Skip to main content
24 votes

Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?

A first remark This same phenomenon of 'control' qubits changing states in some circumstances also occurs with controlled-NOT gates; in fact, this is the entire basis of eigenvalue estimation. So not ...
Niel de Beaudrap's user avatar
19 votes
Accepted

Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?

Imagine you have an eigenvector $|u\rangle$ of $U$. If you have a state such as $|1\rangle|u\rangle$ and you apply controlled-$U$ to it, you get out $e^{i\phi}|1\rangle|u\rangle$. The phase isn't ...
DaftWullie's user avatar
  • 59.4k
14 votes
Accepted

Who discovered the phase kickback trick?

The phase kickback trick appears in this paper: Richard Cleve, Artur Ekert, Chiara Macchiavello, Michele Mosca. Quantum Algorithms Revisited. Proceedings of the Royal Society of London A, 454(1969):...
John Watrous's user avatar
  • 6,137
6 votes
Accepted

Does phase kickback require the system to be in the eigenstate?

The first circuit equality fails when $|\psi_k\rangle$ is not an eigenstate of $U$. A simple way to see this is to set the control qubit to $|1\rangle$. In this case, the RHS circuit is equivalent to $...
Adam Zalcman's user avatar
6 votes
Accepted

Can we use Hadamard test to estimate phases?

So QPE using $\mathcal{O}(1/\epsilon)$ queries to $U$ outputs an estimate of the eigenphase $\theta$ given a corresponding eigenvector with additive error and $\Omega(1)$ probability. The method using ...
dylan7's user avatar
  • 324
5 votes

How to understand intuitively the quantum gate phase kickback?

Part of the problem people usually have here is a sort-of-classical intuition. Because you're trying to describe the action of a gate such as controlled-$U$, we divide it up as "if the control ...
DaftWullie's user avatar
  • 59.4k
4 votes

What is Quantum Phase Estimation in Shor's Algorithm?

Phase estimation is the process by which you are given a controlled-$U$ unitary, and a state that you are promised is an eigenvector of $U$ with eigenvalue $e^{2\pi ix/2^t}$, then you can use a $t$-...
DaftWullie's user avatar
  • 59.4k
4 votes

Question about the phase kickback in the phase estimation algorithm

The phase is applied to the overall wave function $|\phi\rangle$, therefore you can factor the phase to any individual qubit. For example if we have a wave function as a result of a controlled ...
Sam Palmer's user avatar
4 votes
Accepted

Phase kickback without controlled operator

In the case where $f(0)=f(1)=1$, $U_f$ is just the controlled-Identity gate. This is still a controlled gate, though a trivial one. And, as one should expect, the phase we get is the eigenvalue of $\...
Norbert Schuch's user avatar
3 votes
Accepted

In the Bernstein-Vazirani circuit, if the secret is all 0s, would the oracle just be nothing?

Yes, this is because the initial state will be starting as $|\psi_{init} \rangle = |000\cdots0\rangle = |0\rangle^{\otimes 8} $. Then you apply a layer of Hadamard gates, follow by the Oracle ...
KAJ226's user avatar
  • 13.9k
3 votes

Unitary Operator impact on both - the Control Qubit and the Target Register in Shor's Algorithm

The key idea in understanding phase kickback is that phase factors do not belong to one register or the other, but instead belong to terms in superposition and are shared by the registers. For example,...
Adam Zalcman's user avatar
3 votes

Is the $|-\rangle$ state the only one that can do the trick for Grover's algorithm?

Remember that the states $|+\rangle$ and $|-\rangle$ form a basis. That means that any state $|\psi\rangle$ that you use can be written as $$ |\psi\rangle=\alpha|+\rangle+\beta|-\rangle. $$ By ...
DaftWullie's user avatar
  • 59.4k
3 votes
Accepted

Is the $|-\rangle$ state the only one that can do the trick for Grover's algorithm?

It would still work if you rotated the $|−\rangle$ state, you refer to, by an angle with cosine less than $\frac{1}{L}$ where $L$ is the square root of the size of the search space (which is a power ...
Learner's user avatar
  • 336
3 votes
Accepted

Can we use quantum phase estimation to estimate the phase of an arbitrary single-qubit state?

No, it cannot be done because if it could be done, you could distinguish between arbitrarily many linearly dependent states -- which is not possible. Another way to say the same thing is that if you ...
FreeAssange's user avatar
2 votes

Show that these two expressions for the oracle transformation are equivalent

You are given a quantum circuit for $U$ compiled into the H/CNOT/T gateset. Derive a controlled version of $U$ by adding a control qubit $q$, replacing every H with a controlled H, every CNOT with a ...
Craig Gidney's user avatar
  • 38.8k
2 votes
Accepted

Show that these two expressions for the oracle transformation are equivalent

If we express the action of $O_x$ on the basis $\mid i, \pm \rangle$ instead of $\mid i , b \rangle$ \begin{eqnarray*} O_x \mid i , + \rangle = \mid i , + \rangle\\ O_x \mid i , - \rangle = (-1)^{x_i}...
AHusain's user avatar
  • 3,663
2 votes

Who discovered the phase kickback trick?

As to "who discovered/invented the quantum phase estimation algorithm," in his 2011 lecture at Keio University describing the linear equations algorithm, at about the 18 minute mark Lloyd claims that ...
Mark Spinelli's user avatar
2 votes
Accepted

Resources and references about phase kickback trick

The IBM Qiskit text has a section on this that you may find useful. For something more "academic" Mermin Sections 4.2-4.5, Schuld and Petruccione Sections 3.5.1-3.5.3, or Nielson and Chuang ...
ryanhill1's user avatar
  • 2,503
2 votes
Accepted

Why does the phase of the eigenstate get kicked up to the ancilla qubit?

Here is a basic example of a two system that might help you to see this better. Suppose I have these two circuits: Circuit 1: Which put the "Controlled qubit" in the state $|1\rangle$ and ...
KAJ226's user avatar
  • 13.9k
2 votes
Accepted

In Shor's algorithm, how can we guarantee that each controlled-U will kickback to the same eigenvalue?

I understand, that cU has multiple eigenvalues with a factor s. How can be guaranteed that each of the controlled-Us will kickback the same eigenvalue? Or, why it is not important? All the $U$s in ...
DaftWullie's user avatar
  • 59.4k
2 votes
Accepted

How to decompose a multi-target controlled gate?

An example of constructing (with help of Qiskit) a controlled version of some simple 4x4 unitary matrix: $$ U = \begin{pmatrix} \mathrm{e}^{i g_1} & 0 & 0 & 0 \\ 0 & \mathrm{e}^{i g_2}...
Davit Khachatryan's user avatar
2 votes

What is Quantum Phase Estimation in Shor's Algorithm?

If you want to learn more generally about phase estimation, the book I am helping write has a whole chapter on it, Learn Quantum Computing with Python and Q# (chapter 8). Ping me here or via email ...
Dr. Sarah Kaiser's user avatar
2 votes

Does phase kickback require the system to be in the eigenstate?

Disclaimer: please do not consider this to be a definite answer as it is not fully rigorous. I am just giving you some evidence or clues from one practical example. You can write the state $|\psi_k\...
Martin Vesely's user avatar
2 votes
Accepted

Why is this interpretation of phase kickback incorrect?

Physically, there is no difference between measuring the left qubit in the state $|0\rangle$ or $|1\rangle$. As stated in a previous answer, if the left qubit is measured in $|1\rangle$ the phase $e^{...
Andrés Ruiz's user avatar
2 votes
Accepted

How to derive phase kickback probabilities

The norm-squared of a complex $a \in \mathbb{C}$ number is calculated by multiplying that number with it's complex conjugate. So, for: $$ a = \frac{1 + e^{2 \pi i \phi}}{2}, $$ you get: $$ \begin{...
diemilio's user avatar
  • 595
1 vote
Accepted

Understanding phase kickback caused by the CNOT gate

There are two ways to see this. simply factor the sum you have: $$\frac{1}{2}(|00\rangle - |01\rangle - |10\rangle + |11\rangle) \\ = \frac{1}{2}(|0\rangle - |1\rangle)(0\rangle - |1\rangle) \\ =|-\...
Lior's user avatar
  • 1,220
1 vote

In the Bernstein-Vazirani circuit, if the secret is all 0s, would the oracle just be nothing?

With the Quantum Oracle $|x \rangle \xrightarrow{f_s} (-1)^{s\cdot x} |x \rangle$ (between the Hadamard sandwich) for the BV algorithm, with ($n$-bit) input secret bits $s$, here is what it does: $|00\...
Sandipan Dey's user avatar
1 vote

In Shor's algorithm, how can we guarantee that each controlled-U will kickback to the same eigenvalue?

Have you seen this document? https://qiskit.org/textbook/ch-algorithms/shor.html Note that in Shor's algorithm, we use the quantum computer as a subroutine to essentially find the period of the ...
KAJ226's user avatar
  • 13.9k
1 vote

controlled-Z rotation gates in symmetrical fashion

Exchanging the two qubits swaps the basis states $|01\rangle \leftrightarrow |10\rangle$, but keeps $|00\rangle$ and $|11\rangle$ unchanged. Suppose you have a gate whose action on the computational ...
Adam Zalcman's user avatar
1 vote

controlled-Z rotation gates in symmetrical fashion

For the mathematical explanation, check here: Why is the action of controlled-Z unaltered by exchanging target control qubits? Maybe it would help you to see CZ in a different (symmetric) notation, ...
luciano's user avatar
  • 5,813

Only top scored, non community-wiki answers of a minimum length are eligible