Archives for April 1st, 2017 | Open System

In the last post, SU(1,1) parametrization, we have concluded that the group SU(1,1) has the shape of a doughnut. Its elements can be uniquely parametrized by an angle $0\leq\theta<2\pi$ and a complex number $z$ inside of the unit disk: $|z|<1.$

Thus SU(1,1) can be parametrized using the Cartesian product of the circle and the (interior of) the disk. But parametrization is one thing, and the actual construction is a different thing. Today we will see how every SU(1,1) matrix can be uniquely decomposed into a product of disk parametrized positive part, and circle parametrized unitary part.

We already know that every SU(1,1) matrix $A$ is of the form

(1) $\begin{equation*}A=\begin{bmatrix}\lambda&\mu\\ \bar{\mu}&\bar{\lambda}\end{bmatrix},\end{equation*}$

with

(2) $\begin{equation*}|\lambda|^2-|\mu|^2=1.\end{equation*}$

We also know (see Eqs. (11,14,15) from SU(1,1) parametrization) that it determines $0\leq\theta<2\pi,$ and $|z|<1$ so that

(3) $\begin{equation*}|\lambda|=\frac{1}{\sqrt{1-|z|^2}},\end{equation*}$

(4) $\begin{equation*}\lambda=|\lambda|e^{i\theta},\end{equation*}$

(5) $\begin{equation*}\mu=\bar{\lambda}\bar{z}.\end{equation*}$

Thus, writing it explicitly, $A$ is of the form:

(6) $\begin{equation*} A=\begin{bmatrix}\frac{e^{i\theta}}{\sqrt{1-|z|^2}}&\frac{\bar{z}e^{-i\theta}}{\sqrt{1-|z|^2}}\\ \frac{ze^{i\theta}}{\sqrt{1-|z|^2}}&\frac{e^{-i\theta}}{\sqrt{1-|z|^2}}\end{bmatrix}.\end{equation*}$

But then, as you can easily check, $A$ can be written as a product (so called polar decomposition)

(7) $\begin{equation*}A=P(z)U(\theta),\end{equation*}$

where

(8) $\begin{equation*} P(z)=\begin{bmatrix}\frac{1}{\sqrt{1-|z|^2}}&\frac{\bar{z}}{\sqrt{1-|z|^2}}\\ \frac{z}{\sqrt{1-|z|^2}}&\frac{1}{\sqrt{1-|z|^2}}\end{bmatrix},\end{equation*}$

(9) $\begin{equation*}U(\theta)=\begin{bmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{bmatrix},\end{equation*}$

with $0\leq\theta<2\pi,$ and $|z|<1.$

The matrix $P(z)$ is evidently Hermitian: $P(z)=P(z)^*.$ It is also positive. Hermitian matrix is positive when all its eigenvalues are positive. In two dimensions there are two eigenvalues. Their product is the determinant, and so it is equal to 1. Thus either both eigenvalues are positive or both are negative. But the trace of $P(z)$ , which is the sum of eigenvalues, is positive $\mathrm{Tr}(P(z))=2/\sqrt{1-|z|^2}$ is positive, therefore both eigenvalues must be positive, and so $P(z)$ is positive.

Exercise: calculate the two eigenvalues of $P(z)$

In fact every positive matrix from SU(1,1) is of the form $P(z)$ for some $z.$ Since suppose $A$ is in SU(1,1) and is positive. We write it in the form as in Eq. (6). Positive matrix must be Hermitian, therefore $e^{i\theta}$ must be real. This happens only for $\theta=0$ or $\theta=\pi.$ But $e^{i\pi}=-1,$ so for $\theta=\pi$ we would have $\mathrm{Tr}(A)=-2/\sqrt{1-|z|^2}<0,$ which is impossible for a positive matrix. Therefore $\theta=0$ and $A=P(z).$

The matrix $U=U(\theta)$ is evidently unitary, $UU^*=I.$ In fact, every unitary matrix from SU(1,1) must be of the type $U(\theta).$ Why? Suppose $A$ in SU(1,1) is unitary. Write it in the form $A=P(z)U(\theta).$ Suppose $AA^*=I.$ Then $P(z)U(z)U(z)^*P(z)^*.$ But $U(z)$ is unitary and $P(z)$ is Hermitian, so $P(z)^2=I.$ Thus squares of the eigenvalues of $P(z)$ are both equal to $1,$ and since $P(z)$ is positive, both must be equal to 1. Therefore $P(z)=I,$ and so $A=U(\theta).$