# Canonical ensemble: correlation function for quadratic potential energy

In this paper I can't understand the following

A given system has the potential energy $U(x_1,x_2,x_3)=k_1 x_1^2 + k_2 (x_2-x_1)^2 + k_3 x_3^2$.
Since the energy is quadratic, the correlation matrix is given in terms of the Hamiltonian matrix $H$ by $<x_i x_j> = k_BT (H^{−1})_{ij} \quad$ with $\quad H_{ij}=\frac{\partial^2 U}{\partial x_i \partial x_j} \quad$

• shouldn't you also use kinetic energy in the partition function? – tonydo Jun 20 '16 at 10:55
• @tonydo That would only give you a multiplicative constant – valerio Jun 20 '16 at 11:01
• I should write the sum as an integral instead maybe, I'll try a bit later – David Jun 20 '16 at 11:14
• Ok so I used integrals instead as the variables are continuous (edit). I get something that looks like the expected result but I still miss something. – David Jun 20 '16 at 15:38
• I have something which can relate your edit with the $(H^{-1})_{ij}$ matrix. Why did you erase it? If it was wrong, post a correct version and that maybe help with my calculus (which may be wrong, of course ^^") – VictorSeven Jun 20 '16 at 16:35

A system is in a heat bath of temperature T so we work with the canonical ensemble. We consider $N$ degrees of freedom $x_1, x_2, ..., x_N$ and $x$ is the vector $(x_1~ x_2 ~ ... ~ x_N)^T$. The potential energy is quadratic so it can be expressed as a function of its second derivatives:

$U=\sum_{i,j} x_i ~H_{i,j} ~x_j = x^T H x ~~$ with $H_{i,j}=\frac{\partial^2 U}{\partial x_i \partial x_j}$.

$H$ is assumed invertible. The partition function $Z$ in the canonical ensemble is:

$Z= \int (d^Nx) ~ e^{ -\frac{1}{2} \beta x^T H x} ~ = ~ \sqrt{\frac{(2 \pi)^N}{\det(H)}}~~$ with $~~\beta=\frac{1}{k_BT}$,

where we used the multivariate Gaussian formula.

$<x_i x_j> = \frac{1}{Z} \int (d^Nx) ~ x_i x_j e^{ -\frac{1}{2} \beta x^T H x}$

$= \frac{1}{Z} (-\frac{2}{\beta})\frac{\partial}{\partial H_{i,j}}\int (d^Nx) ~ e^{ -\frac{1}{2} \beta x^T H x}$

$= -2k_BT \frac{1}{Z} \frac{\partial Z}{\partial H_{i,j}}$

$= -2k_BT \frac{\partial \ln(Z)}{\partial H_{i,j}}$

We plug-in the expression of $Z$ that we found above:

$<x_i x_j>= k_BT \frac{\partial }{\partial H_{i,j}}(\ln(\det(H))$

$H$ is invertible so we use Jacobi's formula:

$\frac{\partial }{\partial H_{i,j}}(\ln(\det(H)) = Tr(H^{-1} \frac{\partial H}{\partial H_{i,j}})$

$\frac{\partial H}{\partial H_{i,j}}$ is a matrix for which the element $\{i,j\}$ is $1$ and all other elements are $0$. In other words:

$E\equiv \frac{\partial H}{\partial H_{i,j}}$ with $E_{k,l}=\delta_{k,i} \delta_{l,j}.$

$Tr(H^{-1} \frac{\partial H}{\partial H_{i,j}}) = Tr(H^{-1}E)$

$(H^{-1}E)_{k,l}=\sum_m H^{-1}_{k,m}E_{m,l}=\sum_m H^{-1}_{k,m}\delta_{m,i}\delta_{j,l} = H^{-1}_{k,i} \delta_{j,l}$

$Tr(H^{-1} E)=\sum_n (H^{-1}E)_{n,n}=\sum_n H^{-1}_{n,i} \delta_{j,n} = H^{-1}_{j,i} = H^{-1}_{i,j}$

and

$<x_i x_j> = k_B T H^{-1}_{i,j}$