- #1
AxiomOfChoice
- 531
- 1
Suppose you start with a system of N particles identified by position vectors r_1, r_2, \ldots, r_N and masses \mu_1, \mu_2, \ldots, \mu_N. Then (quantum mechanically) the kinetic energy operator for this system is given by (assuming \hbar = 1)
<br /> T = \sum_{i = 1}^N -\frac{\Delta_i}{2\mu_i},<br />
where \Delta_i = \Delta_{r_i} is the Laplacian in the variable r_i. Now, let's make the change of variables \eta_i = r_i - r_N. According to the book I have, the kinetic energy in the new variables looks like
<br /> T' = \sum_{i = 1}^{N-1} -\frac{\Delta_i}{2\mu_i'} + \sum_{i < j} \nabla_i \cdot \nabla_j,<br />
where \Delta_i = \Delta_{\eta_i}, \nabla_i = \nabla_{\eta_i}, and
<br /> \frac{1}{\mu'_i} = \frac{1}{\mu_i} + \frac{1}{\mu_N}.<br />
I simply do not see how this is true. It seems that, if one applied the chain rule, one obtains \Delta_{\eta_i} = \Delta_{r_i}; is that not true?
<br /> T = \sum_{i = 1}^N -\frac{\Delta_i}{2\mu_i},<br />
where \Delta_i = \Delta_{r_i} is the Laplacian in the variable r_i. Now, let's make the change of variables \eta_i = r_i - r_N. According to the book I have, the kinetic energy in the new variables looks like
<br /> T' = \sum_{i = 1}^{N-1} -\frac{\Delta_i}{2\mu_i'} + \sum_{i < j} \nabla_i \cdot \nabla_j,<br />
where \Delta_i = \Delta_{\eta_i}, \nabla_i = \nabla_{\eta_i}, and
<br /> \frac{1}{\mu'_i} = \frac{1}{\mu_i} + \frac{1}{\mu_N}.<br />
I simply do not see how this is true. It seems that, if one applied the chain rule, one obtains \Delta_{\eta_i} = \Delta_{r_i}; is that not true?
