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I have been looking to find a derivation showing that any 2D special unitary operator can be written as: $$ \hat{U}= \begin{bmatrix} \cos(\theta)& e^{i\gamma}\sin(\theta)\\ -e^{-i\gamma}\sin(\theta)& \cos(\theta) \end{bmatrix}. $$ I have not had any luck thus far, but my reasoning until now went like this:
I define a unitary $U$:

$$ U = \left[\begin{matrix}U_{1} & U_{2} \\U_{3} & U_{4} \end{matrix}\right] \label{UUunit} $$

Since we defined the above matrix as unitary, one can say $UU^\dagger=UU^{-1}\Rightarrow U^\dagger =U^{-1}$:

$$ U^\dagger = \left[\begin{matrix}U_{1}^* & U_{3}^* \\U_{2}^* & U_{4}^*\end{matrix}\right] = U^{-1} = \frac{1}{det(U)}\left[\begin{matrix}U_{4} & -U_{2} \\-U_{3} & U_{1}\end{matrix}\right] $$

We restrict $U$ with $det(U)=U_1U_4 - U_2U_3 =1$.
The above equation then gives the relation $U_1 = U_4^*$ and $U_2 = -U_3^*$, while from $UU^\dagger = 1$ we obtain three different identities

$$ |U_{1}|^{2} + |U_{2}|^{2} = 1 \\ |U_{3}|^{2} + |U_{4}|^{2}= 1 \\ U_{1} U_{3}^* + U_{2} U_{4}^*= 0 \\ $$ Tuplets of numbers $\{U_i,U_j\}$ that satisfy the above relations can be parametrized as $U_i = e^{i\mu_i}\cos(\theta)$ and $U_j = e^{i \gamma_j}\sin(\theta)$ (for $\gamma,\theta, \mu \in \mathbb{R}$).\ Applying all the constraints found until now, one gets: $$ U_1 = e^{i\mu} \cos(\theta)\\ U_4 = e^{-i\mu} \cos(\theta)\\ U_2 = e^{i\gamma} \sin(\theta)\\ U_3 = -e^{-i\gamma} \sin(\theta)\\ $$

My problem with this is the presence of 3 angles $\mu, \gamma, \theta$; while the general description requires only two. What am I missing?

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    $\begingroup$ Have you considered that the overall operator is only specified up to a global phase? $\endgroup$
    – Philip
    Apr 25, 2021 at 10:25
  • $\begingroup$ Do you mean there is some way I could extract the third angle $\mu$ by making it become some global shift? $\endgroup$
    – Oti
    Apr 26, 2021 at 8:50

2 Answers 2

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Surely you recognize $\operatorname{diag}[e^{i\phi},e^{-i\phi}]$ is unitary unimodular but not of the $\hat U$ form.

Your derivation is fine, but, to avoid confusion, call the U you found V, and your independent parameter γ by γ-μ, instead. Thus, you found the unitary, unimodular
$$ V= \begin{bmatrix} e^{i\mu} \cos(\theta)& e^{i(\gamma-\mu)}\sin(\theta)\\ -e^{-i(\gamma-\mu)}\sin(\theta)& e^{-i\mu} \cos(\theta) \end{bmatrix}= M\hat{U}M, $$ for $$ M= \begin{bmatrix} e^{i\mu/2} & 0\\ 0& e^{-i\mu/2} \end{bmatrix}, $$ unitary and unimodular as well. In your group SU(2), which has 3 parameters, μ specifies the orientation of the x and y axes you choose on that plane.

It should be instructive to read up on sundry parameterizations of unitary matrices (set φ=0).


Note addressing your comment, arguably:

From your language, it appears you are thinking of some type of Bloch matrix "effectively" invariant under a rotation preserving the z-axis, $$ M^\dagger \hat U M =\begin{bmatrix} \cos(\theta)& e^{i(\gamma-\mu)}\sin(\theta)\\ -e^{-i(\gamma-\mu)}\sin(\theta)& \cos(\theta) \end{bmatrix} ~, $$ where I am using your original, not the above, shifted parameter! That is, the $\hat U$ I'm handling above is this very matrix, so your original $\hat U= V M^{-2}$. If you are using your $\hat U$ to dot on a lower component of a two-spinor to represent a most general 2-spinor (qubit), then you are allowed this rotation at the very start, and μ is fictitious/redundant, given the arbitrariness of the absolute phase.

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  • $\begingroup$ Hi, I am not sure I understand, but from your V how would I go around finding the unitary in the same form as the one presented in the original post? $\endgroup$
    – Oti
    Apr 26, 2021 at 8:49
  • $\begingroup$ $\hat U$, M, and V are all unitary, related as above. You can invert M and take it to the left, around V. But the original unitary matrix is not the most general SU(2) group element. You found it. $\endgroup$ Apr 26, 2021 at 10:31
  • $\begingroup$ I am sure that is the unitary I am supposed to find. I am using it to time evolve a density operator: does this information help? $\endgroup$
    – Oti
    Apr 27, 2021 at 12:07
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As Cosmas Zachos has pointed out, the general form for a 2D special unitary operator you give in your post is in fact not the most general matrix. You won't have any luck proving this, because it is not true. The matrix you did find is the most general.

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