I have a problem with the conditional Hamiltonian. In the original article on HHL (p.3) they wrote that applying the conditional Hamiltonian correspond to: $$ \sum_{\tau=0}^{T-1}|\tau\rangle\langle\tau|\otimes e^{2i\pi A\frac{\tau}{T}}$$
Where $T=2^t$ the number of qubits in the clock register.

And I saw an implementation in this article (p.50), for a 2 qubits register they apply 2 gates $e^{i\pi A}$ and $e^{i\pi A/2}$.

What I don't understand, is that it doesn't correspond to the sum above which have 4 terms, but to this one (I change the index of the sum): $$\sum_{\tau=1}^{2^{t-1}}|\tau\rangle\langle\tau|\otimes e^{2i\pi A\frac{\tau}{2^t}}$$.

Did I miss something ?

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    $\begingroup$ Could you say what page number it is in the review article? $\endgroup$ – AHusain Jul 19 '19 at 1:10
  • $\begingroup$ @AHusain I added it. $\endgroup$ – lufydad Jul 19 '19 at 9:13

Be careful! They don't apply $e^{i\pi A}$ and $e^{i\pi A/2}$. They apply $$ |0\rangle\langle 0|\otimes I\otimes I+|1\rangle\langle 1| \otimes I\otimes e^{i\pi A} $$ and $$ I\otimes |0\rangle\langle 0|\otimes I+I\otimes |1\rangle\langle 1|\otimes e^{i\pi A/2}, $$ i.e. controlled versions of the gates, controlled off two different qubits.

So, consider the 4 possible values of the first two registers: $|00\rangle, |01\rangle, |10\rangle$ and $|11\rangle$. On the third register, these give you respectively, $I,e^{i\pi A/2},e^{i\pi A},e^{i3\pi A/2}$, as required.


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