# Apply the conditional Hamiltonian evolution (HHL)

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 ?

• Could you say what page number it is in the review article? Jul 19, 2019 at 1:10
• @AHusain I added it. Jul 19, 2019 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.