Angular Momentum r p = L = r i r + + r - PowerPoint PPT Presentation

Angular Momentum � r × � p = � L = � r × � i � ∇ ∂ r + ˆ ∂θ + ˆ � r ∂ ∂ ∂ θ 1 1 ∇ = ˆ φ r sin θ ∂φ r ˆ = cos θ cos φ ˆ x + cos θ sin φ ˆ y − sin θ ˆ θ z ˆ φ = − sin φ ˆ x + cos φ ˆ y � � r × ˆ r × ˆ � � r ∂ ∂ ∂ θ 1 1 = r × ˆ ∂ r + � ∂θ + � � φ L r sin θ ∂φ i r � � ˆ ∂θ − ˆ � φ ∂ ∂ 1 = θ sin θ ∂φ i y ) ∂ � � = ( − sin φ ˆ x + cos φ ˆ ∂θ i � 1 ∂ − (cos θ cos φ ˆ x + cos θ sin φ ˆ y − sin θ ˆ z ) sin θ ∂φ � � � − sin φ ∂ ∂θ − cos θ sin θ cos φ ∂ = L x i ∂φ � � � cos φ ∂ ∂θ − cos θ sin θ sin φ ∂ = L y i ∂φ � ∂ = L z ∂φ i

Angular Momentum = L ± L x ± i L y � � � ( − sin φ ± i cos φ ) ∂ ∂θ − (cos φ ± i sin φ ) cot θ ∂ = ∂φ i ± � e ± i φ � � ∂θ ± i cot θ ∂ ∂ = ∂φ � 2 e i φ � � � − e − i φ � � � ∂θ + i cot θ ∂ ∂ ∂θ − i cot θ ∂ ∂ = L + L − ∂φ ∂φ − � 2 e i φ � e − i φ � � ∂ 2 ∂ 2 − 1 ∂ = ∂φ − i cot θ ∂θ 2 − i sin 2 θ ∂θ∂φ i cot θ ( − i ) e − i φ � � ∂θ − i cot θ ∂ ∂ ∂φ i cot θ e − i φ � �� ∂θ∂φ − i cot θ ∂ 2 ∂ 2 ∂φ 2 − � 2 � � ∂θ + cot 2 θ ∂ 2 ∂ 2 ∂θ 2 + cot θ ∂ ∂φ 2 + i ∂ = ∂φ L 2 L + L − − i [ L y , L x ] + L 2 z = L + L − + L 2 = z − � L z − � 2 � � ∂θ + cot 2 θ ∂ 2 ∂ 2 ∂ 2 ∂θ 2 + cot θ ∂ ∂φ 2 + i ∂ ∂φ 2 + 1 ∂ = ∂φ + i ∂φ − � 2 � � ∂ 2 1 ∂ � sin θ ∂ � 1 = + sin 2 θ sin θ ∂θ ∂θ ∂φ 2

Angular Momentum − � 2 � � ∂ 2 ∂ sin θ ∂ L 2 1 � � 1 = + sin 2 θ ∂φ 2 sin θ ∂θ ∂θ eigenfunctions Y m ℓ ( θ , φ ) no Y m for half-integer ℓ ℓ � � − � 2 d r 2 d 1 � � � + L 2 � + V ( r ) ψ = E ψ 2 µr 2 dr dr K rad = p 2 K rot = I 2 I = µr 2 2 µ , 2 I ,

Diatomic Molecule m 1 r 1 = m 2 r 2 ( m 1 + m 2 ) r 1 = m 2 ( r 1 + r 2 ) , ( m 1 + m 2 ) r 2 = m 1 ( r 1 + r 2 ) m 2 2 ( r 1 + r 2 ) 2 m 2 1 ( r 1 + r 2 ) 2 m 1 r 2 1 + m 2 r 2 = 2 = m 1 ( m 1 + m 2 ) 2 + m 2 I ( m 1 + m 2 ) 2 m 1 m 2 ( r 1 + r 2 ) 2 = µr 2 = m 1 + m 2 H ψ = L 2 2 I ψ = E ψ ψ = Y m ℓ ( θ , φ ) E = � 2 ℓ ( ℓ +1) 2 I

O 2 Diatomic Molecule (6 . 58 × 10 − 16 eV s ) 2 (3 × 10 8 m/s ) 2 � 2 2 2 � 2 c 2 E rot = M O r 2 = ≈ 16 M p c 2 r 2 8 × 938 × 10 6 eV (10 − 10 m ) 2 40 × 10 − 16 eV 2 ≈ 8 × 10 8 eV 10 − 20 5 × 10 − 3 eV ≈ f vib = 5 × 10 13 Hz hf vib = 2 π 6 . 58 × 10 − 16 eV s × 5 × 10 13 1 /s E vib = 0 . 2 eV = E vib ≫ E rot

Diatomic Molecule