Question:
Does the angular momentum operator commutes with the Hamilitonian in the isotropic potential?
Solution:
The Hamilitonian operator is $ \hat{H} = \frac{\hat{p}^2}{2m} + \hat{V} $.
$ \begin{eqnarray*} [L_k, H] &=& [L_k, \frac{\hat{p}^2}{2m} + \hat{V}] \\ &=& \frac{1}{2m}[L_k, \hat{p}^2] + [L_k, \hat{V}] \\ \end{eqnarray*} $
Again, in order not to clutter the derivation, let's solve a subproblem first.
$ \begin{eqnarray*} & & [L_a, \hat{p}^2] \\ &=& [L_a, p_b p_b] \\ &=& [L_a, p_b] p_b + p_b [L_a, p_b] \\ &=& \epsilon_{abc} p_c p_b + p_b \epsilon_{abc} p_c \\ &=& \epsilon_{abc} p_c p_b + p_b \epsilon_{abc} p_c \\ &=& 2\epsilon_{abc} p_b p_c \\ &=& 0 \end{eqnarray*} $
The last equality comes from the fact that the sum cancel each other! Suppose $ a = 1 $, then the non-vanishing terms of the sum is just $ p_2 p_3 - p_3 p_2 $, which is simply 0. Looking backward, if I go just a little further with my question 3, I would have reach the same result. Now let's solve the potential side
$ \begin{eqnarray*} & & [L_a, x_b x_b] \\ &=& [L_a, x_b] x_b + x_b [L_a, x_b] \\ &=& \epsilon_{abc} x_c x_b + x_b \epsilon_{abc} x_c \\ &=& \epsilon_{abc} x_c x_b + x_b \epsilon_{abc} x_c \\ &=& 2\epsilon_{abc} x_b x_c \\ &=& 0 \end{eqnarray*} $
The reasoning is identical, so the commutator is just 0.
Does the angular momentum operator commutes with the Hamilitonian in the isotropic potential?
Solution:
The Hamilitonian operator is $ \hat{H} = \frac{\hat{p}^2}{2m} + \hat{V} $.
$ \begin{eqnarray*} [L_k, H] &=& [L_k, \frac{\hat{p}^2}{2m} + \hat{V}] \\ &=& \frac{1}{2m}[L_k, \hat{p}^2] + [L_k, \hat{V}] \\ \end{eqnarray*} $
Again, in order not to clutter the derivation, let's solve a subproblem first.
$ \begin{eqnarray*} & & [L_a, \hat{p}^2] \\ &=& [L_a, p_b p_b] \\ &=& [L_a, p_b] p_b + p_b [L_a, p_b] \\ &=& \epsilon_{abc} p_c p_b + p_b \epsilon_{abc} p_c \\ &=& \epsilon_{abc} p_c p_b + p_b \epsilon_{abc} p_c \\ &=& 2\epsilon_{abc} p_b p_c \\ &=& 0 \end{eqnarray*} $
The last equality comes from the fact that the sum cancel each other! Suppose $ a = 1 $, then the non-vanishing terms of the sum is just $ p_2 p_3 - p_3 p_2 $, which is simply 0. Looking backward, if I go just a little further with my question 3, I would have reach the same result. Now let's solve the potential side
$ \begin{eqnarray*} & & [L_a, x_b x_b] \\ &=& [L_a, x_b] x_b + x_b [L_a, x_b] \\ &=& \epsilon_{abc} x_c x_b + x_b \epsilon_{abc} x_c \\ &=& \epsilon_{abc} x_c x_b + x_b \epsilon_{abc} x_c \\ &=& 2\epsilon_{abc} x_b x_c \\ &=& 0 \end{eqnarray*} $
The reasoning is identical, so the commutator is just 0.
No comments:
Post a Comment