Why should the inverse phase estimation in HHL change the ket $|b\rangle$?

Question

In HHL the forward phase estimation allows to set the phases of eigenvalues. It does not change the input ket $|b\rangle$ as far as this is a combination of the eigenvalues of the operator (and because this operator can be changed to the phase operator and the control phase produces phase kickback for the control qubit).

Why should the inverse phase estimation step change the ket $|b\rangle$? It still is the same control-phase operator, and it thus should again produce phase kickback.

Answer 1

Maybe abstracting a bit the steps might help understanding the gist of it.

The quantum phase estimation algorithm can be seen in general as a gate decomposition implementing (approximately) an isometry $\mathcal U_U$ such that $$\mathcal U_U|u_j\rangle=|u_j\rangle|\phi_j\rangle,$$ for some orthonormal basis $\{|u_j\rangle\}_j$, and with $\phi_j$ the (phases corresponding to the) eigenvalues of some fixed beforehand unitary $U$, $U|u_j\rangle=e^{2\pi i\phi_j}|u_j\rangle$.

It clearly follows that the "inverse quantum phase estimation", meaning $\mathcal U_U^\dagger$, maps $|u_j\rangle|\phi_j\rangle\to |u_j\rangle$.

So in the case of HHL,

1. you start with a state of the form $\sum_j c_j |u_j\rangle$
2. the QPE, $\mathcal U_U$, transforms the above into $\sum_j c_j |u_j\rangle|\phi_j\rangle$,
3. the "controlled rotation step" effectively amounts to a diagonal mapping of the form $$\sum_j c_j |u_j\rangle|\phi_j\rangle \to \sum_j c_j \alpha_j |u_j\rangle|\phi_j\rangle$$ for some set of coefficients $\alpha_j$ (which are related to the eigenvalues etc etc, but I don't care about this here),
4. the inverse QPE step $\mathcal U_U$ now gives $$\sum_j c_j \alpha_j |u_j\rangle|\phi_j\rangle \to \sum_j c_j \alpha_j |u_j\rangle.$$

So in summary, you can think of the combination of QPE-controlledRotation-inverseQPE as a way to implement the mapping $|u_j\rangle\to \alpha_j |u_j\rangle$. Of course, you might wonder why we couldn't just directly do this from the beginning, but note that we don't actually know the states $|u_j\rangle$, which in HHL are the eigenstates of $A$. Using the QPE effectively allows you to apply a unitary diagonal in the basis $\{|u_j\rangle\}$ without needing to know the base itself.