What is the effect of wiping the right half of Fourier Transform bins? I'm trying to change the pitch of a signal using a Fourier Transform (FFT) followed by an Inverse Fourier Transform (IFFT).
I've found many examples, some of which zero out the right half of the real and imaginary bins before changing the pitch. For example, if the signal was $8192$ bins, the real and imaginary parts from $4096$ to $8192$ are set to $0$. This seems to make the math for pitch changing easier, but reduces the volume by one half. This seems to be corrected by doubling the magnitude.
I'm wondering what effect wiping the right half of the bins has on the final signal, apart from reducing the volume. I am confused why these bins exist if they can be wiped without affecting the final signal too much.
The discrete Fourier transform (DFT) transforms a sequence of generally complex-valued numbers to another sequence of generally complex-valued numbers. If the input to the DFT is real-valued - as is the case for a sampled audio signal - then the right half of the DFT is redundant. If $X[k]$ is the length $N$ DFT of a real-valued length $N$ sequence $x[n]$, the following holds:
$$X[k]=X^*[N-k],\qquad k=0,1,\ldots, N-1\tag{1}$$
where $^*$ denotes complex conjugation.
For even $N$, the values $X[0],X[1],\ldots,X[N/2]$ completely represent the original sequence $x[n]$, so the DFT coefficients $X[N/2+1],\ldots,X[N-1]$ can be discarded. For odd $N$, the coefficients $X[0],\ldots,X[(N-1)/2]$ are sufficient.
Note that there are efficient routines for computing the DFT of real-valued input sequences, which only compute the non-redundant DFT coefficients.