# Quantum Simulation of Schrodinger's equation

While studying some quantum mechanics from Neilsen's book on quantum computing and came across following

because x and p are conjugate variables related by a quantum Fourier transform: $U_{FFT}xU_{FFT}^{\dagger}=p$

A single FFT will take $x$ to $p$, then why is there a $U^\dagger$ ? It seems like I'm missing something basic here...

-
that's always the way matrices transform. A vector $v$ transforms according to $Uv$, but a matrix $A$ transforms according to $UAU^\dagger$. –  Martin May 19 '14 at 9:18
@Martin how can one write a finite dimensional matrices for x and p –  Kishor Bharti May 19 '14 at 9:21
the momentum operator is the same as: $\hat{P}= \sum_i p_i|p_i><p_i|$ So clearly if the eigenstate of position is the fourier transform of momentum, you will need a fourier transform and it's complex conjugate on either side of that equation to get position. –  Mew May 19 '14 at 9:22
@KishorBharti You don't, of course - but the statememt is not limited to finite dimensions, and x and p are unbounded operators. Note that an operator is always defined by its action, i.e. $v$ a vector in the domain of $x$, then $xv$ is another such vector. The transformation of this equation would be $UxU^{\dagger}Uv$. –  Martin May 19 '14 at 9:24