Inquantum computing, thequantum phase estimation algorithm is aquantum algorithm to estimate the phase corresponding to an eigenvalue of a givenunitary operator. Because the eigenvalues of a unitary operator always haveunit modulus, they are characterized by their phase, and therefore the algorithm can be equivalently described as retrieving either the phase or the eigenvalue itself. The algorithm was initially introduced byAlexei Kitaev in 1995.^{[1][2]: 246}

Phase estimation is frequently used as a subroutine in other quantum algorithms, such asShor's algorithm,^[2]: 131 thequantum algorithm for linear systems of equations, and thequantum counting algorithm.

The algorithm operates on two sets of qubits, referred to in this context asregisters. The two registers contain${\displaystyle n}$ and${\displaystyle m}$ qubits, respectively. Let${\displaystyle U}$ be aunitary operator acting on the${\displaystyle m}$-qubit register. The eigenvalues of a unitary operator have unit modulus, and are therefore characterized by their phase. Thus if${\displaystyle |\psi ⟩}$ is aneigenvector of${\displaystyle U}$, then${\displaystyle U|\psi ⟩=e^{2\pi i\theta }|\psi ⟩}$ for some${\displaystyle \theta \in \mathbb{R}}$. Due to the periodicity of the complex exponential, we can always assume.

The goal is producing a good approximation for${\displaystyle \theta }$ with a small number of gates and a high probability of success. Thequantum phase estimation algorithm achieves this assuming oracular access to${\displaystyle U}$, and having${\displaystyle |\psi ⟩}$ available as aquantum state. This means that when discussing the efficiency of the algorithm we only worry about the number of times${\displaystyle U}$ needs to be used, but not about the cost of implementing${\displaystyle U}$ itself.

More precisely, the algorithm returnswith high probability an approximation for${\displaystyle \theta }$, within additive error${\displaystyle \epsilon }$, using${\displaystyle n=O(\log (1/\epsilon ))}$ qubits in the first register, and${\displaystyle O(1/\epsilon )}$controlled-U operations. Furthermore, we can improve the success probability to${\displaystyle 1-\mathrm{\Delta }}$ for any${\displaystyle \mathrm{\Delta }>0}$ by using a total of${\displaystyle O(\log (1/\mathrm{\Delta })/\epsilon )}$ uses of controlled-U, and this is optimal.^[3]

The initial state of the system is:

${\displaystyle |{\mathrm{\Psi }}_0⟩=|0{⟩}^{\otimes n}|\psi ⟩,}$

where${\displaystyle |\psi ⟩}$ is the${\displaystyle m}$-qubit state that evolves through${\displaystyle U}$. We first apply then-qubit Hadamard gate operation${\displaystyle H^{\otimes n}}$ on the first register, which produces the state:$${\displaystyle |{\mathrm{\Psi }}_1⟩=(H^{\otimes n}\otimes I_m)|{\mathrm{\Psi }}_0⟩=\frac{1}{2^{\frac{n}{2}}}(|0⟩+|1⟩{)}^{\otimes n}|\psi ⟩=\frac{1}{2^{n/2}}\sum _{j=0}^{2^n-1}|j⟩|\psi ⟩.}$$Note that here we are switching between binary and${\displaystyle n}$-ary representation for the${\displaystyle n}$-qubit register: the ket${\displaystyle |j⟩}$ on the right-hand side is shorthand for the${\displaystyle n}$-qubit state${\displaystyle |j⟩\equiv \underset{\ell =0}{\overset{n-1}{⨂}}|j_{\ell }⟩}$, where${\displaystyle j=\sum _{\ell =0}^{n-1}{j}_{\ell }{2}^{\ell }}$ is the binary decomposition of${\displaystyle j}$.

This state${\displaystyle |{\mathrm{\Psi }}_1⟩}$ is then evolved through the controlled-unitary evolution${\displaystyle U_C}$ whose action can be written as$${\displaystyle U_C(|k⟩\otimes |\psi ⟩)=|k⟩\otimes (U^k|\psi ⟩),}$$ for all${\displaystyle k=0,...,2^n-1}$. This evolution can also be written concisely as$${\displaystyle U_C=\sum _{k=0}^{2^n-1}|k⟩⟨k|\otimes U^k,}$$ which highlights its controlled nature: it applies${\displaystyle U^k}$ to the second register conditionally to the first register being${\displaystyle |k⟩}$. Remembering the eigenvalue condition holding for${\displaystyle |\psi ⟩}$, applying${\displaystyle U_C}$ to${\displaystyle |{\mathrm{\Psi }}_1⟩}$ thus gives$${\displaystyle |{\mathrm{\Psi }}_2⟩\equiv U_C|{\mathrm{\Psi }}_1⟩=\left(\frac{1}{2^{n/2}}\sum _{k=0}^{2^n-1}{e}^{2\pi i\theta k}|k⟩\right)\otimes |\psi ⟩,}$$ where we used${\displaystyle U^k|\psi ⟩=e^{2\pi ik\theta }|\psi ⟩}$.

To show that${\displaystyle U_C}$ can also be implemented efficiently, observe that we can write${\displaystyle U_C=\prod _{\ell =0}^{n-1}{C}_{\ell }(U^{2^{\ell }})}$, where${\displaystyle C_{\ell }(U^{2^{\ell }})}$ denotes the operation of applying${\displaystyle U^{2^{\ell }}}$ to the second register conditionally to the${\displaystyle \ell }$-th qubit of the first register being${\displaystyle |1⟩}$. Formally, these gates can be characterized by their action as$${\displaystyle C_{\ell }(U^k)(|j⟩\otimes |\psi ⟩)=|j⟩\otimes (U^{j_{\ell }k}|\psi ⟩).}$$This equation can be interpreted as saying that the state is left unchanged when${\displaystyle j_{\ell }=0}$, that is, when the${\displaystyle \ell }$-th qubit is${\displaystyle |0⟩}$, while the gate${\displaystyle U^k}$ is applied to the second register when the${\displaystyle \ell }$-th qubit is${\displaystyle |1⟩}$. The composition of these controlled-gates thus gives$${\displaystyle \prod _{\ell =0}^{n-1}{C}_{\ell }(U^{2^{\ell }})(|j⟩\otimes |\psi ⟩)=|j⟩\otimes \left(U^{\sum _{\ell =0}^{n-1}{j}_{\ell }{2}^{\ell }}|\psi ⟩\right)=U_C,}$$ with the last step directly following from the binary decomposition${\displaystyle j=\sum _{\ell =0}^{n-1}{j}_{\ell }{2}^{\ell }}$.

From this point onwards, the second register is left untouched, and thus it is convenient to write${\displaystyle |{\mathrm{\Psi }}_2⟩=|{\tilde{\mathrm{\Psi }}}_2⟩\otimes |\psi ⟩}$, with${\displaystyle |{\tilde{\mathrm{\Psi }}}_2⟩}$ the state of the${\displaystyle n}$-qubit register, which is the only one we need to consider for the rest of the algorithm.

The final part of the circuit involves applying the inversequantum Fourier transform (QFT)${\displaystyle \mathcal{Q}\mathcal{F}\mathcal{T}}$ on the first register of${\displaystyle |{\mathrm{\Psi }}_2⟩}$:$${\displaystyle |{\tilde{\mathrm{\Psi }}}_3⟩={\mathcal{Q}\mathcal{F}\mathcal{T}}_{2^n}^{-1}|{\tilde{\mathrm{\Psi }}}_2⟩.}$$The QFT and its inverse are characterized by their action on basis states as It follows that

${\displaystyle |{\tilde{\mathrm{\Psi }}}_3⟩=\frac{1}{2^{\frac{n}{2}}}\sum _{k=0}^{2^n-1}{e}^{2\pi i\theta k}\left(\frac{1}{2^{\frac{n}{2}}}\sum _{x=0}^{2^n-1}{e}^{\frac{-2\pi ikx}{2^n}}|x⟩\right)=\frac{1}{2^n}\sum _{x=0}^{2^n-1}\sum _{k=0}^{2^n-1}{e}^{-\frac{2\pi ik}{2^n}\left(x-2^n\theta \right)}|x⟩.}$

Decomposing the state in the computational basis as$|{\tilde{\mathrm{\Psi }}}_3⟩=\sum _{x=0}^{2^n-1}{c}_x|x⟩,$ the coefficients thus equal$${\displaystyle c_x\equiv \frac{1}{2^n}\sum _{k=0}^{2^n-1}{e}^{-\frac{2\pi ik}{2^n}(x-2^n\theta )}=\frac{1}{2^n}\sum _{k=0}^{2^n-1}{e}^{-\frac{2\pi ik}{2^n}\left(x-a\right)}{e}^{2\pi i\delta k},}$$where we wrote${\displaystyle 2^n\theta =a+2^n\delta ,}$ with${\displaystyle a}$ is the nearest integer to${\displaystyle 2^n\theta }$. The difference${\displaystyle 2^n\delta }$ must by definition satisfy${\displaystyle 0⩽|2^n\delta |⩽\frac{1}{2}}$. This amounts to approximating the value of${\displaystyle \theta \in [0,1]}$ by rounding${\displaystyle 2^n\theta }$ to the nearest integer.

The final step involves performing ameasurement in the computational basis on the first register. This yields the outcome${\displaystyle |y⟩}$ with probability$${\displaystyle Pr(y)=|c_y{|}^2={\left|\frac{1}{2^n}\sum _{k=0}^{2^n-1}{e}^{\frac{-2\pi ik}{2^n}(y-a)}{e}^{2\pi i\delta k}\right|}^2.}$$ It follows that${\displaystyle \mathrm{Pr}(a)=1}$ if${\displaystyle \delta =0}$, that is, when${\displaystyle \theta }$ can be written as${\displaystyle \theta =a/2^n}$, one always finds the outcome${\displaystyle y=a}$. On the other hand, if${\displaystyle \delta \ne 0}$, the probability reads$${\displaystyle \mathrm{Pr}(a)=\frac{1}{2^{2n}}{\left|\sum _{k=0}^{2^n-1}{e}^{2\pi i\delta k}\right|}^2=\frac{1}{2^{2n}}{\left|\frac{1-e^{2\pi i{2}^n\delta }}{1-e^{2\pi i\delta }}\right|}^2.}$$ From this expression we can see that${\displaystyle Pr(a)⩾\frac{4}{{\pi }^2}\approx 0.405}$ when${\displaystyle \delta \ne 0}$. To see this, we observe that from the definition of${\displaystyle \delta }$ we have the inequality${\displaystyle |\delta |⩽\frac{1}{2^{n+1}}}$, and thus:^{[4]: 157 [5]: 348}

We conclude that the algorithm provides the best${\displaystyle n}$-bit estimate (i.e., one that is within${\displaystyle 1/2^n}$ of the correct answer) of${\displaystyle \theta }$ with probability at least${\displaystyle 4/{\pi }^2}$. By adding a number of extra qubits on the order of${\displaystyle O(\log (1/ϵ))}$ and truncating the extra qubits the probability can increase to${\displaystyle 1-ϵ}$.^[5]

Consider the simplest possible instance of the algorithm, where only${\displaystyle n=1}$ qubit, on top of the qubits required to encode${\displaystyle |\psi ⟩}$, is involved. Suppose the eigenvalue of${\displaystyle |\psi ⟩}$ reads${\displaystyle \lambda =e^{2\pi i\theta }}$,${\displaystyle \theta \in [0,1)}$. The first part of the algorithm generates the one-qubit state$|\varphi ⟩\equiv \frac{1}{\sqrt{2}}(|0⟩+\lambda |1⟩)$. Applying the inverse QFT amounts in this case to applying aHadamard gate. The final outcome probabilities are thus${\displaystyle p_{\pm }=|⟨\pm |\varphi ⟩{|}^2}$ where$|\pm ⟩\equiv \frac{1}{\sqrt{2}}(|0⟩\pm |1⟩)$, or more explicitly,$${\displaystyle p_{\pm }=\frac{|1\pm \lambda {|}^2}{4}=\frac{1\pm \cos (2\pi \theta )}{2}.}$$ Suppose${\displaystyle \lambda =1}$, meaning${\displaystyle |\varphi ⟩=|+⟩}$. Then${\displaystyle p_{+}=1}$,${\displaystyle p_{-}=0}$, and we recover deterministically the precise value of${\displaystyle \lambda }$ from the measurement outcomes. The same applies if${\displaystyle \lambda =-1}$.

If on the other hand${\displaystyle \lambda =e^{2\pi i/3}}$, then${\displaystyle p_{\pm }=[1\pm \cos (2\pi /3)]/2}$, that is,${\displaystyle p_{+}=1/4}$ and${\displaystyle p_{-}=3/4}$. In this case the result is not deterministic, but we still find the outcome${\displaystyle |-⟩}$ as more likely, compatibly with the fact that${\displaystyle 2/3}$ is closer to 1 than to 0.

More generally, if${\displaystyle \lambda =e^{2\pi i\theta }}$, then${\displaystyle p_{+}\ge 1/2}$ if and only if${\displaystyle |\theta |\le 1/4}$. This is consistent with the results above because in the cases${\displaystyle \lambda =\pm 1}$, corresponding to${\displaystyle \theta =0,1/2}$, the phase is retrieved deterministically, and the other phases are retrieved with higher accuracy the closer they are to these two.

• Shor's algorithm
• Quantum counting algorithm
• Parity measurement

• Quantum algorithms

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