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I need to understand how one can think of a function as a vector (in Hilbert space, more specifically) so I can follow along QM texts. I've read this question's answers, but they were uninspiring to me. I've also read this introduction but found it too short to really grasp the concept.

I've been recommended Cohen-Tannoudji's QM book. Do you have any other recommendations or explanations to offer?

My goal is not only to understand the concept, but use it. Also, for the interested, I'm a chemistry major - mathematics is poorly taught to chemistry majors, in my experience.

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Are you familiar with the definition of a vector space? It helps to change the way you think of vectors rather than changing the way you think of functions. For example, it is perfectly natural to think of an element of $\mathbb R$ as a vector in the infinite-dimensional $\mathbb Q$-vector space $\mathbb R$. – Alex Becker May 28 '12 at 17:20
I've had a linear algebra course - nothing fancy. I remember having to verify if X or Y were vector spaces using certain criteria, but these are lost to memory. I'm taking a look at wikipedia now. – CHM May 28 '12 at 17:22
If you are willing to take on a challenge after you review the definition of vector spaces and vector space isomorphisms: a)Verify that the set $Funct(X,F)$ of functions from $X$ to a field $F$ is a vector space under pointwise addition and $(\lambda\cdot f)(x)=\lambda f(x)$. b) The set $\prod_{x\in X} F$ of vectors with positions indexed by $X$ is a vector space under pointwise addition and vector scaling. c) Show $Funct(X,F)\cong \prod_{x\in X} F$ by the isomorphism given in my solution below. – rschwieb May 29 '12 at 12:12

If your working definition of "vector" is "an ordered list of numbers", then you can start there.

Let's consider $v=(A,B,C)$ where $A,B,C$ are all real numbers. This gadget would be a vector for us. How could one look upon it as a function? Well we could name the positions #1, #2, and #3, and say $v(1)=A, v(2)=B, v(3)=C$. Therefore the domain (things for $i$ in $v(i)$) is $1,2,3$ and the output is determined by whatever is in that position.

Now let's think of a sequence of real numbers $a_1, a_2,\dots$. If I dress them up: $v=(a_1,a_2,\dots)$ this is also a (very long) vector. To think of it as a sequence we can pull the same trick of using the $i$'th position to record the output for $i$, and so $v(i)=a_i$ has given you a function from $\mathbb{N}$ into $\mathbb{R}$. Really, $a_i$ was already in functional notation, and you probably wouldn't need to introduce $v(i)$, but I'm making a point here :)

But why should we stop with a finite set or $\mathbb{N}$? Could we use $\mathbb{R}$ as subscripts for a vector? Sure, except now that the inputs for $i$ are not discrete anymore, it becomes impossible for us to physically write a vector with continuum many positions. But we can still imagine it. Just imagine you have a bunch of positions, each labeled with a real number $r$, and the whole bunch are bracketed by parentheses (). In the $r$'th spot, you put the output of $v$, and so we would write $v(r)=Q$, where $Q$ is that particular output.

Really, order isn't even necessary. Given any function from a set $X$ into a set $Y$ you just imagine a bunch of positions, one for each thing in $X$, surrounded by parentheses. In each position you are allowed to fill in a thing from $Y$. Let's call the whole vector $f$. What is $f$'s value at $x\in X$? Well, whatever value of $y$ is in the $x$th position of course! We denote this with $f(x)=y$.

In this way $f$ can be thought of as a very long vector with entries indexed by $X$, and each entry contains the value of $Y$ which is the output.

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This is a tricky way to think about things, because there are a lot of things one does with "a list of components" that don't work in this situation. I once read a textbook that advised forgetting everything you know about finite-dimensional linear algebra, on the grounds it's just as likely to lead you wrong as it is to help in the infinite-dimensional case. – Hurkyl May 28 '12 at 21:00
@Hurkyl I can't imagine what you are referring to "not working", but I would appreciate elaboration on what you have in mind. This description is completely standard in analysis and topology, where the set of functions from $X$ to $Y$ is sometimes denoted $Y^X$, since the functions are thought of as elements of this product space. If $X$ and $Y$ have addition and/or multiplication operations, then pointwise addition/multiplication of functions behave exactly as you would guess for the "vectors". – rschwieb May 28 '12 at 23:05
Things like the fact that the vectors with a $1$ in one slot and $0$ elsewhere don't form a basis, there are limitations on how you can put numbers in slots, two different "lists" of numbers can still be the same vector, very few linear transformations can be written in the corresponding matrix form, et cetera. – Hurkyl May 28 '12 at 23:58
(To explain the third point, we mod out by negligible functions in interesting function spaces like $L^2$) – Hurkyl May 29 '12 at 0:03
@hurkyl As for the basis thing, that is not very important. For example in the algebra of formal power series $F[[x]]$, you have the same problem that the powers of $x$ are not a basis. For the second comment, I do not understand how you could have two different lists which were the same vector. By definition, they differ on a coordinate, and therefore are unequal vectors. Linear transformations with infinite matrices are still possible. – rschwieb May 29 '12 at 0:08

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