# Tag Info

### 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 ...
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### 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 ...
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### 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):...
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### 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 ...
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### 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,...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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}...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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) \\ =|-\...
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### 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, ...
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