Why should a Fourier transform not be a change of basis?

  • #26
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It doesn't make sense to do ##\vec{x} - A\vec{x}## if ##A## is a change of basis matrix. It only makes sense if ##A## is a linear operator. In that expression, one must use the same basis for ##\vec{x}##, ##A## and ##A\vec{x}##.
Every matrix is a linear operator.
 
  • #27
atyy
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Every matrix is a linear operator.
No. It has to transform correctly under the change of basis.
 
  • #28
etotheipi
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Every matrix is a linear operator.
I think we should perhaps say that every linear transformation ##T## has a matrix representation ##\mathcal{M}(T)## (assuming we deal with finite vector spaces only). Just like a vector is not equal to its coordinate matrix w.r.t. some basis, a linear transformation is not equal to its coordinate matrix w.r.t. two bases.
 
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  • #29
atyy
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It changes a lot more than the basis. It is a mapping from one space to an entirely different space.
The way to see the Fourier transform heuristically as a change of basis is to treat the function ##\phi(x)## that is being Fourier transformed as the representation of the vector, ie. ##\phi(x)## are the coordinates of a vector, with ##x## being the index for the coordinates. Its Fourier transform ##\tilde{\phi}(k)## is the representation of the same vector in different coordinates.

The technical issue is that the representation of the basis vectors is not normalizable, but heuristically it works out as follows. Suppose one were working in 2D, then orthonormal basis vectors would have representation ##[1,0]## and ##[0,1]##, ie. each basis vector should have a peak for one index, and zero for other indices. For a continuous index, one would get a Dirac delta with a peak at only one ##x## (since ##x## is the index). In the position representation, the position basis vectors are ##\delta(x-y)## (where ##y## is a continuous index) and the momentum basis vectors are ##\text{exp}(ikx)## (where k is a continuous index).

So suppose the vector we want to represent is ##|\Phi\rangle##, then in the position representation, we can write it heuristically (without being careful about normalization and signs) using either a sum of position basis vectors or momentum basis vectors as follows:
##\langle x |\Phi\rangle = \phi(x) = \int{\phi(y) \delta(x-y)dy} = \int{\tilde{\phi}(k) \text{exp}(ikx)} dk##
 
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  • #30
martinbn
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This was already said, but because the thread was revived I will repeat it. To illustrate the problem with the fact that there are two different spaces one needs to look at the simplest case, periodic functions in one varible. The Fourier transform maps ##L^2(\mathbb R/\mathbb Z)## to ##l^2(\mathbb Z)##. So it would be hard to view this as a change of basis.
 
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  • #31
Saw
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The way to see the Fourier transform heuristically as a change of basis is to treat the function ϕ(x) that is being Fourier transformed as the representation of the vector, ie. ϕ(x) are the coordinates of a vector, with x being the index for the coordinates. Its Fourier transform ϕ~(k) is the representation of the same vector in different coordinates.
⟨x|Φ⟩=ϕ(x)=∫ϕ(y)δ(x−y)dy=∫ϕ~(k)exp(ikx)dk
I am glad you used this term, "heuristically", because that was one of the objectives of the OP: to confirm that the view of the FT as something akin to a change of basis can help to solve the problem of better understanding its mechanics, mechanics which consist precisely of solving problems by changing "perspective" or "representation system".

It is fine for me if the FT is not, strictly speaking, a change of basis. As FactChecker also said, the analogy may fly

as long as it is not taken too literally in advanced operations.
Yes, analogies cannot be taken too far. They only work within their domain of applicability, which is given by their practical purpose. In this case, it is solving problems from another "angle". (By the way, it is said that the angle between time and frequency domains is 90 degrees...) If this other angle is another "space" instead of a "basis", that simply means that we need a generalized term that, at least for this purpose, covers the two things: "representation system", for example?

A different thing is that the second scope is, yes, understanding well why "change of basis" is not technically accurate. Here I still have questions:

* Does this apply only to the continuous case with infinite interval?

* The basis functions are not square-integrable, they don't belong to the same vector space. I see, although I still have to assimilate it...

*
The technical issue is that the representation of the basis vectors is not normalizable,
This is linked, I understand, to the fact that the product of one basis vector (function) with itself is not 1 but infinity. But it would be 1 if you could normalize by dividing by the same infinite time interval over which you have integrated... Physics4Fun gave some solutions by means of which you could divide by infinity. In any case, I tend to think that this should not be an insurmountable problem.

* Then there is the issue of the subtraction not being possible, but I did not grasp if in the end this objection was maintained.

* Latest objection:

The Fourier transform maps ##L^2(\mathbb R/\mathbb Z)## to ##l^2(\mathbb Z)##.
Could you elaborate on that?
 
  • #32
Stephen Tashi
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Is the terminology "Fourier transform" unambiguous? Is it to be a map from a vector space to possibility different vector space of coefficients? - or is it to be a map from a vector space of coefficients into itself?

On the one hand we can define a transformation ##c_f(v)## from a vector space ##V## into the (in general different) vector space ##C## whose elements are coefficients of vectors in ##V## in some basis ##B_f##. (e.g. ##V## can be an n-dimensional vector space real valued functions defined on [0,1] and ##C## can be the vector space of n-tuples of real numbers)

On the other hand we can consider a transformation ##T## from ##C## into itself defined as follows: To find ##T(c)## let ##v## be the vector in ##V## who coefficients in the basis ##B_t## are the n-tuple ##c##. Then ##T(c)## is defined as ##c_f(v)## which is the n-tuple of coefficients for ##v## in a different basis ##B_f##.

Does "Fourier transform" refer to a function like ##c_f(v)## or does it refer to a function like ##T##?
 
  • #33
Infrared
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The Fourier transform maps ##L^2(\mathbb R/\mathbb Z)## to ##l^2(\mathbb Z)##. So it would be hard to view this as a change of basis.
I think you're thinking of Fourier series. A function doesn't need to be periodic to have a Fourier transform.
 
  • #34
atyy
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This is linked, I understand, to the fact that the product of one basis vector (function) with itself is not 1 but infinity. But it would be 1 if you could normalize by dividing by the same infinite time interval over which you have integrated... Physics4Fun gave some solutions by means of which you could divide by infinity. In any case, I tend to think that this should not be an insurmountable problem.
If you extend the notion of "basis", you may not need the property that "basis functions" be square-integrable. The Dirac delta function, which is a "basis function", is not even a proper function, and cannot be multiplied by itself. In quantum mechanics, the quantum state must be square-integrable so that we can calculate probabilities. Since not all "basis functions" are square integrable, they can be "basis functions" but not quantum states. Take a look at rigged Hilbert spaces to see how it may be possible to extend the notion of "basis". Here is one of the previously linked references and a couple of others about rigged Hilbert spaces.

http://galaxy.cs.lamar.edu/~rafaelm/webdis.pdf
Quantum Mechanics in Rigged Hilbert Space Language
Rafael de la Madrid Modino

https://arxiv.org/abs/quant-ph/0502053
The role of the rigged Hilbert space in Quantum Mechanics
R. de la Madrid

https://arxiv.org/abs/1411.3263
Quantum Physics and Signal Processing in Rigged Hilbert Spaces by means of Special Functions, Lie Algebras and Fourier and Fourier-like Transforms
Enrico Celeghini, Mariano A. del Olmo

Celeghini and del Olmo state the interesting point that for a rigged Hilbert space, some bases may be countably infinite, whereas others may be uncountably infinite. This is in contrast to Hilbert space where all bases have the same cardinality.

Karen Smith's notes are also interesting as she discusses ideas that even for Hilbert spaces (without going to rigged Hilbert spaces), there are different notions of what a basis is.

http://www.math.lsa.umich.edu/~kesmith/infinite.pdf
Bases for Infinite Dimensional Vector Spaces
Karen E. Smith
 
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  • #35
martinbn
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I think you're thinking of Fourier series. A function doesn't need to be periodic to have a Fourier transform.
I am talking about Fourier transform in general. Fourier series is a special case. If you have a locally compact abelian group ##G##, its dual ##\widehat{G}## is as well, and the Fourier transform takes a function with domain the group ##f:G\rightarrow\mathbb C##, to a function on the dual group ##\widehat{f}:\widehat{G}\rightarrow\mathbb C##. In the case of ##G=\mathbb R##, the dual is (noncannonically) isomorphic to itself. In the case of ##G=\mathbb R/\mathbb Z##, the dual is ##\widehat{G}=\mathbb Z##.

https://en.wikipedia.org/wiki/Pontryagin_duality
 
  • #36
Infrared
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I agree that is true, but it looks to me like the locally compact group considered in the OP and throughout the throughout the thread is ##(\mathbb{R},+),## so I'm not sure why the Fourier transform on a circle is directly relevant.

Working over ##\mathbb{R}##, you can definitely view the Fourier transform as map from a space to itself, e.g. use the space of Schwartz functions. I still agree it doesn't look like a change of basis.
 
  • #37
atyy
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The other heuristic that suggests looking at the Fourier transform as something like a change of basis is the fact that it diagonalizes differentiation. Wikipedia's article on Spectral Theory makes the interesting comment:

"There have been three main ways to formulate spectral theory, each of which find use in different domains. After Hilbert's initial formulation, the later development of abstract Hilbert spaces and the spectral theory of single normal operators on them were well suited to the requirements of physics, exemplified by the work of von Neumann.[5] The further theory built on this to address Banach algebras in general. This development leads to the Gelfand representation, which covers the commutative case, and further into non-commutative harmonic analysis.

The difference can be seen in making the connection with Fourier analysis. The Fourier transform on the real line is in one sense the spectral theory of differentiation qua differential operator. But for that to cover the phenomena one has already to deal with generalized eigenfunctions (for example, by means of a rigged Hilbert space). On the other hand it is simple to construct a group algebra, the spectrum of which captures the Fourier transform's basic properties, and this is carried out by means of Pontryagin duality."
 
  • #38
martinbn
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I agree that is true, but it looks to me like the locally compact group considered in the OP and throughout the throughout the thread is ##(\mathbb{R},+),## so I'm not sure why the Fourier transform on a circle is directly relevant.

Working over ##\mathbb{R}##, you can definitely view the Fourier transform as map from a space to itself, e.g. use the space of Schwartz functions. I still agree it doesn't look like a change of basis.
Well, I was thinking about Fourier in general, and the circle case makes the point obvious. I also think that it doesn't look like a change of bases. Even more, I am not sure how standard the terminology about position and momentum basis is, but it is somewhat misleading. The spaces in question have countable bases, while these position and momentum "bases" are uncountable.
 
  • #39
atyy
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In some cases, the energy basis is a true basis that is countable, while position and momentum "bases" are generalized "bases" that are uncountable. See the rigged Hilbert spaces references above. It is interesting that even for vector spaces, whether the basis is countable or not depends on the definition, and may differ between a Fourier basis and a Hamel basis.
 
  • #40
martinbn
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In some cases, the energy basis is a true basis that is countable, while position and momentum "bases" are generalized "bases" that are uncountable. See the rigged Hilbert spaces references above. It is interesting that even for vector spaces, whether the basis is countable or not depends on the definition, and may differ between a Fourier basis and a Hamel basis.
The rigged Hilbert spaces do not change anything about the fact that the position "basis" is uncountable. I didn't see where in the references it is called a basis.
 
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  • #43
atyy
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I didn't see where in the references it is called a basis.
Take a look at post $34.
 
  • #44
martinbn
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Take a look at post $34.
I did, it desn't change anything. Refferences to rigged Hilbert spaces will not change the fact that the exponentals are uncountable and the spaces are seperable.
 

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