2
$\begingroup$

Given the following quantum state:

$\frac{1}{2}(|0000\rangle + |0100\rangle + |1000\rangle + |1100\rangle)$

How do I apply a QFT (given by the formula below) to that state in superposition?

$QFT_n|j\rangle = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\omega_{N}^{j \times k} |k\rangle$ where $N = 2^n$

$\endgroup$
2
$\begingroup$

QFT on any Superposition (Linear Combination of Basis States) can be applied using Linearity.

$$QFT_n|\psi\rangle = \sum_{k=0}^{2^n-1}a_kQFT_n|k\rangle$$

Hence $QFT_4|\psi\rangle$ where $|\psi\rangle = \frac{1}{2}(|0000\rangle + |0100\rangle + |1000\rangle + |1100\rangle)$ is $$QFT_4(\frac{1}{2}(|0000\rangle + |0100\rangle + |1000\rangle + |1100\rangle)) \\ = \frac{1}{2}(QFT_4|0000\rangle + QFT_4|0100\rangle + QFT_4|1000\rangle + QFT_4|1100\rangle) \\ = \frac{1}{2}(QFT_4|0\rangle + QFT_4|4\rangle + QFT_4|8\rangle + QFT_4|12\rangle) \\ = \frac{1}{2}(\frac{1}{4}\sum_{k=0}^{15}\omega_N^{k\times 0}|k\rangle + \frac{1}{4}\sum_{k=0}^{15}\omega_N^{k\times 4}|k\rangle + \frac{1}{4}\sum_{k=0}^{15}\omega_N^{k\times 8}|k\rangle + \frac{1}{4}\sum_{k=0}^{15}\omega_N^{k\times 12}|k\rangle ) \\ =\frac{1}{8}\sum_{k=0}^{15}(\omega_N^{k\times 0}+\omega_N^{k\times 4}+\omega_N^{k\times 8}+\omega_N^{k\times 12})|k\rangle $$

Here $\omega_N = e^{\frac{i2\pi}{2^4}} = e^{\frac{i\pi}{8}}$, therefore $\omega_N^0 = 1$, $\omega_N^4 = e^{\frac{i\pi}{2}} = i$, $\omega_N^8 = e^{i\pi}=-1$ and $\omega_N^{12} = e^{\frac{i3\pi}{2}}=-i$.

Thus $QFT_4|\psi\rangle$ is

$$ QFT_n|\psi\rangle = \frac{1}{8}\sum_{k=0}^{15}(\omega_k^{k\times 0}+\omega_k^{k\times 4}+\omega_k^{k\times 8}+\omega_k^{k\times 12})|k\rangle \\ = \frac{1}{8}\sum_{k=0}^{15}(1^{k}+i^{k}+(-1)^{k}+(-i)^{k})|k\rangle $$

This sum $(1^{k}+i^{k}+(-1)^{k}+(-i)^{k})$ is $4$ when $k$ divides 4 otherwise its 0.

Thus $QFT_4|\psi\rangle$ is

$$QFT_n|\psi\rangle = \frac{1}{8}\sum_{k'=0}^{3}(4)|4k'\rangle \\ = \frac{1}{2}(|0\rangle + |4\rangle + |8\rangle + |12\rangle) \\ = \frac{1}{2}(|0000\rangle + |0100\rangle + |1000\rangle + |1100\rangle) $$

Here I noticed that our initial state is an eigenvector of $QFT_4$ with an eigenvalue $1$. If I noticed this before I could have directly written the answer.

Nonetheless, I hope this helps.

$\endgroup$
1
  • 1
    $\begingroup$ Very well explained, thank you! $\endgroup$
    – user12136
    Aug 2 '20 at 19:00

Your Answer

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