Now, let's analyze the QPE algorithm when the target wires are prepared in the state $\vert \phi \rangle$.

--- ***Exercise P.4.2.*** Show that the state at *Step 2* of the QPE circuit is $$ \left( \frac{1}{2^{t/2}}\right) \sum_{\lambda} c_{\lambda} \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_t)_\lambda} \vert 1 \rangle \right) \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_{t-1}\theta_t)_\lambda}\vert 1 \rangle \right)\cdots \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_1\theta_{2}\cdots\theta_t)_\lambda} \vert 1 \rangle \right) \otimes \vert \psi_{\lambda} \rangle . $$

Solution.

We start at the initial state $$ \vert 0\rangle^{\otimes t} \vert \phi \rangle, $$ where $$ \vert \phi \rangle = \sum_{\lambda} c_{\lambda} \vert \psi_{\lambda} \rangle. $$ Recall from Exercise P.4.1 that $$ U\vert \phi \rangle = \sum_\lambda c_{\lambda} e^{2 \pi i (0.\theta_1\theta_2\cdots \theta_t)_\lambda} \vert \psi_\lambda \rangle. $$ At *Step 1*, applying the Hadamard transform to the estimation wires results in $$ \begin{align} H^{\otimes t} \vert 0\rangle^{\otimes t} \vert \phi \rangle &= \left( \frac{1}{2^{t/2}}\right) \left( \vert 0 \rangle + \vert 1 \rangle \right) ^{\otimes t} \otimes \vert \phi \rangle \\ &= \left( \frac{1}{2^{t/2}}\right) \left( \vert 0 \rangle + \vert 1 \rangle \right) ^{\otimes t} \otimes \sum_{\lambda} c_{\lambda} \vert \psi_{\lambda} \rangle \\ &= \left( \frac{1}{2^{t/2}}\right) \sum_{\lambda} c_{\lambda} \left( \vert 0 \rangle + \vert 1 \rangle \right) ^{\otimes t} \otimes \vert \psi_{\lambda} \rangle . \end{align} $$ Next, the controlled-$U$ is applied between the first qubit and the target wires $2^{t-1}$ times to obtain $$ \left( \frac{1}{2^{t/2}}\right) \sum_{\lambda} c_{\lambda} \left( \vert 0 \rangle + U^{2^{t-1}} \vert 1 \rangle \right) \left( \vert 0 \rangle + \vert 1 \rangle \right)^{\otimes (t-1)} \otimes \vert \psi_{\lambda} \rangle $$ $$ \begin{align} &= \left( \frac{1}{2^{t/2}}\right) \sum_{\lambda} c_{\lambda} \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_t)_\lambda} \vert 1 \rangle \right) \left( \vert 0 \rangle + \vert 1 \rangle \right)^{\otimes (t-1)} \otimes \vert \psi_{\lambda} \rangle \end{align} $$ where the controlled-$U^{2^{t-1}}$ generates the associated phase of $e^{2\pi i 0.\theta_t}$ for each of the eigenvectors in the superposition. Similarly, on the second qubit, we obtain $$ \left( \frac{1}{2^{t/2}}\right) \sum_{\lambda} c_{\lambda} \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_t)_\lambda} \vert 1 \rangle \right) \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_{t-1}\theta_t)_\lambda}\vert 1 \rangle \right) \left( \vert 0 \rangle + \vert 1 \rangle \right)^{\otimes (t-2)} \otimes \vert \psi_{\lambda} \rangle . $$ We continue in this manner, applying the controlled-$U$ and then kicking back the phases to the estimation wires, up until the last qubit, where $U$ is applied only once. At the end, we obtain $$ \left( \frac{1}{2^{t/2}}\right) \sum_{\lambda} c_{\lambda} \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_t)_\lambda} \vert 1 \rangle \right) \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_{t-1}\theta_t)_\lambda}\vert 1 \rangle \right)\cdots \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_1\theta_{2}\cdots\theta_t)_\lambda} \vert 1 \rangle \right) \otimes \vert \psi_{\lambda} \rangle. $$

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--- The state obtained in the previous exercise $$ \left( \frac{1}{2^{t/2}}\right) \sum_{\lambda} c_{\lambda} \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_t)_\lambda} \vert 1 \rangle \right) \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_{t-1}\theta_t)_\lambda}\vert 1 \rangle \right)\cdots \left( \vert 0 \rangle + e^{2 \pi i(0.\theta_1\theta_{2}\cdots\theta_t)_\lambda} \vert 1 \rangle \right) \otimes \vert \psi_{\lambda} \rangle \tag{1} \label{eq:step2} $$ looks very similar to what we obtained after *Step 2* when the target wires were prepared with an eigenvector (rather than a superposition of eigenvectors that we prepared here). The difference here is that the entire estimation register is also now in a superposition. Using our knowledge about what the state on the estimation register looks like, what results can we expect? --- ***Exercise P.4.3.*** Apply the inverse Fourier transform to the state above. Show that the probability of obtaining the state $|\theta_\lambda \rangle$ associated with the phase $\theta_\lambda$ corresponding to the eigenvalue $e^{2 \pi i \theta_\lambda}$ and the eigenvector $\vert \psi_\lambda \rangle$ is $\vert c_\lambda \vert^2$, upon measuring the estimation wires after *Step 3*.

Solution.

Since the Fourier transform is a linear transformation, applying the QFT$^\dagger$ to Eq. ($\ref{eq:step2}$) will transform each of the values in the superposition, giving us the following state $$ \sum_{\lambda} c_{\lambda} \vert \theta_1\theta_2\cdots \theta_t \rangle_\lambda \otimes \vert \psi_{\lambda} \rangle . $$ Here $\vert \theta_1\theta_2\cdots \theta_t \rangle_\lambda$ are the bits that represent $(0.\theta_1\theta_2\cdots \theta_t)_\lambda$ or $\theta_\lambda$. Since we obtain these phases on the estimation wires in superposition, the probability of observing $\theta_\lambda$ is $|c_{\lambda}|^2$.

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--- This extends to phases that do not have a $t$-bit binary expansion as well. Instead of one approximation, our state is a superposition of approximations. However, as we saw in earlier nodes, if the number of estimation wires is not large enough, the success probability will be lower (though, still never less than $40\%$). Given that we are already obtaining a superposition of phases, we may need more and more qubits to get more precision, as well as satisfactory results.

Codercise P.4.1 - Eigenphases of the T-gate

Codercise P.4.2 - Eigenphases of a two-qubit gate

Codercise P.4.3 - Experimenting with more unitaries

Phase kickback for non-eigenstates

QPE for non-eigenstates

Finding the ground state energy