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

#### Haorong Wu

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?

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#### George Jones

Staff Emeritus
Gold Member
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
$U \left ( \Delta t \right ) = \left (1-\frac 2 N \sin^2 \left ( \frac {\Delta t} 2 \right ) \right ) I -2 i \sin \left ( \frac {\Delta t} 2 \right ) \vec r \cdot \vec \sigma$
Writing $\vec{r} = r \hat{r}$ gives
$$\sin \left( \frac{\theta}{2} \right) = 2r \sin \left( \frac{\Delta t}{2} \right) .$$

#### tnich

Homework Helper
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

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!

#### Haorong Wu

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!

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

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