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Let $p(x)$ and $f(x)$ be sufficiently smooth functions and $D=\frac{d}{dx}$. It is easy to show that $$e^{p(x)D}f(x)=f(e^{p(x)D}x).\tag{1}$$

If $p(x)=a \in \mathbb{R}$ , we have the shift operator as $e^{aD}f(x)=f(x+a)$.

I appreciate it if you could help me to get answers for the following questions:

First, if $p=D$, can we say that $$e^{DD}f(x)=f(x+D)~?\tag{2}$$

Update: The first question is answered below. No, LHS is a function and RHS is an operator.

Second, what is the relationship between $e^{p(x+D)D}f(x)$ and $e^{p(x)D}f(x)$? For instance:

$$e^{p(x+D)D}f(x)-e^{p(x)D}f(x)=(e^{p(x+D)D}-e^{p(x)D})f(x)= \qquad ?\tag{3}$$

$$\frac{e^{p(x+D)D}}{e^{p(x)D}}=e^{p(x+D)D} e^{-p(x)D}=\qquad ?\tag{4}$$

or is it possible to somehow simplify $e^{p(x+D)D}f(x)$?


Update: The problem above seems to be ambiguious. I decided to rephrase it as follows:

Assume that $p(x)$ ($p:\mathbb{R} \to \mathbb{R}$) in known numerically, is it possible to find a matrix representation of operator $p(x+D)D$? or is it possible to define $p(x+D)D$ at all?

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    $\begingroup$ I would think that $e^{DD} f(x)$ would work out to be function $\sum_{n=0}^\infty f^{(2n)}(x)/n!$, not an operator like $f(x + D)$. $\endgroup$ Apr 12, 2022 at 20:21
  • $\begingroup$ Good point Michael. $e^{DD}f(x)$ is a function but $f(x+D)$ is an operator. Thanks! $\endgroup$
    – Mirar
    Apr 12, 2022 at 20:32

2 Answers 2

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People are liable to make mistakes in Heaviside operator calculus manipulations, as they forget symbols on the right. Here is a trick to reduce your $e^{DD}$ to the Weierstrass transform, namely convolution with a Gaussian.

Adapt the identity $$ e^{u^2}= \frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty} \! dy ~ e^{-y^2/4} e^{-uy}, $$
to the formal one, $$ e^{D^2}= \frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty} \! dy ~ e^{-y^2/4} e^{-yD}, $$ so, then, effectively, a linear combination of your starting Lagrange translation operator.

You then have $$ e^{D^2} f(x)= \frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}\!\! \! dy ~ e^{-y^2/4} e^{-yD} f(x)\\ = \frac{1}{\sqrt{4\pi}}\int_{-\infty}^{\infty}\!\! \! dy ~ e^{-y^2/4} f(x-y). $$

So your (2) is not even wrong.

You may learn more here and links therein...


(Soft) Addendum on comments When $[p,D]\neq 0$, you are not in Kansas anymore, and Lagrange's rewriting of the Taylor expansion fails dramatically and insidiously. Indeed, occasionally you see c-number arguments of functions shifted by operators, as in deformation quantization, but such operators commute with everything relevant and behave like c-numbers in that context. Your case, by contrast, is a recipe for grief.

Your proposed $\phi(x,t)=e^{t p(x+D)D} f(x)$, s.t. $\phi(x,0)= f(x)$ and $\phi(x,1)= e^{p(x+D) D }f(x)$ and $\partial_t \phi(x,t)= p(x+D)D \phi (x,t)$, is true, and is a standard Hausdorff move in CBH expansion procedures, as Lie algebra books cover. But I am not quite sure what you are really up to, so I couldn't comment. Your (4) is ill-defined, as the numerator and denominator don't commute.

As you are exploring your way, make sure your constructions agree with the known case in $e^{txD} f(x) = e^{t{d\over d\ln x}} f(e^{})= f(xe^t) $.

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  • $\begingroup$ Thank you Cosmas. I found this answer helpful. I am aware of the Weierstrass transform and am mainly concerened with $p(x+D)$ operator. Based on the link above and the links therein, I see that $e^{p(x+D)D}f$ is the solution of $\frac{d}{dt}\phi(x,t)=p(x+D)D\phi(x,t)$, with $\phi(x,0)=f(x)$. Is my understanding correct? $\endgroup$
    – Mirar
    Apr 13, 2022 at 4:48
  • $\begingroup$ I mean the solution at $t=1$. I am wondering whether it would be possible to find a differential equation that $p(x+D)f$ is its solution (?) $\endgroup$
    – Mirar
    Apr 13, 2022 at 4:54
  • $\begingroup$ I updated the question @Cosmas. The intention is defining $p(x+D)D$ if $p(x)$ is known. $\endgroup$
    – Mirar
    Apr 13, 2022 at 17:35
  • $\begingroup$ If the power series of p(x) is known, p(x+D) is well defined and messy. You want a "normal-ordered" form with all D s to the right? Like $(x+D)^2= x^2+1 +2xD +D^2$ ? $\endgroup$ Apr 13, 2022 at 18:00
  • $\begingroup$ Yes, that is correct. The data that I have ($p$ and $f$) are numerically known. If I know the action of $p(x+D)D$ I can compute $e^{p(x+D)D}f$. What if we represent $p$ locally with cubic/quadratic spline? The highest degree is 3 or 2. Will it simplify the calculation? $\endgroup$
    – Mirar
    Apr 13, 2022 at 18:20
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  1. In OP's equations it is ambiguous how far to the right the operator $D\equiv\frac{d}{dx}$ is acting. See e.g. this Phys.SE post for a similar ambiguity.

  2. Example. OP's eq. (1) is not true in general as an operator identity where $D$ acts all the way to the right. Take e.g. $p(x)=a$ and $f(x)=x^2$ in eq. (1). Then$^1$ $$LHS ~=~e^{p(x)D}f(x)~=~e^{aD}x^2~=~ (x+a)^2e^{aD}$$ while $$RHS~=~f(e^{p(x)D}x)~=~(e^{aD}x)^2~=~ (x+a)(x+2a)e^{2aD}$$

  3. Example. OP's eq. (2) is not true in general as an operator identity where $D$ acts all the way to the right. Take e.g. $f(x)=x$ in eq. (2). Then $$LHS ~=~e^{D^2}f(x)~=~e^{D^2}x~=~ (x+2D)e^{D^2}$$ while $$RHS ~=~f(x+D)~=~x+D$$

  4. Let's introduce a full stop notation "." to indicate how far $D$ acts to the right.

  5. Example. OP's eq. (1) is not true in general as a function identity if we insert a full stop "." to the right of each side. Take e.g. $p(x)=a$ and $f(x)=x^2$ in eq. (1). Then $$LHS ~=~e^{p(x)D}f(x).~=~e^{aD}x^2.~=~ (x+a)^2$$ while $$RHS~=~f(e^{p(x)D}x).~=~(e^{aD}x)^2.~=~ (x+a)(x+2a)$$

  6. We can formulate a correct version of OP's eq. (1) as follows: $$\begin{align} e^{p(x)D}f(x). ~=~&e^{p(x)D}f(x)e^{-p(x)D}. \cr ~=~&e^{p(x)D}f(x)e^{-p(x)D} \cr ~=~&f(e^{p(x)D}xe^{-p(x)D}) \cr ~=~&f(e^{p(x)D}xe^{-p(x)D}.) \cr ~=~&f(e^{p(x)D}x.) \end{align} \tag{1}$$

  7. Example. OP's eq. (2) is not true in general as a function identity if we insert a full stop "." to the right of each side. Take e.g. $f(x)=x^2$ in eq. (2). Then $$LHS~=~e^{D^2}f(x). ~=~e^{D^2}x^2.~=~ x^2+D^2x^2.~=~x^2+2$$ while $$RHS ~=~f(x+D).~=~(x+D)^2.~=~ (x+D)x.~=~x^2+1$$
    It doesn't help if we put the full stop in a different place, e.g. $$RHS ~=~f(x+D.)~=~(x+D.)^2~=~x^2$$

  8. There are similar issues with many of OP's other formulas.

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$^1$ Notation. Note that one often implicitly identify a function, say, $p: x\mapsto p(x)$ with its corresponding left multiplication operator $m_p: g(x) \mapsto (m_p g)(x):=p(x)g(x)$, or a value $p(x)$ of the function.

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  • $\begingroup$ In my understanding, if a differential operator acts on a real function ($f:\mathbb{R}\to\mathbb{R}$), it produces a function. In your example, $$e^{aD}x^{2}=(1+aD+\frac{1}{2}a^{2}D^{2}+...)x^{2}=x^{2}+2ax+a^{2}+ 0 + 0 + ... = (x+a)^{2}$$ Similarly, RHS is equal to $(x+a)^{2}$ because $e^{aD}$ acting on $x$, generates $x+a$. I am sorry, but I do follow your line of reasoning in this answer. Are you trying to say that the problem above is not correctly posed? $\endgroup$
    – Mirar
    Apr 13, 2022 at 4:13
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 13, 2022 at 6:57
  • $\begingroup$ Thank you @Qmechanic. Unfortunately I am not familiar with the full stop "." notation. Do you mind pointing me to a reference? Additionally, do you believe that the problem, defined as it is, is ambiguous? Do you have any suggestion for improvement? $\endgroup$
    – Mirar
    Apr 13, 2022 at 7:20
  • $\begingroup$ Is the full stop notation unclear? $\endgroup$
    – Qmechanic
    Apr 13, 2022 at 10:40
  • $\begingroup$ Yes. And the main question is whether I should modify the question description to avoid ambiguity? $\endgroup$
    – Mirar
    Apr 13, 2022 at 11:11

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