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Qubits and quantum evolution

  1. Feb 25, 2015 #1
    Hello,

    I have a problem where I'm given the following

    [itex]H=-\frac{\hbar\Omega}{2}\sigma_x\quad\quad\quad\textrm{and}\quad\quad\quad\Psi(0)=\left|0\right\rangle\quad[/itex]
    Where

    [itex]\sigma_x=\begin{pmatrix}0 & 1\\1&0\end{pmatrix}\quad\quad\quad\textrm{and}\quad\quad\quad\left|0\right\rangle=\begin{pmatrix}1\\0\end{pmatrix}[/itex]
    And in general

    [itex]\Psi(t)=\textrm{exp}\left[-i\frac{H}{\hbar}t\right]\Psi(0)[/itex]
    So

    [itex]\Psi(t)=\textrm{exp}\left[i\frac{\Omega t}{2}\sigma_x\right]\left|0\right\rangle[/itex]
    The problem is I need to get from here to

    [itex]\Psi(t)=\textrm{exp}\left[i\frac{\Omega t}{2}\sigma_x\right]\left|0\right\rangle=\begin{pmatrix}\textrm{cos}(\Omega t/2)\,\,\,&i\textrm{sin}(\Omega t/2)\\i\textrm{sin}(\Omega t/2)\,\,\,&\textrm{cos}(\Omega t/2)\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}\\\\\\
    \quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,=\begin{pmatrix}cos(\Omega t/2)\\isin(\Omega t/2)\end{pmatrix}
    [/itex]

    I can't work out how to get to this cos and sine matrix. I tried this

    [itex]\Psi(t)=\textrm{exp}\left[i\frac{\Omega t}{2}\sigma_x\right]\left|0\right\rangle=\left\lbrace\textrm{cos}\left(\frac{\Omega t}{2}\sigma_x\right)+i\textrm{sin}\left(\frac{\Omega t}{2}\sigma_x\right)\right\rbrace\left|0\right\rangle\\\\
    \quad\quad\quad\quad\quad\quad\quad\quad\quad=\left\lbrace\textrm{cos}\begin{pmatrix}0 & \frac{\Omega t}{2}\\\frac{\Omega t}{2} & 0\end{pmatrix}+i\textrm{sin}\begin{pmatrix}0 & \frac{\Omega t}{2}\\\frac{\Omega t}{2} & 0\end{pmatrix}\right\rbrace\begin{pmatrix}1\\0\end{pmatrix}[/itex]

    Beyond this I cannot see how to get from here to

    [itex]\begin{pmatrix}\textrm{cos}(\Omega t/2)\,\,\,&i\textrm{sin}(\Omega t/2)\\i\textrm{sin}(\Omega t/2)\,\,\,&\textrm{cos}(\Omega t/2)\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}[/itex]
    Any help would be really appreciated
     
  2. jcsd
  3. Feb 25, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    There might be more than one way to approach the problem, but I would proceed thus. You can calculate the exponential of a matrix easily if the matrix is diagonal. For
    $$
    A = \left( \begin{array}{cc} a_1 & 0 \\ 0 & a_2 \end{array} \right)
    $$
    $$
    \exp(A) = \left( \begin{array}{cc} e^{a_1} & 0 \\ 0 & e^{a_2} \end{array} \right)
    $$
    Can you find the rotation that brings you from the z-basis to the x-basis?
     
  4. Feb 26, 2015 #3
    Thanks for responding,

    So based on what you've said for the exponential of a matrix the expression in my original post would become

    [itex]
    \Psi(t)=\textrm{exp}\left[i\frac{\Omega t}{2}\sigma_x\right]\left|0\right\rangle=\textrm{exp}\left[\begin{pmatrix}0 & i\frac{\Omega t}{2}\\i\frac{\Omega t}{2} & 0\end{pmatrix}\right]\left|0\right\rangle=\begin{pmatrix}0 & \textrm{exp}\left[i\frac{\Omega t}{2}\right]\\\textrm{exp}\left[i\frac{\Omega t}{2}\right] & 0\end{pmatrix}\left|0\right\rangle\\\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\begin{pmatrix}0 & \textrm{cos}\left(\frac{\Omega t}{2}\right)+i\textrm{sin}\left(\frac{\Omega t}{2}\right)\\\textrm{cos}\left(\frac{\Omega t}{2}\right)+i\textrm{sin}\left(\frac{\Omega t}{2}\right) & 0\end{pmatrix}\left|0\right\rangle

    [/itex]

    is that correct? It doesn't seem much closer to the intended end result.

    When you asked about the rotation are you referring to the relationships [itex]\sigma_y\sigma_z=-i\sigma_x[/itex] etc.?
     
    Last edited: Feb 26, 2015
  5. Feb 26, 2015 #4

    DrClaude

    User Avatar

    Staff: Mentor

    Calculating the exponential of a matrix by taking the exponential of the elements only works for a diagonal matrix. You can show this by considering the series expansion of the exponential function.

    You do not have a diagonal matrix, this is why I said you need to look at the rotation matrix that will bring you from the z basis to the x basis, in which ##\sigma_x## is diagonal.
     
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