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$U_{a,b}=\sum^{d-1}_{x=0}\omega^{bx}|x+a\rangle\langle x|$,$\omega=e^{\frac{2\pi i}{d}}$,$a,b\in\{0,1,2,...,d-1\}$ Can someone please give me the pic of the quantum circuit?

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  • $\begingroup$ $ a, b $ are like some fixed parameters? And you are looking for a generic circuit that performs $ U_{a,b} $ for all possible values of $ a $ and $ b $? $\endgroup$
    – tsgeorgios
    Oct 29, 2020 at 10:32
  • $\begingroup$ Yes,exactly. It's easy to perform it on 2-dimendision system,but I don't how to extend to high-dimensional $\endgroup$ Oct 29, 2020 at 13:55

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Since we are talking about a unitary operation on qubits, i assume $ d = 2^n $ where $ n $ is the number of qubits.

We define the unitary operations $ V_{a} = \sum_{x=0}^{d - 1} | x + a \rangle \langle x| $ and $ D_{b} = \sum_{x=0}^{d - 1} \omega_d^{bx} | x \rangle \langle x| $.

Notice that we can write $ U_{a, b} = V_a \cdot D_b $.

In the Fourier basis (see here) we have $ QFT^{\dagger} \cdot V_a \cdot QFT = D_{-a} $ since for a basis state $ |x \rangle $ \begin{align*} V_a \cdot QFT \cdot |x \rangle &= V_a \cdot \frac{1}{\sqrt{d}} \sum_{k=0}^{d-1} \omega_d^{kx} |k \rangle \\ & = \frac{1}{\sqrt{d}} \sum_{k=0}^{d-1} \omega_d^{kx} |k + a\rangle \\ & = \frac{1}{\sqrt{d}} \sum_{k=0}^{d-1} \omega_d^{(k - a)x} |k \rangle \\ & = \omega_d^{-ax} QFT \cdot |x \rangle \end{align*} and so $QFT^{\dagger} \cdot V_a \cdot QFT \cdot |x \rangle = D_{-a} \cdot |x \rangle $ which proves the previous identity.

We conclude that is enough to implement $ D_b $ and then $ U_{a, b} = QFT \cdot D_{-a} \cdot QFT^{\dagger} \cdot D_b $

But since $ \omega_d^{b x} = \omega_d^{b \sum_{k=0}^{n-1} x_k 2^k} = \prod_{k=0}^{n-1} \omega_d^{b 2^k x_k} $, it holds that

$$ D_b |x \rangle = \omega_d^{b x} |x \rangle = \Big( \prod_{k=0}^{n-1} \omega_d^{b 2^k x_k} \Big) |x_{n-1} .. x_1 \rangle = \otimes_{k=0}^{n-1} \Big( \omega_d^{b 2^k x_k} |x_k \rangle \Big) = \otimes_{k=0}^{n-1} P_k |x_k \rangle $$

where $ P_k $ are the single qubit phase gates $P_k = \begin{bmatrix} 1 & 0 \\ 0 & e^{i\frac{2 \pi b \cdot 2^k}{d}} \end{bmatrix}$

If you want to implement this in Qiskit, see QFT and Phase Gate and be careful with Qiskit convention that $q_0$ is the least significant qubit.

Here is the circuit for $ n = 3, a, b = 1 $ made in IBM Q: enter image description here For different values of $a, b$ just multiply then angles of $ U1 $ gates in $ D_{-a}, D_b $ by $a, b$ accordingly.

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  • $\begingroup$ I have one question: How to get $QFT^{\dagger} \cdot V_a \cdot QFT = D_{-a}$ by $QFT^{\dagger} \cdot V_a \cdot QFT \cdot |x \rangle = D_{-a} \cdot |x \rangle$ as it is calculation of matrices and vectors, not numbers. $\endgroup$ Mar 2, 2021 at 8:42
  • $\begingroup$ Since the two matrices agree on every basis state $ | x \rangle $ they must be equal. $\endgroup$
    – tsgeorgios
    Mar 2, 2021 at 9:52

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