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In a chapter of Quantum Computation and Quantum Information by Nielsen and Chuang (10th edition) about quantum phase estimation I get a little confused. Namely:

Before applying inverse QFT our quantum state is: $$ \frac{1}{2^{t/2}}(|0\rangle + e^{2\pi i 2^{t-1} \phi}|1\rangle)(|0\rangle + e^{2\pi i 2^{t-2} \phi}|1\rangle)...(|0\rangle + e^{2\pi i 2^{0} \phi}|1\rangle) $$

This later is said to be equivalent to $$ \frac{1}{2^{t/2}}(|0\rangle + e^{2\pi i 0.\phi_t}|1\rangle)(|0\rangle + e^{2\pi i 0.\phi_t\phi_{t-1}}|1\rangle)...(|0\rangle + e^{2\pi i 0.\phi_t\phi_{t-1}...\phi_0}|1\rangle), $$ where $0.\phi_t\phi_{t-1}...\phi_0 = \phi_t * 2^{-1} + ... + \phi_0 * 2^{-t}$. I don't really get the transition, don't we need some $2^{-t}$ in the exponent to get from $2\pi i 2^{t-1} \phi$ to $2\pi i 0.\phi_t$ etc.

Also, I have read about quantum phase estimation in Qiskit documentation and Wikipedia and there the result after applying inverse QFT is clearly stated to be $2^t \phi$ whereas Nielsen and Chuang seem to imply that we get directly $\phi$.

Is it just a mistake in the book or am I missing something?

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The first part of the Quantum Phase Estimation (QPE) is the implementation of this circuit (pg. 222 of Mike and Ike)

enter image description here

(There should be $+$ between $|0\rangle$ and $e^{2 \pi i 2^{n} }|1\rangle$ )

This gives you the state the first equation in your question:

$$ \frac{1}{2^{t/2}}(|0\rangle + e^{2\pi i 2^{t-1} \phi}|1\rangle)(|0\rangle + e^{2\pi i 2^{t-2} \phi}|1\rangle)...(|0\rangle + e^{2\pi i 2^{0} \phi}|1\rangle) $$

To see how this is equivalent to the second question, let's suppose that $\phi$ can be written exactly as $t$ bits, and so $$\phi = 0.\phi_1 \phi_2 \cdots \phi_t = \dfrac{\phi_1}{2^1} + \dfrac{\phi_2}{2^2} + \cdots + \dfrac{\phi_t}{2^t} = \sum_{i=1}^{t} \dfrac{\phi_i}{2^i} $$

Now note the following:

\begin{align} 2^k \phi &= \phi_1 \phi_2 \cdots \phi_k.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} \\ &=\phi_1 \phi_2 \cdots \phi_k + 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} \end{align}

And therefore

\begin{align} e^{2\pi i 2^k \phi} &= e^{2\pi i (\phi_1 \phi_2 \cdots \phi_k + 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} )} \\ &=e^{2\pi i \phi_1 \phi_2 \cdots \phi_k} e^{2\pi i 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} }\\ &= 1 \cdot e^{2\pi i 0.\phi_{k+1} \phi_{k+2} \cdots \phi_{t} } \hspace{0.75 cm} \textrm{since} \ \ e^{2\pi i \cdot n} = 1\ \ \forall n \in \mathbb{N} \end{align}

Thus that prompts the result:

$$ \frac{1}{2^{t/2}}(|0\rangle + e^{2\pi i 0.\phi_t}|1\rangle)(|0\rangle + e^{2\pi i 0.\phi_t\phi_{t-1}}|1\rangle)...(|0\rangle + e^{2\pi i 0.\phi_t\phi_{t-1}...\phi_0}|1\rangle) $$

that you see (formula 5.21 in Mike and Ike). Now, in term of QFT part, remember what the QFT transformation does, which is the following:

$$QFT: |\phi_n\rangle \otimes |\phi_{n-1} \otimes \rangle \otimes \cdots \otimes |\phi_1\rangle \rightarrow \dfrac{|0\rangle + e^{2\pi i 0.\phi_n}|1\rangle}{\sqrt{2}} \otimes \dfrac{|0\rangle + e^{2\pi i 0.\phi_{n-1}\phi_n}|1\rangle}{\sqrt{2}} \otimes \cdots \otimes \dfrac{|0\rangle + e^{2\pi i 0.\phi_1 \cdots \phi_{n-1} \phi_n}|1\rangle}{\sqrt{2}}$$

Which again is the state that you have at the end of the circuit above. Thus, by performing the inverse of QFT, we get back $|\phi_n\rangle \otimes |\phi_{n-1} \otimes \rangle \otimes \cdots \otimes |\phi_1\rangle$. Therefore, upon measurement on each of the qubit in the First Register, you will be able to exact $\phi_n, \cdots, \phi_1$ which allow you to reconstruct the phase $\phi$.

Hopefully I was able to help and provide some clarifications.

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  • $\begingroup$ Thank you for this, so just to make sure, we always have $\theta \in [0,1)$ right? (which now actually makes sense given that our eigenvalue is defined as $e^{2\pi i \theta}$) $\endgroup$
    – puma
    Oct 30 '21 at 0:06
  • $\begingroup$ No problem. yes, $\theta \in [0,1)$. $\endgroup$
    – KAJ226
    Oct 30 '21 at 20:15

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