# Re-writing kinetic energy term by using Euler's homogeneous function theorem

I've been reading Tuckerman's statistical mechanics textbook, in Feynman's path-integral chapter, the book introduced a method to convert from thermodynamic to virial energy estimators by using Euler's homogeneous function theorem:

$\alpha(x_1, \cdots, x_P) = \frac{1}{2} m \omega_P^2 \sum_{k=1}^{P} (x_k - x_{k+1})^2$ the function $\alpha$ is a homogeneous function of degree 2. Hence, applying Euler's theorem, which we can write as:

$\alpha(x_1, \cdots, x_P) = \frac{1}{2} \sum_{k=1}^{P}x_k \frac{\partial \alpha}{\partial x_k}$.

I know the general theorem: $n f(x_1, \cdots, x_P) = \sum_{k=1}^{P} x_k \frac{\partial f}{x_k}$

To prove this is correct, I first let $p=1$, then we arrive $\alpha = \frac{1}{2}m \omega_P^2 (x_1^2 - x_1 x_2 )$, which isn't same as the outcome of Euler's theorem? I'm quite confused about this part, thanks in advance for any help!

• What do you mean "include"? $\frac12m\omega_P^2$ is a multiplicative constant — it does not involve any of the variables. It will be there, on both sides of your equation. Feb 11 '17 at 15:30
• What about trying the dim 1 (i.e $p=1$ case), just to check ? Feb 11 '17 at 16:55
• @Thomas, when $P=1$, $\alpha(x_1, \cdots, x_P) = \frac{1}{2} m \omega_P^2 (x_1^2 - x_1x_2)$ instead, it doesn't reveal its original expression, I think it's because of this $x_{k+1}$ term being troublesome Feb 11 '17 at 23:22

The confusion here is purely because I ignored the physical meaning of $P$ in Feynman's path-integrals. In which case, the resemblance of the partition function to that of a classical cyclic polymer chain of $P$ points to coin the term classical isomorphism and to exploit the isomorphism between the classical and approximate quantum partition functions. In regard to our current question, $P \ge 2$ must be devised in order to form a cyclic polymer, otherwise, we only construct a single particle (reveals classical system with single point).
When $P=2$, $\alpha = \frac{1}{2} m \omega_P^2 [(x_1 - x_2)^2 + (x_2 - x_3)^2] = \frac{1}{2} m \omega_P^2 [(x_1 - x_2)^2 + (x_2 - x_1)^2] = m \omega_P^2(x_1 - x_2)^2$, then the theorem can be easily proved.