A two dimensional square box with a corner length L, the energy level of a particle with a mass h^2/(8mL^2)(n_1^2+n_2^2) . What is the quantum number of the first, second and third degenerate energy levels?

h^2/(8mL^2)(n_1^2+n_2^2)

1 Answer
Jun 14, 2018

I don't know what you mean by corner length... I assume it is just an L xx L square, because that is what the given energy expression implies.


Degenerate energy levels are those for which the total energy is the same, i.e. if

n = n_1 + n_2
n_1, n_2 = 1,2,3,4,5, . . .

and

E_(n_1n_2) = h^2/(8mL^2)(n_1^2 + n_2^2)

then we want all combinations of increasing n_1^2 + n_2^2 for which there exist more than one possibility for the same E_(n_1n_2).

The quantum number we want is n -= n_1+n_2.

You should be working these out as trial and error, but here are the first 15 combinations, degenerate or not:

ul(n_1" "" "n_2" "" "n_1^2+n_2^2" "" "E_(n_1n_2)" "" "" ""degeneracy"
1" "" "color(white)(/)1" "" "2" "" "" "" "color(white)(/)2h^2//8mL^2" "" "1
1" "" "color(white)(/)2" "" "5" "" "" "" "color(white)(/)5h^2//8mL^2" "" "color(blue)(bb2)
2" "" "color(white)(/)1
2" "" "color(white)(/)2" "" "8" "" "" "" "color(white)(/)8h^2//8mL^2" "" "1
1" "" "color(white)(/)3" "" "10" "" "" "color(white)(/.)10h^2//8mL^2color(white)(/)" "color(blue)(bb2)
3" "" "color(white)(/)1
2" "" "color(white)(/)3" "" "13" "" "" "color(white)(/.)13h^2//8mL^2color(white)(/)" "color(blue)(bb2)
3" "" "color(white)(/)2
1" "" "color(white)(/)4" "" "17" "" "" "color(white)(/.)17h^2//8mL^2color(white)(/)" "color(blue)(bb2)
4" "" "color(white)(/)1
3" "" "color(white)(/)3" "" "18" "" "" "color(white)(/.)18h^2//8mL^2color(white)(/)" "1
2" "" "color(white)(/)4" "" "20" "" "" "color(white)(/.)20h^2//8mL^2color(white)(/)" "color(blue)(bb2)
4" "" "color(white)(/)2
3" "" "color(white)(/)4" "" "25" "" "" "color(white)(/.)25h^2//8mL^2color(white)(/)" "color(blue)(bb2)
4" "" "color(white)(/)3

Clearly, the first three degenerate levels, which are highlighted, have:

  1. n = n_1 + n_2 = 1 + 2 = bb3
  2. n = n_1 + n_2 = 1 + 3 = bb4
  3. n = n_1 + n_2 = 2 + 3 = bb5

Note that even if n is the same for two combinations of n_1 and n_2, they might NOT be degenerate levels. For example, 2+3 = 5, and 1+4 = 5 as well, but 2^2 + 3^2 = 13 and 1^2 + 4^2 = 17, so those levels are NOT the same energy.

Also, at most you can only have a 2-fold degenerate energy because there are only two permutations you can have of two numbers.