What is the functional form for a ket vector in the position basis?

Question

$\newcommand{\ket}[1]{|#1\rangle}$ $\newcommand{\bra}[1]{\langle#1}$

This is a question that has long confused me, what is the actual functional form of a ket vector, specifically in the position basis? I am guessing the answer is that the ket vector is too abstract to have a functional form except perhaps under specific circumstances, but let me try to explain my confusion. In Shankar section 1.10, he describes a function being expanded as a series of kets as such:

Let us denote by $f_n(x)$ the discrete approximation to $f(x)$ that concides with it at $n$ points and vanishes in between. Let us now interpret the order $n$-tuple {$f_n(x_1)$,$f_n(x_2)$,...,$f_n(x_n)$} as components of a ket $\ket{f_n}$ in a vector space $V^n(R)$: $$\ket{f_n}\leftrightarrow \begin{bmatrix} f_n(x_{1}) \\ f_n(x_{2}) \\ \vdots \\ f_n(x_{n}) \end{bmatrix}$$ The basis vectors in this space are: $$\ket{x_i}\leftrightarrow \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \\ 0 \\ \vdots \\ 0 \\ \end{bmatrix}$$ corresponding to the discrete function which is unity at $x=x_i$ and zero elsewhere. (...) Try to imagine a space containing $n$ mutually perpendicular axes, one for each point $x_i$. Along each axis is a unit vector $\ket{x_i}$. The function $f_n(x)$ is represented by a vector whose projection along the $I$th direction is $f_n(x_i)$: $$\ket{f_n}=\sum_{i=1}^n f_n(x_i)\ket{x_i}$$

This discussion seems to imply that a ket $\ket{x}$ is either a kronecker delta or more realistically $\ket{x}=\int dx^\prime\delta(x-x^\prime)|x'\rangle$. The reason I am skeptical of this is that this would require an uncountably infinite number of Dirac deltas to define the entirety of a physical position space, as Shankar clearly relates each point in space with a distinct ket.

I am also skeptical given that $\int dx^\prime\delta(x-x^\prime)$ is a functional and therefore should live in the dual space, which is the space of bras (although I do understand there is a one to one correspondence between the two, although I am not sure how to see it explicitly in this case). I know this discussion often includes the idea of a "Rigged Hilbert Space", such as here, however I do not fully follow the discussion being had. Is Shankar's discussion here purely superficial and not meant to represent the underlying mathematics?

To further my confusion, I can imagine a wavefunction defined on a finite interval. We can expand this function in terms of a power series of polynomials. In this case, we would almost certainly associate $\ket{x}$ with $\ket{x_n}=x^n$ as the function would be properly expanded as: $$f(x)=\sum_{n=0}^{\infty} a_n x^n=\sum \ket{x}\bra{x}\ket{f}$$ in which clearly $a_n=\bra{x}\ket{f}$ and therefore $\ket{x}=x^n$, as the polynomials now form the basis (or at least a linear combination of polynomials form basis states, i.e. Laguerre polynomials). This differs from the previous interpretation of kets as Dirac deltas.

Any discussion or resources you could point me to about this would be greatly appreciated!

Answer 1

The Hilbert space for a $1D$ particle without spin is $\mathcal H = L^2(\mathbb R)$. For every $x\in \mathbb R$, $|x\rangle$ is not an element of $\mathcal H$, so it is not a proper ket (in elementary QM courses, this is often mentioned as the fact that it is not normalizable).

However, in the rigged Hilbert space formalism, we can make sense of it as a generalized ket and equations like $\langle \psi|x\rangle = \psi^*(x)$ and $\mathbb I = \int \text d x |x\rangle\langle x|$ are true. (Note that there is one $|x\rangle $for each $x\in\mathbb R$.) I think Shankar is trying to justify heuristically the fact that shouldn't worry to much about this subtleties (at least at first) and that it is perfectly ok to use $|x\rangle$ as just a regular ket.

Of course, to make it mathematically sound, you need a rigged Hilbert space $\Phi\subset \mathcal H \subset \Phi^\times$, in which case, for all $x\in\mathbb R$, you can define a continous antilinear functional on $\Phi$ by : $$\forall \psi \in\Phi, \langle \psi|x\rangle = \psi^*(x) = \int \text dx'\psi^*(x')\delta(x'-x)$$

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Now, about your idea of taking polynomials as a basis for $\mathcal H =L^2([a,b])$ : this is definitely possible, but it would be very confusing to write those ket $|x\rangle$ (as this is usually understood to be what I described above). The functions $\psi_n(x) = x^n$ do span a dense subset, but they are not orthonormal so you do not have the closure relation $\sum_n |\psi_n\rangle\langle \psi_n| = \mathbb I$. By applying the Gram-Schmidt orthonormalization process, you will obtain an orthonormal basis $\big\{|n\rangle\big\}$ of polynomial wave functions.

Answer 2

Shankar's discussion is not

purely superficial and not meant to represent the underlying mathematics.

He tells you precisely that kets are vectors; so you should never have an equation with open kets on one side and pure functions (numbers) on the other. In that sense, several equations after your "skeptical" point don't make any sense, but I don't know what you are after.

Following the discrete paradigm, here are some correct equations, possibly helpful to some of your confusion, $$ |f\rangle = \int\!\!dx~~f(x)|x\rangle \equiv \int\!\!dx~~\langle x|f\rangle |x\rangle ~~~\leadsto \\ f(x')=\langle x'|f\rangle=\int\!\!dx~~f(x)\langle x'|x\rangle= \int\!\!dx~~f(x) \delta(x-x'). $$

If you want to picture $|x\rangle$, think of an infinite-dimensional vector, whose merging entries (their intervals/rungs have shrunk to a continuum) correspond to the value of x. So $|137.4848\rangle$ is a vector empty everywhere except at the 137.4848 location/step, where the component is infinite.