strangerep said:
OK, good. But did you also study the theory of Fourier transforms in the more general context of distributions? From your other remarks, it seems not.
Then you need to study distribution theory at a more advanced level. The integral I gave above is ##2\pi \delta(q-q')##.
It does have meaning as a distribution. (Sorry, but this is kinda basic.)
(Sigh.) It has the properties of the Dirac delta distribution. Again, this is elementary.
I know this, of course. But any regular function (of x, say) which grows at infinity no faster than a polynomial, (i.e., a so-called function of "slow growth"), qualifies as a distribution (since when integrated against a Schwartz function, the result is finite). Such functions can of course be multiplied, and the product also grows no faster than a polynomial, hence also qualifies as a distribution. However, if one takes their respective Fourier transforms (which are still distributions), these cannot be simply multiplied (but rather must be convolved).
Hi strangerep,
I now understand more precisely what you are talking about. And all I can say is that IMHO you are mislead. I'm going to try to explain why.
If I’m correct, you are interpreting the expression ##\int \exp(2 \pi i q x) \cdot \exp(-2 \pi i q' x) \cdot dx## as ##\mathcal{F}1(q'-q)## with ##\mathcal{F}## the Fourier transform and 1 the constant function equal to 1. This interpretation is, in a way, possible but you seem to forget that this is
an abuse of notation. How do you define the Fourier transform of a tempered distribution ##T##? By showing that there is one and only one distribution (##\mathcal{F}T##) such that:
##\mathcal{F}T(\phi)= T(\mathcal{F}\phi)##
for all rapidly decreasing functions ##\phi##.
There exists no integral definition for it. Now we both agree that it is correct that if ##T## is regular (associated with the function I will call ##\hat{T}## even if generally we intentionally do not make the distinction) and
if the function ##\hat{T}## has a Fourier transform, ##\mathcal{F}T## is also a regular distribution, and is associated with the function ##t \mapsto \int \exp(-2 \pi i t x) \cdot \phi(x) \cdot dx##. This is not the case for function 1 !
We tend to use abuses of notations for their evocative properties but we all know that they are always tricky somewhere. One may write :
##\delta(t) = \int \exp(-2 \pi i t x) \cdot 1 \cdot dx##
but one should keep in mind that in distribution context, ##\int \exp(-2 \pi i t x) ... dx## is purely symbolic. There is no « actual » exponential function involved here. Just as we all know that ##\frac{\partial ...}{\partial ...}## is a single symbol and certainly not a fraction.
Now if we leave Fourier transforms and choose some ##t##, what can we say about the function ##E_t : x \mapsto \exp(2 \pi i t x)##? It is not even a member of ##\mathcal{L}_2(\mathbb{R})## so it cannot be a test function. Can we interpret it as a tempered distribution ? We know that we can, as the distribution :
##E_t(\phi) = \int \exp(2 \pi i t x) \cdot \phi(x) \cdot dx##
But this expression only makes sense
if ##\phi## is rapidly decreasing function. So according to distribution theory, an expression like :
##E_q(x \mapsto \exp(-2 \pi i q’ x)) = \int \exp(2 \pi i t q) \cdot \exp(-2 \pi i q’ x) \cdot dx##
just does not make sense and it has nothing to do with any Fourier transform. Similarly if one considers ##|q\rangle## to be an abstract generalized vector, distribution theory says ##\langle q|q’\rangle## is not a meaningful expression and Fourier transform is of no help here.
I said and I repeat that if we want
a definition of orthonormality we cannot rely on a particular family of generalized eigenvectors (unlike you do when considering only Dirac deltas or only complex exponentials). We must think of a undefinite family, potentially containing weird elements (for example Dirac combs or whatever you want) and define concepts that can apply to all families (or at least enough families). This is precisely what Mr Carvi is trying to do and (I suppose) why he has felt necessary to write an article about it. This article would be totally pointless if the problem could be solved as easily as you pretend, right? It might be, but I seriously doubt it and the previous fact should incite to caution.