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General form for 2 x 2 unitary matrices

  1. Mar 23, 2015 #1
    I'm trying to show that any unitary matrix may be written in the form [itex]\begin{pmatrix}e^{i\alpha_1}\cos{\theta} & -e^{i\alpha_2}\sin{\theta}\\ e^{i\alpha_3}\sin{\theta} & e^{i\alpha_4}\cos{\theta}\end{pmatrix}[/itex]

    Writing the general form of a unitary matrix as
    [itex]U=\begin{pmatrix} u_{11} & u_{12}\\ u_{21} & u_{22}\end{pmatrix}[/itex]

    gives

    [itex]U^{\dagger}U=
    \begin{pmatrix}u_{11}^* & u_{21}^*\\u_{12}^* & u_{22}^*\end{pmatrix}\begin{pmatrix} u_{11} & u_{12}\\ u_{21} & u_{22}\end{pmatrix}=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}[/itex]

    [itex]\Longrightarrow |u_{11}|^2+|u_{21}|^2=1 \:\:,\:\: |u_{12}|^2 + |u_{22}|^2=1\\

    \Longrightarrow |u_{11}|=\cos(\theta) \:\:,\:\: |u_{21}|=\sin(\theta) \:\:,\:\: |u_{12}|=\cos(\varphi) \:\:,\:\: |u_{22}|=\sin(\varphi)[/itex]
    for some [itex]\theta , \varphi[/itex]

    I'm not really sure where to go from here.
     
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  3. Mar 23, 2015 #2

    vanhees71

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    Don't be too quick :-). First write down all equations, and not only the diagonal ones.
     
  4. Mar 23, 2015 #3

    Orodruin

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    You have so far only used the conditions on the diagonal. You need to also fulfill the off-diagonal elements equal to zero.

    Suggestion: Call ##|u_{22}| =\cos\varphi## instead of the 12 element. It makes no difference but will simplify your math slightly.
     
  5. Mar 24, 2015 #4
    Ok so for the orthogonal elements:
    [itex]u_{11}^*u_{12}+u_{21}^*u_{22}=0[/itex]
    [itex]u_{12}^*u_{11}+u_{22}^*u_{21}=0[/itex]
     
  6. Mar 24, 2015 #5

    Orodruin

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    So what do you get when you insert into these the things you have already found? (Note that they are both equivalent by simple complex conjugation.)
     
  7. Mar 24, 2015 #6
    [itex]\cos(\theta)\sin(\varphi)+\sin(\theta)\cos(\varphi)=0[/itex]
    using your suggestion of [itex]u_{22}=\cos(\varphi)[/itex]
     
  8. Mar 24, 2015 #7

    Orodruin

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    Well, almost, you forgot the phases. Once you have taken them into account, I suggest using some trigonometric identities.
     
  9. Mar 24, 2015 #8
    After using the product formula and some simplification I get

    [itex]\frac{1}{2}\big{[}\sin(\theta+\varphi)(e^{i(\alpha_1+\alpha_2)}+e^{i(\alpha_3+\alpha_4)})+\sin(\theta-\varphi)(e^{i(\alpha_3+\alpha_4)}-e^{i(\alpha_1+\alpha_2)})\big{]}=0[/itex]
     
  10. Mar 24, 2015 #9

    Orodruin

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    A hint: Start by identifying the phase of both terms to be equal (note that this puts a condition on the phases! and that it is also a matter of choice of phases whether you put the phases equal or equal modulo π).
     
  11. Mar 24, 2015 #10

    Strilanc

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    In addition to adding the must-equal-zero-off-diagonal constraints, I recommend breaking ##|\mu_x|^2## into ##(a_x + b_x i)(a_x - b_x i)##.

    Alternatively, you could take the approach of "Well, each column and row must have their 2-norm equal to 1. So if topleft has squared amplitude |a|=x then bottom-left and top-right must be |b|=|c|=1-x...".
     
  12. Mar 24, 2015 #11
    Heres everything I've written up now. Does this look ok?

    [itex] U=\left(\!\begin{array}{cc}u_{11} & u_{12}\\u_{21} & u_{22}\end{array}\!\right)[/itex]

    By unitarity we have

    [itex]U^{\dagger}U=\left(\!\begin{array}{cc}u_{11}^* & u_{21}^*\\u_{12}^* & u_{22}^*\end{array}\!\right)
    \left(\!\begin{array}{cc}u_{11} & u_{12}\\u_{21} & u_{22}\end{array}\!\right)=\left(\!\begin{array}{cc}1 & 0\\0 & 1\end{array}\!\right)[/itex]

    [tex]\Longrightarrow |u_{11}|^2+|u_{21}|^2=1 \:\:,\:\: |u_{12}|^2 + |u_{22}|^2=1[/tex]

    [tex]\Longrightarrow |u_{11}|=\cos(\theta) \:\:,\:\: |u_{21}|=\sin(\theta) \:\:,\:\: |u_{12}|=\sin(\varphi) \:\:,\:\: |u_{22}|=\cos(\varphi)
    [/tex]
    [tex]UU^{\dagger}=
    \left(\!\begin{array}{cc}u_{11} & u_{12}\\u_{21} & u_{22}\end{array}\!\right)\left(\!\begin{array}{cc}u_{11}^* & u_{21}^*\\u_{12}^* & u_{22}^*\end{array}\!\right)=\left(\!\begin{array}{cc}1 & 0\\0 & 1\end{array}\!\right)[/tex]
    [tex]\Longrightarrow |u_{11}|^2+|u_{12}|^2=1 \:\:,\:\: |u_{21}|^2+|u_{22}|^2=1.[/tex]
    From these relations we get
    [tex]\cos^2(\theta)+\sin^2(\varphi)=1 \:\:,\:\: \sin^2(\theta)+\cos^2(\varphi)=1[/tex]

    [tex]\Longrightarrow \cos^2(\theta)-\sin^2(\theta)=\cos^2(\varphi)-\sin^2(\varphi)=0[/tex]
    [tex]\Longleftrightarrow \frac{1}{2}\big{[}(1+\cos(2\theta))-(1-\cos(2\theta))\big{]}=\frac{1}{2}\big{[}(1+\cos(2\varphi))-(1-\cos(2\varphi))\big{]}
    [/tex]
    [tex]\Longleftrightarrow \cos(2\theta)=\cos(2\varphi)\Longleftrightarrow \theta=\varphi\pm k\pi \:\:\:, k\in \mathbb{Z}[/tex]
    [tex]|u_{11}|=\cos(\theta)\Longrightarrow u_{11}=\pm e^{i\alpha_1}\cos(\theta).[/tex]
    Likewise, [tex]u_{12}=\pm e^{i\alpha_2}\sin(\theta) \:\:,\:\: u_{21}=\pm e^{i\alpha_3}\sin(\theta) \:\:,\:\: u_{22}=\pm e^{i\alpha_4}\cos(\theta)[/tex]
    So we may now write the general form for a unitary operation acting on 1 qubit as
    [tex]\left(\!\begin{array}{cc}e^{i\alpha_1}\cos(\theta) & -e^{i\alpha_2}\sin(\theta)\\e^{i\alpha_3}\sin(\theta) & e^{i\alpha_4}\cos(\theta)\end{array}\!\right)[/tex]
     
  13. Mar 24, 2015 #12

    Strilanc

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    I think you mixed up ##\theta## and ##\varphi## halfway through.
     
  14. Mar 24, 2015 #13

    Orodruin

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    You are missing one condition on the phases. The way you have written it down, the matrix is not necessarily unitary. A general unitary 2x2 matrix has 4 real parameters.
     
  15. Mar 24, 2015 #14
    So do I need to eliminate one of the alphas?
     
  16. Mar 24, 2015 #15

    Orodruin

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    Yes, but you cannot just do it randomly, you must do it in a way that ensures that the result is unitary.
     
  17. Mar 24, 2015 #16
    Soo check the unitarity condition again with my new U to get another set of equations?
     
  18. Mar 24, 2015 #17

    Orodruin

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    That would be one way of doing it yes, but you can already get the condition from the contributions to the off diagonal elements in the product having the same phase.

    Edit: Also, you should really not need to use the ##UU^\dagger## condition. Everything you need is already present from the ##U^\dagger U = 1## condition.
     
  19. Mar 25, 2015 #18
    Ok so using [itex]UU^{\dagger}=I[/itex] I get for the orthogonal parts

    [itex]e^{i(\alpha_1-\alpha_3)}\cos(\theta)\sin(\theta)-e^{i(\alpha_2-\alpha_4)}\sin(\theta)\cos(\theta)=0[/itex]

    Which gives

    [itex]\alpha_1-\alpha_2-\alpha_3+\alpha_4=0[/itex]

    doing this for the other orthogonal row and collumn vector gives the same condition
     
  20. Mar 25, 2015 #19

    Orodruin

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    Yes, so this is the final piece of the puzzle you need to reduce the number of real parameters of the general 2x2 unitary matrix to four.
     
  21. Mar 25, 2015 #20
    It seems to me that this should be the easy part but im not really sure what to do here
     
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