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I have the following complex vector in $\mathbb{C}^2$: Vec= [[ 0.89741876+0.j] [-0.33540402+0.28660724j]]

I'm trying to implement a $U_3$ gate to prepare this state. The general $U_3$ gate is defined as $$ U_3(\gamma,\beta,\delta) = \begin{bmatrix} \cos\left(\frac{\gamma}{2}\right) & -e^{i\delta} \sin\left(\frac{\gamma}{2}\right) \\ e^{i\beta} \sin\left(\frac{\gamma}{2}\right) & e^{i(\delta + \beta)} \cos\left(\frac{\gamma}{2}\right) \end{bmatrix}\\ $$ To solve for the parameters, I have $$ \cos\left(\frac{\gamma'}{2}\right) = g_0\qquad e^{i\beta'}\sin\left(\frac{\gamma'}{2}\right) = g_1 $$ Where $g_0$ is the real entry of Vec , and g_1 is the complex one. Then, I can get the solutions

$$ \gamma'=2\cos^{-1}(g_0),\qquad e^{i\beta'}=\pm\frac{g_1}{\sqrt{1-g_0^2}}\qquad\beta'=-i\log\frac{g_1}{\sqrt{1-g_0^2}}+k\pi,\ \text{where}\ k\in\mathbb{Z}$$ $\delta'$ could be any real value. With the values specified in Vec, we can find that $\gamma' = 0.913825694123124 , \beta_1'= 2.434484141132717 , \beta_2'=5.5760767947225105$ (differ by $1*\pi$). Therefore, they are two different vectors, as could be seen on the Bloch sphere: (the blue vector corresponds to $\beta_2'$, the purple one is $\beta_1'$)enter image description here The second solution doesn't quite work as I checked the values using Mathematica, but it looks like the blue vector might also contribute to the rotation as I tried to simulate the time-evolution of some arbitrary vector. Should I keep the blue vector, or how can I get rid of it? Thanks a lot for the help!!!

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    $\begingroup$ Note that that $\gamma'' = -\gamma'$ is also a valid angle for $\cos(\frac{\gamma}{2}) = g_{0}$, so that $\sin(\gamma'') = -\sin(\gamma')$; this $-1$ then cancels out the $k\pi$ in your $\beta'$. Or, they are both $-1$ and then it's a global phase - which you can't get with your $U_{3}$, you'd need to introduce an extra phase factor $e^{i\alpha'}$ $\endgroup$
    – JSdJ
    Mar 23 at 8:47
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    $\begingroup$ Also, $g_{0}$ is not generally referred to as the real entry - just the 'top' entry, or maybe the $|0\rangle$-coefficient. Likewise for $g_{1}$:) $\endgroup$
    – JSdJ
    Mar 23 at 8:49
  • $\begingroup$ @JSdJ Thanks so much, that's really helpful:) $\endgroup$
    – ZR-
    Mar 23 at 17:18
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When we want to use Bloch sphere formalism it's better to use its conventions. So, $\theta \in [0, \pi]$ and $\varphi \in [0, 2\pi)$:

$$|\psi \rangle = \cos{\left( \frac{\theta}{2} \right)} |0\rangle + e^{i\varphi}\sin{\left( \frac{\theta}{2} \right)} |1\rangle$$

The chosen $\gamma$ and $\beta$ will correspond to $\theta$ and $\varphi$ if we chose them the same way: $\gamma \in [0, \pi]$ and $\beta \in [0, 2\pi)$. If we don't choose them that way then we should do some extra steps to find the corresponding Bloch angles for the visualization.

If we use the convention for the Bloch angles the $\pm$ sign is not nessasry (in the convention we always have $\sin{\left( \frac{\gamma}{2} \right)} > 0$):

$$e^{i \beta '} = \frac{e^{i \beta '} \sin{\left( \frac{\gamma}{2} \right)}}{\sin{\left( \frac{\gamma}{2} \right)}} = \frac{g_1}{\sqrt{ 1 - \cos^2{\left( \frac{\gamma}{2} \right)}}} = \frac{g_1}{\sqrt{1 - g_0^2}}$$

Without the $\pm$ cases we will obtain only one solution. The $-$ case assumes that $\sin{\left( \frac{\gamma}{2} \right)} < 0$ and in that case the estimated $\beta$ and $\gamma$ will not directly correspond to $\varphi$ and $\theta$ (some extra steps are needed in order to find them).

Also for the calculated $\gamma ' \approx 0.91$, the $\sin{\left( \frac{\gamma '}{2} \right)} > 0$ and again $-$ sign can be disregarded.

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    $\begingroup$ Thanks so much!! That helps a lot:) $\endgroup$
    – ZR-
    Mar 23 at 17:01

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