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

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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?
 

George Jones

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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

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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?
 
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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!
 
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!
 

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