Cubic Potential Well

Cubic Poth well V(x,y,z) = { 0, 0 < x < L, 0 < y < L, 0 < z < L { infinity, otherwise V(x,y,z) = V(x) + V(y) + V(z) V(x) = { 0, 0 < x < L { infinity, otherwise Ex = (n_x^2 * pi^2 * hbar^2) / (2 * m * L^2) psi_n(x) = { sqrt(2/L) * sin(n_x * pi * x / L), 0 <= x <= L { 0, otherwise V(y) = { 0, 0 < y < L { infinity, otherwise Ey = (n_y^2 * pi^2 * hbar^2) / (2 * m * L^2) psi_n(y) = { sqrt(2/L) * sin(n_y * pi * y / L), 0 <= y <= L { 0, otherwise V(z) = { 0, 0 < z < L { oo, otherwise Ez = (n_z^2 * pi^2 * hbar^2) / (2 * m * L^2) psi_n(z) = { sqrt(2/L) * sin(n_z * pi * z / L), 0 <= z <= L { 0, otherwise E = Ex + Ey + Ez = (pi^2 * hbar^2 / (2 * m * L^2)) * (n_x^2 + n_y^2 + n_z^2) En_x n_y n_z = (pi^2 * hbar^2 / (2 * m * L^2)) * (n_x^2 + n_y^2 + n_z^2) psi_n_x n_y n_z (x, y, z) = { (2/L)^(3/2) * sin(n_x * pi * x / L) * sin(n_y * pi * y / L) * sin(n_z * pi * z / L), 0 <= x, y, z <= L { 0, otherwise The energy of ground state E111 = (3 * pi^2 * hbar^2) / (2 * m * L^2) => Non-degenerate. psi111 = { (2/L)^(3/2) * sin(pi * x / L) * sin(pi * y / L) * sin(pi * z / L), 0 <= x, y, z <= L { 0, otherwise The energy of first excited state- (let m2 = 2) E112 = E121 = E211 = (6 * pi^2 * hbar^2) / (2 * m * L^2) Corresponding wave function psi_n1n2n3 -> psi112 -> psi121 -> psi211 psi112 = { (2/L)^(3/2) * sin(pi * x / L) * sin(pi * y / L) * sin(2 * pi * z / L), 0 <= x, y, z <= L { 0, otherwise psi121 = { (2/L)^(3/2) * sin(pi * x / L) * sin(2 * pi * y / L) * sin(pi * z / L), 0 <= x, y, z <= L { 0, otherwise psi111 = { (2/L)^(3/2) * sin(pi * x / L) * sin(pi * y / L) * sin(pi * z / L), 0 <= x, y, z <= L { 0, otherwise These eigen fn have same energy. All are Linearly independent state having same eigen values. E112 = E121 = E211 = (6 * pi^2 * hbar^2) / (2 * m * L^2) → 3-fold degenerate. Enx ny nz = ((nx^2 + ny^2 + nz^2) * pi^2 * hbar^2) / (2 * m * L^2) Energy (nx ny nz) Degeneracy 1 (1,1,1), (1,1,1), (1,1,1) 1 (3 * pi^2 * hbar^2) / (2 * m * L^2) 2 (1,1,2), (1,2,1), (2,1,1) 3 (6 * pi^2 * hbar^2) / (2 * m * L^2) 3 (1,2,2), (2,1,2), (2,2,1) 3 (9 * pi^2 * hbar^2) / (2 * m * L^2) 4 (1,1,3), (1,3,1), (3,1,1) 3 (12 * pi^2 * hbar^2) / (2 * m * L^2) 5 (2,2,2) 1 (15 * pi^2 * hbar^2) / (2 * m * L^2) 6 (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) 6 (18 * pi^2 * hbar^2) / (2 * m * L^2) Question 4.95 (4) Sum of energies of third and fourth level E3 + E4 = (9+16) * (pi^2 * hbar^2) / (2 * m * L^2) = (10 * pi^2 * hbar^2) / (m * L^2) (a) Question 4.95 (5) Degeneracy of the fourth level = 3 (proven) Question 4.14 V(x,y) = { 0, 0 < x < a, 0 < y < a { infinity, otherwise = V(x) + V(y) So, V(x) = { 0, 0 < x < a { infinity, otherwise V(y) = { 0, 0 < y < a { infinity, otherwise Enx * Eny = Enx + Eny = (nx^2 * pi^2 * hbar^2) / (2 * ma^2) + (ny^2 * pi^2 * hbar^2) / (2 * ma^2) = (nx^2 + ny^2) * (pi^2 * hbar^2) / (2 * ma^2) nx = ny = 1,2,3... The energy of first excited state → E12 = E21 = E = (1+4) * (pi^2 * hbar^2) / (2 * ma^2) = 5 * (pi^2 * hbar^2) / (2 * ma^2) (option-c) 2-fold of degeneracy n1^2 + n2^2 = 5 (n1, n2) = (1,2) (n1, n2) = (2,1) = 2-fold of degeneracy Ques.5Page-68 En1n2n3 = (n1^2 + n2^2 + n3^2) * (pi^2 * hbar^2) / (2 * ma^2) E = 14 * (pi^2 * hbar^2) / (2 * ma^2) ∴ n1^2 + n2^2 + n3^2 = 14 option-d full fill this above condition (option-d) is right