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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? $\endgroup$ Commented Jun 20, 2019 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
    Commented Jun 20, 2019 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
    Commented Jun 20, 2019 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. $\endgroup$ Commented Jun 20, 2019 at 7:07
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    $\begingroup$ Great, then you can accept the answer :) $\endgroup$
    – smapers
    Commented Jun 27, 2019 at 6:25

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