# General Special Unitary matrix in 2D

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?

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

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).

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.
• $\hat U$, 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. Apr 26, 2021 at 10:31