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In the paper "Compiling basic linear algebra subroutines for quantum computers" here, the introduction seems to imply that the subroutines covered will touch on matrix addition and multiplication, but the subroutines in question concern the product of matrix exponentials, and the sum of exponentials (i.e for A, B Hermitian, $e^{i(A+B)t}$).

Can it be inferred that after these results are achieved that there is some way to get $(A+B)$ from $e^{i(A+B)t}$ ?

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Yes, see for instance this stack exchange question: Using the HHL algorithm to compute $A|b\rangle$ instead of $A^{−1}|b\rangle$ . It is discussed that if you have access to an exponential $e^{iX \otimes H\tau}$, then you can implement $H$. To see this, note that $$ e^{iX\otimes H \tau} |0\rangle |\psi\rangle = (\mathbb{I} + i\tau X \otimes H + O(\tau^2)) |0\rangle |\psi\rangle = |0\rangle |\psi\rangle + i\tau |1\rangle H|\psi\rangle + O(\tau^2). $$ Now measure the first register. If the outcome is 1, then we get the state $|1\rangle H|\psi\rangle + O(\tau)$. For $\tau$ sufficiently small this approximates $H|\psi\rangle$.

So in your case, rather than summing $A$ and $B$, you could sum $X \otimes A$ and $X\otimes B$ with $X$ the Pauli X gate. Then you can use the exponential $e^{iX \otimes (A+B)t}$ to implement $A+B$ using the above approach.

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  • $\begingroup$ That was my question. I'm not sure how that relates, can you elaborate? $\endgroup$ – IntegrateThis Jun 20 at 6:18
  • $\begingroup$ The question I referred to exactly answers your question: gives access to $e^{iHt}$, how to implement $H$? I elaborated my reply. $\endgroup$ – smapers Jun 20 at 6:31
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    $\begingroup$ It stands for the identity matrix. The first equality follows from the series expansion of the matrix exponential. Please don't worry about asking :) $\endgroup$ – smapers Jun 20 at 6:51
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    $\begingroup$ This answer is nice but the original question was "is it possible to get $A+B$", not "is it possible to apply $A+B$". It may be a mistake from @IntegrateThis though, let's wait for a clarification. $\endgroup$ – Nelimee Jun 20 at 7:07
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    $\begingroup$ Great, then you can accept the answer :) $\endgroup$ – smapers Jun 27 at 6:25

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