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

[Infrared]

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.

 

[atyy]

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."

 

[martinbn]

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.

 

[atyy]

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.

 

[martinbn]

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.

 

[atyy]

I didn't see where in the references it is called a basis.

https://www.amazon.com/dp/047116433X/?tag=pfamazon01-20
https://www.amazon.com/dp/1107028728/?tag=pfamazon01-20

 

[martinbn]

https://www.amazon.com/dp/047116433X/?tag=pfamazon01-20
https://www.amazon.com/dp/1107028728/?tag=pfamazon01-20

In over 1400 pages there isn't a single mention of rigged Hilbert spaces, or any of the equivalent names.

 

[atyy]

I didn't see where in the references it is called a basis.

Take a look at post $34.

 

[martinbn]

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.