Inner product of signals (or wavefunction)

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SUMMARY

The discussion centers on calculating the inner product of two signals, X and Y, using Fourier transforms. The user proposes using the Fast Fourier Transform (FFT) to obtain P = fft(X) and Q = fft(Y), and then computing their inner product as conj(P)*Q. However, it is clarified that while signals can be treated as vectors in Hilbert space, the inner product must be expressed as an integral of wave functions, and the proposed method does not yield a valid inner product. The convolution theorem is mentioned as a related concept.

PREREQUISITES
  • Understanding of Hilbert space and its vectors
  • Knowledge of Fourier transforms and their applications
  • Familiarity with the Fast Fourier Transform (FFT) algorithm
  • Basic concepts of wave functions and their projections
NEXT STEPS
  • Study the mathematical foundations of inner products in Hilbert space
  • Learn about the convolution theorem and its implications in signal processing
  • Explore the concept of rigged Hilbert spaces for advanced applications
  • Investigate the relationship between time-domain and frequency-domain representations of signals
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Physicists, signal processing engineers, and anyone interested in the mathematical foundations of wave functions and their applications in signal analysis.

wang7022
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Dear All:

Any idea for the following interesting question:

As we know we can calculate inner product of two wave functions A and B
as <A|B>. here both A and B are vector in hilbert space. here we may use
fourier transform to get momentum representation of A and B, and get same
result.

Then let's apply this idea in this way, if we have two real signals, such as X and Y and I want to get their inner product because signals could be consider as vector in hilbert space also. Now I use Fourier transform to get P = fft(X) and Q = fft(Y).
then get their inner product = conj(P)*Q.
now may I condiser this result as an inner product of signal X and Y?

Thanks for comments and I still don't have clear idea in this problem so just let's discuss it.
 
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wang7022 said:
As we know we can calculate inner product of two wave functions A and B
as <A|B>. here both A and B are vector in hilbert space. here we may use
fourier transform to get momentum representation of A and B, and get same
result.
Many things are confused here. ##|A \rangle## and ##|B \rangle## are vectors in Hilbert space, indeed, and thus are not wave functions. They become wave functions when projected on position "eigenstates" (*),
$$
\psi_A(x) = \langle x | A \rangle
$$
Likewise, one can obtained momentum-space wave functions,
$$
\phi_A(p) = \langle p | A \rangle
$$

While one can go from ##\psi_A(x)## to ##\phi_A(p)## and vice-versa from Fourier transforms, don't forget that ##\langle A | B \rangle## involves an integral when expressed in term of wave functions:
$$
\langle A | B \rangle = \int \psi_A^*(x) \psi_B(x) \, d x = \int \phi_A^*(p) \phi_B(p) \, d p
$$

wang7022 said:
Then let's apply this idea in this way, if we have two real signals, such as X and Y and I want to get their inner product because signals could be consider as vector in hilbert space also. Now I use Fourier transform to get P = fft(X) and Q = fft(Y).
then get their inner product = conj(P)*Q.
now may I condiser this result as an inner product of signal X and Y?
No.

This does however remind me of the convolution theorem.

(*) I don't want to get into the mathematical weeds here. Look up "rigged Hilbert space" to learn more.
 

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