Write all the unitary vectors of $\mathbb{C}^2$ in terms of real parameters

Question

Write all the unitary vectors of $\mathbb{C}^2$ in terms of real parameters. After so, write all the orthonormal basis of $\mathbb{C}^2$

I thought that for the first part, a vector $\vec{u}=(x_1,x_2)$ is unitary if $|x_1|^2+|x_2|^2=1$, but I don't know how the book goes from here to directly put that the solutions are $$x_1=e^{i\alpha}\cos\theta$$ $$x_2=e^{i\beta}\sin \theta$$ Also clueless for the second part, I'm guessing I can do $\vec{u}\cdot\vec{x}=0$ for a generic vector $\vec{x}=(x_3,x_4)$

Answer 1

If $$|x_1|^2+|x_2|^2=1,$$ you have a solution in terms of real numbers of the form $$x_1=\cos\theta\\x_2=\sin\theta,$$ where $\theta \in\mathbb R$. But $x_1,x_2\in\mathbb C$, so you take the magnitudes ($|\sin\theta|$ and $|\cos\theta|$) and multiply by unit vectors in $\mathbb C$. Therefore $$x_1=e^{i\alpha'}|\cos\theta|\\x_2=e^{i\beta'}|\sin\theta|$$ Here $\alpha'$ and $\beta'$ are any random real numbers. Note that $e^{i\pi}=-1$, so if $\cos\theta<0$, then $$|\cos\theta|=e^{i\pi}\cos\theta$$ This $\pi$ factor can be then absorbed into $\alpha=\alpha'+\pi$. A similar approach is valid for $\beta$

EDIT: Based on the comments, I think I should clarify a couple of points:

1. Both $x_1$ and $x_2$ are complex numbers. Each complex number is equivalent to a pair of real numbers. So a vector in $\mathbb C^2$ has four components.
2. Let $x_1=r_1e^{i\alpha}$ and $x_2=r_2e^{i\beta}$. These are complex number written in the polar form. $e^{i\alpha}$ is a unit vector in $\mathbb C$. So you multiply a unit vector by the magnitude:$$x_1=|x_1|e^{i\alpha}$$
3. Then $$|\vec u|=(r_1\cos\alpha)^2+(r_1\sin\alpha)^2+(r_2\cos\beta)^2+(r_2\sin\beta)^2=r_1^2+r_2^2=1$$ If you look at the last equation, both $r_1^2$ and $r_2^2$ are between $0$ and $1$. So if I choose $r_1=|\sin\theta|$ then I can write $r_2=|\cos\theta|$. With these, you get the form in the first question.

For the second part, this question might provide some insight. It is a lot of calculation anyway. $$\vec u\cdot\vec v=(\cos\theta e^{i\alpha},\sin\theta e^{i\beta})\cdot(\cos\gamma e^{i\mu},\sin\gamma e^{i\nu})=\cos\theta\cos\gamma e^{i(\alpha-\mu)}+\sin\theta\sin\gamma e^{i(\beta-\nu)}$$ It's easy to see that $\vec u\cdot\vec u=1$. Now let's look for an orthogonal base. The number of vectors in such a base is $2$ (two complex numbers = 4 real numbers). Since $\vec u\cdot\vec v=0$, the six parameters and two constraints (real and imaginary parts are zero) yield 4 independent variables. There are several cases.

• $\cos\theta=0$. This yields $\sin\theta =1$ (the $-1$ case is just a change of phase in the exponent). That means $\sin\gamma=0$. In this case, the basis is $((0,e^{i\beta}),(e^{i\mu},0))$. You can choose any $\beta$ and $\nu$
• $\cos\gamma=0$. Similar case as before, with the same basis $((e^{i\alpha},0),(0,e^{i\nu}))$, in different order.
• $\cos\theta\cos\gamma\ne 0$. Then $$\frac{\sin\theta\sin\gamma}{\cos\theta\cos\gamma}e^{i(\beta-\nu-\alpha+\mu)}=-1$$. This means two things. One, $\tan\theta\tan\gamma=\pm1$, and two $\beta-\nu-\alpha+\mu=n\pi$. Odd $n$ values correspond to $\tan\theta\tan\gamma=1$ and even ones to $\tan\theta\tan\gamma=-1$.

Answer 2

Let $x_1 = x_{1r}+ix_{1i}$ and $x_2 = x_{2r}+ix_{2i}$ $$|x_{1r}|^2 + |x_{1i}|^2+|x_{2r}|^2+|x_{2i}|^2 = 1$$

Let without loss of generality, $$|x_{1r}|^2 + |x_{1i}|^2 = cos^2\alpha_1$$ Hence $$|x_{2r}|^2+|x_{2i}|^2 = sin^2\alpha_1$$

Hence, $$x_{1r} = cos\alpha_1 \times cos\alpha_2$$ $$x_{1i} = cos\alpha_1 \times sin\alpha_2$$ $$x_{2r} = sin\alpha_1 \times cos\alpha_3$$ $$x_{2i} = sin\alpha_1 \times sin\alpha_3$$

Hence any unit vector is of the the above form for some $\alpha_1,\alpha_2,\alpha_3$.

Orthonormal basis is same as $2 \times 2 $ orthonormal matrix: $$A^{-1} = A^H$$ $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

$$A^{-1} = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$

$$A^H = = \begin{bmatrix} a^* & c^* \\ b^* & d^* \end{bmatrix}$$

Now setting $$A^{-1} = A^H$$, we have,

$$a = d^*, b = -c^*$$

Hence columns of the below matrix with $$|a|^2+|b|^2 = 1$$(general form of $a,b$ from first part of this answer) is your set of all orthonormal basis.

$$A = \begin{bmatrix} a & b \\ -b^* & a^* \end{bmatrix}$$

Answer 3

$\mathbb{C}^2$ is isomorphic to tessarines via isomorphism:

$\left(x_1,x_2\right)\to \frac{x_1+x_2}{2}+j\frac{x_2-x_1}{2} $

Representing $x_1$ and $x_2$ as complex numbers we get:

$\left(a_1+i b_1,a_2+i b_2\right)\to\frac{1}{2} j \left(a_2+i b_2-\left(a_1+i b_1\right)\right)+\frac{1}{2} \left(a_1+a_2+i b_1+i b_2\right)$

Expanding,

$\left(a_1+i b_1,a_2+i b_2\right)\to\frac{a_1+a_2}{2}+\frac{b_1+b_2}{2}i+\frac{a_2-a_1}{2}j+\frac{b_2-b_1}{2} ij$

To get the modulus (magnitude), we have to take the squares of all components of this vector with positive sign except the coefficient of hyperbolic unity $j$:

$\left|\left(a_1+i b_1,a_2+i b_2\right)\right|=\sqrt{\left(\frac{a_1+a_2}{2}\right)^2+\left(\frac{b_1+b_2}{2}\right)^2-\left(\frac{a_2-a_1}{2}\right)^2+\left(\frac{b_2-b_1}{2}\right)^2}$

So, the unit vectors are those for which $\left(\frac{a_1+a_2}{2}\right)^2+\left(\frac{b_1+b_2}{2}\right)^2-\left(\frac{a_2-a_1}{2}\right)^2+\left(\frac{b_2-b_1}{2}\right)^2=1$