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

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data
    The problem is as follows. I have two spins, [itex]m_S[/itex] and [itex]m_I[/itex]. The first spin can either be [itex]\uparrow[/itex] or [itex]\downarrow[/itex] , 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 [itex]\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)[/itex]

    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

    [itex]\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)[/itex]

    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. jcsd
  3. Mar 12, 2014 #2
    Distribute out the tensor product. [itex]\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>[/itex]. Then apply the controlled operation to every basis state.
     
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