1.
This is just an English issue. It seems to me you’re misreading the sentence
Zettili: To free state vectors from coordinate meaning, Dirac introduced what was to become an invaluable notation in quantum mechanics.
This is the same use of the word “free” as in “in order to free up some resources”. Although I’ll disagree that this notation is “invaluable”, let me phrase the sentence in an essentially equivalent manner:
Me: Dirac introduced an invaluable notation to emphasize the coordinate-independent notion of a state-vector.
(And recall that by a state-vector, Zettili just means an element of the Hilbert space under consideration in a given problem).
2 (see edit below).
It means evaluate the function $\psi:\Bbb{R}^3\times\Bbb{R}\to\Bbb{C}$ on the element $(\vec{r},t)$ of its domain to get the complex number $\psi(\vec{r},t)$.
There isn’t a true bra $\langle \vec{r},t|$ anyway, i.e this doesn’t live in the dual of the Hilbert space, but rather lives in a larger space of tempered distributions. So, there’s really no point trying to force everything into a Bra or a Ket when it doesn’t belong.
3 and 4.
In general if you have a linear map $\lambda:H\to\Bbb{C}$ and a linear operator $A:H\to H$, then these can be composed to give you $\lambda\circ A:H\to\Bbb{C}$. If you now want to apply this on an element $\psi\in H$, you get the complex number $(\lambda\circ A)(\psi):=\lambda(A(\psi))$. This is literally elementary function notation and definition of composition.
Anyway, if you want to do all sorts of symbolic gymnastics, then here we go step by step:
- given an element $\phi\in H$, we can define a linear functional $\Phi:H\to\Bbb{C}$ as $\Phi(\psi):= \langle\phi,\psi\rangle$.
- In QM, people don’t write $\Phi$ (or whatever other name you can come up with), they would instead prefer to write $\langle \phi|$, and call it “bra of $\phi$”.
- given an element $\psi\in H$, don’t leave it naked, instead enclose in half a right-angled bracket, $|\psi\rangle$, called the “ket of $\psi$”.
So, instead of writing $\Phi(\psi)$, we now write $\langle \phi|(|\psi\rangle)$. Observe that this is just following the syntax of
\begin{align}
\text{function_name}(\text{name of point in the domain of the function}).
\end{align}
Now, stop writing parentheses, to get $\langle\phi||\psi\rangle$. Now, delete the extra line in the middle, to get $\langle\phi|\psi\rangle$.
This is half the reason why people like the notation. For the other half, we go as follows:
- as we have already declared our “bra notation”, instead of writing $\Phi\circ A$ we now write $\langle\phi|\circ A$.
- composition symbols are annoying in Physics, so write $\langle\phi|A$ instead and call this the “*action of the operator $A$ on the bra of $\phi$”.
But you have to remember that when you apply this on an element of $H$, by definition it means you apply $A$ first, then you apply the linear functional, i.e
\begin{align}
\bigg(\langle\phi|\circ A\bigg)(|\psi\rangle) := (\langle\phi|)\bigg(A(|\psi\rangle)\bigg).\tag{$*$}
\end{align}
Finally, in keeping with our tradition above of removing parentheses and vertical lines, both sides (which are equal by definition of function composition) are just written as $\langle\phi|A|\psi\rangle$.
Once again, the equality in $(*)$ is literally nothing more than basic definitions of how functions work. For any sets $A,B,C$ and any functions $f:A\to B$ and $g:B\to C$, we can define $g\circ f:A\to C$ by setting for each $a\in A$, $(g\circ f)(a):= g(f(a))$.
Nothing at all about self-adjointness here.
Some extra remarks (can be skipped on first reading).
If $H\neq \Bbb{C}$, and $A:H\to H$ and $\lambda:H\to\Bbb{C}$ are linear, then the composition $A\circ\lambda$ is not even defined. That’s why we don’t write $A$ before the $\lambda$.
In math there’s the notion of the transposed operator or dual operator or adjoint operator (closely related to Hilbert space adjoints but not exactly the same), whereby given a linear map $T:V\to W$ between two vector spaces, we get an induced map $T’$ (also denoted $T^t$ or $T^*$ (again, not exactly the same as the Hilbert space adjoint)), which is a linear map $T’:W^*\to V^*$, defined by setting $T’(\omega):=\omega\circ T$. In other words, we don’t write $T(\omega)$ or $T\circ\omega$ because that’s nonsense, but instead, we define a new operator $T’$ with correct domains and targets so that $T’(\omega):=\omega\circ T$ is a sensible definition.
Edit to 2: Thanks to @Ghoster for the comment.
In order to maintain a clean distinction between abstract Hilbert spaces and concrete ones like $L^2(\Bbb{R})$, then I should explain the above slightly differently. But the tldr is the following:
once you express the abstract ket suitably as a function, the corresponding bra $\langle x|$ just acts like evaluation as described above.
More properly, start with an abstract Hilbert space $\mathscr{H}$ (modelling a 1D quantum system say). The position observable is supposed to be described by a self-adjoint operator $X:\mathscr{H}\to\mathscr{H}$ (warning: lots of domain issues of operators on infinite-dimensional which I’m ignoring). Then, one has (by the spectral theorem in multiplication operator form) a unitary isomorphism $\Xi:\mathscr{H}\to L^2(\Bbb{R})$. This unitary isomorphism $\Xi$ gives us several concrete interpretations:
for each abstract $\psi\in\mathscr{H}$, we get a function $\Xi(\psi)\equiv \psi_{\Xi}\in L^2(\Bbb{R})$. We can call this the “position-space representation of $\psi$”.
for each operator $A:\mathscr{H}\to\mathscr{H}$ (self-adjoint or not) we get an induced operator $A_{\Xi}:L^2(\Bbb{R})\to L^2(\Bbb{R})$ defined by $A_{\Xi}:= \Xi\circ X\circ \Xi^{-1}$. We call $A_{\Xi}$ the position-space representation of the operator $A$.
applying the previous bullet point to $A=X$ the position operator itself we get an operator $X_{\Xi}:L^2(\Bbb{R})\to L^2(\Bbb{R})$. The definition of $\Xi$ and $X$ together are designed to give us a multiplication operator, i.e $(X_{\Xi}(f))(x):=x\cdot f(x)$, i.e the “position representation” ($X_{\Xi}$) of the position operator ($X$) simply multiplies each function by the position coordinate.
lastly, for each $x_0\in\Bbb{R}$, we can define a linear map $\mathscr{H}\to\Bbb{C}$ by sending $\psi$ to $\psi_{\Xi}(x_0)$. People often call this $\langle x_0|$ “bra of $x_0$” but technically it doesn’t live in $\mathscr{H}^*$. This definition is summarized symbolically as follows:
\begin{align}
\langle x_0|\psi\rangle=\psi_{\Xi}(x_0).
\end{align}
Now, we can consider the momentum observable as well, i.e a self-adjoint $P:\mathscr{H}\to\mathscr{H}$, and we get a (different) unitary isomorphism $\Pi:\mathscr{H}\to L^2(\Bbb{R})$. So, now we can speak of “momentum-space representations” of various quantities. So, now we have things like $\psi, A, \psi_{\Xi},A_{\Xi}, \psi_{\Pi}, A_{\Pi}$, and these are all a-priori different things, even though $\psi_{\Xi}$ and $\psi_{\Pi}$ both live in the same space $L^2(\Bbb{R})$ and $A_{\Xi},A_{\Pi}$ are both operators $L^2(\Bbb{R})\to L^2(\Bbb{R})$.
Just so we’re on the same page, observe (third bullet point above) that $X_{\Xi}$ and $P_{\Pi}$ are the same operator $L^2(\Bbb{R})\to L^2(\Bbb{R})$, namely they’re both the operator which takes a function and multiplies by its argument. But carefully observe that $P_{\Xi}$ (position representation of momentum operator) is NOT the same as $P_{\Pi}$ (momentum representation of momentum operator) and so
\begin{align}
P_{\Xi}\neq X_{\Xi}= P_{\Pi}.
\end{align}
If you want to figure out how to get from “position representation” to “momentum representation”, then we need to understand how $\Pi\circ\Xi^{-1}:L^2(\Bbb{R})\to L^2(\Bbb{R})$ (which is a unitary isomorphism) behaves. It turns out this is precisely the Fourier transform (modulo some constant factors involving $\hbar$), i.e to get from the function $\psi_{\Xi}$ to $\psi_{\Pi}$, you take its Fourier transform. So, we can now write equations like
\begin{align}
\langle p|\psi\rangle&= \psi_{\Pi}(p)= (\Pi\circ\Xi^{-1})(\psi_{\Xi})(p)=\frac{1}{(2\pi \hbar)^{1/2}}\int_{\Bbb{R}}e^{-\frac{ipx}{\hbar}}\psi_{\Xi}(x)\,dx = \frac{1}{(2\pi \hbar)^{1/2}}\int_{\Bbb{R}}e^{-\frac{ipx}{\hbar}}\langle x|\psi\rangle\,dx
\end{align}
Actually, up till now I didn’t actually discuss what truly makes $X$ a position operator and $P$ a momentum operator, and where the Fourier transform comes from. For this, we can first define $P$ to be the operator such that $P_{\Xi}:=-i\hbar\frac{d}{dx}$. This seems ad-hoc but it’s actually very natural once you realize that this differential operator generates a $1$-parameter unitary group of translations $f\mapsto f(\cdot -x_0)$ (and afterall in classical mechanics, momentum is defined to be the observable which generates translations in space), then one can easily check that the commutation relations hold (say on a nice space of functions like Schwartz space). Then, the role of the Fourier transform is more obvious. To more properly see the interrelations, one should also look at Stone’s theorem and the Stone-Von-Neumann theorem; see this answer for more details (but can definitely be skipped on first reading).
