# In quantum search algorithm, how to interpret the effect of U(t)？

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• Haorong Wu
In summary: R_{\hat n} \left ( \theta \right ) = \cos \left ( \frac \theta 2 \right ) I - i \sin \left ( \frac \theta 2 \right ) \hat n \cdot \vec \theta##. Then in this case, ##\sin \left ( \frac \theta 2 \right ) = 2 \sin \left ( \frac {\Delta t} 2 \right )##. However, the resulting equation does not follow the definition of rotation by ##\theta##, leading to confusion and potential error.
Haorong Wu
TL;DR Summary
In quantum search algorithm, how to interpret the effect of U(t) as a rotation on the Bloch sphere?
In Nielsen's QCQI, in page 259, it reads,

$$U \left ( \Delta t \right ) = \left ( \cos^2 \left ( \frac {\Delta t} 2 \right ) - \sin ^2 \left ( \frac {\Delta t} 2 \right ) \vec \psi \cdot \hat z \right ) I \\ -2 i \sin \left ( \frac {\Delta t} 2 \right ) \left ( \cos \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi + \hat z} 2 + \sin \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi \times \hat z} 2 \right ) \cdot \vec \sigma$$ where ##U \left ( \Delta t \right )## is a operation of a Hamiltonian, ##\Delta t## is the time interval, ##\vec \psi## is the initial state.

Well, it seems complicated. But with ##\vec r = \cos \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi + \hat z} 2 + \sin \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi \times \hat z} 2 ## and ## \vec \psi \cdot \hat z = \frac 2 N -1 ##
where ##N## is the number of the elements in the search space, it would be simplified to ##U \left ( \Delta t \right ) = \left (1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right ) \right ) I -2 i \sin \left ( \frac {\Delta t} 2 \right ) \vec r \cdot \vec \sigma##.

Then the book reads, ##U \left ( \Delta t \right ) ## is a rotation on the Bloch sphere about an axis of rotation ##\vec r## and through an angle ##\theta## defined by ##\cos \left ( \frac {\theta} 2 \right ) = 1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right ) ##.

My problem is, the definition of the rotation by ##\theta## about any ##\hat n## axis is ## R_{\hat n} \left ( \theta \right ) = \cos \left ( \frac \theta 2 \right ) I - i \sin \left ( \frac \theta 2 \right ) \hat n \cdot \vec \theta##. Then in this case, ##\sin \left ( \frac \theta 2 \right ) = 2 \sin \left ( \frac {\Delta t} 2 \right ) ##.

Then ##\sin^2 \left ( \frac \theta 2 \right ) + \cos^2 \left ( \frac \theta 2 \right ) \neq 1##.

Where have I made a mistake?

Haorong Wu said:
Then in this case, ##\sin \left ( \frac \theta 2 \right ) = 2 \sin \left ( \frac {\Delta t} 2 \right ) ##.

I don't understand this.

Note that ##\vec{r}## is not a unit vector in
Haorong Wu said:
##U \left ( \Delta t \right ) = \left (1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right ) \right ) I -2 i \sin \left ( \frac {\Delta t} 2 \right ) \vec r \cdot \vec \sigma##
Writing ##\vec{r} = r \hat{r}## gives
$$\sin \left( \frac{\theta}{2} \right) = 2r \sin \left( \frac{\Delta t}{2} \right) .$$

Haorong Wu
Haorong Wu said:
Summary: In quantum search algorithm, how to interpret the effect of U(t) as a rotation on the Bloch sphere?

In Nielsen's QCQI, in page 259, it reads,

$$U \left ( \Delta t \right ) = \left ( \cos^2 \left ( \frac {\Delta t} 2 \right ) - \sin ^2 \left ( \frac {\Delta t} 2 \right ) \vec \psi \cdot \hat z \right ) I \\ -2 i \sin \left ( \frac {\Delta t} 2 \right ) \left ( \cos \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi + \hat z} 2 + \sin \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi \times \hat z} 2 \right ) \cdot \vec \sigma$$ where ##U \left ( \Delta t \right )## is a operation of a Hamiltonian, ##\Delta t## is the time interval, ##\vec \psi## is the initial state.

Well, it seems complicated. But with ##\vec r = \cos \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi + \hat z} 2 + \sin \left ( \frac {\Delta t} 2 \right ) \frac {\vec \psi \times \hat z} 2 ## and ## \vec \psi \cdot \hat z = \frac 2 N -1 ##
where ##N## is the number of the elements in the search space, it would be simplified to ##U \left ( \Delta t \right ) = \left (1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right ) \right ) I -2 i \sin \left ( \frac {\Delta t} 2 \right ) \vec r \cdot \vec \sigma##.

Then the book reads, ##U \left ( \Delta t \right ) ## is a rotation on the Bloch sphere about an axis of rotation ##\vec r## and through an angle ##\theta## defined by ##\cos \left ( \frac {\theta} 2 \right ) = 1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right ) ##.

My problem is, the definition of the rotation by ##\theta## about any ##\hat n## axis is ## R_{\hat n} \left ( \theta \right ) = \cos \left ( \frac \theta 2 \right ) I - i \sin \left ( \frac \theta 2 \right ) \hat n \cdot \vec \theta##. Then in this case, ##\sin \left ( \frac \theta 2 \right ) = 2 \sin \left ( \frac {\Delta t} 2 \right ) ##.

Then ##\sin^2 \left ( \frac \theta 2 \right ) + \cos^2 \left ( \frac \theta 2 \right ) \neq 1##.

Where have I made a mistake?
it has taken me a while to unpack all of the notation. I think the issue is the definitions of ##\vec \psi ## and ##\hat z##. On p. 259 these are given as:
##\vec \psi = (2 \alpha \beta, 0,\alpha^2 - \beta^2)##
##\hat z = (0, 0, 1)##
With these definitions, I get the same result as in the book.
Edit: Also, is ##\vec r## a unit vector?

Last edited:
Haorong Wu
George Jones said:
I don't understand this.

Note that ##\vec{r}## is not a unit vector in

Writing ##\vec{r} = r \hat{r}## gives
$$\sin \left( \frac{\theta}{2} \right) = 2r \sin \left( \frac{\Delta t}{2} \right) .$$
Thanks, George. I made a mistake when I assumed that ##\vec r## is normalized.

In fact, ##\left | \vec r \right | =\sqrt {\alpha ^2 \beta ^2 + \alpha ^4 \cos ^2 \frac {\Delta t} 2}##, and the result is consistent with ##\sin^2 \left ( \frac \theta 2 \right ) + \cos^2 \left ( \frac \theta 2 \right ) = 1##

Thanks!

tnich said:
it has taken me a while to unpack all of the notation. I think the issue is the definitions of ##\vec \psi ## and ##\hat z##. On p. 259 these are given as:
##\vec \psi = (2 \alpha \beta, 0,\alpha^2 - \beta^2)##
##\hat z = (0, 0, 1)##
With these definitions, I get the same result as in the book.

Yes, after calculation, I found out that I made a mistake when I assume ##\vec r## is normalized which is not.

Thanks!

## 1. What is U(t) in the context of quantum search algorithm?

U(t) refers to the unitary operator used in the quantum search algorithm, which is responsible for performing the search operation on the quantum state. It is a time-dependent operator that is applied repeatedly to the initial state to amplify the amplitude of the target state and decrease the amplitude of the other states.

## 2. How does U(t) affect the outcome of the quantum search algorithm?

U(t) plays a crucial role in the success of the quantum search algorithm. It is responsible for amplifying the amplitude of the target state, making it more likely to be measured at the end of the algorithm. The number of times U(t) is applied also affects the efficiency of the algorithm, with more iterations leading to a higher probability of finding the target state.

## 3. Can U(t) be interpreted in a classical way?

No, U(t) cannot be interpreted classically as it is a quantum operator that operates on a quantum state. It is a fundamental part of quantum mechanics and cannot be explained using classical concepts.

## 4. How is the effect of U(t) measured in the quantum search algorithm?

The effect of U(t) is measured by the probability of finding the target state after applying the operator multiple times. In the ideal case, the probability of measuring the target state should approach 1 as the number of iterations increases. This indicates that U(t) is effectively amplifying the amplitude of the target state.

## 5. Are there any alternatives to using U(t) in the quantum search algorithm?

Yes, there are alternative approaches to performing quantum search without using U(t). For example, the Grover's algorithm uses a different set of quantum gates to achieve the same goal. However, U(t) is the most commonly used approach and has been proven to be the most efficient for searching unstructured databases.

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