7
$\begingroup$

Context:

On the 5th page of the paper Quantum circuit design for solving linear systems of equations (Cao et al, 2012) there's this circuit:

enter image description here

enter image description here

enter image description here


Schematic:

A brief schematic of what's actually happening in the circuit is:

enter image description here


Question:

Cao et al.'s circuit (in Figure 4) is specifically made for the matrix:

$$A = \frac{1}{4} \left(\begin{matrix} 15 & 9 & 5 & -3 \\ 9 & 15 & 3 & -5 \\ 5 & 3 & 15 & -9 \\ -3 & -5 & -9 & 15 \end{matrix}\right)$$

whose eigenvalues are $\lambda_1 = 2^{1-1}=1,\lambda_2 = 2^{2-1}=2,\lambda_3 = 2^{3-1}=4$ and $\lambda_4 = 2^{4-1} = 8$ and corresponding eigenvectors are $|u_i\rangle = \frac{1}{2}\sum_{j=1}^{4}(-1)^{\delta_{ij}}|j\rangle^C$. In this case since there are $4$ qubits in the clock register, the $4$ eigenvalues can be represented as states of the clock register itself (no approximation involved) i.e. as $|0001\rangle$ (binary representation of $1$), $|0010\rangle$ (binary representation of $2$), $|0100\rangle$ (binary representation of $4$) and $|1000\rangle$ (binary representation of $8$).

After the first Quantum phase estimation step, the circuit's (in Fig. 4) state is

$$|0\rangle_{\text{ancilla}} \otimes \sum_{j=1}^{j=4} \beta_j |\lambda_j\rangle |u_j\rangle$$

Everything is fine till here. However, after this, to produce the state

$$\sum_{j=1}^{j=4} \beta_j |u_j\rangle_I |\lambda_j\rangle^C ((1-C^2/\lambda_j^2)^{1/2}|0\rangle + C/\lambda_j|1\rangle)$$

it seems necessary to get to the state $$\sum_{j=1}^{j=4} \beta_j |u_j\rangle_I |\frac{2^{3}}{\lambda_j}\rangle^C\otimes |0\rangle_{\text{ancilla}}$$

That is, the following mappings seem necessary in the $R(\lambda^{-1})$ rotation step:

$$|0001\rangle \mapsto |1000\rangle, |0010\rangle \mapsto |0100\rangle, |0100\rangle \mapsto |0010\rangle \ \& \ |1000\rangle \mapsto |0001\rangle$$

which implies that the middle two qubits in the clock register need to be swapped as well as the two end qubits.

But, in the circuit diagram they seem to be swapping the first and third qubit in the clock register! That doesn't seem reasonable to me. In the paper (Cao et al.) claim that the transformation they're doing using their SWAP gate is

$$\sum_{j=1}^{j=4} \beta_j |u_j\rangle_I |\lambda_j\rangle^C\otimes |0\rangle_{\text{ancilla}} \mapsto \sum_{j=1}^{j=4} \beta_j |u_j\rangle_I |\frac{2^{4}}{\lambda_j}\rangle^C\otimes |0\rangle_{\text{ancilla}}$$

According to their scheme, $|1000\rangle \to |0010\rangle$ (see the third page). This scheme doesn't make sense to me because the state $|0001\rangle$ (representing the eigenvalue $1$) would have to be transformed to $|2^4/1\rangle$. But the decimal representation of $16$ would be $|10000\rangle$ which is a 5-qubit state! However, our clock register has only $4$ qubits in total.

So, basically I think that their SWAP gates are wrong. The SWAPs should actually have been applied between the middle qubits and the two end qubits. Could someone verify?


Supplementary question:

This is not a compulsory part of the question. Answers addressing only the previous question are also welcome.

@Nelimee wrote a program (4x4_system.py in HHL) in QISKit to simulate the circuit (Figure 4) in Cao et al's paper. Strangely, his program works only if the SWAP gate is applied between the middle two qubits but not in between the two end qubits.

The output of his program is as follows:

<class 'numpy.ndarray'>
Exact solution: [-1  7 11 13]
Experimental solution: [-0.84245754+0.j  6.96035067+0.j 10.99804383+0.j 13.03406367+0.j]
Error in found solution: 0.16599956439346453

That is, in his program only the mapping $|0100\rangle \mapsto |0010\rangle$ takes place in the clock register in the $R(\lambda^{-1})$ step. There's no mapping $|1000\rangle \mapsto |0001\rangle$ taking place.

Just in case anyone figures out why this is happening please let me know in the comments (or in an answer).

$\endgroup$
3
$\begingroup$

I don't see the need for the swap gate either. Although I also don't think you need the set of swap gates that you're wanting. Remember that the standard implementation of the Fourier transform outputs the binary string $j\in\{0,1\}^4$ where the eigenvalues are of the form $2\pi j/16$ but in reverse order, so the least significant bit is at the top, and the most significant bit is at the bottom. Thus, $|1000\rangle$ already corresponds to $j=1$, and so the rotation that you want is with an angle $8\pi/2^m$, and $|0100\rangle$ corresponds to $j=2$, requiring angle $4\pi/2^m$, and so on.

However, I have to emphasise that what I see here is something of a fudge. It seems to be that the authors have very much built in the knowledge of the eigenvalues - not just that they're of the form $2\pi j/16$, but specifically that they are 1,2,4,8. It looks to me like, if one had a matrix that also contained a $j=4$ term, for example, that the inverse would not be correctly calculated because the angle of rotation actually created (LHS) $$ \frac{8\pi}{2^m}+\frac{4\pi}{2^m}=\frac{12\pi}{2^m}\neq\frac{8\pi}{3\cdot 2^m} $$ does not match the required angle for the inverse (RHS). Perhaps that was already clear to you, but it was not clear to me in trying to understand the circuit in order to answer your question!

In terms of testing with QISKit, it's not something that I use myself. However, can I suggest trying a different $|b\rangle$ as input? It is possible (perhaps not likely, however) that some of what's going on is masked slightly because the $|b\rangle$ chosen in the paper is an equally weighted superposition of the 4 eigenvectors. I'd want to chose something with different amplitudes for each eigenvector, just to be sure.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.