# Integral Notations in Quantum Mechanics [duplicate]

I've been learning about Quantum Dynamics, time evolution operators, etc. I am confused about the notation used in integrals. Normally I am used to integrals written in this way (with $$dx$$ on the right side):
$$\int f(x)dx$$
In this manner of notation, I can easily see the integrand as it is sandwiched by the integral sign and the $$dx$$. However, I often see integrals written in this way (with $$dx$$ beside the integral sign): $$\int dx f(x)$$ Is this notation not ambiguous? This is especially confusing for me when used in products, as I cannot identify what is the integrand sometimes. For example, I don't understand which is true in the following (when evaluating time evolution operator): \begin{align}\left(\int ^t_{t_0} dt' H(t')\right)^2\stackrel{?}{=}\int ^t_{t_0} H(t') dt'\int ^t_{t_0} H(t'')dt''\\\stackrel{?}{=}\int ^t_{t_0} dt' H(t')\int ^t_{t_0} dt'' H(t'')\\\stackrel{?}{=}\int ^t_{t_0} dt'\int ^t_{t_0} dt'' H(t') H(t'') \end{align}

The last line is especially confusing for me as I'm not sure if the integrand changes. Could I please get clarification for these different notations? Is there a reason for such notation? (If I'm not wrong, it is to group the integrals and the integrands in separate places for convenience? I'm not sure if it sacrifices clarity for this though.)

EDIT: There is also an issue of when operators are involved: \begin{align}\int dx \hat{F}(x) \hat{G}(x)\stackrel{?}{=}\int \hat{F}(x)dx\hat{G}(x)\\\stackrel{?}{=}\int \hat{F}(x)\hat{G}(x)dx\end{align} How do you know which operator is in the integrand? And assuming the general case where $$\hat{F}$$ and $$\hat{G}$$ do not commute, you cannot write the integral with $$\hat{G}(x)$$ on the left of the integral. How is this not ambiguous?

## marked as duplicate by knzhou, Kyle Kanos, Jon Custer, Martin, GiorgioPMar 21 at 18:33

• – Qmechanic Mar 21 at 3:20
• I added another section to my answer, that you may find interesting. – DanielSank Mar 21 at 5:34
• Yes, it is indeed an intriguing point, thank you for your help – Hexiang Chang Mar 21 at 7:12

I started seeing $$\int dx f(x)$$ in my freshman year of undergraduate. It's pretty common and the more you learn about integration the more it makes sense.

Now, regarding this part:

\begin{align}\left(\int ^t_{t_0} dt' H(t')\right)^2\stackrel{?}{=}\int ^t_{t_0} H(t') dt'\int ^t_{t_0} H(t'')dt''\\\stackrel{?}{=}\int ^t_{t_0} dt' H(t')\int ^t_{t_0} dt'' H(t'')\\\stackrel{?}{=}\int ^t_{t_0} dt'\int ^t_{t_0} dt'' H(t') H(t'') \end{align}

All of the equals signs there are correct. Integrals factor like this: $$\int dx \int dy \, f(x) \, g(y) = \left( \int dx \, f(x) \right) \left( \int dy \,g(y) \right) \, ,$$ which is all you did there. In fact, these are all the same: \begin{align} \int \int dx \, dy \, f(x) g(y) &= \int dx \int dy \, f(x) g(y) \\ &= \int dx \, dy \, f(x) g(y) \\ &= \left( \int dx \, f(x) \right) \left( \int dy \, g(y) \right) \\ &= \left( \int dx \, f(x) \right) \left( \int dx \, g(x) \right) \\ \end{align} Note, however, that you cannot factor something like this: $$\int_0^t f(t') \left( \int_0^{t'} dt'' f(t'') \right) dt'$$ because the limit of the second integral depends on the first integral's integration variable. You can, however, write it as $$\int_0^t dt' f(t') \int_0^{t'} dt'' f(t'') \, .$$

## Operators

There is also an issue of when operators are involved: \begin{align}\int dx \hat{F}(x) \hat{G}(x)\stackrel{?}{=}\int \hat{F}(x)dx\hat{G}(x)\\\stackrel{?}{=}\int \hat{F}(x)\hat{G}(x)dx\end{align}

There's really no difference. The key is to remember that the $$dx$$ really doesn't mean anything other than to remind you which variable(s) in the integrand is being integrated. By convention we tend to write the $$dx$$ either at the front or at the end. I've never seen it written in the middle like that. I think everyone would know what you mean, but putting the $$dx$$ is the middle of the integrands runs the risk that a reader won't notice them.

How do you know which operator is in the integrand?

Ok that's a good question! It really comes down to the fact that notation has to be clear. If you use the symbol $$x$$ to denote both an integration variable and a not-integrated variable, that's just asking for trouble. It also shouldn't ever happen because integration variables are consumed by the integral, so they can't be referred to anywhere else in an equation. For example, this makes no sense: $$g(x) = \int_0^1 \sin(x) dx$$ because there's no "free" $$x$$ on the right hand side.

And assuming the general case where $$\hat{F}$$ and $$\hat{G}$$ do not commute, you cannot write the integral with $$\hat{G}(x)$$ on the left of the integral. How is this not ambiguous?

Well, you certainly would not write $$\int dx \hat F(x) \hat G(x) \neq \left( \int dx \hat F(x) \right) \hat G(x) \, .$$ That just makes no sense.

## A speech about functions, integrals, and notation

A function $$f$$ is a well-defined thing independent of any specific choice of variable. A function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ takes one real number to another one. It is, therefore, completely unambiguous to write an integral as $$\int_0^1 f$$ with no $$dx$$. If $$f$$ is the sine function, then we can write e.g. $$\int_0^1 \sin$$ with absolutely no ambiguity. So why then do we so often write things like $$\int_0^1 \sin(x) \, dx$$? Well, consider a slightly more complicated function like the function $$f$$ defined by the equation $$f(x) = \sin(x)/x$$. How would we write this without variables? Well, we'd name the inversion function $$\text{inv}$$ defined by $$\text{inv}(x) = 1/x$$, and then we could say $$f = \sin \cdot \, \text{inv}$$ and we would write the integral as $$\int_0^1 \sin \cdot \, \text{inv} \, .$$ That's a perfectly clear representation, but I think it's less common in practice for three reasons:

1. It's cumbersome to have to name every function. Imagine having to write $$(\sin \circ \, \text{square}) \cdot \, \text{inv}$$ instead of $$\sin(x^2)/x$$.

2. We often solve integrals by "variable transformation", and for some people it's easier to see what transformations to make if we represent the functions by their action on their variables.

3. If you want to evaluate a multi-dimensional integral, at some point you have to use Fubini's theorem and that's only really possibly once the integrals are expressed as nested one-dimensional integrals over separate variables.

Still, even with those three points, I do think that especially for gaining a better understanding of integration (e.g. the change of variables formula) it can be helpful to practice the "no $$dx$$ notation". For example, I found that it was a key piece of my understanding how probability distributions transform under a change of variables. The "no $$dx$$ notation" also makes a lot of sense for people with experience in programming because it has strong ties to the notion of a type system.

• I think it's also important to point out that this form makes the notion of an integral as a functional more apparent, ie an integral is a function that takes a function and returns a number. – Aaron Mar 21 at 5:52
• by the way, the function f(x) = sinx(x) / x is called sinc. But the point stands for anything more complicated. Good explanation overall. – Chieron Mar 21 at 13:35
• @Chieron yeah good point. I should have picked a different function that doesn't already have a short name :-) – DanielSank Mar 21 at 16:12

The notation is not ambiguous; it's purely convention. The correspondence is

$$\left( \int_{t_1}^{t_2} dt H(t) \right) \left( \int_{t'_1}^{t'_2}dt’ H(t',t) \right) \iff \int_{t_1}^{t_2}\int_{t'_1}^{t'_2}H(t)H(t',t)dt' dt.$$

That is instead of evaluating "inside out" we evaluate the integrals from right to left.

If you square an integral as

$$\left(\int ^t_{t_0} dt' H(t')\right)^2$$

you should know that in general these two integrals don't talk to one another, except in very special cases. That is, they are completely separate entities.

• An example of such a very special case would be $$\left(\int_{\Bbb R}\exp -x^2 dx\right)^2=\int_{\Bbb R}\exp -x^2 dx\int_{\Bbb R}\exp -y^2 dy\\=\int_{\Bbb R^2}\exp -(x^2+y^2)dxdy=\int_0^{2\pi} d\theta\int_0^\infty r\exp -r^2 dr,$$where we convert a product of two integrals into a double integral so we can transform the variables. – J.G. Mar 21 at 11:48