Quantum algorithm for linear systems of equations (HHL09): Step 1 - Number of qubits needed

Question

This is a continuation of Quantum algorithm for linear systems of equations (HHL09): Step 1 - Confusion regarding the usage of phase estimation algorithm

Questions (contd.):

Part 2: I'm not exactly sure how many qubits will be needed for the Step 1 of the HHL09.

In Nielsen and Chuang (section 5.2.1, 10th anniversary edition) they say:

Thus to successfully obtain $\varphi$ accurate to $n$-bits with probability of sucess at least $1-\epsilon$ we choose

$$t=n+\lceil { \log(2+\frac{1}{2\epsilon})\rceil}$$

So, say we want an accuracy of $90\%$ i.e. $1-\epsilon = 0.9 \implies \epsilon = 0.1$ and a precision of $3$-bits for $\frac{\lambda_j t}{2\pi}$ or $\lambda_j$ we'd need

$$t = 3 + \lceil { \log_2(2+\frac{1}{2 (0.1)})\rceil} = 3 + 3 = 6$$

Apart from that, since $|b\rangle$ can be represented as a sum of $N$ linearly independent eigenvectors of a $N\times N$ dimensional matrix $A$, we'd need minimum $\lceil{\log_2(N)\rceil}$ qubits to produce a vector space having at least $N$ - dimensions. So, we need $\lceil{\log_2(N)\rceil}$ for the second register.

Now, for the first register we not only $\lceil{\log_2(N)\rceil}$ qubits won't be sufficient to represent the $N$ eigenvalues $|\lambda_j\rangle$, that is because we'll need more bits for representing each $|\lambda_j\rangle$ precisely upto $n$-bits.

I guess we should again use the formula $$n+\lceil { \log(2+\frac{1}{2\epsilon})\rceil}$$ in this case. If we want each eigenvalue $|\lambda_i\rangle$ to be represented with a $3$-bit precision and $90\%$ accuracy then we'd need $6\times \lceil{\log_2(N)\rceil}$ for the first register. Plus, one more qubit which is needed for the ancilla.

So, we should need a total of $(6+1)\lceil{\log_2(N)\rceil}+1$ qubits for Step 1 of the HHL09 algorithm. That's quite a lot!

Say we want to solve a $2\times 2$ linear equation system such that $A$ is Hermitian that itself would require $7\lceil{\log_2(2)\rceil}+1 = 8$ qubits! In case $A$ is not Hermitian we'd need even more qubits. Am I right?

However, in this^{[$\dagger\dagger$]} paper on page 6 they claim that they used the HHL09 algorithm to estimate the pseudoinverse of $A$ which of size ~$200\times 200$. In that paper, $A$ is defined as:

$$A := \begin{pmatrix} W - \gamma \Bbb I_d & P \\ P & 0 \end{pmatrix}$$

where $P$,$W$ and $\Bbb I_d$ are all $d\times d$ matrices.

In the H1N1 related simulated Lloyd et al. have claimed to have made, $d = 100$. And they further claim that they used the HHL09 algorithm to estimate the pseudo-inverse of $A$ (which is of size $200\times 200$). That would need a minimum of $7\lceil{\log_2(200)\rceil}+1 = 7(8)+1 = 57$ qubits to simulate. I have no idea how they could possibly do that using the current quantum computers or quantum computer simulations. As far as I know, IBM Q Experience at present supports ~$15$ qubits (that too it isn't as versatile as their $5$-qubit version).

Am I missing something here? Does this Step 1 actually require a lesser number of qubits than what I have estimated?

[$\dagger\dagger$]: A Quantum Hopfield Neural Network Lloyd et al. (2018)

Answer 1

Calculation of the inverse of an $N\times N$ matrix can be done by applying HHL with $N$ different $\vec{b}_i$ (specifically, HHL is applied $N$ times, once for each computational basis vector used as the $\vec{b}_i$).

In each case, phase estimation has to be done for an $N \times N$ matrix.

The number of qubits required for phase estimation is written on page 249 of the 10th anniversary edition of N&C:

"The quantum phase estimation procedure uses two registers. The first register contains $t$ qubits."

"The second register [...] contains as many qubits as is necessary to store $|u\rangle$", where $|u\rangle$ is an $N$-dimensional vector.

So you are correct that we would need $6$ qubits for the first register, and $\log N=8$ qubits for the second register.

This is 14 qubits in total to do the phase esitmation part of each HHL iteration involved in calculating the inverse of a matrix. 14 qubits is well within the capabilities of a laptop.