In the rigged Hilbert space approach to continuous variable quantum mechanics, what does a general bra-ket or ket-bra rigorously denote?

Question

I'm trying to rigorously understand Dirac notation in continuous variable quantum mechanics. I'm told the most common approach is via rigged Hilbert spaces. This means we have: $$\Phi \subset H \cong H^* \subset \Phi^* \text{ and } \Phi \subset H \cong H^\times \subset \Phi^\times$$ where:

• $H$ is the Hilbert space of square integrable functions $\mathbb{R} \to \mathbb{C}$
• $\Phi$ is Schwartz space, which I'll take to be the space of quantum states
• $(-)^*$ denotes the dual space of linear functionals
• $(-)^\times$ denotes the antidual space of antilinear functionals

In Dirac notation, bras are exactly the space $\Phi^*$ of linear functionals, and kets are exactly the space $\Phi^\times$ of antilinear functionals.

My question: for a general bra $\langle a |$ and ket $|b\rangle$, rigorously speaking, what do $\langle a|b \rangle$ and $| b \rangle \langle a |$ denote? Can they be well-defined for all $\langle a |$ and $|b\rangle$? If not, why not?

Partial answers: I can see that for certain $\langle a |$ and $|b\rangle$ these expressions could be given a rigorous definition, e.g.:

• If the antilinear functional $| b \rangle \in \Phi^*$ actually corresponds to a state in $\Phi$ (i.e. is of the form $\langle -, f_b \rangle$ for $f_b \in \Phi$, where $\langle -,- \rangle$ is the inner product on $H$), then $\langle a|b \rangle$ could be defined as applying the linear functional $\langle a |$ to the state $f_b$.
• If both $\langle a |$ and $|b\rangle$ correspond to elements of $H$ (i.e. are of the form $\langle f_a, - \rangle$ and $\langle -, f_b \rangle$ for $f_a, f_b \in H$), then $\langle a | b \rangle$ could be defined as the inner product $\langle f_a, f_b \rangle$ in $H$.
• In this thesis from de la Madrid, equation (3.5.39) suggests that for 'position basis vectors' ${}_X\langle x |$ and $| y \rangle_X$ we can define ${}_X\langle x | y \rangle_X$ as the Dirac delta 'function' $\delta_{x-y}$, i.e. the functional mapping $f \mapsto f(x-y)$. This confuses me a lot -- I'm expecting every braket $\langle a|b \rangle$ to be somehow equivalent to a scalar in $\mathbb{C}$, and I can't see how this would be.
• Below this question, @Albatross gives a very useful answer and further comments, including saying "there is certainly no such prescription which is well-defined for arbitrary $\langle a |$ and $|b\rangle$, because if $\langle a | = \delta_x$ and $| b \rangle = \overline{\delta}_x$, then $\langle a|b \rangle = \delta_0$" -- I don't see why this should be true.

Any help appreciated!

Answer 1

This confuses me a lot -- I'm expecting every braket $\langle a|b \rangle$ to be somehow equivalent to a scalar in $\mathbb{C}$, and I can't see how this would be.

Based on your points 3,4 it seems you want to define/understand $\langle b|a\rangle$ by its numerical value. However, this is not always possible, because in some cases (continuous eigenvalues $a,b$) this expression appears in integrals, and there it behaves as delta distribution on the space over which $b$ runs, concentrated at $a$, which is sometimes denoted $\delta_a(b)$. This is known to be not representable by any function of $a,b$. At least at $a=b$, there is no finite value we can assign, and it is doubtful whether we can assign any finite value meaningfully even at points $a\neq b$.

For example, take position operator with "eigenvalues" $x'$ and corresponding position kets $|x'\rangle$.

What is then $\langle x|x'\rangle$ for various combinations of values $x,x'$? Usually in physics, we do not care about the actual numerical value of this object, as its proper use is in integrals, where we integrate formal product of $\langle x|x'\rangle$ with some other object dependent on $x$, over $x$.

Assuming any $|\psi\rangle$ can be expressed as

$$ |\psi\rangle = \int |x\rangle \langle x |\psi\rangle dx, $$ this does not require $\langle x|\psi\rangle$ to always, for all $x$ and $\psi$, to attain finite numerical value.

For example, for $|\psi\rangle = |x'\rangle$, we expect:

$$ |x'\rangle = \int |x\rangle \langle x |x'\rangle dx. $$ If this is to hold for any $|x'\rangle$, the object has to behave as delta distribution, and thus obey the distributional equation $$ \langle x|x'\rangle = \delta(x-x'). $$

This equation means we can't represent the object $\langle x|x'\rangle$ using functions of $x,x'$. It acts as delta distribution, as linear functional on functions:

$$ \int \langle x|x'\rangle \psi(x) dx = \psi(x'). $$ We can also try to regard it as an "integral kernel" representation of the linear functional $\langle x|$ in $x'$ coordinates, since its action on $|\psi\rangle$ can be expressed as integral of $\langle x|x'\rangle \psi(x')$ over $x'$:

$$ \langle x|\psi\rangle = \int \langle x|x'\rangle \psi(x') dx' . $$

But since this representation behaves as delta distribution, it does not really have a value at all points $x,x'$. Instead, it has (after integration) an effect on functions, or on kets. Often such effect of a linear functional can be expressed using a set of values (a numerical representation of the linear functional, to be used in integral), but for singular distributions such as delta, this is not possible.

Answer 2

Check out Gel'fand & Vilenkin's Generalised Functions, Vol IV, they have a section on rigged Hilbert spaces.

They state that given a linear operator $A$ on a topological vector space $V$, then a linear functional $F$ on $V$ is called a generalised eigenvector with eigenvalue $k$ when:

$F(Av) = k(Fv)$

for all $v \in V$. It then turns out that $e^{-ikx}$ are generalised eigenvectors for the translation operator on the Schwartz space of rapidly decaying functions. Furthermore, the Plancheral equality shows that the set of all generalised eigenvectors is complete. They also have a generalised spectral theory: they show that any self-adjoint or unitary operator defined on a nuclear space has a complete set of generalised eigenvectors.

I realise that this doesn't answer your specific question, but hopefully the answer is in Gel'fand and Vilenkin.

Answer 3

The question is fundamentally misled: If there was an direct and unambiguous way to translate Dirac notation bijectively into rigorous mathematics, there would be functional analysis texts doing that. The manipulations physicists do with Dirac notation are so much "simpler" than many functional-analytic proofs that it is simply impossible for this to translate to rigorous mathematics in the fashion the question asks for. The general rigorous map demanded in the question does not exist.

Secondly, I dispute the claim that rigged Hilbert spaces are the "most common" way to make sense of Dirac notation. Rather, they are the most common excuse to not think too hard about it - "eh, rigged Hilbert spaces exist, I'm sure Dirac notation makes sense".

Here is a brief dictionary with which you can replace most of the Dirac notation by ordinary functional analysis on Hilbert spaces, no rigging of the space necessary:

Our fundamental starting point is the spectral theorem for unbounded self-adjoint operators, by which such an operator $A$ has a spectral measure decomposition $A = \int a\mathrm{d}E_A(a)$ and a multiplication representation on $L^2(\mathbb{R})$ where it acts as the multiplication operator.

Now, we fix the Hilbert space to always be $L^2(\mathbb{R})$, and interpret the isomorphisms to the multiplication representation of some operator $A$ simply as a "basis change" on this space, i.e. a map from the space to itself.

• Inner products $\langle a \vert b\rangle$ where $a$ and $b$ are "eigenvectors" of self-adjoint operators $A$ and $B$ are meant to the integral kernels of the transforms between the multiplication representations on $L^2(\mathbb{R}^n)$ of those operators. E.g. "$\langle x\vert p\rangle = \mathrm{e}^{\mathrm{i}px}$" for position and momentum is rigorously just the statement that the complex exponential is the kernel of the Fourier transform (which goes from the multiplication representation of position to that of momentum and vice versa).

  Note that this perfectly explains things like the $\langle a\vert a'\rangle = \delta(a -a')$ property: Indeed, the integral kernel of the identity transformation is the $\delta$.

• Outer products $\lvert a\rangle \langle a \rvert$ where $a$ are the "eigenvectors" of a self-adjoint operator $A$ are rigorously simply the spectral measure $\mathrm{d}E_A(a)$.

• Outer products $\lvert a\rangle \langle b\rvert$ where $a$ and $b$ are the "eigenvectors" of a self-adjoint operator $A$ mean there is a switch from the multiplication representation of $B$ to that of $A$ (or vice versa).

As an example, let's use this to interpret the following funny representation of the identity operator: $$ 1 = \int \lvert x\rangle\langle x\rvert \mathrm{d}x = \int \lvert x\rangle \langle x\rvert \mathrm{d}x\int \lvert p\rangle\langle p\rvert \mathrm{d} p = \int \mathrm{e}^{\mathrm{i}xp}\lvert x \rangle\langle p\rvert \mathrm{d}x\mathrm{d}p, \tag{1}$$ which, with the "definitions" that the position wavefunction of a state $\lvert \psi\rangle$ is $\psi(x) = \langle x\vert \psi\rangle$ and the momentum wavefunction is $\psi(p) = \langle p\vert \psi\rangle$ can be used to show that the position and momentum wavefunctions are Fourier transforms of each other at the physical level of rigor (sandwich between $\langle x'\rvert$ and $\lvert \psi\rangle$).

With the dictionary above, eq. (1) is really nothing more than the statement that the Fourier transform changes between the multiplication representation of the position and the momentum operator. This, in turn, is mathematically nothing more than knowing that the Fourier transform exchanges multiplication and differentiation. If your instinct is to say it feels like the physical argument based on eq. (1) shows something more from less, then that's because all the knowledge e.g. about position and momentum being multiplication and differentiation etc. is hidden in the way the physicist "knows" $\langle x\vert p\rangle = \mathrm{e}^{\mathrm{i}px}$, which in the end will end up being some variant of the Stone-von Neumann theorem.

The reasoning via Dirac notation is very good at hiding this kind of complexity, but there is almost always a solution based in "normal" functional analysis instead of trying to micking the literal sequence of equations via rigged Hilbert spaces.