Infinite dimensional vector spaces (in quantum mechanics)

Question

I'm reading Shankar's "principles of quantum mechanics" and I'm extremely confused at section 1:10, which is about generalizing the statements made about vector spaces and operators to infinite dimensions. I don't find it as straightforward as simply "changing sums to integrals". I simply don't understand how to generalize from a discrete finite basis to a continuous basis. What does linear independence mean in a set of continuously indexed set of vectors? How do we construct a basis set? Also, it says that a function $f$ can be thought of as a vector that gets projected onto a basis (position basis) such that if $|f\rangle$ is the vector, then $f(x)$ is not exactly the function itself, but its projection in the $|x\rangle$ basis, but then uses both interchangably at some points. Is it not possible to talk about the vector itself without projecting it onto a basis in this case? Finally, what can I read to learn more about this that at the same or at a similar level to Shankar's QM book?

Answer 1

The "$|x \rangle $ basis" does not exist (in the Hilbert space), because no function exists that satisfies its defining property (the delta "function" does not exist). The changing sum to integral analogy is atrocious. Because for a basis of a Hilbert space the basis expansion is still given by (infinite) summation.

The author is doing the classic theoretical physics thing, where the reader is lured into a make believe world. It is probably best to accept it simply as a heuristic explanation of something that requires more advanced mathematical knowledge.

This whole thing can be mathematically explained by the usage of the theory of distributions and the theory of rigged Hilbert spaces (which requires some advanced knowledge of functional analysis unfortunately). An economic introduction to the theory of distributions and a review of the rigged Hilbert space theory can be found in the book "Mathematical Methods in Physics" by Blanchard and Brüning. But unfortunately this book assumes that the reader knows some basic measure/integration theory. It also contains most of the other mathematical background needed to truly understand quantum mechanics (for example Hilbert spaces).

Classical introductory texts to QM are "Modern Quantum Mechanics" by Sakurai and "Introduction to Quantum Mechanics" by Griffiths. I would recommend either of them over Shankar, but they both suffer from similar lack of mathematical rigor.

The only book (that i know of) on QM that keeps it real is "Quantum Theory for Mathematicians" by Hall. But like the title says it requires some math knowledge and also does not tackle many important topics of QM.

Answer 2

You can think about the Fourier transform as being a basis expansion in a continuous basis. For example, if you write $$ f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{isx}\int_{-\infty}^{\infty}f(y)e^{-isy}dy ds $$ If you think of the integrals as sums, it really is a basis expansion in exponentials: $$ \sum_{s} \frac{e^{isx}}{\sqrt{2\pi}}\left\langle f,\frac{e^{isx}}{\sqrt{2\pi}}\right\rangle = \int_{s=-\infty}^{\infty}\frac{e^{isx}}{\sqrt{2\pi}}\left\langle f,\frac{e^{isx}}{\sqrt{2\pi}}\right\rangle ds $$ The inner product is an integral, and the sum is a continuous sum, which becomes an integral. Why would you want such an expansion? Well, the exponentials are the eigenfunctions of $\frac{1}{i}\frac{d}{dx}$. Indeed, $$ \frac{1}{i}\frac{d}{dx} \frac{e^{isx}}{\sqrt{2\pi}}= s \frac{e^{isx}}{\sqrt{2\pi}}. $$ This isn't entirely correct because the so-called eigenfunctions don't belong to $L^2(\mathbb{R})$. However, integrals over any small interval in the eigenvalue parameter do belong to $L^2(\mathbb{R})$. For example, the following is in $L^2(\mathbb{R})$: $$ \int_a^b \frac{e^{isx}}{\sqrt{2\pi}}ds = \frac{1}{\sqrt{2\pi}}\frac{e^{ibx}-e^{iax}}{ix} $$ So these wave packs cancel in the right way to give you an actual $L^2(\mathbb{R})$ object. You can derive this mathematically, or you can use the uncertainty principle to conclude that $e^{isx}$ cannot be a valid state. So, these continuous expansions in Physics relate to uncertainty, because the objects in such a continuous expansion are not physically realizable due to the uncertainty principle. But, you can still expand using these objects, and the integral "sums" of such objects do result in physically realizable states. Physics and Mathematics complement each other beautically in this context.