consistently ADD and CARRY) is also useful in its own right.

A good rite of passage in seeing if you "really get it" is if you can derive this from the expression $\frac{I}{I - e^{\frac{d}{dx}}}$ and explain why that makes sense as a summation formula. (Hint: you should first try to be absolutely clear as to why $e^{\frac{d}{dx}}[f] = f(x+1)$ )