How to find the square root of a matrix

Question

Could anyone help me to understand how the author was able to take the square root of the following matrix:

$$
\rho  = {1 \over Z}\left( {\matrix{
   {e^{ - 2\beta B} } & 0 & 0 & 0  \cr 
   0 & {\cosh 2\beta \delta } & { - e^{i\theta } \sinh 2\beta \delta } & 0  \cr 
   0 & { - e^{ - i\theta } \sinh 2\beta \delta } & {\cosh 2\beta \delta } & 0  \cr 
   0 & 0 & 0 & {e^{2\beta B} }  \cr 

 } } \right)
$$

to be in the following form:

$$
\sqrt \rho   = {1 \over {\sqrt Z }}\left( {\matrix{
   {e^{ - \beta B} } & 0 & 0 & 0  \cr 
   0 & {\cosh \beta \delta } & { - e^{i\theta } \sinh \beta \delta } & 0  \cr 
   0 & { - e^{ - i\theta } \sinh \beta \delta } & {\cosh \beta \delta } & 0  \cr 
   0 & 0 & 0 & {e^{\beta B} }  \cr 

 } } \right)
$$

Answer 1

You already got the simple math answer from @Doug M 's comment. But, beyond that, you are meant to recognize the familiar physics structure of it.

That is, the evident factorization to two commuting matrices, one of them diagonal,
$$ \rho = {1 \over Z}\left( {\matrix{ {e^{ - 2\beta B} } & 0 & 0 & 0 \cr 0 & {\cosh 2\beta \delta } & { - e^{i\theta } \sinh 2\beta \delta } & 0 \cr 0 & { - e^{ - i\theta } \sinh 2\beta \delta } & {\cosh 2\beta \delta } & 0 \cr 0 & 0 & 0 & {e^{2\beta B} } } } \right)= {1 \over Z}\left( {\matrix{ {e^{ - 2\beta B} } & 0 & 0 & 0 \cr 0 & 1 & 0 & 0 \cr 0 & 0& 1 & 0 \cr 0 & 0 & 0 & {e^{2\beta B} } } } \right) \left( {\matrix{ 1 & 0 & 0 & 0 \cr 0 & {\cosh 2\beta \delta } & { - e^{i\theta } \sinh 2\beta \delta } & 0 \cr 0 & { - e^{ - i\theta } \sinh 2\beta \delta } & {\cosh 2\beta \delta } & 0 \cr 0 & 0 & 0 & 1 } } \right). $$ The first diagonal factor is trivial to take the square root of, diag$(e^{-\beta B}, 1,1,e^{\beta B})/\sqrt{Z}$.

So all you need to do is recognize the middle (unimodular) 2×2 block as a phased- equivalent hyperbolic rotation, $$ M= \begin{pmatrix} {\cosh 2\beta \delta } & { - e^{i\theta } \sinh 2\beta \delta } \\ { - e^{ - i\theta } \sinh 2\beta \delta } & {\cosh 2\beta \delta } \end{pmatrix}= \begin{pmatrix} 1&0\\ 0 & -e^{-i\theta}\end{pmatrix} \begin{pmatrix} {\cosh 2\beta \delta } & \sinh 2\beta \delta \\ \sinh 2\beta \delta & {\cosh 2\beta \delta } \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & - e^{ i\theta }\end{pmatrix}. $$

But you know the hyperbolic rotation matrix angles (rapidity) add upon composition, as you may verify here through the standard double-angle hyperbolic function identities, so the middle hyperbolic rotation resolves to just $$ \begin{pmatrix} {\cosh \beta \delta } & \sinh \beta \delta \\ \sinh \beta \delta & {\cosh \beta \delta } \end{pmatrix} ^2, $$ and, of course, you may insert the two phases in-between these two factor matrices to preserve the equivalence in the product.

This is a standard calculation you are expected to be repeating again and again, by now in your head, as you go on, ever mindful of the interlocking pieces...