3
$\begingroup$

I'm trying to work through the proof that the quantum Fourier transform can be described by a unitary operator, i.e $F^{\dagger}F=\mathbb{1}$, where $$ F=\frac{1}{\sqrt{N}}\sum_{j,k=0}^{N-1}\exp\left(\frac{2\pi ijk}{N}\right) |k\rangle\langle j |$$ and $$F^\dagger=\frac{1}{\sqrt{N}}\sum_{j,k=0}^{N-1}\exp\left(\frac{-2\pi ijk}{N}\right)|j\rangle\langle k |.$$

Together they give : $$F^{\dagger}F=\frac{1}{N}\sum_{j',k'j,k=0}^{N-1}\exp\left[\frac{2\pi i(j'k'-jk)}{N}\right]|j\rangle\langle k| k'\rangle\langle j' |.$$

Using $\langle k |k'\rangle = \delta_{k',k}$ and $\exp\left[\frac{2\pi i(j'-j)}{N}\right]=\delta_{j',j}$, The expression reduces to

$$F^{\dagger}F=\frac{1}{N}\sum_{j=0}^{N-1}|j\rangle\langle j|=\frac{1}{N} \mathbb I.$$

My question is: Shouldn't the factor of $\frac{1}{N}$ vanish somewhere? Or is the definition of a unitary operator $U^\dagger U \propto \mathbb{1}$ rather than $=\mathbb{1}$ ?

$\endgroup$
4
$\begingroup$

The error in the OP's question comes from the second Kronecker delta, which is not correct (furthermore, the sum over $k$ is not dealt with...). After using the first Kronecker delta $\delta_{k\,k'}$, one has to use the identity $$\frac{1}{N}\sum_{k=0}^{N-1}\exp{[\frac{2\pi i k(j'-j)}{N}]}=\delta_{j'\,j},$$ which directly gives that $F^\dagger F =1$.

One can easily check that the $1/N$ has to be included in the identity, since one is summing complex numbers of modulus one. Or equivalently, one trivialy sees that for $j=j'$, $$\frac{1}{N}\sum_{k=0}^{N-1} 1 = 1.$$

| cite | improve this answer | |
$\endgroup$
-1
$\begingroup$

I don't believe it's true that $\langle k | k' \rangle = \delta_{k',k}$. If I recall correctly, the whole point of the $\frac{1}{\sqrt{N}}$ factor is to normalize the momentum eigenstates, so we should have $\langle k | k' \rangle = N\delta_{k',k}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm sorry, in my efford to edit my question I edit yours instead. I apologise for my mistake, I reverted it back to your original answer. $\endgroup$ – Constandinos Damalas Dec 14 '14 at 18:05
  • $\begingroup$ @PhotonicBoom No worries. $\endgroup$ – Alessandro Power Dec 14 '14 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.