Physical Chemistry Week 12

Properties of spherical harmonics Λ 2 𝑌 𝑙𝑚 =−𝑙 𝑙+1 𝑌 𝑙𝑚 Energy: 𝐻 =− ℏ 2 2𝑚 ⋅ Λ 2 𝑟 2 , 𝐻 𝑌 𝑙𝑚 = ℏ 2 2𝐼 ⋅𝑙 𝑙+1 𝑌 𝑙𝑚 , 𝐸 𝑙 = 𝑙 𝑙+1 ℏ 2 2𝐼 Square of angular momentum ( 𝐿 2 ): Classical: 𝐸= 𝐿 2 /2𝐼; quantum: 𝐿 2 =− ℏ 2 Λ 2 𝐿 2 𝑌 𝑙𝑚 =𝑙 𝑙+1 ℏ 2 𝑌 𝑙𝑚 , 𝐿 2 𝑙 =𝑙 𝑙+1 ℏ 2

Angular momentum ( 𝐿 ) Classical: 𝐿 = 𝑟 × 𝑝 ; quantum: 𝐿 = 𝑟 × 𝑝 𝑟 =𝑥 𝑒 𝑥 +𝑦 𝑒 𝑦 +𝑧 𝑒 𝑧 , 𝑝 = 𝑝 𝑥 𝑒 𝑥 + 𝑝 𝑦 𝑒 𝑦 + 𝑝 𝑧 𝑒 𝑧 𝑝 𝜇 = ℏ 𝑖 𝜕 𝜕𝜇 , 𝜇=𝑥,𝑦,𝑧 ‘×’ means cross product 𝑎 × 𝑘 𝑏 =𝑘 𝑎 × 𝑏 , 𝑎 × 𝑎 = 0 , 𝑎 × 𝑏 =− 𝑏 × 𝑎 , 𝑎 × 𝑏 + 𝑐 = 𝑎 × 𝑏 + 𝑎 × 𝑐 𝑒 𝑥 × 𝑒 𝑦 = 𝑒 𝑧 , 𝑒 𝑦 × 𝑒 𝑧 = 𝑒 𝑥 , 𝑒 𝑧 × 𝑒 𝑥 = 𝑒 𝑦

Continued 𝐿 &= 𝑥 𝑒 𝑥 +𝑦 𝑒 𝑦 +𝑧 𝑒 𝑧 × ℏ 𝑖 𝜕 𝜕𝑥 𝑒 𝑥 + 𝜕 𝜕𝑦 𝑒 𝑦 + 𝜕 𝜕𝑧 𝑒 𝑧 &= ℏ 𝑖 𝑥 𝜕 𝜕𝑦 −𝑦 𝜕 𝜕𝑥 𝑒 𝑧 + 𝑦 𝜕 𝜕𝑧 −𝑧 𝜕 𝜕𝑦 𝑒 𝑥 + 𝑧 𝜕 𝜕𝑥 −𝑥 𝜕 𝜕𝑧 𝑒 𝑦 𝐿 𝑥 = ℏ 𝑖 𝑦 𝜕 𝜕𝑧 −𝑧 𝜕 𝜕𝑦 𝐿 𝑦 = ℏ 𝑖 𝑧 𝜕 𝜕𝑥 −𝑥 𝜕 𝜕𝑧 𝐿 𝑧 = ℏ 𝑖 𝑥 𝜕 𝜕𝑦 −𝑦 𝜕 𝜕𝑥

Commutation relations 𝐿 𝑥 , 𝐿 𝑦 &=− ℏ 2 𝑦 𝜕 𝜕𝑧 −𝑧 𝜕 𝜕𝑦 𝑧 𝜕 𝜕𝑥 −𝑥 𝜕 𝜕𝑧 − 𝑧 𝜕 𝜕𝑥 −𝑥 𝜕 𝜕𝑧 𝑦 𝜕 𝜕𝑧 −𝑧 𝜕 𝜕𝑦 &=− ℏ 2 𝑦 𝜕 𝜕𝑥 +𝑦𝑧 𝜕 2 𝜕𝑧𝜕𝑥 −𝑦𝑥 𝜕 2 𝜕 𝑧 2 − 𝑧 2 𝜕 2 𝜕𝑦𝜕𝑥 +𝑧𝑥 𝜕 2 𝜕𝑦𝜕𝑧 & + ℏ 2 𝑧𝑦 𝜕 2 𝜕𝑥𝜕𝑧 − 𝑧 2 𝜕 2 𝜕𝑥𝜕𝑦 −𝑥𝑦 𝜕 2 𝜕 𝑧 2 +𝑥 𝜕 𝜕𝑦 +𝑥𝑧 𝜕 2 𝜕𝑧𝜕𝑦 &= ℏ 2 𝑥 𝜕 𝜕𝑦 −𝑦 𝜕 𝜕𝑥 &=𝑖ℏ 𝐿 𝑧 𝐿 𝑥 , 𝐿 𝑦 =𝑖ℏ 𝐿 𝑧 , 𝐿 𝑦 , 𝐿 𝑧 =𝑖ℏ 𝐿 𝑥 , 𝐿 𝑧 , 𝐿 𝑥 =𝑖ℏ 𝐿 𝑦

Continued 𝐿 2 , 𝐿 𝑧 = 𝐿 𝑥 2 , 𝐿 𝑧 + 𝐿 𝑦 2 , 𝐿 𝑧 + 𝐿 𝑧 2 , 𝐿 𝑧 𝐿 2 , 𝐿 𝑧 = 𝐿 𝑥 2 , 𝐿 𝑧 + 𝐿 𝑦 2 , 𝐿 𝑧 + 𝐿 𝑧 2 , 𝐿 𝑧 𝐿 𝑧 2 , 𝐿 𝑧 = 𝐿 𝑧 3 − 𝐿 𝑧 3 =0 • 𝐿 𝑥 2 , 𝐿 𝑧 &= 𝐿 𝑥 2 𝐿 𝑧 − 𝐿 𝑥 𝐿 𝑧 𝐿 𝑥 + 𝐿 𝑥 𝐿 𝑧 𝐿 𝑥 − 𝐿 𝑧 𝐿 𝑥 2 &= 𝐿 𝑥 𝐿 𝑥 , 𝐿 𝑧 + 𝐿 𝑥 , 𝐿 𝑧 𝐿 𝑥 &=−𝑖ℏ 𝐿 𝑥 𝐿 𝑦 + 𝐿 𝑦 𝐿 𝑥 𝐿 𝑦 2 , 𝐿 𝑧 =𝑖ℏ 𝐿 𝑦 𝐿 𝑥 + 𝐿 𝑥 𝐿 𝑦 𝐿 2 , 𝐿 𝑧 =0

Continued 𝐻 , 𝐿 𝑧 = 𝐿 2 2𝐼 , 𝐿 𝑧 =0 These commutation relations confirm that 𝑌 𝑙𝑚 are the eigenfunctions for 𝐻 , 𝐿 2 and 𝐿 𝑧 , i.e. 𝐻 𝑌 𝑙𝑚 = 𝑙 𝑙+1 ℏ 2 2𝐼 𝑌 𝑙𝑚 𝐿 2 𝑌 𝑙𝑚 =𝑙 𝑙+1 ℏ 2 𝑌 𝑙𝑚 𝐿 𝑧 𝑌 𝑙𝑚 =𝑚ℏ 𝑌 𝑙𝑚

Schrödinger equation for hydrogen atom Some notations Mass: nucleus 𝑚 𝑁 , electron 𝑚 𝑒 , total 𝑚 CM = 𝑚 𝑁 + 𝑚 𝑒 , reduced 𝜇= 𝑚 𝑒 𝑚 𝑁 𝑚 𝑒 + 𝑚 𝑁 Position vector: nucleus 𝑟 𝑁 , electron 𝑟 𝑒 , centre of mass (CM) 𝑅 = 𝑚 𝑒 𝑟 𝑒 + 𝑚 𝑁 𝑟 𝑁 𝑚 𝑒 + 𝑚 𝑁 , electron relative to nucleus 𝑟 = 𝑟 𝑒 − 𝑟 𝑁 If 𝑅 = 𝑅 𝑥 𝑒 𝑥 + 𝑅 𝑦 𝑒 𝑦 + 𝑅 𝑧 𝑒 𝑧 , ∇ 𝑅 2 = 𝜕 2 𝜕 𝑅 𝑥 2 + 𝜕 2 𝜕 𝑅 𝑦 2 + 𝜕 2 𝜕 𝑅 𝑧 2 (similar for ∇ 𝑟 2 ) Classical momentum: nucleus 𝑝 𝑁 = 𝑚 𝑁 𝑟 𝑁 , electron 𝑝 𝑒 = 𝑚 𝑒 𝑟 𝑒 , CM 𝑝 CM = 𝑚 CM 𝑅 , electron relative to nucleus 𝑝 𝜇 =𝜇 𝑟 Vector without arrow means modulus 𝑟= 𝑟 , etc O positron 𝑚 𝑁 𝑚 𝑒 electron 𝑟 𝑒 𝑟 𝑁 𝑟 = 𝑟 𝑒 − 𝑟 𝑁

Continued Classical energy: 𝐸= 𝑝 𝑒 2 2 𝑚 𝑒 + 𝑝 𝑁 2 2 𝑚 𝑁 − 𝑒 2 𝑟 = 𝑝 CM 2 2 𝑚 CM + 𝑝 𝜇 2 2𝜇 − 𝑒 2 𝑟 , CM motion and relative motion are separated Quantum Hamiltonian: 𝐻 =− ℏ 2 2 𝑚 CM ∇ 𝑅 2 − ℏ 2 2𝜇 ∇ 𝑟 2 − 𝑒 2 𝑟 S.E. − ℏ 2 2 𝑚 CM ∇ 𝑅 2 − ℏ 2 2𝜇 ∇ 𝑟 2 − 𝑒 2 𝑟 Φ 𝑟 , 𝑅 =𝐸Φ 𝑟 , 𝑅

Separation of 𝑟 and 𝑅 Let Φ 𝑟 , 𝑅 =𝜒 𝑅 𝜓 𝑟 − ℏ 2 2 𝑚 CM ∇ 𝑅 2 𝜒 𝜓− ℏ 2 2𝜇 𝜒 ∇ 𝑟 2 𝜓− 𝑒 2 𝑟 𝜒𝜓=𝐸𝜒𝜓 divide both sides by 𝜒𝜓: − ℏ 2 2 𝑚 CM ∇ 𝑅 2 𝜒 𝜒 − ℏ 2 2𝜇 ∇ 𝑟 2 𝜓 𝜓 − 𝑒 2 𝑟 =𝐸 Thus − ℏ 2 2 𝑚 CM ∇ 𝑅 2 𝜒 𝜒 = 𝐸 CM and − ℏ 2 2𝜇 ∇ 𝑟 2 𝜓 𝜓 − 𝑒 2 𝑟 = 𝐸 𝑒 where 𝐸 CM + 𝐸 𝑒 =𝐸

Equation of 𝜒 − ℏ 2 2 𝑚 CM ∇ 𝑅 2 𝜒= 𝐸 CM 𝜒 This is just a free particle moving in 3D space Plane wave solution 𝜒 𝑅 =𝐴 𝑒 𝑖 𝑘 CM ⋅ 𝑅 Wave vector: 𝑘 CM with modulus 2 𝑚 CM 𝐸 CM ℏ and along the direction of 𝜐 𝑝

Equation of 𝜓 − ℏ 2 2𝜇 ∇ 𝑟 2 𝜓− 𝑒 2 𝑟 𝜓= 𝐸 𝑒 𝜓 − ℏ 2 2𝜇 ∇ 𝑟 2 𝜓− 𝑒 2 𝑟 𝜓= 𝐸 𝑒 𝜓 ∇ 𝑟 2 can be expressed in spherical coordinate system located at nucleus ∇ 𝑟 2 = 𝜕 2 𝜕 𝑟 2 + 2 𝑟 𝜕 𝜕𝑟 + Λ 2 𝑟 2 From now on, we will omit the subscript of 𝐸 𝑒 for simplicity Let 𝜓 𝑟 =𝑅 𝑟 𝑌 𝜃,𝜙 − ℏ 2 2𝜇 𝑟 2 𝜕 2 𝑅 𝜕 𝑟 2 + 2 𝑟 𝜕𝑅 𝜕𝑟 𝑌− 𝑒 2 𝑟𝑅𝑌−𝐸 𝑟 2 𝑅𝑌− ℏ 2 2𝜇 𝑅 Λ 2 𝑌=0

Continued divide both sides by 𝑅𝑌: − ℏ 2 𝑟 2 2𝜇𝑅 𝜕 2 𝑅 𝜕 𝑟 2 + 2 𝑟 𝜕𝑅 𝜕𝑟 − 𝑒 2 𝑟−𝐸 𝑟 2 − ℏ 2 2𝜇𝑌 Λ 2 𝑌=0 Thus − ℏ 2 2𝜇𝑌 Λ 2 𝑌=𝐴 − ℏ 2 𝑟 2 2𝜇𝑅 𝜕 2 𝑅 𝜕 𝑟 2 + 2 𝑟 𝜕𝑅 𝜕𝑟 − 𝑒 2 𝑟−𝐸 𝑟 2 =−𝐴

Equation of 𝑌 Rearrange − ℏ 2 2𝜇𝑌 Λ 2 𝑌=𝐴 as Λ 2 𝑌=− 2𝜇𝐴 ℏ 2 𝑌 𝑌= 𝑌 𝑙 𝑚 𝑙 𝜃,𝜙 , − 2𝜇𝐴 ℏ 2 =−𝑙 𝑙+1 𝐴= 𝑙 𝑙+1 ℏ 2 2𝜇

Equation of 𝑅 − ℏ 2 𝑟 2 2𝜇𝑅 𝜕 2 𝑅 𝜕 𝑟 2 + 2 𝑟 𝜕𝑅 𝜕𝑟 − 𝑒 2 𝑟−𝐸 𝑟 2 =− 𝑙 𝑙+1 ℏ 2 2𝜇 i.e. − ℏ 2 2𝜇 𝜕 2 𝜕 𝑟 2 + 2 𝑟 𝜕 𝜕𝑟 + 𝑙 𝑙+1 ℏ 2 2𝜇 𝑟 2 − 𝑒 2 𝑟 𝑅=𝐸𝑅