0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You need to proofread what you wrote. $\endgroup$ – Cosmas Zachos Sep 3 at 19:48
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.