Translation Operators on Smooth Functions
by Eric Thrane		3/12/2006

In this notebook we're going to look at some properties of the translation operator: T(Δ).   The translation operator is defined as follows:

	T(Δ) f(x) = f(x-Δ)

It just shifts a function to the right by an amount equal to Δ.  In quantum mechanics we argue that the translations in position space are generated by the momentum operator:

	T(Δ) = \(TraditionalForm\`e\^\(i\ Δ\ p\)\)
	
When we're working in the position space representation, (with spatial wavefunctions), we know that the momentum operator p is represented by i (∂_x).  Thus, the position space representation of T(Δ) is:

	T(Δ) = \(TraditionalForm\`e\^\(\(-Δ\)\ ∂\_x\)\) = (1 - Δ (∂_x) +(1/(2!)) (((Δ^2)( (∂_x)))^2) + ...)
	
I was daydreaming about this the other day, and it occured to me that the derivation has nothing to do with quantum mechanics...the same reasoning applies to finding a translation operator that operates on the set of smooth, (but not necessarily normalizable,) functions.  Let's see how this derivation works...

Consider a smooth function f(x).  To translate f  some infinitessimal distance ϵ it is clear that:

	f(x-ϵ) = f(x) - ϵ (∂_x) f(x) = (1 - ϵ (∂_x)) f(x)

(It's just the start of Taylor series for small ϵ.)  Now suppose we want to do a whole bunch of infinitessimal translation that add up to a finite translation of length Δ.  Specifically let's break up the single translation by Δ into N translations by (Δ/N)and then send N->∞:

	f(x-Δ) = (((1 - (Δ/N) (∂_x)))^N) f(x)		N->∞

But as we send N->∞, this just becomes:

	f(x-Δ) = \(TraditionalForm\`e\^\(\(-Δ\)\ ∂\_x\)\)f(x)
	
And so in the language of group theory, we says i (∂_x)is the generator of translations, since exponentiating it multiplied by -i Δ gives us an operator that translates functions by Δ.

Notice, we did not have to talk about quantum mechanics or even normalizability.  The only tacit assumption we made is that the every derivative of f(x) is finite, and I think that can even probably be relaxed for many cases where the infinity can be characterized by something like a delta-function.  All of this is pretty cool because it means that high school student with some calculus experience can learn about continuous groups, generators, and operators...no quantum mechanics required.

Let's start with a normalizable Gaussian function.

f[x_] := (E^(-(x^2)))

Plot[{f[x]}, {x, -3, 3}];

Δ is the amount we will translate by.

Δ = 2;

g[x_] := f[x]
\(For[a = 1, 
      a < 50, \(a++\), 
	\(g[x_] = 
          g[x] + \((\(-Δ\))\)\^a*
              D[f[x], {x, a}]\/\(a!\);\)
]\)

D[f[x].{x,a}] takes the a-th derivative with respect to x of f[x].  The first fifty terms in this series ought to do the trick!

Plot[{g[x]}, {x, -1, 5}];

Bingo: the function has been translated by 2.  Now let's try a non-normalizable function.

F[x_] := (x^2)

Plot[{F[x]}, {x, -3, 3}];

Δ = 5;

G[x_] := F[x]
\(For[a = 1, 
      a < 50, \(a++\), 
	\(G[x_] = 
          G[x] + \((\(-Δ\))\)\^a*
              D[F[x], {x, a}]\/\(a!\);\)
]\)

Plot[{G[x]}, {x, 2, 8}];

So this method appears to work for non-normalizable functions as well.

Keywords: translations, translation operator, nth-order derivative.