# Circuit for the inverse quantum Fourier transform

• Haorong Wu
In summary: After the exchange, ##R_k^{-1}= R_k^{\dagger}=\begin{pmatrix} 1 & 0 \\ 0 & e^{-2 \pi i/2^k} \end{pmatrix} ##. Then everything works out.
Haorong Wu
Homework Statement
Give a quantum circuit to perform the inverse quantum Fourier
Relevant Equations
The quantum Fourier transform is defined as ##\left | j \right > =\frac 1 {\sqrt {2^n}} \sum_{j=0}^{2^n-1} e^{2 \pi ijk / 2^n} \left | k \right >##, and it is equal to ##\left | j_1 , j_2 , \dots , j_n \right > \rightarrow \frac { \left ( \left | 0 \right > + e^{2 \pi i 0.j_n} \left | 1 \right > \right ) \left ( \left | 0 \right > + e^{2 \pi i 0. j_{n-1} j_n} \left | 1 \right > \right ) \dots \left ( \left | 0 \right > + e^{2 \pi i 0. j_1 j_2 \dots j_n} \left | 1 \right > \right ) } {2^{n/2}}##
First, the inverse quantum Fourier transform is ##\left | k \right > =\frac 1 {\sqrt {2^n}} \sum_{j=0}^{2^n-1} e^{-2 \pi ijk / 2^n} \left | j \right >##, and it is equal to ##\left | k_1 , k_2 , \dots , k_n \right > \rightarrow \frac { \left ( \left | 0 \right > + e^{-2 \pi i 0.k_n} \left | 1 \right > \right ) \left ( \left | 0 \right > + e^{-2 \pi i 0. k_{n-1} k_n} \left | 1 \right > \right ) \dots \left ( \left | 0 \right > + e^{-2 \pi i 0. k_1 k_2 \dots k_n} \left | 1 \right > \right ) } {2^{n/2}}##
where I merely change the signs in the exponentials.

Second, since the quantum circuits are reversible, then I think if I reverse the quantum Fourier transform circuit, then I should have the circuit for the inverse quantum Fourier transform as I did in the picture:

Then, I tried to calculate the outcome of the circuit:
## \left | k_n \dots k_1 \right > \\ \rightarrow \left | k_n \dots k_2 \right > \frac {\left | 0 \right > + e^{2\pi i 0.k_1} \left | 1 \right >} {\sqrt 2} \\ \rightarrow \left | k_n \dots k_3 \right > \frac {\left | k_2 , 0 \right > + e^{2\pi i 0.k_1 k_2} \left |k_2, 1 \right >} {\sqrt 2} =\left | k_n \dots k_2 \right > \frac {\left | 0 \right > + e^{2\pi i 0.k_1 k_2} \left | 1 \right >} {\sqrt 2} \\ \rightarrow \left | k_n \dots k_3 \right > \frac {\left | 0 \right > + e^{2\pi i 0.k_2} \left | 1 \right >} {\sqrt 2} \frac {\left | 0 \right > + e^{2\pi i 0.k_1 k_2} \left | 1 \right >} {\sqrt 2} \\ \dots \\ \rightarrow \frac { \left ( \left | 0 \right > + e^{2\pi i 0.k_n} \left | 1 \right > \right ) \left ( \left | 0 \right > + e^{2\pi i 0. k_{n-1} k_n } \left | 1 \right > \right ) \dots \left ( \left | 0 \right > + e^{2\pi i 0.k_1 k_2 \dots k_n} \left | 1 \right > \right ) } {2^{n/2}}##
which does not match the previous equation. The signs in the exponentials are positive instead of negative.

I think there may be some possible mistakes:
1. I got the wrong inverse quantum Fourier transform, or
2. though quantum circuits are reversible, merely reversing the circuit of quantum Fourier transform will not yield the proper circuit for inverse quantum Fourier transform, or
3. (most probably) I miscalculate the changes in the circuits.

Also, I know that Hadamard gate can be expressed as ## \left | k \right > \rightarrow \frac {\left | 0 \right > + e^{2 \pi i 0.k} \left | 1 \right > } {\sqrt 2} = \frac {\left | 0 \right > + e^{-2 \pi i 0.k} \left | 1 \right > } {\sqrt 2}## where the negative sign appears in the exponential. Then how can I get other negative signs out? Should I change ##R_k=\begin{pmatrix} 1 & 0 \\ 0 & e^{2 \pi i/2^k} \end{pmatrix} ## to ##R_k=\begin{pmatrix} 1 & 0 \\ 0 & e^{-2 \pi i/2^k} \end{pmatrix} ##? If so, I believe the origin circuit is right for the inverse quantum Fourier transform.

Thanks!

Haorong Wu said:
Should I change ##R_k=\begin{pmatrix} 1 & 0 \\ 0 & e^{2 \pi i/2^k} \end{pmatrix} ## to ##R_k=\begin{pmatrix} 1 & 0 \\ 0 & e^{-2 \pi i/2^k} \end{pmatrix} ##? If so, I believe the origin circuit is right for the inverse quantum Fourier transform.
Yes, I believe so. To reverse the circuit, you would need to reverse the action of each gate.

Haorong Wu
tnich said:
Yes, I believe so. To reverse the circuit, you would need to reverse the action of each gate.
Hi, tnich. Sorry to get back to you this late.

I forgot to exchange the inputs and outputs of the ##R_k## gates. After the exchange, ##R_k^{-1}= R_k^{\dagger}=\begin{pmatrix} 1 & 0 \\ 0 & e^{-2 \pi i/2^k} \end{pmatrix} ##. Then everything works out.

Thanks!

## 1. What is a circuit for the inverse quantum Fourier transform?

The inverse quantum Fourier transform (IQFT) is a mathematical operation used in quantum computing to convert a quantum state into its corresponding classical state. A circuit for the IQFT is a series of quantum gates that perform this operation on a given quantum state.

## 2. How does the circuit for the inverse quantum Fourier transform work?

The circuit for the IQFT works by applying a series of Hadamard gates and controlled phase gates to the input quantum state. These gates manipulate the quantum state in a way that allows for the transformation into its classical state.

## 3. What is the purpose of the circuit for the inverse quantum Fourier transform?

The circuit for the IQFT is used in quantum algorithms, such as Shor's algorithm for factoring large numbers, to convert the quantum state into a classical state that can be measured and used to obtain the desired result.

## 4. Are there any limitations to the circuit for the inverse quantum Fourier transform?

One limitation of the circuit for the IQFT is that it requires a large number of quantum gates, which can lead to errors and decrease the overall efficiency of the quantum algorithm. Additionally, the circuit may not work for certain types of quantum states, such as entangled states.

## 5. How is the circuit for the inverse quantum Fourier transform different from the circuit for the quantum Fourier transform?

The circuit for the IQFT is the inverse of the circuit for the quantum Fourier transform (QFT). While the QFT converts a classical state into a quantum state, the IQFT converts a quantum state back into its classical state. The two circuits use different combinations of gates to achieve their respective operations.

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