# If the discretized $x$ and $p$ are conjugate variables, why are there two $U$s in $U_{FFT}xU_{FFT}^{\dagger}=p$?

While studying some quantum mechanics from Nielsen's book on quantum computing, specifically the simulation of the single-particle Schrödinger equation using a quantum computer, I came across following statement:

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

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

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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
Because you use fast fourier transform you can use two transforms at the price of only $\frac{2\log(2N)}{N}$ compared to standard fourier transform. – Jannick Oct 2 '15 at 21:52