Fourier Transform: the group perspective

May 30, 2011

Fourier Transform: the group perspective

Filed under: math, physics, science — vivisheksudhir @ 9:41 pm

After another substantially long break, here I am. And this post (and hopefully a follow-up on the same theme) is something I have been planning to pen for a while now, except that time has not been kind to me. Unfortunately though, today I decided to clear the accumulating pile of papers on my desk, and came across my own random notes on this mildly interesting perspective on a classic mathematical workhorse.

Just for fun, let us think of some functions living in the Hilbert space $\mathcal{H}(\mathbb{R})$ , i.e the normed vector space on the real field consisting of functions $f:\mathbb{R} \rightarrow \mathbb{C}$ , equipped with the standard inner product,

$(f,g) \equiv \int_{-\infty}^{+\infty} f^{*}(x)g(x)\, dx$ ,

for any two functions $f,g \in \mathcal{H}$ . Now, consider the transformations on elements of the underlying field,

$x \rightarrow \hat{t}_a x = x+a$ ,

i.e., translations in the complex field. Corresponding to the set of translation operators $\hat{t}_a$ on $\mathbb{R}$ , let us postulate some operators $\hat{T}_a : \mathcal{H} \rightarrow \mathcal{H}$ , which are defined by their action,

$\hat{T}_a f(x) \equiv f(\hat{t}_a x) = f(x+a)$ .

Evidently, we have the following properties (henceforth, dropping the hats on top of operators):

$t_a t_b = t_{a+b}$
$t_0 t_a = t_a = t_a t_0$
$(t_a)^{-1} = t_{-a}$

which are consequently mapped onto the operator on $\mathcal{H}$ , via the following isomorphisms:

$T_a T_b = T_{a+b}$
$T_0 T_a = T_a = T_a T_0$
$(T_a)^{-1} = T_{-a}$ .

The essential point now is that the set of operators $\lbrace t_a \rbrace$ form an abelian group with the identity element $t_0$ ; and the set of operators $\lbrace T_a \rbrace$ furnish a representation of the group on the Hilbert space $\mathcal{H}$ . What else can we say about our representation? Lets see. From the structure of the inner product, we see that, for any two $f,g \in \mathcal{H}$ ,

$(T_a f,T_a g) = (f,g)$ ,

i.e., the $T_a$ are unitary operators; that is to say, the representation is unitary.

The other (almost trivial, but consistently overlooked) thing to note for the translation group is that it is a continuous group, i.e., the parameter $a$ can take any value (in particular, ones that are infinitesimally close to each other), and still the elements $t_a$ are in the group. Thus, the elements of the group form a continuum. [This continuum on which the group elements become points, is the group manifold, and opens the door to the study of groups as topological objects, a la Pontryagin.] Also, intuitively, it is obvious that any element $t_b$ of the group, can be constructed by starting from some other element $t_a$ and then composing several infinitesimal translation operations; this means that the group is connected as well. Owing to the way in which the representation of the group is constructed, all these properties of the group, specifically continuity and connectedness, are inherited by the representations $T_a$ as well.

At this point, a trained mathematician will see whats coming – an application of Stone’s theorem. But since I write this in the spirit of a physicist, let me try and make it a bit more transparent.

Since the group is continuous and connected, it is intuitive (in a hand-wavy sense) to see that an infinitesimal translation by $\delta a$ is affected by the operator,

$T_{\delta a} = T_0 + i\,\delta a\, \tau$ ,

where $\tau$ is another operator. This statement just says that since the translation is small, it must be very close to the identity translation, but corrected by an infinitesimal term proportional to the translation itself. The seemingly artificial factor of the imaginary $i$ exists for a reason, viz: since we know $T_a$ is unitary (i.e., that $T_a^\dagger = T_{-a}$ ), the unitarity must be manifest in the above relation, atleast to $O(\delta a)$ ; and the $i$ factor makes this explicit.

Now, a finite translation by $a$ , can be built up by dividing $a$ into $n$ small pieces of length $\delta a$ , so that $\delta a = a/n$ , and then applying these small translations one after the other. Mathematically,

$T_a = \lim_{n \rightarrow \infty} \left( T_0 + i\frac{a}{n}\tau \right)^n$ .

This limit can be evaluated in a standard manner, to give the general result,

$T_a = \exp(i\,a\, \tau)$ .

This *is* the statement of Stone’s Theorem, that a unitary representation (in our case, $T_a$ ) of a one-parameter group can be expressed as a one-parameter (in our case, $a$ ) exponential of a hermitian operator (in our case, $\tau$ ). This extends in an obvious manner to groups paramerised by more than a single variable.

But what the hell is $\tau$ now? Formally, we can invert the last expression above, to yield,

$i\tau = \left[\frac{dT_a}{da}\right]_{a=0}$ .

That is to say, $\tau$ , is the operator that enables us to “explore” the neighbourhood of the identity operator on the group manifold.

As for every operator equation, this equation only has a discernible meaning when it is converted into an equation for some function $f \in \mathcal{H}$ ; doing this,

$i \tau f(x) = \left[ \frac{d}{da}f(x+a) \right]_{a=0} = \left[ \frac{d}{da} \left( f(x) + a f'(x)+... \right) \right]_{a=0}= f'(x)$ .

Since this holds for arbitrary $f(x)$ , it must be that,

$\tau = -i \frac{d}{dx}$ ,

so that,

$T_a = \exp\left( a \frac{d}{dx} \right)$ ,

which is a priori obvious if we had thought about the functional representation of the Taylor expansion. For a physicist of course, the generator $\tau$ is nothing but the momentum operator familiar from elementary quantum mechanics.

But this convoluted method of finding the infinitesimal operator (in our case, $\tau$ ), the so-called generator of the group, is valuable since it is general, and applies to most topological groups. The key of course, is to spot the group’s identity element and then setup an equation which expresses the generator as the infinitesimal change about that identity.

Typically, atleast for simple groups like this, an explicit form of the generator(s) is all one needs – in essence, that closes the study of the group, since everything that can be said about it, follows from the properties of the generator(s).

So lets go just one step further in that direction; since we have an operator, $\tau$ , an obvious question one could ask is: are there some functions in $\mathcal{H}$ , that remain invariant to being acted on by the operator? That is, is there a set of (eigen)functions ${u_k (x)}$ , that satisfy,

$\tau u_k (x) = k u_k (x)$ .

Since we know the explicit form of $\tau$ , this is a differential equation, with the solution, $u_k(x) = e^{ikx}$ , which is the Fourier kernel. Now, since $\tau$ is hermitian (implied by the unitarity of $T_a$ ), we have,

orthogonality: $(u_k(x), u_m(x)) = \delta(k-m)$
completeness: $(u_k(x), u_k (y)) = \delta(x-y)$

In particular, completeness means that for any $f \in \mathcal{H}$ ,

$f(x) = \int (u_k, f) u_k(x) dk$ ,

which, given that $u_k(x) = e^{ikx}$ , is the Fourier theorem.

Having recovered the Fourier transform via this mechanism, i.e., as the invariant eigenspace of the unitary representation of the translation group, it is worth looking back to make an obvious observation: the eigenfunctions of the translation operator must, by definition, be left alone by translations, i.e., they must be periodic functions; and that is exactly the result. This is of course a re-assuring sanity check.

Note:

The keen mathematician would point out that the eigenfunctions $u_k$ are not really even $L^2$ and hence not strictly elements of a Hilbert space. This quirky, but pain-in-the-ass, mathematical fact *is* the reason for physical effects like the wave-particle duality and hence of the Heisenberg uncertainty principle.

Nothing can be done about this, except to conclude that the translation operators do not have a “nice” representation in the ordinary Hilbert space – one has to add some additional structure to it, so that things become rigorous. This structure is a weakening of the $L^2$ condition, and hence a weakening of the definition of the inner product. One considers a set of test functions $\Phi \subseteq \mathcal{H}$ , with respect to which the inner product is defined; and $\Phi$ is restricted in such a way (for eg, the Schwartz space, consisting of functions which decrease rapidly, along with all their derivatives) that one can have a finite norm for all $f \in \mathcal{H}$ , with respect to this new test set. This approach is known as “rigging” the Hilbert space [arXiv:quant-ph/0502053v1], and the final product is the Gelfand triple.

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Comments (2)

2 Comments »

hahaha…

Comment by sandeep — May 30, 2011 @ 10:10 pm

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- what is this comment supposed to mean??
  
  Comment by vivisheksudhir — June 10, 2011 @ 8:21 am
  
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