In the context of Quantum Mechanics and Hilbert spaces, I understand that a function can be interpreted as $\psi(x) = \langle x \vert \psi \rangle$ in the position basis, and things like $$\int_a^b|\psi(x)|^2dx$$make sense interpreting $x$ as a label. But if I think of $\psi(x)$ as an element of $L^2$ and expand $\psi(x)$ in a discrete basis, like $\psi(x)=\sum_a(\phi_a(x),\psi(x))\phi_a(x)$, now the label is $a$ and what does $x$ means now and why the inner product is still the same and does not make any reference to $a$?
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$\begingroup$ Illustrate by Fourier analyzing a periodic function, and then resolving a function in quantum oscillator eigenfunctions. Can you now see the map from natural to real numbers and sharpen your question? $\endgroup$– Cosmas ZachosCommented Apr 22, 2022 at 10:38
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$\begingroup$ probably closely related: physics.stackexchange.com/a/359982/50583 $\endgroup$– ACuriousMind ♦Commented Apr 22, 2022 at 12:13
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2 Answers
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When you represent a periodic function by a Fourier series you do pretty much the same thing: you represent it by sines instead of δ-functions.
For your specific example, you have two alternate resolutions of the identity/ completeness relations, $$ 1\!\!1= \int\!\! dx ~|x\rangle\langle x| \\ 1\!\!1= \sum_a |a\rangle \langle a| $$ where, importantly, $\langle a|x \rangle = \phi_a^* (x)$, your change of basis "matrix".
You then have $$ \psi(x)=\langle x|\psi\rangle = \sum_a \langle x|a\rangle \langle a|\psi\rangle = \sum_a \phi_a(x) ~~\psi_a, ~~\hbox{where}\\ \psi_a=\langle a|\psi\rangle= \int\!\! dx ~ \phi^*_a(x)~ \psi(x), $$ by the above completeness relation. In your expression, you used the same x as a variable, and as a dummy integration variable in your inner product--a disastrous practice.
It is crucial to test-drive this equivalence with the Hermite functions $\psi_n(x)=\langle x|n\rangle$, where the discrete index is the natural number identifying the quantum oscillator energy level, if you have not already done so.
Useful link as per your comment.
answered Apr 22, 2022 at 11:47
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$\begingroup$ An important point is also that $\psi(x) \in \mathbb C$ is a number while $|a\rangle, |\psi\rangle \in \mathcal H$ are states on the Hilbert space. However, $|x\rangle$ does (strictly speakin) not exist as a state on the Hilbert space. Only the "bra" $\langle x|$ exists as a linear map on the Hilbert space ($\langle x| : \mathcal H \to \mathbb C$). This is the reason why they must be treated differently, even though they are similar in prinicple. One would write $|\psi\rangle = \sum_a \psi_a |\phi_a\rangle$ as an equality on $\mathcal H$, but not $|\psi\rangle = \int dx \psi(x) |x\rangle$. $\endgroup$– CreamCommented Apr 22, 2022 at 12:30
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$\begingroup$ Thanks a lot for your answers. This was my first question in this forum. I need to "digest" all these. My goal is to find out how you can do QM without the rigged Hilbert space, delta functional, etc. I know that it has to be with the spectral theorem and that we need to restate our equations in terms of projection operators, which are well-defined in Hilbert space. I need to study. It is not that I don't like the rigged Hilbert space; it is because I want to know why so many books don't talk about it explicitly, but used all the time. $\endgroup$– PabloCommented Apr 22, 2022 at 12:48
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$\begingroup$ You say: "In your expression, you used the same x as a variable, and as a dummy integration variable in your inner product--a disastrous practice" ¿How can I write the same expression correctly? $\endgroup$– PabloCommented May 26, 2022 at 22:25
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$\begingroup$ $\psi(x)=\sum_a(\phi_a(y),\psi(y))~~\phi_a(x)$, but it's still a clumsy expression... $\endgroup$ Commented May 27, 2022 at 13:33
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The expression $$p(a,b)=\int_a^b|\psi(x)|^2dx$$ is the probability that you will find a particle in the interval $[a,b]$. So it is a measurement of position.
The expression $$\psi(x)=\sum_c(\phi_c(x),\psi(x))\phi_c(x)$$ is writing down a function in terms of other functions, which doesn't change its value anywhere on the interval $[a,b]$ so it won't affect $p(a,b)$. Each $c$ is just a label, like the $x$: it's just a different way of labeling the same function.
