# Chaitanya's Random Pages

## September 14, 2010

### My Six Favourite Formulas – #3

Filed under: mathematics — ckrao @ 1:01 pm

Continuing the look at my favourite formulas, behold the following: $\displaystyle \left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x}) = f(\mathbf{ x+a})$

Here $f$ is an analytic scalar-valued function on some domain of $\mathbb{R}^n$. This is in fact the Taylor series in disguise, in which knowledge of all derivatives of a function at one point enables us to calculate its value at a different point. It may more commonly be seen in operator form as: $\displaystyle f(\mathbf{ x+a}) = f({\mathbf x}) + ({\mathbf a}.\nabla)f({\mathbf x}) + \frac{({\mathbf a}.\nabla)^2}{2!}f({\mathbf x}) + \frac{({\mathbf a}.\nabla)^3}{3!}f({\mathbf x}) + \ldots$

or in scalar form as: $\displaystyle f(x+a) = f(x) + af'(x) + \frac{a^2}{2!}f''(x) + \frac{a^3}{3!}f'''(x) + \ldots$

The exponent ${\mathbf a}.\nabla$ $\left(= a_1 \frac{\partial}{\partial x_1} + \ldots + a_n \frac{\partial}{\partial x_n}\right)$ represents a directional derivative operator (a tangent vector), which when applied to a function gives its rate of change in the ${\mathbf a}$ direction: $\displaystyle \left({\mathbf a}.\nabla\right) g({\mathbf x}) = \left[\frac{d}{dt} g({\mathbf x} + t{\mathbf a}) \right]_{t=0}$

Similarly $\displaystyle \left({\mathbf a}.\nabla\right)^k$ is the k’th order directional derivative.

The beautiful compact formula at the top of this post reveals another use of the exponential function apart from what we may be used to seeing. It takes us from from the infinitesimal (local) world of tangent vectors to the macroscopic (global) world of translations. Applying the exponential of the tangent vector in the direction ${\mathbf a}$ is equivalent to a translation by ${\mathbf a}$. $({\mathbf a}.{\mathbf \nabla})$ is also known as a generator of translation. In the theory of Lie algebras and groups, the exponential map generalises this concept and takes us from a Lie algebra (the space of tangent vectors at the identity element) to its corresponding Lie group. In quantum mechanics the operator $e^{-iHt/\hbar}$ generates the evolution of $\psi(x,0)$ into $\psi(x,t)$ according to the time-dependent Schrödinger equation $\displaystyle i\hbar \frac{\partial \psi}{\partial t} = H\psi$ in the same way.

1. Non-rigorous justification

One intuitive way of seeing why the Taylor series formula holds is to regard $f({\mathbf x}+{\mathbf a})$ as a large number $n$ of successive infinitesimal translations by ${\mathbf a}/n$ and then to approximate each infinitesimal translation by its linearisation $\displaystyle f({\mathbf x} + {\mathbf a}/n) \approx Tf({\mathbf x}) := \left(1 + \frac{{\mathbf a}. {\mathbf \nabla}}{n}\right)f({\mathbf x}).$

Then $\displaystyle \begin{array}{lcl} f({\mathbf x} + {\mathbf a})&=& f({\mathbf x} + n.{\mathbf a}/n)\\&=& \lim_{n\rightarrow \infty}T^nf({\mathbf x})\\&=&\lim_{n\rightarrow \infty} \left(1 + \frac{{\mathbf a}. {\mathbf \nabla}}{n}\right)^nf({\mathbf x})\\&=& \left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x}).\end{array}$

2. Fourier transform “proof”

If we assume the Fourier transform of $f$ exists (defined by $\displaystyle F({\mathbf k}) = \frac{1}{(2\pi)^{n/2}}\int_{{\mathbb R}^n} f({\mathbf x}) e^{-i{\mathbf k}.{\mathbf x}}\ d{\mathbf x}$), another way of understanding the Taylor series formula is by using the following two properties of Fourier transform pairs: $\displaystyle \left({\mathbf a}.\nabla\right)f({\mathbf x}) \leftrightarrow i({\mathbf a}.{\mathbf k})F({\mathbf k})$ $\displaystyle f({\mathbf x} + {\mathbf a}) \leftrightarrow e^{i {\mathbf a}.{\mathbf k}}F({\mathbf k})$

Using the power series definition of the exponential function, we can use repeated applications of the first property to write $\displaystyle e^{{\mathbf a}.\nabla}f({\mathbf x}) \leftrightarrow e^{i {\mathbf a}.{\mathbf k}}F({\mathbf k})$

and our formula results by comparing this with the second Fourier transform property above. This equation also shows why in physics momentum (proportional to the wave number ${\mathbf k}$) is said to be the generator of translation.

3. Proof of the formula by solving a partial differential equation

Fix the vector $\mathbf{a}$ and let $g({\mathbf x},t) = \left[e^{t{\mathbf a}.\nabla}\right]f({\mathbf x})$. Partially differentiating both sides with respect to $t$ gives $\displaystyle \begin{array}{lcl}\frac{\partial g}{\partial t} ({\mathbf x},t)&=&\frac{\partial }{\partial t}\left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x})\\&=&\frac{\partial }{\partial t}\sum_{k=0}^{\infty}\frac{\left(t{\mathbf a}.{\mathbf \nabla}\right)^k}{k!}f({\mathbf x})\\&=&\sum_{k=1}^{\infty}\frac{t^{k-1}\left({\mathbf a}.{\mathbf \nabla}\right)^k}{(k-1)!}f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)\sum_{k=1}^{\infty}\frac{t^{k-1}\left({\mathbf a}.{\mathbf \nabla}\right)^{k-1}}{(k-1)!}f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)\sum_{k=0}^{\infty}\frac{t^k\left({\mathbf a}.{\mathbf \nabla}\right)^k}{k!}f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)\left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x})\\&=&\left({\mathbf a}.{\mathbf \nabla}\right)g({\mathbf x},t).\end{array}$

This is a first order partial differential equation which may be solved by the method of characteristics. Introduce a parameter $u$ so that ${\mathbf x}$ and $t$ are functions of $u$. Then by the chain rule, $\displaystyle \frac{dg}{du}={\mathbf \nabla}g.\frac{d{\mathbf x}}{du} + \frac{\partial g}{\partial t}\frac{dt}{du} \quad \left(= \sum_{i=1}^n \frac{\partial g}{\partial x_i} \frac{dx_i}{du} + \frac{\partial g}{\partial t}\frac{dt}{du}\right).$

From before, $\frac{\partial g}{\partial t}=\left({\mathbf a}.{\mathbf \nabla}\right)g = \left({\mathbf \nabla}g\right).{\mathbf a}$, so $\displaystyle \frac{dg}{du} = {\mathbf \nabla}g.\frac{d{\mathbf x}}{du} +\left({\mathbf \nabla}g\right).{\mathbf a}\frac{dt}{du} = {\mathbf \nabla}g.\frac{d}{du}\left({\mathbf x} + t{\mathbf a} \right).$

This equation becomes 0 if $\frac{d}{du}\left({\mathbf x} + t{\mathbf a} \right) = 0$, or equivalently, $g$ is constant if ${\mathbf x} + t{\mathbf a}$ is constant. We conclude that $\displaystyle g({\mathbf x}, t) = \Psi({\mathbf x} + t{\mathbf a})$

where $\Psi$ is a function. We know that $g({\mathbf x},t=0) = e^0 f({\mathbf x}) = f({\mathbf x})$, and so $\Psi({\mathbf x}) = \Psi({\mathbf x} + 0{\mathbf a}) = f({\mathbf x})$ for all ${\mathbf x}$.  Hence we have shown $g({\mathbf x},t) = \Psi({\mathbf x}+t{\mathbf a}) = f({\mathbf x}+t{\mathbf a})$, and it is easily verified that this satisfies $\frac{\partial g}{\partial t} ({\mathbf x},t) = \left({\mathbf a}.{\mathbf \nabla}\right)g({\mathbf x},t).$ By setting $t$ equal to 1 we conclude that $\displaystyle \left[e^{{\mathbf a}.\nabla}\right]f({\mathbf x}) = f(\mathbf{ x+a}).$  Advertisements

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