How to give a closed form to $e^{a(x) \frac{d}{dx} + b(x)I}[f]$ in physicists style abuse of notation?

Question

In Quantum Mechanics we have the famous time evolution result (here $a$ is a constant)

$$ e^{a \frac{d}{dx}}[f] = f(x+a) $$

Which is an abuse of notation but makes sense due to Taylor's Theorem.

In this answer I show we can give a closed form to $e^{a(x) \frac{d}{dx}}$ whereas if we can find a constant $r$ and function $q$ so that:
$$a(x) = \frac{q(r)}{q'(x)} $$

Then $$ e^{a(x) \frac{d}{dx}} [f] = f(q^{-1}(q(r)+q(x)))$$. As an example if $q = \ln(x)$ and $r=2$ then: $e^{\ln(2) x \frac{d}{dx}} = f(2x)$ (letting $q(x)=x, r=a$ we also can prove the quantum mechanics result above as special case of this)

Now we also know that $e^{b(x)I}[f] = e^{b(x)}f$ where $I$ is the identity operator.

So with this in place I am curious if we can generally speaking give some kind of closed form for a generic exponential of a first order linear differential operator:

$$ e^{a(x) \frac{d}{dx} + b(x)I}[f] $$

My question is "what should this evaluate to?" and to put some boundaries on it, can we expect a general formula which involves finitely many functions $w_1(x) ... w_k(x), c_1(x), ... c_k(x)$ and finitely many non-negative real numbers $d_1 ... d_k$ such that

$$ e^{a(x) \frac{d}{dx} + b(x)I}[f] = w_1(x)f^{(d_1)}(c_1(x)) + w_2(x)f^{(d_2)}(c_2(x)) + ... w_k(x)f^{(d_k)}(c_k(x)) $$

Where $f^{(d_k)}$ indicates the $d_k$ fractional derivative of $f$?

Some Ideas:

My first intuition when working with this abuse of notation is to factor it via integration factors (and at this point this is more symbol shuffling than math, I can hardly give a definition of what any of this means):

$$ e^{a \frac{d}{dx} + bI} = e^{a(x) e^{-\int \frac{a}{b}} \left( e^{\int \frac{a}{b}} I \right)' } $$

But this doesn't necessarily help since I don't have any tools at the moment for evaluating $ e^{\frac{d}{dx} \left(g(x) I \right)} $.

Even something as simple as $e^{\frac{d}{dx}(2I)}$ we know to be $f(x+2)$. But how does one arrive at $f(x+2)$ from $2f(x)$ which is the interpretation of $2I$ or $e^2f$ which is $e^{2I}$. It's just absolutely not clear to me how to proceed here.

Another part of the trouble is that $a \frac{d}{dx}$ and $bI$ don't necessarily commute and because they do not commute I don't feel comfortable making the either of the jumps $e^{a \frac{d}{dx} + bI} = e^{a \frac{d}{dx}} \circ e^{bI}$ or $e^{bI} \circ e^{a \frac{d}{dx}}$. (Actually we know for fact both those jumps are wrong from our earlier example)

One simplification. If $c$ is a number and we did hypothetically know what $e^{\frac{d}{dx}(a(x)I)}$ was then $e^{c \frac{d}{dx}(a(x)I}$ would obviously just be the $c$ iterate of this linear operator. This comes in handy as $e^{\frac{d}{dx}(2I)} = e^{2 \frac{d}{dx}}$ then must be applying $e^{\frac{d}{dx}}$ twice. Of course $f \rightarrow f(x+1)$ twice is $f(x+2)$.

Answer 1

Note that essentially if you want to define $u(x,t)=(e^{t(a(x)\partial_x+b(x)I)}f)(x)$ for $t\geq 0$, then what you want is that $u$ solves the partial differential equation $$ \partial_tu(x,t)=a(x)\partial_xu(x,t)+b(x)u(x,t), $$ with initial data $$ u(x,0)=f(x). $$ This is why PDE are relevant in your question. If you can manage to prove that for a given $f\in E$, with $E$ some function space (e.g. a Banach space) the above Cauchy problem for the PDE admits a unique solution $u$ in $C^0([0,1];E)$ (or in a subspace of it), then the natural way of defining $e^{a(x)\partial_x+b(x)I}f$ is simply by $u(\cdot,1)$ (this discussion applies in general to any differential operator, not just for $a\partial_x+b$). In your specific case, the above PDE is easily solvable in a very general sense (as @LL_3.14 wrote in the comments) if $a,b$ are regular enough (especially $a$), using the method of characteristics (see, e.g., L.C. Evans “Partial Differential Equations”, §3.2.2).

You might be interested in semigroup theory (in particular, strongly continuous semigroups) and its applications to partial differential equations, which deals with questions similar to what you asked. I don’t know much about the general theory, but I found these introductory lecture notes (see esp. 1.4 and chapter 3), and also L.C. Evans “Partial Differential Equations” has some material in section 7.4. See also this post for why semigroups are of interest in PDE.

Answer 2

In physics, the operator $-i d/dx$ is a essential self-adjoint operator acting on complex functions in a Hilbert space $L^2(\mathbb R)$ with integrable derivative squared absolutely $\int|\partial_x f|^2\ dx < \infty$.

This is a dense linear subspace in the Hilbert space. It follows that there exists a maximal extension of this primary domain of defininition of $d/dx$, given by the inverse Fourier transform of functions, square integrable under the map $f(k) \to k f(k)$

$$f \in \text{dom} (-i d/dx) \quad \text{iff} \quad k\ \mathit F(f)(k) \in \mathit L^2(\mathbb R)$$

As a selfadjoint operator, $i d/dx $ has real eigenvalues, can be diagonalized (by Fourier transform) and its exponential with an imaginary factor is a unitary operator

$$ e^{i a (-i d/dx)} f(x) =f(x+a)$$ by Fourier transform.

Restricted to complex anyalytic functions, square integrable over the real line, this forth and back Fourier transform indeed generates the Taylor expansion.

But the exponential is much more than that trival application.

Its an unitary operator with domain the complete Hilbert space.

Taking the standard orthogonal basis from the oscillator problem $f_n = H_n(x)e^{-x^2} $ one concludes, that its the translation operator for any vector in Hilbert space.

Caveat: On an intervall, this is true only for functions with its translate inside the interval. In case of boundaries present, the derivative operator needs boundary conditions in order to construct a selfadjoint operator.

Then its exponential may be one one of the infintely many representations of the unitary translation group on the unit circle with combinations of transfer and reflection with phase changes on the point $\phi = \pm \ \pi$.

Comes the question, what is $a(x)(-i\partial x) + b(x)$.

b is symmetric. The differential operator is not symmetric on a Hilbert space with respect to the Lebesgue measure in $\mathit L^2(\mathbb R)$.

But if $a$ is positiv and absolute continuous, one may consider the differential as acting in a Hilbert space $\mathit L^2( .. ,\frac{dx}{a(x)} )$, such that

$$\int f^*(x) ( -i a(x) \partial_x g(x) \frac{dx}{a(x)} = \int ( -i a(x) \partial_x a(x) \ f)* g(x) \frac{dx}{a(x)} $$

This is the standard case if the coordinate map is changed to orthogonal curvilinear coordinates and the second order derivatives in the euclidean Laplacian are transformed to products of pairs of adjoint operators.

Standard example is the system of spherical coordinate, where the operators $$ dr \partial_r r^2\partial_r , \qquad(\mathbb R_+, r^2 \ dr) $$ $$\partial_\theta \sin \theta \partial_\theta, \qquad ( (0,\pi), \sin\theta \ d\theta) $$ and $$-i\partial_{\phi}, \qquad ((-\pi,\pi), d\phi )$$

are selfadjoint with real spectrum, but only the third is a the generator of a unitary group. Radial translations don't make much sense as translations along the meridians. Instead the combined products of eigenfunctions constitute the series of all integer dimensional unitary representation spaces for the unitary representation of the rotation group.

Bilinear combinations like $L_x +- i L_y$ in spherical coordinates often constitute pairs of ladder operators, such that their exponentials generate tensor products of n-eigenstates over all of the Hilbert space, that are used to represent mixed states.

The translation group in euclidean space is intimately interwoven with Fourier transform.

This algebra of trivialities shaping the free quantum theory breaks down if non translation invariant measures in Hilbert spaces must be used.

So in the 1930ties, quantum theory degenerated to representation theory of symmetry groups.

With gravitation and curved 3-spaces in a dynamic geometry, even this field is of no use anymore.

Answer 3

This problem was compeletely solved by Viskov in 1997, see

O.V. Viskov. Expansion in powers of a noncommutative binomial. Tr. Mat. Inst. Steklova, 216:70–75, 1997.

But the paper remained unkown until 2015. I've very recently written some notes on this theorem here, section 5. A special case of the theorem is the following:

$$ \exp\left(\left(f(x) \frac{\partial}{\partial x} + g(x)\right)t\right) = e^\gamma (e^\varphi)^{\frac{\partial}{\partial x}} $$ where $$ \gamma(t, x) = \int_0^t g(\alpha(\tau, x)) \text d\tau, \quad\text{and}\quad \varphi(t, x) = \int_0^t f(\alpha(\tau, x)) \text d\tau, $$ and $\alpha$ is the unique solution to the Cauchy problem $$ \frac{\partial\alpha(t, x)}{\partial t} = f(\alpha(t, x)), \quad \alpha(0, x) = x. $$ In fact in this special case, we can workout $$ \varphi(t, x) = \alpha(t,x) - x. $$ When I write $(e^\varphi)^D$ I mean it in the sense of the binomial series $$ A^B = \sum_{k=0}^\infty (A-1)^k \binom{B}{k}, $$ And this notation carries the property $$ (e^A)^B = \sum_{n=0}^\infty \frac1{n!} A^n B^n. $$ (Note that if $A$ and $B$ don't commute, we can swap them nor can write $e^{AB}$ in the latter equation). Therefore $$ \exp\left(\left(f(x) \frac{\partial}{\partial x} + g(x)\right)t\right) q(x) = e^{\gamma(t,x)} q(x + \varphi(t, x)) = e^{\gamma(t,x)} q(\alpha(t, x)). $$ For the proof, you should refer to the notes I mentioned, as they correct Viskov's proof (and probably also because Viskov's paper is in russian).

Answer 4

Here are several references that give slightly different perspectives and elaborations on`the 'disentanglement' of the operator

$$e^{t(q(x) + g(x)\partial_x)}:$$

"Evolution operator equations: integration with algebraic and finite difference methods. Applications to physical problems in classical and quantum mechanics and quantum field theory" by Dattoli, Ottaviani, Torre, and Vazquez

"Combinatorial Models of Creation-Annihilation" by Blasiak and Flajolet

"Boson Normal Ordering via Substitutions and Sheffer-type Polynomials" by Blasiak, Horzela, Penson, Duchamp, and Solomon

Commutation Relations, Normal Ordering, and Stirling Numbers by Mansour and Schork

I also posted a semi-organized draft of notes "The Creation Op $\mathfrak{D}= q(z) +g(z)\partial_z$: Scaled flow and operator identities" (pdf) on this in my blog post "A Creation Op, Scaled Flows, and Operator Identities" that has some info not contained in the other refs above.

Answer 5

So it seems that

$$ \frac{v(x)f(x) \circ Q^{-1}(Q(x)+Q(a))}{v(x)} = e^{Q(a)O_{u,v}}$$

Where $O_{u,v} = \frac{u(x)}{v(x)}\frac{d}{dx}\left[v(x)I \right] $ and $Q(x) = \int \frac{1}{u(x)}$ .

So as a concrete example. If we let $v(x)=1$ and $u(x)=1$ then $Q(x) = x$ so we assert:

$$ \frac{f\circ (x+a)}{1} = e^{a \frac{d}{dx}}$$

If we let $u(x)=1$ and $v(x)=e^{x}$ then:

$$ \frac{e^x f(x) \circ (x+a) }{e^x} = e^{ae^{-x}\frac{d}{dx}[{e^x}I]} $$

I.E.

$$ e^a f(x) = e^{a\frac{d}{dx} + aI} $$

The argument for this is a bit verbose and involves basically pattern matching on some examples to create a nice generalization. I will update the answer with this in time. Let $u,v$ be the functions from the integration factor trick above recovers the general solution.

On a side note, it's very interesting that any arbitary $c(x)f(d(x))$ can likely be expressed as the exponential of some differential operator by finding function $Q$, constant $a$, and function $v$ such that:

$$ \begin{matrix} Q(d(x)) = Q(x) + Q(a) \\ \frac{v(Q^{-1}(Q(x)+Q(a)))}{v(x)} = c(x) \end{matrix} $$

Then $e^{ \frac{Q(a)}{Q'(x)v(x)} \frac{d}{dx} \left(v(x) I \right)} = c(x)f(d(x))$