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In the paper "Compiling basic linear algebra subroutines for quantum computers" here, they discuss (page 2 bottom right) using the HHL algorithm to multiply a vector by a matrix. However, after having read the HHL09 paper, what is being estimated is the state $|x \rangle = A^{-1} |b \rangle$, where $A |x \rangle = |b \rangle$ for some Hermitian matrix $A$ of dimension $n$, and vectors $b, x$ of size $n$, where $b$ and $A$ are known.

Is it possible to use the HHL algorithm to compute $A |b \rangle$ knowing $A$ and $b$. For instance, running the algorithm in reverse seems not possible (?), because not all steps in the algorithm are unitary.

glS
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IntegrateThis
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2 Answers2

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Note: the graphics have been generated with the LaTeX code available here. Credits to @Niel de Beaudrap.

Yes it is possible!

The HHL algorithm can be schematically depicted as

HHL algorithm drawing

Let's split down the parts:

  1. The first part aims at computing an approximation of the eigenvalues of $H$, $H = A$ if $A$ is hermitian, else $H = \begin{pmatrix} 0 & A \\ A^\dagger & 0 \end{pmatrix}$.
  2. The second part invert these eigenvalues and adds a phase proportional to these eigenvalues in front of your state (see the original HHL09 paper for the exact expression).
  3. The third part uncompute the first part, leaving you with a state like $$ \sum_{j=1}^{N}\beta_j \vert u_j \rangle \left( \sqrt{1 - \frac{C^2}{\lambda_j^2}} \vert 0 \rangle + \frac{C}{\lambda_j}\vert 1 \rangle \right). $$
  4. The measure project the resulting state. If the measure succeed (measured value is $1$), then this means that your state is now $$ \sum_{j=1}^{N}\beta_j \vert u_j \rangle \frac{C}{\lambda_j}\vert 1 \rangle = \left(\sum_{j=1}^{N}C\frac{\beta_j}{\lambda_j} \vert u_j \rangle \right) \vert 1 \rangle $$ which is the solution of the linear system considered.

Everything apart being equal, if you do not invert the eigenvalues in step 2, you will end up with a state like $$ \sum_{j=1}^{N}\beta_j \vert u_j \rangle \left( \sqrt{1 - C^2\lambda_j^2} \vert 0 \rangle + C\lambda_j\vert 1 \rangle \right). $$ for step 3 and a result like $$ \left(C\sum_{j=1}^{N}\beta_j\lambda_j \vert u_j \rangle \right) \vert 1 \rangle $$ if your measurement succeeded.

This is the state you are looking for: the result of the operation $A\times b$, scaled by $C$ in order to be of unit-norm.

Adrien Suau
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    Isn't this rather circular since you're assuming you have $U$? Is it not the case that $U|b\rangle=A|b\rangle$? So it's a basic assumption that you can already compute that product. – DaftWullie Jun 18 '19 at 08:08
  • Or is $U=e^{iHt}$? You could still extract the action fairly easily with measurement on an ancilla, up to some higher order terms. – DaftWullie Jun 18 '19 at 08:11
  • $U = e^{iHt}$ for different values of $t$ (as I said, the drawing is schematic, in the real algorithm you apply controlled $U$, $U^2$, ...) so you are right, $U\vert b \rangle = e^{iHt} \vert b \rangle = C\sum_{j=1}^{N}\beta_je^{i\lambda_j t} \vert u_j \rangle$. We don't have the result, but something close. The phase estimation step is here to change this $e^{i\lambda_jt}$ in $\lambda_j$. – Adrien Suau Jun 18 '19 at 08:32
  • But why not simply apply $e^{iX\otimes H\delta t}$ on $|0\rangle|b\rangle$? If $\delta t$ is small, and you find the first qubit to be in the $|1\rangle$ state, you've got $H|b\rangle$ with high probability. – DaftWullie Jun 18 '19 at 10:40
  • @DaftWullie what is X in that case? – IntegrateThis Jun 18 '19 at 18:00
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    The standard Pauli X matrix – DaftWullie Jun 18 '19 at 18:19
  • @DaftWullie I do not understand why applying $e^U$ on $\vert b \rangle$ gives $U\vert b\rangle$ and not $e^U \vert b \rangle$, but if that's the case then of course, your version is better. Post it as an answer. – Adrien Suau Jun 19 '19 at 07:37
  • @Nelimee In the HHL paper they rotate conditioned on $|\tilde{\lambda} \rangle$ after performing a Fourier transform, so in your version you would simply ignore the rotation? I'm not sure where you obtained your second last equation. – IntegrateThis Jun 20 '19 at 00:09
  • They do not rotate conditioned on $\vert \overline{\lambda}\rangle$, they rotate conditioned on $\vert \frac{1}{\overline{\lambda}}\rangle$. In this version, I simply ignore the "inverting $\overline{\lambda}$" part. – Adrien Suau Jun 26 '19 at 11:09
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Let's assume that you have a Hermitian matrix $$ H=\left(\begin{array}{cc} 0 & A^\dagger \\ A & 0 \end{array}\right). $$ Let $|b\rangle$ be the state that we want to apply $A$ to, extended to work on the space that $H$ acts on. So, our aim is to implement $H|b\rangle$.

Let $X$ be the standard Pauli $X$ matrix. If we implement a unitary evolution $$ U=e^{-i X\otimes H\delta t}, $$ then the action of this on an input state $|0\rangle|b\rangle$ can be expressed through a Taylor expansion, assuming $\delta t\|H\|\ll 1$: $$ U|0\rangle|b\rangle=|0\rangle|b\rangle-i\delta t |1\rangle(H|b\rangle)-\frac{\delta t^2}{2}|0\rangle(H^2|b\rangle)+\ldots. $$ So, if we measure the ancilla qubit in the standard basis, then with probability $\delta t^2\|H|b\rangle\|^2$, we find the ancilla to be in state $|1\rangle$, and the other register is in the desired state $H|b\rangle$, up to an accuracy $O(\delta t^2)$ (coming from the $O(\delta t^4)$ term in the Taylor expansion). If you don't get the $|1\rangle$ answer to your measurement, then to accuracy $O(\delta t^2)$, you've stil got the original state $|0\rangle|b\rangle$, so you can just repeat until success.

By that simple method, the overall error is very bad - you have to repeat $O(1/\delta t^2)$ times so that the overall error becomes $O(1)$. I'm sure this could be improved by replacing the $X$ with a cycle operation on a larger dimensional Hilbert space. However, note that this method is very similar to part of the HHL algorithm (where they do the controlled-rotation from the phase estimation register onto an ancilla) and, at least there, it is claimed that it can be improved using amplitude amplification. So I assume (without having looked at the details) that something similar is possible here.

DaftWullie
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    Ok, did not think about the Taylor expansion of the exponential, you are right it is simpler to implement than HHL. – Adrien Suau Jun 20 '19 at 07:02