0
$\begingroup$

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

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!

$\endgroup$
4
  • 1
    $\begingroup$ possible duplicate of physics.stackexchange.com/questions/364208/… $\endgroup$ Jul 13 at 21:46
  • 1
    $\begingroup$ The weakness here is $\vert x_n\rangle\to x^n$, which doesn’t make sense in the way the basis of eigenstates of $\hat x$ is constructed. In fact, $f(x)$ is interpreted as a number (the function evaluated at $x$) not a vector in the Hilbert space. The vector would be $\vert f\rangle$, whereas the component of the vector along basis state $\vert x\rangle$ is $\langle x\vert f\rangle=f(x)\in \mathbb{C}$. $\endgroup$ Jul 13 at 21:52
  • $\begingroup$ Thank you for sharing the previous link. Does this imply then that $|x\rangle$ is just the real number line? Or at least is that an appropriate way to think of it? $\endgroup$
    – Schoppe
    Jul 13 at 22:07
  • $\begingroup$ Not quite. The argument $x$ is a specific value on the real line but $\vert x\rangle$ is a vector, so that $\mathbb{I}=\int dx \vert x\rangle \langle x\vert $ generalizes to the continuous case the sum $\sum_k \vert k\rangle\langle k\vert$ in the discrete case. $\endgroup$ Jul 14 at 0:10
4
$\begingroup$

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 : $$\forall \psi \in\Phi, \langle \psi|x\rangle = \psi^*(x) = \int \text dx'\psi^*(x')\delta(x'-x)$$


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.

$\endgroup$
2
  • 1
    $\begingroup$ Thank you! This is exactly the type of discussion and answer I was looking for. Do you know of any resources which discuss this in detail? The idea of using a non-renormalizable "object" which is not even a part of the Hilbert space as a basis for the Hilbert space always left me confused. Griffiths talks about it momentarily when discussing momentum eigenstates, but then offers no clarification afterwards. How has this problem traditionally been dealt with, meaning before the use of a Rigged Hilbert Space? $\endgroup$
    – Schoppe
    Jul 14 at 12:56
  • 1
    $\begingroup$ For resources on Rigged Hilbert spaces, you could start with the explanations and resources given in the answers here $\endgroup$ Jul 14 at 12:58
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ The point I was trying to make is that you can form a basis using the polynomials such that a function $f(x)$ can be expressed as $f(x)=\sum a_n x^n$ where the set of basis vectors are {$1,x,x^2,x^3,...$}. We say that $|x|rangle$ forms a basis for position space, so I was trying to draw an equivalence between these two notions. If $|x\langle$ corresponds to a Dirac delta function, this still doesnt clarify how, because a Dirac delta is not a function but a functional, it belongs to the dual space and not the vector space. $\endgroup$
    – Schoppe
    Jul 14 at 1:24
  • $\begingroup$ The Dirac delta function operates like a function here, not a functional. It is the wave function of the position ket, as Dirac stresses in his book. $\endgroup$ Jul 14 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.