I was studying quantum mechanics and the meaning of the bra-ket notation. We define the following ket that belongs to the Hilbert space:
$$|\psi\rangle \in \mathcal{H} \tag{1}$$
We define two vectors $|\alpha\rangle$ and $|\beta\rangle$ as;
$$
|\alpha\rangle\rightarrow
\begin{align}
\boldsymbol{a} &=
\begin{bmatrix}
x_{1} \\
x_{2} \\
\vdots \\
x_{m}
\end{bmatrix}
\end{align}
$$
$$
|\beta\rangle\rightarrow
\begin{align}
\boldsymbol{b} &=
\begin{bmatrix}
b_{1} \\
b_{2} \\
\vdots \\
b_{m}
\end{bmatrix}
\end{align}
$$
The classical way of describing the inner product between two vectors is given as;
$$
\langle\alpha|\beta\rangle = a_1^{\ast}b_1+a_2^{\ast}b_2+…+a_N^{\ast}b_N \tag{2}
$$
Now, 'Introduction to Quantum Mechanics' by David Griffiths states that vectors in quantum mechanics, for the most part, are functions that live in infinite-dimensional spaces. The previous notation can be awkward to use, because the inner product, just defined, does not always exist for $N\rightarrow\infty$.
To represent a physical state the wavefunction $\Psi$ must be normalized:
$$\int_{-\infty}^{\infty}|\Psi|^2dx=1$$
The set of all square-integrable functions are defined as:
$$
f(x) \;\; \text{such that} \;\; \int_a^b|f(x)|^2dx<\infty \tag{3}
$$
This constitutes a much smaller vector space. Mathematicians call it $L^2(a,b)$, while physicists call it Hilbert space. In another book, 'Quantum Mechanics: A modern development by Leslie E. Ballentine', I have seen the definition of the Hilbert space defined for the set of vectors ${\psi_i}$ as:
$$\lim_{i\to\infty}=||\psi_i-\chi||=0 \tag{4}$$
Where $\psi_i=\sum_n^{i}c_n\phi_n$. So Hilbert space for vectors is defined as the linear combination of an infinite set of vectors that converge towards a vector $\chi$. Here $\phi_n: n=1,2,…$ so we can construct a vector $\psi_n=\sum_nc_n\phi_n$ as a linear combination of an infinite set of functions. This must mean that the constructed vector is 'a vector of infinite elements', compliant with vectors $|\alpha\rangle$ and $|\beta\rangle$, or more generally that $\psi$ is a vector in the basis of an infinite set of vectors and thus a vector of infinite dimension. This is how I understand the Hilbert space of vectors.
The function (3) defines the norm of a $\textit{function}$, so why is this applicable to the $\textit{vectors}$ as well? Also, why does this equation make sure that the inner product of equation (2) converges? Lastly, how does (4) show the same as (3)?
I want to point out that I know how to show that eq. (3) is a vector space, but I struggle to see how this vector space describes the vector given in eq. (1). The concept of Hilbert space confuses me a lot and I struggle to find a satisfactory explanation.
Edit: Maybe we can only say that the Hilbert space defined as in eq. (3) describes that the inner-product converges in eq. (2), but then why is eq. (1) true -It does not describe an inner product?
Maybe the definition of Hilbert space for functions is completely seperated from the definition of Hilbert space for vectors. This is my interpretation thus far.
Best Answer
First, let's make it clear that a Hilbert space is not a particular vector space, but is rather any vector space that has a well-defined inner product and is complete w.r.t. the norm derived from it. Complete here means that you're allowed to take the limit of any sequence of vectors whose consequitive term difference tends to the zero vector, i.e. one whose elements get closer and closer (such a sequence is called fundamental).
If you think about it, a sequence whose elements get arbitrarily close (the closeness being measured by the norm: $\|a_{n} - a_{m}\|\to 0$) is a perfect candidate for one that has a limit. Unfortunately, for the infinite-dimentional case, not every fundamental sequence ought to have a limit, hence we need the definition of "completeness".
That said, $L^2(a,b)$ is a particular vector space which happens to fulfill the criteria for being a Hilbert space. Simpler examples include every finite-dimentional vector space - they're Hilber spaces, too, with the inner product defined by your eq. 2 (w.r.t. the particular basis you've chosen to represent your vectors as columns).
Let's look at your questions now.
I think you're accustomed to vectors being matrices of size $n\times 1$. That's true, but is only part of the story. More generally, the term vector describes an element of any vector space. When you get to know infinite-dimensional spaces, functions can become vectors, too - even if you cannot write them as finite column matrices!
Any set in which you can take (finite) linear combinations is a vector space, and its elements are called vectors. E.g. for $f,g:(a,b)\to\mathbb R$ we can always get another function $af+bg$ for any real $a,b$, hence they form a vector space and we cal $f,g$ vectors.
The two equations - 2 and 3 - are unrelated. They refer to different vector spaces. For finite-dimensional spaces, an inner product can be defined with your eq. 2, while for the function space $L^2$, the inner product is $\left\langle f,g\right\rangle :=\int_{a}^{b}f(x)g(x)\text{dx}$ (if your functions are complex-valued, you take $f^*$).
You ask why does the integral $\int_{a}^{b}fg\text{dx}$ converge, given that $f,g$ are square-integrable, and that is a very reasonable question. An easy answer uses the fact that $|ab|\le \frac 1 2 (a^2 + b^2)$ (expand $0\le(a+b)^2$ to see this): $$\int_{a}^{b}fg\le\int_{a}^{b}\left|fg\right|\le\frac{1}{2}\left(\int_{a}^{b}f^{2}+\int_{a}^{b}g^{2}\right)<\infty$$ (Omiting the x argument here for brevity). The last inequality follows from the square integrability of $f$ and $g$.
Actually, it does not. Your eq. 4 defines the limit of the sequence $\psi_i$, not a vector space. Ballentine constructs a Hilbert space formally (read "artificially") from a countable set of "abstract" elements $\psi_i$ which we'll consider an ortonormal basis.
Given all the $\psi_i$, Balentine constructs a vector space $V$ of all formal finite linear combinations of $\psi_i$. It is an inner product space by your eq. 2: two finite combinations of $\psi_i$s are multiplied component-wise, and this induces a norm on every such vector. This space, though, is not yet Hilbert, because it is not complete (remember that a Hilber space is one that is both complete and has an inner product). For example, the sequence with terms $v_{n}=\sum_{i=1}^{n}\frac{1}{i}\psi_{i}$ is fundamental ($\|v_n-v_{m}\|=\sum_{n+1}^m\frac 1 {i^2}\to 0$ sufficiently fast), but has no limit in $V$ (each next term is linearly independent of the span of all previous terms, so its limit would be an infinite linear combination which does not exists in $V$ by construction). Hence Balentine includes "artificially" all such limit points, calling the resulting space $\mathcal{H}$, to make it complete. Thus we get "artificially" some infinite linear combinations, apart from the finite ones (this process is called completion, here's a wiki article on it).
What you can ask now is, how the heck is this space (constructed from a countably many $\psi_i$s with your eq. 4) related to the square-integrable functions $f:(a,b)\to \mathbb R$ (your eq. 3)? It is a very good question, allow me to answer it in the next section.
I think you're confused because you don't see how the space Balentine constructs from the $\psi_i$s is the same as the space of square-integrable functions. It is not apriori clear what functions are the $\psi_i$s, and the question is highly non-trivial.
First you need to realize that the elements of the $\mathcal{H}$ described above (from Balentine's book) $v_{n}=\sum_{i=1}^{n}c_{i}\psi_{i}$ are completely determined by the coeffiecients $c_i$ in front of the $\psi_i$s. That is, its elements are in fact square-integrable sequences, and this space is called $l^2$. To get a high-level overview about why this space - from your eq. 2 - is the same as the one described above, allow me to throw some links:
You don't need to read through all these, just make sure to understand the logic flow:
And if you're interested in what square-integrable functions from $L^2$ correspond to the $\psi_i$s from the artificially built $l^2$, look closely at the answer by Jess Madnick in the first question linked. It may give you an intuition about how such a big random space like $L^2$ may be constructed from countably many functions only.
I'm not sure whether you expected such an answer, it's just that the truth is not so simple :) As the comments to your question say, you can try reading some books on these topics, just try not to get too distracted from your main goal. I had a similar experience with differential geometry and classical mechanics, and there's a risk you get aside from your target if you dive too deep in the details. I hope my answer gives you a way to move forward without needing to spend another month reading functional analysis books (which is a wonderful experience on its own, don't get me wrong :)).