Integrate $dx$ over interval $a\le x \le b$ instead of just $b-a$

Question

In the Wikipedia article on the wave function it's stated that the probability of a spin-less particle in 1D space being found in the interval $a\le x \le b$ at time $t$, where $x$ is the position, given the particle's wave function $\psi (x,t)$, is defined as the integral: $$ P_{a\le x \le b}(t)=\int_a^b dx |\psi (x,t)|^2 $$ Sticking to the math part instead of the physics part, I don't understand why there's an integral. I know that $\int dx=\int 1dx=x+C$, which means that the integral part of this function just evaluates to $b-a$. Why didn't they just write that?

Of course, given that Wikipedia isn't the greatest source of information, I am assuming that the math on Wikipedia is accurate.

Answer 1

This is a matter of notation. What they mean is $\int\limits_a^b |\psi (x,t)|^2 \mathbb{d}x$.

Such notation is particularly usefull when dealing with multiple integral: $\int \mathbb{d}x \int \mathbb{d}y \int \mathbb{d}z f(x,y,z)$ is a bit cleaner then $\int\int\int f(x,y,z) \mathbb{d}x \mathbb{d}y \mathbb{d}z$ if we try to understand which limits refer to a particular variable under integration.

Another point for this notation from differential geometry point of view is the fact that $f(x)\mathbb{d}x$ is a differential 1-form, which is the function $f(x)$ and the 1-form $\mathbb{d}x$ multiplied; here the multiplication of forms and functions is commutative simply by notation (much as multiplying a vector by a scalar).

Answer 2

Let's take a specific situation as an example.

We don't take for granted that $|\psi(x)|^{2}=1$ more that in some scenario (like for example the quantum mechanical particle in a box of width $[0,a]$) $\psi(x)=B \sin \left( \frac{n \pi x}{a} \right) $ for some constants $B, n$ and $a$.

In this instance an integral exists such that $$\int_{0}^{a}B^{2}\sin^{2}\left(\frac{n \pi x}{a} \right)^{2} dx=1$$

From which can be readily computed that the constant $B=\frac{\sqrt{2}}{a}$

Answer 3

It looks like you left out some information. A couple of lines above on the Wikipedia page, it was stated that $$\left|\Psi(x,t)^2\right| = \Psi(x,t)^*\Psi(x,t) = \rho(x,t),$$ and that this is interpreted as the probability density that the particle is at $x$. Thus $$P_{a\leq x\leq b} (t) = \int_a^bdx\left|\Psi(x,t)\right|^2 = \int_a^b \rho(x,t)\,dx.$$ Presumably, $\rho(x,t) \neq 1$. Meaning therefore $$\int_a^b\rho(x,t)\,dx \neq \int_a^b 1\,dx \neq b-a.$$