- #1
alebruna
- 6
- 0
- TL;DR Summary
- Example of a quantum free particle, understanding the observables necessary to describe the system
I have a question about an example about the choice of the operators needed to describe a system, the text is reported below:
"3D systems with ##H = (p_1^2+ p_2^2+ p_3^2)/(2m)## but no potential. Classically, the number of degrees of freedom is 6 corresponding to the six canonical variables xi and pi. The maximum number of observables corresponding to mutually commuting operators is three. In view of fundamental commutators:
$$[L_j ,L_m] = i\hbar\epsilon_{jmn} L_n\ ,\ \ \ [L_j , x_m] = i\hbar\epsilon_{jmn} x_n\ ,\ \ \ [L_j , p_m] = i\hbar\epsilon_{jmn} p_n $$
$$[L_j , x_2] = 0\ ,\ \ \ [L_j , p_2] = 0\ ,\ \ \ [L_j ,L_2] = 0 $$
with ##x^2 =\sum_i x_i^2 ## and ##p^2 = \sum_i p_i^2##, there are many possibilities
i) ##x_1, x_2, x_3##,
ii) ##p_1, p_2, p_3##,
iii) ##x_1, x_2, p_3##,
iv) ##L_3, L^2,H##
v) ##p_i, L_i, H \propto p_2##,
vi)## x_k, L_k, x_2 ##,
The choice of angular-momentum triplet ##L_1##, ##L_2## and ##L_3## is excluded since ##[L_n,L_m] = i\epsilon_{nmk} L_k##. The fact that ##H## commutes with various operators represents an important information. These conserved quantities are:
$$p_1,\ p_2,\ p_3,\ L_i ,\ i = 1, 2, 3 ,\ L^2 = L_1^2+ L_2^2+ L_3^2$$
in addition to ##H##. Considering such constants of motion, the more advantageous choices should be cases (ii) and (iv) because the three observables correspond to constants of motion in addition to be mutually commuting operators."
I have a question, why are only 3 observables needed to describe a system of 6 degrees of freedom?
"3D systems with ##H = (p_1^2+ p_2^2+ p_3^2)/(2m)## but no potential. Classically, the number of degrees of freedom is 6 corresponding to the six canonical variables xi and pi. The maximum number of observables corresponding to mutually commuting operators is three. In view of fundamental commutators:
$$[L_j ,L_m] = i\hbar\epsilon_{jmn} L_n\ ,\ \ \ [L_j , x_m] = i\hbar\epsilon_{jmn} x_n\ ,\ \ \ [L_j , p_m] = i\hbar\epsilon_{jmn} p_n $$
$$[L_j , x_2] = 0\ ,\ \ \ [L_j , p_2] = 0\ ,\ \ \ [L_j ,L_2] = 0 $$
with ##x^2 =\sum_i x_i^2 ## and ##p^2 = \sum_i p_i^2##, there are many possibilities
i) ##x_1, x_2, x_3##,
ii) ##p_1, p_2, p_3##,
iii) ##x_1, x_2, p_3##,
iv) ##L_3, L^2,H##
v) ##p_i, L_i, H \propto p_2##,
vi)## x_k, L_k, x_2 ##,
The choice of angular-momentum triplet ##L_1##, ##L_2## and ##L_3## is excluded since ##[L_n,L_m] = i\epsilon_{nmk} L_k##. The fact that ##H## commutes with various operators represents an important information. These conserved quantities are:
$$p_1,\ p_2,\ p_3,\ L_i ,\ i = 1, 2, 3 ,\ L^2 = L_1^2+ L_2^2+ L_3^2$$
in addition to ##H##. Considering such constants of motion, the more advantageous choices should be cases (ii) and (iv) because the three observables correspond to constants of motion in addition to be mutually commuting operators."
I have a question, why are only 3 observables needed to describe a system of 6 degrees of freedom?
Last edited:
