# Independent conditions of unitary matrix, $U U^{\dagger}=1$ and $U^{\dagger}U = 1$

I am trying to show that, for a simple 2x2 complex matrix $$\begin{equation} U = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \end{equation}$$ there will be only 4 real constraints for $$U$$ to be unitary. I will denote the conjugate transpose by $$\dagger$$, for example the complex transpose of a matrix $$A$$ is $$A^{\dagger}$$. I will also use $$*$$ to denote complex conjugation of a complex number, for example $$a^{*}$$ is the complex conjugate of complex number $$a$$. I first carry out the calculation $$\begin{equation} UU^{\dagger}=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a^{*} & c^{*} \\ b^{*} & d^{*} \end{bmatrix} = \begin{bmatrix} |a|^{2}+|b|^{2} & a c^{*} + b d^{*} \\ a^{*}c+b^{*}d & |c|^{2} + |d|^{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{equation}$$ which leads to the constraints \begin{align} & |a|^{2}+|b|^{2} = 1 \\ & a c^{*} + b d^{*} = 0 \\ & a^{*}c+b^{*}d = 0 \\ & |c|^{2} + |d|^{2} = 1. \end{align} The second and third are same, and each of them gives 2 real constraints. The first and the fourth both give 1 real constraint. So in total the above four equations give 4 real constraints.

I know that $$UU^{\dagger} = U^{\dagger}U$$ and so $$U^{\dagger}U=1$$ should give no more constraints. But still I calculate it as $$\begin{equation} U^{\dagger}U= \begin{bmatrix} a^{*} & c^{*} \\ b^{*} & d^{*} \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} |a|^{2}+|c|^{2} & a^{*} b + c^{*} d \\ ab^{*}+c d^{*} & |b|^{2} + |d|^{2} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{equation}$$ which leads to the constraints \begin{align} & |a|^{2}+|c|^{2} = 1 \\ & a^{*} b + c^{*} d = 0 \\ & ab^{*}+c d^{*} = 0 \\ &|b|^{2} + |d|^{2} = 1. \end{align} Now it seems to me that it is not obvious why the constraints obtained from $$U^{\dagger}U = 1$$ are independent of those obtained from $$U U^{\dagger}=1$$. However, we know that $$U U^{\dagger}=1$$ should imply $$U^{\dagger}U = 1$$. Could anyone give me some suggestions? I think I just miss some simple points, but currently haven't figured out the key points. Thanks a lot!

• Does † mean transpose? And what does a∗ refer to? Please clarify your notations. – Soumyadwip Chanda Sep 12 at 6:13
• @SoumyadwipChanda $\dagger$ is common notation for the conjugate transponse and ${a}^*$ is common notation for the conjugate of the complex number $a$. – Gae. S. Sep 12 at 6:16
• Yeah, these are physicist notations – LL 3.14 Sep 12 at 9:40
• @SoumyadwipChanda sorry for the notation issue, by $\dagger$ I mean conjugate transpose and $*$ means complex conjugation of a complex number. I have edited the post and so hopefully this will clarify. – ocf001497 Sep 12 at 15:14
• @Gae.S. thanks for helping me clarify – ocf001497 Sep 12 at 15:18

Since $$ac^\ast=-bd^\ast$$, $$|a|^2|c|^2=|b|^2|d|^2$$, i.e. $$|a|^2-|a|^2|d|^2=|d|^2-|a|^2|d|^2$$, whence $$|a|=|d|$$. But $$|b|^2-|c|^2=|d|^2-|a|^2=0$$, so$$ab^\ast=-bd^\ast\cdot(b/c)^\ast=-|b|^2(d/c)^\ast=-|c|^2(d/c)^\ast=-cd^\ast.$$