Simultaneous diagonalization of potential and kinetic energy

I am trying to prove that the matrix expression of the potential energy (Hessian matrix from a Taylor expansion in several variables of the potential) is diagonal considering small oscillations, when written in normal coordinates. But, the accounts do not seem to work out.

Consider a potential of a system of N degrees of freedom $$V(q_{1},...,q_{N})=v(q_{i})$$ and defining $$\vec{q}-\vec{q}_{min}=\vec{\eta}$$, where $$\vec{q}_{min}$$ is a minimum of stable potential:

$$V_{ij}=\frac{\partial V}{\partial q_{i}\partial q_{j}}(\vec{q}=\vec{q}_{min}).$$

The transformation between the generalized coordinates $$q_{i}$$ and the normal coordinates $$\xi _{i}$$ is:

$$\vec{q}=A\vec{\xi }+\vec{q}_{min} \Longleftrightarrow q_{i}=\sum_{j}^{}A_{ij}\xi_{j} + q_{i}^{min}$$

where $$A$$ is the matrix formed by the eigenvectors of the system:

$$\det(V-\lambda ^{(\alpha )}T)=0.$$

So I tried to write $$V_{ij}$$ in terms of $$\xi$$ :

$$\frac{\partial V}{\partial \xi _{i} }=\frac{\partial V}{\partial q_{i}} \frac{\partial q_{i}}{\partial \xi _{i}}=\frac{\partial V}{\partial q_{i}}\sum_{j}^{}A_{ij}\delta _{ij}=A_{ii}\frac{\partial V}{\partial q_{i}}$$

$$\frac{\partial^{2} V}{\partial \xi _{i}^{2} }=\frac{\partial }{\partial \xi _{i}} (A_{ii}\frac{\partial V}{\partial q_{i}})=A_{ii} \frac{\partial [\frac{\partial V}{\partial q_{i}}] }{\partial q_{i}} \frac{\partial q_{i}}{\partial \xi_{i}}=\frac{\partial^{2} V}{\partial q _{i}^{2} }A_{ii}^{2}$$

but for cross derivatives I get a nonzero result:

$$\frac{\partial V}{\partial \xi _{j}\partial \xi _{i}}=\frac{\partial V}{\partial \xi _{j}}(A_{ii}\frac{\partial V}{\partial q_{i}})=A_{ii}\frac{\partial[\frac{\partial V}{\partial q_{i}}] }{\partial q_{k}}\frac{\partial q_{k}}{\partial \xi_{j} }=A_{ii} \frac{\partial ^{2}V}{\partial q_{i} \partial q_{k}}\sum_{j}^{}A_{ij}\delta _{kj}=A_{ik}A_{ii} \frac{\partial ^{2}V}{\partial q_{i} \partial q_{k}}\neq 0$$ taking into account that we can exchange the index $$k$$ for $$j$$.

What could I do wrong? Where did I go wrong?

\begin{align*} &\text{with the potential energy ~V=V(\boldsymbol q)~ and the kinetic enegry ~T=\frac 12\, \boldsymbol{\dot q}\cdot \boldsymbol{\dot q} ~}\\ &\text{ you obtain this non linear copuled differential equations}\\\\ &\boldsymbol{\ddot{q}}=\underbrace{\left[\frac{\partial}{\partial \boldsymbol q}\left(\frac{\partial V}{\partial \boldsymbol{q}}\right)\right]}_{\boldsymbol K}\,\boldsymbol q=\boldsymbol{K(\boldsymbol q)}\,\boldsymbol{q}\tag 1 \end{align*} where

• $$~\boldsymbol q~$$ are the $$n$$ generalized coordinates
• $$~\boldsymbol K~$$ $$n\times n~$$ symmetric matrix

substitute $$~\boldsymbol q=\boldsymbol A\,\boldsymbol{\xi}+\boldsymbol{\xi}_{\text{min}}$$ into Eq. (1) you obtain: (where $$~\boldsymbol A~$$ is the matrix of the eigen vectors) \begin{align*} &\boldsymbol{A}\,\boldsymbol{\ddot{\xi}}=\left[\boldsymbol{K}\left(\boldsymbol A\,\boldsymbol{\xi}+\boldsymbol{\xi}_{\text{min}}\right)\right]\, \left(\boldsymbol A\,\boldsymbol{\xi}+\boldsymbol{\xi}_{\text{min}}\right)\overset{\text{linear}}{=} \boldsymbol K(\boldsymbol q=\boldsymbol\xi_{\text{min}})\,\boldsymbol A\,\boldsymbol\xi\\\\ &\boldsymbol{\ddot{\xi}}=\underbrace{\boldsymbol{A}^{-1}\,\boldsymbol{K}_L\,\boldsymbol A}_{\boldsymbol D}\,\boldsymbol\xi\tag 2\\ &\text{where}\\ &\boldsymbol D=\text{Diag}\left[\lambda_1~,\lambda_2\ldots~,\lambda_n\right]\\ &\text{and}\\ &\boldsymbol{K}_L=\boldsymbol{K}\bigg|_{\boldsymbol q=\boldsymbol\xi_{\text{min}}} \end{align*} thus Eq. (2) is a linear uncoupled differential equation

You were messy some of the indexes. First, remind that each column vector is normalized eigen vector of matrix $$\bf{V}$$, therefore $$\bf{V} \hat{A}_k = \lambda_k \hat{A}_k$$. $$\hat{A}_k$$ is the $$k_{th}$$ column vectoe of mtrix $$\bf{k}$$, in terms of index form:

$$\sum_j V_{ij} A_{jk} = \lambda_k A_{ik}$$

I try to re-write your equations:

$$\frac{\partial V}{\partial \xi_i} = \sum_j \frac{\partial V}{\partial q_j}\frac{\partial q_j}{\partial \xi_i} = \sum_j \frac{\partial V}{\partial q_j}\frac{\partial }{\partial \xi_i} \sum_k A_{jk} \xi_k =\sum_j \frac{\partial V}{\partial q_j} \sum_k A_{jk} \delta_{ik} = \sum_j \frac{\partial V}{\partial q_j} A_{ji}$$

$$\frac{\partial^{2} V}{\partial \xi _{i}^{2} }=\frac{\partial }{\partial \xi _{i}} \sum_j A_{ji}\frac{\partial V}{\partial q_{j}}= \sum_k \frac{\partial q_k}{\partial \xi _{i}} \frac{\partial}{\partial q_k} \left( \sum_j A_{ji}\frac{\partial V}{\partial q_{j}} \right) = \sum_k A_{ki} \frac{\partial}{\partial q_k} \left( \sum_j A_{ji}\frac{\partial V}{\partial q_{j}} \right) \\ = \sum_j\sum_k A_{ki}A_{ji} \frac{\partial^2 V}{\partial q_k \partial q_j} =\sum_j\sum_k A_{ki}A_{ji} V_{kj} = \lambda_i \sum_k A_{ki} A_{ki} = \lambda_i.$$

$$\frac{\partial^{2} V}{\partial \xi _{p} \partial \xi _{q}}=\frac{\partial }{\partial \xi_ {p}} \sum_j A_{jq}\frac{\partial V}{\partial q_{j}}= \sum_k \frac{\partial q_k}{\partial \xi _{p}} \frac{\partial}{\partial q_k} \left( \sum_j A_{jq}\frac{\partial V}{\partial q_{j}} \right) = \sum_k A_{kp} \frac{\partial}{\partial q_k} \left( \sum_j A_{jq}\frac{\partial V}{\partial q_{j}} \right) \\ = \sum_j\sum_k A_{kp}A_{jq} \frac{\partial^2 V}{\partial q_k \partial q_j} = \sum_j\sum_k A_{kp}A_{jq} V_{kj} = \lambda_q \sum_k A_{kp} A_{kq} = 0$$

• Thanks, it has been very helpful, just a question, the eigenvalue equation of the system has the form $V \cdot \vec{a_{k}} = \lambda_ {k} T \cdot \vec {a_ {k}}$ , then the matrix A made up of the column vectors $\vec{a_{k}}$ and the diagonal matrix $D = diag (\lambda_ {1}, .., \lambda_ {k}, ..,\lambda_ {N})$ what is the matrix that is diagonalized in the form $ADA^{-1}$?. Feb 5 '21 at 15:48
• $\bf{D} = \bf{A}^T \bf{V} \bf{A}$. The matrix $\bf{V}$ is diagonalized by orthogonal matrix $\bf{A}$ intp digonal matrix $\bf{D}$. Therefrore, the expansion of potential near the equilibrium point $q_{min}$ will have no cross terms. $V(q) = \sum_i\sum_j V_{ij} q_i q_j = \sum_i D_{ii} \xi_i^2$. This is call dynamical matrix.
– ytlu
Feb 6 '21 at 14:06