{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# #12.3.7 - Quantum Mechanics 1 HW 6 Solutions Leo...

This preview shows pages 1–5. Sign up to view the full content.

Quantum Mechanics - 1: HW 6 Solutions Leo Radzihovsky Paul Martens November 12, 2007 1 Problem 1 Show that H E 1 · E 2 commutes with L , where E 1 , 2 are vector operators. The definition of a vector operator. E (1 or 2) i , L j = ı ijk E k (1) (transforms like a vector under rotations) E (1) i E (2) i , L j = E (1) i E (2) i , L j + E (1) i , L j E (2) i (2) = ı ijk E (1) i E (2) k + E (1) k E (2) i (3) = 0 (4) Because E 1 · E 2 commute with L all functions of E 1 · E 2 commute with L H E 1 · E 2 , L = 0 (5) 2 Problem 2 Show that L z = - ı ( x∂ y - y∂ x ) = - ı ∂ φ relating coordinates x = r sin θ cos φ (6) y = r sin θ sin φ (7) z = r cos θ (8) and derivatives y = ∂r ∂y r + ∂θ ∂y θ + ∂φ ∂y φ (9) x = ∂r ∂x r + ∂θ ∂x θ + ∂φ ∂x φ (10) 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
r = x 2 + y 2 + z 2 (11) z r = cos θ (12) calculating derivatives ∂r ∂x = x r = sin θ cos φ (13) ∂r ∂y = y r = sin θ sin φ (14) θ derivatives can be taken easily by making use of the chain rule and equation (12) ∂x z r = - z r 2 ∂r ∂x = - sin θ ∂θ ∂x = - sin θ ∂θ ∂x (15) solving for ∂θ ∂x . ∂θ ∂x = 1 r cos θ cos φ (16) the calculation for ∂θ ∂y is almost identical ∂θ ∂y = 1 r cos θ sin φ (17) finding φ derivatives y x = tan φ (18) taking derivatives of both sides - tan φ x = sec 2 φ ∂φ ∂x (19) solving for ∂φ ∂x . ∂φ ∂x = - sin φ r sin θ (20) a similar argument can be used for y ∂φ ∂y = cos φ r sin θ (21) Putting all of it together L z - ı = x∂ y - y∂ x = r sin θ cos φ ( y r∂ r + y θ∂ θ + y φ∂ φ ) - r sin θ sin φ ( x r∂ r + x θ∂ θ + x φ∂ φ ) (22) L z = - ı ∂ φ (23) 2
3 Problem 3 Suppose we have three identical bosonic particles constrained to move on a circle in an equilateral configuration and want to find the wavefunction and the energy spectrum. The Hamiltonian of this system is H = L 2 z 2 I (24) where I = 3 mr 2 (25) The eigenstates of this Hamiltonian are the eigenstates of the L z operator. L z | z = z | z (26) φ z ( φ ) = e ı z φ (27) The position of each of the particles are φ 1 = φ (28) φ 2 = φ + 2 π 3 (29) φ 3 = φ + 4 π 3 (30) Because these are bosons, we must require that the full wavefunction be symmetric under permutations. For the rigid structure above this amounts to φ m ( φ φ + 2 π 3 ) = φ m ( φ ). Therefore, z = 3 n , where n Z . φ n ( φ ) = e 3 ınφ (31) E n = 9 2 n 2 2 I (32) 4 Problem 4 Express Y m ’s in Cartesian coords 4.1 Y 0 0 Y 0 0 = 1 4 π (33) 4.2 Y ± 1 1 Y ± 1 1 = 3 8 π sin θ e ± ıφ sin θ cos φ ± ı sin θ sin φ (34) = 3 8 π x ± ıy x 2 + y 2 + z 2 (35) 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4.3 Y 0 1 Y 0 1 = 3 4 π cos θ = 3 π 4 z x 2 + y 2 + z 2 (36) 4.4 Y ± 2 2 Y ± 2 2 = 15 32 π sin 2 θ (cos 2 φ ± ı sin 2 φ ) (37) = 15 32 π sin 2 θ ( cos 2 φ - sin 2 φ ± 2 ı sin φ cos φ ) (38) Y ± 2 2 = 15 32 π x 2 - y 2 ± 2 ıxy x 2 + y 2 + z 2 (39) 4.5 Y ± 1 2 Y ± 1 2 = 15 8 π sin θ cos θ
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 13

#12.3.7 - Quantum Mechanics 1 HW 6 Solutions Leo...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online