# Conditional rotation in the bloch sphere with a 2-qubit system

1. Feb 24, 2014

### Verdict

1. The problem statement, all variables and given/known data
The problem is as follows. I have two spins, $m_S$ and $m_I$. The first spin can either be $\uparrow$ or $\downarrow$ , and the second spin can be -1, 0 or 1.
Now, I envision the situation as the first spin being on the bloch sphere, with up up to and down at the bottom.
What I want to do is as follows:
Given an initial situation $\left|\psi\right> = \left|\psi_1\right> \otimes \left|\psi_2\right> = \left|\uparrow\right> \otimes \frac{1}{\sqrt{3}}\left( \left|-1\right> + \left|0\right>+ \left|1\right> \right)$

I want to rotate m_S around the x-axis by pi/2, followed by a waiting time t. In this waiting time t, I want m_S to rotate around the z-axis, conditional on the state of m_I. If m_I is -1, m_S should rotate clockwise, if it is 0, m_S should not rotate, and if it is 1, m_S should rotate anticlockwise.

After this has happened, I want to perform another rotation, this time around the y-axis. This way, the state of m_S becomes entangled with the state m_I.

3. The attempt at a solution

Now, the rotation part I know how to do, as that can simply be written as

$\left|\psi\right> = R_x (\frac{\pi}{2}) \left|\psi_1\right> \otimes \left|\psi_2\right> = \frac{1}{\sqrt{2}} \left( \left|\uparrow\right> -i \left|\downarrow\right> \right) \otimes \frac{1}{\sqrt{3}}\left( \left|-1\right> + \left|0\right>+ \left|1\right> \right)$

But here I get to the point of the conditional rotation, and I don't know how to proceed. Could anyone help me start with this?

Last edited: Feb 24, 2014
2. Mar 12, 2014

### DonQubito

Distribute out the tensor product. $\sum_i \alpha_i \left|i \right> \otimes \sum_j \beta_j \left|j \right>=\sum_{i,j} \alpha_i \beta_j \left|i\right> \otimes \left|j \right>$. Then apply the controlled operation to every basis state.