Runge-Lenz矢量是守恒量

并不知道这个blog会被我用来记什么，但这个皮肤我捣鼓了挺长时间。最近又在实验室做了一堆家务，不如写点物理的东西。
这是翟荟老师的冷原子物理的第一个作业。

Question

Considering an electron with Coulomb interaction between electron and nucleus, the Hamiltonian is given by

\[\hat{H}=-\frac{\hbar^{2} \nabla_{i}^{2}}{2 m^{*}}-\frac{Z \kappa}{r} \]

and considering the Laplace–Runge–Lene vector defined as

\[\hat{\mathbf{J}}=\frac{1}{2 m^{*}}(\hat{\mathbf{p}} \times \hat{\mathbf{L}}-\hat{\mathbf{L}} \times \hat{\mathbf{p}})-Z_{\kappa} \frac{\mathbf{r}}{r} \]

Show that \([\hat{\mathbf{J}}, \hat{H}]\) = 0. This operator \(\hat{\mathbf{J}}\) and the angular momentum operator \(\hat{I}\) together form the \(SO(4)\) algebra. This only works for the Coulomb potential.

Proof

We ignore the "hat" below. Always keep in mind all observables are operator.
To calculate the commutator \([\bold{J},H]\), we first calculate the following terms:

\[\begin{aligned} \left[(p \times L)_i, p^2\right] &= [\epsilon_{kli} p_k L_l, p_j p_j] \\ &= [\epsilon_{kli} \epsilon_{mnl} p_k r_m p_n, p_j p_j] \\ &= [(\delta_{im}\delta_{kn} - \delta_{in}\delta_{km}) p_k r_m p_n, p_j p_j] \\ &= [p_k r_i p_k, p_j p_j] - [p_k r_k p_i, p_j p_j] \\ &= p_k [r_i, p_j p_j] p_k - p_k [r_k, p_j p_j] p_i \\ &= p_k (2\mathrm{i}\hbar p_i) p_k - p_k (2\mathrm{i}\hbar p_k) p_i \\ &= 0 \end{aligned}\]

\[\begin{aligned} \left[(L \times p)_i, p^2\right] &= [\epsilon_{kli} L_k p_l, p_j p_j] \\ &= [\epsilon_{kli} \epsilon_{mnk} r_m p_n p_l, p_j p_j] \\ &= [r_l p_i p_l, p_j p_j] - [r_i p_l p_l, p_j p_j] \\ &= (2\mathrm{i}\hbar p_l) p_i p_l - (2\mathrm{i}\hbar p_i) p_l p_l \\ &= 0 \end{aligned}\]

\[[p_i, r] = - \mathrm{i}\hbar \frac{r_i}{r} \]

\[[p_i, \frac{1}{r}] = \mathrm{i}\hbar \frac{r_i}{r^3} \]

\[\begin{aligned} \left[\frac{r_i}{r}, p^2\right] &= [r_i,p^2]\frac{1}{r} + r_i \left[\frac{1}{r},p^2 \right] \\ &= 2\mathrm{i}\hbar p_i \frac{1}{r} + r_i \left(\left[\frac{1}{r},p_j \right] p_j + p_j \left[\frac{1}{r},p_j \right] \right) \\ &= 2\mathrm{i}\hbar p_i \frac{1}{r} - \mathrm{i}\hbar \left( \frac{r_i r_j}{r^3} p_j + r_i p_j \frac{r_j}{r^3}\right) \end{aligned}\]

\[\begin{aligned} \left[(p \times L)_i, \frac{1}{r} \right] &= \left[p_k r_i p_k, \frac{1}{r}\right] - \left[p_k r_k p_i, \frac{1}{r} \right] \\ &= \left(\left[p_k, \frac{1}{r}\right] r_i p_k + p_k r_i\left[p_k, \frac{1}{r}\right] \right) - \left(\left[p_k, \frac{1}{r}\right] r_k p_i + p_k r_k\left[p_i, \frac{1}{r}\right] \right) \\ &= \mathrm{i}\hbar \left(\frac{r_k}{r^3} r_i p_k + p_k r_i \frac{r_k}{r^3} \right) - \mathrm{i}\hbar \left(\frac{r_k}{r^3} r_k p_i + p_k r_k \frac{r_i}{r^3} \right) \\ &= \mathrm{i}\hbar \left(\frac{r_k r_i}{r^3}p_k \right) - \mathrm{i}\hbar \frac{1}{r} p_i \end{aligned}\]

\[\begin{aligned} \left[(L \times p)_i, \frac{1}{r} \right] &= \left[r_l p_i p_l, \frac{1}{r}\right] - \left[r_i p_l p_l, \frac{1}{r} \right] \\ &= \left(r_l \left[p_i, \frac{1}{r}\right] p_l + r_l p_i \left[p_l, \frac{1}{r}\right] \right) - \left(r_i \left[p_l, \frac{1}{r}\right] p_l + r_i p_l \left[p_l, \frac{1}{r}\right] \right) \\ &= \mathrm{i}\hbar \left(r_l p_i \frac{r_l}{r^3} - \frac{r_i r_l}{r^3} p_l \right) \end{aligned}\]

\[\left[\frac{r_i}{r}, \frac{1}{r} \right] = 0 \]

Note that

\[\mathrm{i}\hbar \frac{1}{r} p_i = \mathrm{i}\hbar p_i \frac{1}{r} + \mathrm{i}\hbar \left[\frac{1}{r}, p_i \right] = \mathrm{i}\hbar \left(p_i \frac{1}{r} - \mathrm{i}\hbar \frac{r_i}{r^3} \right) \]

\[\mathrm{i}\hbar r_l p_i \frac{r_l}{r^3} = \mathrm{i}\hbar \left(p_i r_l \frac{r_l}{r^3} + \mathrm{i}\hbar \frac{r_i}{r^3} \right) = \mathrm{i}\hbar \left(p_i \frac{1}{r} + \mathrm{i}\hbar \frac{r_i}{r^3} \right) \]

Further, we can verify that \(\left[p_k, \frac{r_k}{r^3} \right] = \left[p_k, r_k \right]\frac{1}{r^3} + r_k \left[p_k, \frac{1}{r^3} \right] = 0\), hence \(\frac{r_k}{r^3}p_k = p_k \frac{r_k}{r^3}\).
So the commutator

\[\begin{aligned} \left[J_i,H\right] &= \left[\frac{1}{2m}\left(p \times L - L \times p \right)_i - Z\kappa\frac{r_i}{r}, \frac{p^2}{2m} - \frac{Z\kappa}{r} \right] \\ &= \frac{1}{(2m)^2} \left(\left[(p \times L)_i, p^2\right] - \left[(L \times p)_i, p^2\right] \right) - \frac{Z\kappa}{2m}\left[\frac{r_i}{r}, p^2\right] \\ &\quad - \frac{Z\kappa}{2m} \left(\left[(p \times L)_i, \frac{1}{r}\right] - \left[(L \times p)_i, \frac{1}{r}\right] \right) + (Z\kappa)^2 \left[\frac{r_i}{r}, \frac{1}{r}\right] \\ &= - \frac{Z\kappa}{2m}\left[\frac{r_i}{r}, p^2\right] - \frac{Z\kappa}{2m} \left[(p \times L)_i, \frac{1}{r}\right] + \frac{Z\kappa}{2m} \left[(L \times p)_i, \frac{1}{r}\right] \\ &\propto \left[\frac{r_i}{r}, p^2\right] + \left[(p \times L)_i, \frac{1}{r}\right] - \left[(L \times p)_i, \frac{1}{r}\right] \\ &= 2\mathrm{i}\hbar p_i \frac{1}{r} - \mathrm{i}\hbar \left( \frac{r_i r_j}{r^3} p_j + r_i p_j \frac{r_j}{r^3}\right) \\ &\quad + \mathrm{i}\hbar \left(\frac{r_k r_i}{r^3}p_k \right) - \mathrm{i}\hbar \frac{1}{r} p_i - \mathrm{i}\hbar \left(r_l p_i \frac{r_l}{r^3} - \frac{r_i r_l}{r^3} p_l \right) \\ &= \left(2\mathrm{i}\hbar p_i \frac{1}{r} - \mathrm{i}\hbar \frac{1}{r} p_i - \mathrm{i}\hbar r_l p_i \frac{r_l}{r^3} \right) + \left(\mathrm{i}\hbar \frac{r_i r_l}{r^3} p_l - \mathrm{i}\hbar r_i p_j \frac{r_j}{r^3}\right) \\ &= 0 \end{aligned}\]

is always zero, meaning the Laplace-Runge-Lenz vector is conserved under Coulomb potential.
当然也有经典的版本，把所有的对易子换成泊松括号(Poisson Bracket)即可。

（选了这课的同学不要照抄作业啊kora