Path integral: uncoupling via staging variables

Question

I am studying the transformation to staging variables for the calculation of path integrals in quantum mechanics, following the scheme presented in the book of Mark Tuckerman "Statistical Mechanics: Theory and Molecular Simulations".

Let us consider a $N$-body quantum system, whose configurations are labelled by $\mathbf{q}\in \mathbb R^N$. Let us define with $\mathbf{q}_k\in \mathbb R^N$, $k=1\cdots P$, a number of $P$ 'replicas' of the system. In the path integral formulation of quantum mechanics, it is known that in the canonical partition function these terms interact via an harmonic bond with shape \begin{equation*} \sum_{k=1}^P(\mathbf{q}_{k+1}-\mathbf{q}_k)^2 , \end{equation*} with boundary conditions $\mathbf{q}_1=\mathbf{q}_{P+1}$. In order to uncouple this quadratic form, it is often convenient to introduce a new set of configurations $\mathbf{u}_k$, denoted 'staging variables' such that \begin{align*} \mathbf{q}_1 &=\mathbf{u}_1 \\ \mathbf{q}_k &= \mathbf{u}_k +\frac{k-1}{k}\mathbf{q}_{k+1}+\frac {1}{k}\mathbf{u}_1 \hspace{5mm} k = 2\cdots P , \end{align*} and these coordinates should allow to uncouple the harmonic bonds to \begin{equation*} \sum_{k=1}^P(\mathbf{q}_{k+1}-\mathbf{q}_k)^2 =\sum_{k=2}^P\frac{k}{k-1}\mathbf{u}_k^2= \sum_{k=2}^P\sum_{j=1}^N\frac{k}{k-1}u_{k,j}^2 , \end{equation*} where I denoted with $u_{k,j}$ the $j^\text{th}$ particle of the $k^\text{th}$ system.

I am trying to prove the last identity but a simple change of variables didn't work out. In the text this is only discussed in a different framework and in the 1d case. Has someone a simple proof? I was also wondering if this could be proven by combining the matrix of the quadratic form with the change of variables' matrix.

Answer 1

I don't think that I've ever seen this spelt out in every detail, but the derivation in Tuckerman, Berne, Martyna and Klein J Chem Phys, 99, 2796 (1993) gives the essential points. The aim is to express a ratio of free-particle density matrices $$ \frac{\rho_0(\mathbf{q}_1,\mathbf{q}_k;(k-1)\epsilon)\,\rho_0(\mathbf{q}_k,\mathbf{q}_{k+1};\epsilon)}{\rho_0(\mathbf{q}_1,\mathbf{q}_{k+1};k\epsilon)} $$ each of which takes a Gaussian form (omitting a lot of physical constants) $$ \rho_0(\mathbf{q},\mathbf{q}';\epsilon) \propto \exp\left[\frac{1}{2\epsilon}(\mathbf{q}-\mathbf{q}')^2\right] $$ in terms of a Gaussian function in the deviation of $\mathbf{q}_k$ from a linearly-interpolated value $$ \mathbf{q}_k^* = \frac{(k-1)\mathbf{q}_{k+1}+\mathbf{q}_1}{k} . $$ The new variables are defined so that $\mathbf{u}_k=\mathbf{q}_k-\mathbf{q}_k^*$. The vectorial nature of the coordinates makes no difference to the algebra, so I'm going to use scalar quantities henceforth to save typing.

The squared distances appearing in the exponents of the density matrices will match up, if we use the identity $$ \frac{(q_1-q_k)^2}{k-1} + (q_k-q_{k+1})^2 - \frac{(q_1-q_{k+1})^2}{k} = \left(\frac{k}{k-1}\right)(q_k-q_k^*)^2 = \left(\frac{k}{k-1}\right)u_k^2 . $$ This can be verified by multiplying it out. Now we sum over $k$ from $2$ to $P$: $$ \sum_{k=2}^P \frac{(q_1-q_k)^2}{k-1} + \sum_{k=2}^P(q_k-q_{k+1})^2 - \sum_{k=2}^P\frac{(q_1-q_{k+1})^2}{k} = \sum_{k=2}^P\left(\frac{k}{k-1}\right)u_k^2 . $$ Redefining the index of the first summation $$ \sum_{k=1}^{P-1} \frac{(q_1-q_{k+1})^2}{k} + \sum_{k=2}^P(q_k-q_{k+1})^2 - \sum_{k=2}^P\frac{(q_1-q_{k+1})^2}{k} = \sum_{k=2}^P\left(\frac{k}{k-1}\right)u_k^2 $$ we see that almost every term cancels with the third summation. There is an extra $k=1$ term in the first sum: $(q_1-q_{2})^2$, which we can put with the second summation, extending the lower limit of that sum to $k=1$. Then, there is an extra $k=P$ term in the third sum, but if we set $P+1\equiv 1$, this term is zero. So we get the final result $$ \sum_{k=1}^P(q_k-q_{k+1})^2 = \sum_{k=2}^P\left(\frac{k}{k-1}\right)u_k^2 . $$ I believe that the premultiplying factors in the combinations of free-particle density matrices all work out OK.