# Matrix Exponential [closed]

i'm working on a paper about quantum entropy and i'm stuck on an analytical step about computing a matrix exponential.

I have the following Hamiltonian $$H=\frac{1}{2}\sum_ip_i^2+\frac{1}{2}\sum_{i,j}x_iK_{ij}x_j$$

so that i get the following wave function $$\psi_0(x_1,x_2,...,x_N)=\frac{1}{2}(\det\Omega)^{1/4}\exp(-x\Omega x/2)$$ where $$\Omega$$ is the square root of K: If $$K=U^TK_DU$$ where $$K_D$$ is diagonal and U is orthogonal, then $$\Omega=U^T\sqrt{K_D}U$$.

To find the density matrix for the N-n oscillators, we trace over the first N ones, so we get : $$\rho(x_{n+1},...,x_N,x'_{n+1},...,x'_N)=\int\prod_i^ndx_i\psi_0(x_1,...,x_n,x_{n+1},...,x_N)\psi_0^*(x_1,...,x_n,x'_{n+1},...,x'_N)$$

The paper solves this integral writing $$\Omega=\left(\begin{matrix} A&B\\ B^T&C\\ \end{matrix}\right)$$ where $$A=n\times n$$ and $$C=(N-n) \times (N-n)$$.

Then on the paper the $$\rho(x_{n+1},...,x_N,x'_{n+1},...,x'_N)$$ is written like $$\rho(x,x')\sim \exp\left(-\frac{x\gamma x+x'\gamma x'}{2} +x\beta x'\right)$$ where $$\gamma=C-\beta$$, $$\beta=\frac{1}{2}B^TA^{-1}B$$ and both x and x' have N-n components.

$$\textbf{How do I get the last result?}$$

I thought that the right way to the solution is to diagonalize $$\Omega$$ but I don't know how because A and C have different ranks.

• You need to proofread what you wrote. – Cosmas Zachos Sep 3 at 19:48

Your formula suggests the use of a block-matrix Gaussian decomposition such as $$\left[\matrix{ A & B\cr C & D}\right]=\left[\matrix{I &BD^{-1} \cr 0 &I }\right] \left[\matrix{A-BD^{-1} C&0\cr 0& D}\right]\left[\matrix{ I &0 \cr D^{-1} C &I}\right],$$ or the one with a lower triangular matrix on the leftmost factor so that $$A^{-1}$$ replaces the $$D^{-1}$$. In these products the $$B$$ and $$C$$ matrices do not have to be square because all the matrix sizes match up.