# Inner product of signals (or wavefunction)

#### wang7022

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 lets apply this idea in this way, if we have two real signals, such as X and Y and I wanna 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 lets discuss it.

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#### DrClaude

Mentor
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:
$$\langle A | B \rangle = \int \psi_A^*(x) \psi_B(x) \, d x = \int \phi_A^*(p) \phi_B(p) \, d p$$

Then lets apply this idea in this way, if we have two real signals, such as X and Y and I wanna 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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