关于一元算子交换性及微积分中微分与积分算子交换条件的技术问询

Hey Joseph, great question—this is exactly the kind of foundational detail that trips up even experienced physicists, so it’s smart to dig into it! Let’s break this down into two parts: the calculus-specific case you’re seeing in your QFT book, and the broader abstract algebra question about unary operators.

第一部分：微积分中微分与积分算子的交换性 #

First off: differential and integral operators are NOT always commutative, but in the standard contexts of quantum field theory (and most physics applications), the conditions for commutativity are almost always satisfied—which is why your book skips the explanation.

The core rule here is the Leibniz integral rule (also called the differentiation under the integral sign theorem), which formalizes when you can swap a partial derivative and an integral. For an integral of the form:
$$I(x) = \int_{a(x)}^{b(x)} f(x, t) dt$$
You can write $\frac{d}{dx}I(x) = \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(x,t) dt + f(x,b(x))\frac{db}{dx} - f(x,a(x))\frac{da}{dx}$.

If your integral has constant bounds ($a(x), b(x)$ are fixed numbers), the boundary terms vanish entirely, leaving you with the clean swap:
$$\frac{d}{dx}\int_{a}^{b} f(x,t) dt = \int_{a}^{b} \frac{\partial}{\partial x}f(x,t) dt$$

Key conditions for this swap to hold: #

• $f(x,t)$ is continuous on a region that includes all $x$ you care about and the entire interval $[a,b]$ (or $[a(x),b(x)]$ for variable bounds).
• The partial derivative $\frac{\partial f}{\partial x}$ is continuous on the same region.
• For variable bounds: $a(x)$ and $b(x)$ are differentiable functions of $x$.

In QFT, we almost always work with smooth (infinitely differentiable) functions that decay rapidly at infinity (like exponential decay), so:

1. The continuity conditions are automatically satisfied.
2. If we’re integrating over all $\mathbb{R}^n$ (common in QFT), the boundary terms at $\pm\infty$ vanish because the function decays to zero.

Quick proof sketch (constant bounds): #

Start with the definition of the derivative:
$$\frac{d}{dx}I(x) = \lim_{h\to0} \frac{1}{h}\left(\int_a^b f(x+h,t)dt - \int_a^b f(x,t)dt\right)$$
Combine the integrals:
$$= \lim_{h\to0} \int_a^b \frac{f(x+h,t)-f(x,t)}{h}dt$$
Since $\frac{\partial f}{\partial x}$ is continuous, the difference quotient $\frac{f(x+h,t)-f(x,t)}{h}$ converges uniformly to $\frac{\partial f}{\partial x}$ on $[a,b]$. By the dominated convergence theorem (a core result in analysis), we can interchange the limit and the integral, giving:
$$= \int_a^b \lim_{h\to0} \frac{f(x+h,t)-f(x,t)}{h}dt = \int_a^b \frac{\partial}{\partial x}f(x,t)dt$$
Which is exactly the swap you’re seeing in the book.

第二部分：抽象代数中一元算子的交换性 #

In the broader sense of abstract algebra, two unary operators $A$ and $B$ commute if applying $A$ then $B$ gives the same result as applying $B$ then $A$—formally, $A(B(v)) = B(A(v))$ for every $v$ in their common domain.

There’s no universal condition for commutativity #

It entirely depends on the operators’ definitions and the space they act on:

• Linear operators: On finite-dimensional vector spaces, linear operators can be represented as matrices. Commutativity here is equivalent to matrix commutativity ($AB = BA$). On infinite-dimensional spaces (like function spaces), commutativity often ties to shared eigenspaces—for example, two self-adjoint linear operators commute if and only if they have a common orthonormal basis of eigenvectors.
• Nonlinear operators: Things get messier. For example, the differential operator $\frac{d}{dx}$ and the squaring operator $S(f) = f^2$ do NOT commute: $\frac{d}{dx}(f^2) = 2f\frac{df}{dx}$, which is not equal to $(\frac{d}{dx}f)^2$. Another example: the operator that takes a function to its absolute value doesn’t commute with differentiation, since $\frac{d}{dx}|f(x)|$ isn’t the same as $|\frac{d}{dx}f(x)|$ (think about $f(x) = x$ at $x=0$).

A quick note on "inverse" operators #

In calculus, the integral operator (over a fixed interval, or the antiderivative) is roughly the inverse of the differential operator. But even here, they don’t commute in the strictest sense—$\frac{d}{dx}\int_a^x f(t)dt = f(x)$, but $\int_a^x \frac{d}{dt}f(t)dt = f(x) - f(a)$. The difference is the constant term, which disappears if we’re working with functions that vanish at the boundary (common in QFT).

备注：内容来源于stack exchange，提问作者Joseph_Kopp