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I have implemented the following 8 qubit QFT circuit similar to the following:

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and loaded the coefficients as follows:

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The output of the QFT is as follows:

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Could anyone help interpret the above result?


With particularly nice values of wavelength,

enter image description here enter image description here enter image description here enter image description here

so the circuit seems to be filtering frequencies in some way, but how to read the "frequency" value from the above result?


Analogous case for discrete fourier transform (300 data points):

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The frequency burst at $X_{18}=120$ can be read straight off the result.

I am unclear what will happen when this exact same case is implemented using quantum Fourier transform on 300 qubits? What will $X_{18}=120$ be manifested in (probability amplitude?) in the output measurement of quantum fourier transform?

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    $\begingroup$ Can you add more details on what you want to do? Which "frequency" do you expect to see? Is it normal that the states are not normalized? What do you mean by "loading the coefficients"? $\endgroup$
    – Tristan Nemoz
    Commented Jun 7, 2023 at 7:11
  • $\begingroup$ @TristanNemoz thank you. I added an analogous case using discrete fourier transform, where the burst "frequency" can be read off the result chart immediately. I guess i am not clear what is the equivalent procedure when using quantum fourier transform? $\endgroup$
    – James
    Commented Jun 7, 2023 at 8:19
  • $\begingroup$ @TristanNemoz another aspect that puzzles me is, how to "load" the sine wave data points into the qubits, $\Psi = a|0\rangle + b|1\rangle$? What numbers should we load $a, b$ with to represent the original sine wave? $\endgroup$
    – James
    Commented Jun 7, 2023 at 8:31
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    $\begingroup$ If you have 300 data points, you don't load them into the amplitudes of 300 qubits. You load them onto the amplitudes of the $2^n$ basis states of $n$ qubits ($n=9$). So you have a very complex circuit to produce the state. $\endgroup$
    – DaftWullie
    Commented Jun 7, 2023 at 9:48
  • $\begingroup$ @DaftWullie thank you, I think I am starting to understand... Say we have $2^8=256$ data points, and have managed to load the waveform into coefficients of the 256 tensored states $\Psi = a |00000000\rangle + b|00000001\rangle + c|00000010\rangle + d|00000011\rangle + ...$ Is this complicated tensored state already pre-constructed prior to us entering the circuit diagram (in the first picture above)? After passing through the sequence of H, CR(pi/2), CR(pi/4), CR(pi/8).. gates, what will be the output coming out of these 8 qubit lines (ie. what's coming out on the right of the diagram)? $\endgroup$
    – James
    Commented Jun 7, 2023 at 10:26

1 Answer 1

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After working through the math, I found that the 3-qubit QFT density transformation is given by the matrix:

enter image description here

(from here)

Extending to $2^8 = 256$ states,

enter image description here enter image description here enter image description here

gives the correct single Fourier spike at the right frequency index.

Therefore the error is likely in circuit construction or density state change implementation.


Algebraic verification for 2 qubit QFT circuit:

enter image description here

Step 0:

$$\sqrt{\frac{1}{4}}[a|00\rangle + b|01\rangle + c|10\rangle + d|11\rangle]$$

Step 1:

$$\sqrt{\frac{1}{8}}[(a+c)|00\rangle + (b+d)|01\rangle + (a-c)|10\rangle + (b-d)|11\rangle]$$

Step 2:

$$\sqrt{\frac{1}{8}}[(a+c)|00\rangle + (b+d)|01\rangle + (a-c)|10\rangle + i(b-d)|11\rangle]$$

Step 3:

$$\sqrt{\frac{1}{16}}[(a+b+c+d)|00\rangle + (a-b+c-d)|01\rangle + (a+ib-c-id)|10\rangle + (a-ib-c+id)|11\rangle]$$

$$=\sqrt{\frac{1}{16}}\begin{bmatrix}a+b+c+d \\ a-b+c-d \\ a+ib-c-id \\ a-ib-c+id \end{bmatrix}$$

Comparing with 2 qubit QFT matrix with $\omega = e^{i\frac{2\pi}{4}}$:

$$\sqrt{\frac{1}{4}}\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & \omega & \omega^2 & \omega^3 \\ 1 & \omega^2 & \omega^4 & \omega^6 \\ 1 & \omega^3 & \omega^6 & \omega^9 \end{bmatrix} = \sqrt{\frac{1}{4}} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\ 1 & -1 & 1 & -1 \\ 1 & -i & -1 & i\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}=\sqrt{\frac{1}{4}}\begin{bmatrix}a+b+c+d\\a+ib-c-id\\a-b+c-d\\a-ib-c+id\end{bmatrix}$$

The $|01\rangle, |10\rangle$ coefficient entries are reversed, which need to be post-processed.


Algebraic verification of 3 qubit QFT circuit:

enter image description here

$$\boxed{\omega^1 = \frac{1+i}{\sqrt2} \\ \omega^2 = i \\ \omega^3 = \frac{i(1+i)}{\sqrt2} \\ \omega^4 = -1 \\ \omega^5 = -\frac{1+i}{\sqrt2} \\ \omega^6 = -i \\ \omega^7 = -\frac{i(1+i)}{\sqrt3} \\ \omega^8 = 1}$$

Step 0:

$$ \begin{bmatrix} 000 & a \\ 001 & b \\ 010 & c \\ 011 & d \\ 100 & e \\ 101& f \\ 110 & g \\ 111 & h \end{bmatrix}$$

Step 1:

$$ \begin{bmatrix} 000 & a+e \\ 001 & b+f \\ 010 & c+g \\ 011 & d+h \\ 100 & a-e \\ 101& b-f \\ 110 & c-g \\ 111 & d-h \end{bmatrix}$$

Step 2:

$$ \begin{bmatrix} 000 & a+e \\ 001 & b+f \\ 010 & c+g \\ 011 & d+h \\ 100 & a-e \\ 101& w^1b-w^1f \\ 110 & c-g \\ 111 & w^1d-w^1h \end{bmatrix}=\begin{bmatrix} 000 & a+e \\ 001 & b+f \\ 010 & c+g \\ 011 & d+h \\ 100 & a+w^4e \\ 101& w^1b+w^5f \\ 110 & c+w^4g \\ 111 & w^1d+w^5h \end{bmatrix}$$

Step 3:

$$ \begin{bmatrix} 000 & a+e \\ 001 & b+f \\ 010 & c+g \\ 011 & d+h \\ 100 & a+w^4e \\ 101& w^1b+w^5f \\ 110 & w^2c+w^2w^4g \\ 111 & w^2w^1d+w^2w^5h \end{bmatrix}=\begin{bmatrix} 000 & a+e \\ 001 & b+f \\ 010 & c+g \\ 011 & d+h \\ 100 & a+w^4e \\ 101& w^1b+w^5f \\ 110 & w^2c+w^6g \\ 111 & w^3d+w^7h \end{bmatrix}$$

Step 4:

$$ \begin{bmatrix} 000 & a+e + c+g \\ 001 & b+f +d+h \\ 010 & a+e -(c+g) \\ 011 & b+f -(d+h)\\ 100 & a+w^4e+w^2c+w^6g \\ 101& w^1b+w^5f + w^3d+w^7h\\ 110 & a+w^4e -(w^2c+w^6g) \\ 111 & w^1b+w^5f -(w^3d+w^7h) \end{bmatrix}=\begin{bmatrix} 000 & a+e + c+g \\ 001 & b+f +d+h \\ 010 & a+e +w^4c+w^4g \\ 011 & b+f +w^4d+w^4h\\ 100 & a+w^4e+w^2c+w^6g \\ 101& w^1b+w^5f + w^3d+w^7h\\ 110 & a+w^4e +w^6c+w^2g \\ 111 & w^1b+w^5f +w^7d+w^3h \end{bmatrix}$$

Step 5:

$$\begin{bmatrix} 000 & a+c+e+g \\ 001 & b+d+f+h \\ 010 & a+w^4c+e + w^4g \\ 011 & w^2b+w^6d+w^2f+w^6h \\ 100 & a +w^4 e + w^2c +w^6g \\ 101& w^1b +w^5f + w^3d +w^7h \\ 110 & a +w^4 e +w^6c + w^2g \\ 111 & w^3b +w^7f +w^1d + w^5h \end{bmatrix}$$

Step 6:

$$\begin{bmatrix} 000 & a+c+e+g + b+d+f+h \\ 001 & a+c+e+g -(b+d+f+h)\\ 010 & a+w^4c+e + w^4g + w^2b+w^2w^4d+w^2f+w^2w^4h \\ 011 & a+w^4c+e + w^4g - (w^2b+w^2w^4d+w^2f+w^2w^4h) \\ 100 & a +w^4 e + w^2c +w^6g + w^1b +w^5f + w^3d +w^7h \\ 101& a +w^4 e + w^2c +w^6g - (w^1b +w^5f + w^3d +w^7h) \\ 110 & a +w^4 e +w^6c + w^2g + w^3b +w^7f +w^1d + w^5h\\ 111 & a +w^4 e +w^6c + w^2g - (w^3b +w^7f +w^1d + w^5h) \end{bmatrix}$$

$$=\begin{bmatrix} 000 & a+c+e+g + b+d+f+h \\ 001 & a+c+e+g +w^4b+w^4d+w^4f+w^4h\\ 010 & a+w^4c+e + w^4g + w^2b+w^6d+w^2f+w^6h \\ 011 & a+w^4c+e + w^4g + w^6b+w^2d+w^6f+w^2h \\ 100 & a +w^4 e + w^2c +w^6g + w^1b +w^5f + w^3d +w^7h \\ 101& a +w^4 e + w^2c +w^6g + w^5b +w^1f + w^7d +w^3h \\ 110 & a +w^4 e +w^6c + w^2g + w^3b +w^7f +w^1d + w^5h\\ 111 & a +w^4 e +w^6c + w^2g + w^7b +w^3f +w^5d + w^1h \end{bmatrix}$$

$$=\begin{bmatrix} 000&1&1&1&... \\ 001&1&w^4&w^{2\times 4}&... \\ 010&1&w^2&w^{2\times 2}&...\\011&1&w^6&w^{2\times 6}&...\\100&1&w^1&w^{2\times 1}&...\\101&1&w^5&w^{2\times 5}&...\\110&1&w^3&w^{2\times 3}&...\\111&1&w^7&w^{2\times 7}&...\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\\e\\f\\g\\h\end{bmatrix}$$

Renaming labels $q_1 \leftrightarrow q_3$ may be done in post-processing.

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