Interpretation of Sakurai's ambiguous statement about wavefunctions and overlap

Question

I wasn't sure if I should've made this matter a question, but I'll give it a try anyway, so I might delete it if anyone finds the question to be of no help or against the guidelines (please let me know).

I've found the following passage from Sakurai a little bit obscure (I don't know if it contains some deep pieces of information, or if it is just a bunch of trivial facts that I'm missing):

Consider the inner product $\langle \beta | \alpha \rangle$. Using the completeness of $|x'\rangle$, we have: $$ \langle \beta | \alpha \rangle = \int dx' \langle \beta | x' \rangle \langle x' | \alpha \rangle = \int dx' \psi_{\beta}^{\ast}(x')\psi_{\alpha}(x')\tag{1.7.6}$$ so $\langle \beta | \alpha \rangle$ characterizes the overlap between the two wavefunctions. Note that we are not defining $\langle \beta | \alpha \rangle$ as the overlap integral; the identification of $\langle \beta | \alpha \rangle$ with the overlap integral follows from our completeness postulate for $|x'\rangle$.

Why does the author talk about overlap between wavefunctions, when you multiply two functions, their product isn't their overlap, unless the words overlap integral here assumes a different meaning from what I've imagined here. Secondly, why does postulating completeness for $|x'\rangle$ allow to pass from $$\text{overlap between wavefunctions} \rightarrow \text{overlap integral} $$ Is this transition hiding some deeper meaning?

Any help is much appreciated, as per usual

Answer 1

Why does the author talk about overlap between wavefunctions, when you multiply two functions, their product isn't their overlap, unless the words overlap integral here assumes a different meaning from what I've imagined here.

No, the words don't have any significantly different meaning than usual. From dictionary dot com: "Overlap: ... extend over and cover a part of..."

The integral is not just the product, it is the integration of the product.

Sakurai is basically saying that the "overlaps" are the regions where both the functions are not zero. If, say, one of the functions was only non-zero for $x>10$ and one of the functions was only non-zero for $x<-572$, then there would be no "overlap" and the integral would be zero. (Of course, there are other ways for the total integral to be zero, but let's not worry about that now.)

Secondly, why does postulating completeness for $|x'\rangle$ allow to pass from $$\text{overlap between wavefunctions} \rightarrow \text{overlap integral} $$ Is this transition hiding some deeper meaning?

No, there is not really any "deeper" meaning. If there is any deeper meaning, it is in the meaning of the "bra." Ultimately, a "bra" (an element of the "dual space") is just a linear function of a "ket" (an element of the Hilbert space). Sakurai is just saying that since we have this nice completeness relation, we can use it as a concrete and effective way of calculating functions like $\langle \beta|\alpha\rangle$.

Answer 2

Assuming both wavefunctions are normalized so that $\int dx |\psi_i(x)|^2=1$ for $i=\{\alpha, \beta\}$, then the inner product $\langle \alpha | \beta \rangle$ is a complex number with a norm between $0$ and $1$. The norm will be zero if the two wavefunctions are orthogonal, and will be $1$ if they are the same. The norm of the inner product is therefore a kind of measure for how well the vectors $|\alpha\rangle$ and $|\beta\rangle$ are aligned in Hilbert space, much like the dot product is a measure for how aligned two vectors are in real 3-dimensional space. The word "overlap" captures this intuition.

Answer 3

Consider two functions, $$f(x):\Bbb R^3\rightarrow \Bbb R$$ $$g(x):\Bbb R^3\rightarrow \Bbb R$$ They may separately have non-zero integral on all of $\Bbb R^3$, i.e.: $$\int_{\Bbb R^3}f(x)\mathrm dx\neq 0 \>\>\>\>\text{and}\>\>\>\>\int_{\Bbb R^3}g(x)\mathrm dx\neq 0$$

However, if the following are true for all $x\in\Bbb R^3$:

$$f(x)\neq0\implies g(x)=0$$ $$g(x)\neq0\implies f(x)=0$$

In words, everywhere that $f(x)$ is non-zero, $g(x)$ is zero, and vice versa, we see that:

$$f(x)\cdot g(x)=0\>\>\>\>\>\>\>\forall x\in\Bbb{R}^3$$

and therefore:

$$\int_{\Bbb R^3}f(x)\cdot g(x) \mathrm dx=0$$

In other words, if the functions have "no overlap" in the sense that they are both zero at all points where the other function is non-zero, the integral of their product is necessarily zero.

Thus, we can consider the integral a measure (in a loose sense) of the degree to which the two functions overlap where they are non-zero.