Codebook: Quantum Fourier Transform

Codebook
1

Ask here about the “Quantum Fourier Transform” Codebook topic from the “Quantum Fourier Transform” module.

2

Hi,
in “QFT circuit for two qubit” there is a calculation for HS and the result is given as a martix (numpy notation): ( [1, 1], [i, -i] ).

I miss the 1 / sqrt(2)

Is this just a typo or do I miss something?

Best regards,
jomu

3

Hi,
in “QFT circuit for two qubit” there are two diagrams for the circuit.
I implemented the first and it looks OK.
This is the code:
qml.Hadamard(wires = 0)
qml.ctrl(qml.S, control=1)(wires=0)
qml.Hadamard(wires = 1)
qml.SWAP(wires=[0,1])
return qml.state()

Then I tried the second but now the systems responds “not OK”.
This is the code:
qml.Hadamard(wires = 1)
qml.ctrl(qml.S, control=0)(wires=1)
qml.Hadamard(wires = 0)
return qml.state()

But when I do a swap at the very first beginning then again it OK:
qml.SWAP(wires=[0,1])
qml.Hadamard(wires = 1)
qml.ctrl(qml.S, control=0)(wires=1)
qml.Hadamard(wires = 0)
return qml.state()

But in this case there is not reduction in the depth of the circuit.

Is there something wrong with the sketch?

Best rergards,
jomu

4

Hi,
in “Properties of the QFT” you write:
“We will look at special functions where the amplitude is non-zero at multiples of
(and zero otherwise) and has no offset”

I tested this with numpy FFT.
When I use an input like:
poly_a = np.array([1, 0, 0, 0,
1, 0, 0, 0,
1, 0, 0, 0,
1, 0, 0, 0 ], dtype=complex)
then the FFT output is periodic as expected.

When I use an input like:
poly_a = np.array([1, 0, 0, 1,
1, 0, 0, 1,
1, 0, 0, 1,
1, 0, 0, 1 ], dtype=complex)
then the FFT output is not periodic any more.

My question: is this “and zero otherwise” a strict requirement for a periodic output?

Best regards,
jomu

5

Hi @jomu, yes there’s a typo. We’re missing the 1/sqrt(2). Good catch!

6

Hi @jomu , I think there might be something wrong with the sketch indeed because the output of the two circuits is different. Let me check and get back to you.

7

Hi @jomu,

Regarding your last question, “zero otherwise” is indeed a strict requirement for FFT to work as expected.

8

Hi Catalina,
thank you so much for all these clarifications and help. I really appreciate it!

Regards,
jomu

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