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Physical Chemistry Week 12
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Properties of spherical harmonics
Ξ 2 π ππ =βπ π+1 π ππ Energy: π» =β β 2 2π β
Ξ 2 π 2 , π» π ππ = β 2 2πΌ β
π π+1 π ππ , πΈ π = π π+1 β 2 2πΌ Square of angular momentum ( πΏ 2 ): Classical: πΈ= πΏ 2 /2πΌ; quantum: πΏ 2 =β β 2 Ξ 2 πΏ 2 π ππ =π π+1 β 2 π ππ , πΏ 2 π =π π+1 β 2
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Angular momentum ( πΏ ) Classical: πΏ = π Γ π ; quantum: πΏ = π Γ π
π =π₯ π π₯ +π¦ π π¦ +π§ π π§ , π = π π₯ π π₯ + π π¦ π π¦ + π π§ π π§ π π = β π π ππ , π=π₯,π¦,π§ βΓβ means cross product π Γ π π =π π Γ π , π Γ π = 0 , π Γ π =β π Γ π , π Γ π + π = π Γ π + π Γ π π π₯ Γ π π¦ = π π§ , π π¦ Γ π π§ = π π₯ , π π§ Γ π π₯ = π π¦
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Continued πΏ &= π₯ π π₯ +π¦ π π¦ +π§ π π§ Γ β π π ππ₯ π π₯ + π ππ¦ π π¦ + π ππ§ π π§ &= β π π₯ π ππ¦ βπ¦ π ππ₯ π π§ + π¦ π ππ§ βπ§ π ππ¦ π π₯ + π§ π ππ₯ βπ₯ π ππ§ π π¦ πΏ π₯ = β π π¦ π ππ§ βπ§ π ππ¦ πΏ π¦ = β π π§ π ππ₯ βπ₯ π ππ§ πΏ π§ = β π π₯ π ππ¦ βπ¦ π ππ₯
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Commutation relations
πΏ π₯ , πΏ π¦ &=β β 2 π¦ π ππ§ βπ§ π ππ¦ π§ π ππ₯ βπ₯ π ππ§ β π§ π ππ₯ βπ₯ π ππ§ π¦ π ππ§ βπ§ π ππ¦ &=β β 2 π¦ π ππ₯ +π¦π§ π 2 ππ§ππ₯ βπ¦π₯ π 2 π π§ 2 β π§ 2 π 2 ππ¦ππ₯ +π§π₯ π 2 ππ¦ππ§ & β 2 π§π¦ π 2 ππ₯ππ§ β π§ 2 π 2 ππ₯ππ¦ βπ₯π¦ π 2 π π§ 2 +π₯ π ππ¦ +π₯π§ π 2 ππ§ππ¦ &= β 2 π₯ π ππ¦ βπ¦ π ππ₯ &=πβ πΏ π§ πΏ π₯ , πΏ π¦ =πβ πΏ π§ , πΏ π¦ , πΏ π§ =πβ πΏ π₯ , πΏ π§ , πΏ π₯ =πβ πΏ π¦
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Continued πΏ 2 , πΏ π§ = πΏ π₯ 2 , πΏ π§ + πΏ π¦ 2 , πΏ π§ + πΏ π§ 2 , πΏ π§
πΏ 2 , πΏ π§ = πΏ π₯ 2 , πΏ π§ + πΏ π¦ 2 , πΏ π§ + πΏ π§ 2 , πΏ π§ πΏ π§ 2 , πΏ π§ = πΏ π§ 3 β πΏ π§ 3 =0 β’ πΏ π₯ 2 , πΏ π§ &= πΏ π₯ 2 πΏ π§ β πΏ π₯ πΏ π§ πΏ π₯ + πΏ π₯ πΏ π§ πΏ π₯ β πΏ π§ πΏ π₯ 2 &= πΏ π₯ πΏ π₯ , πΏ π§ + πΏ π₯ , πΏ π§ πΏ π₯ &=βπβ πΏ π₯ πΏ π¦ + πΏ π¦ πΏ π₯ πΏ π¦ 2 , πΏ π§ =πβ πΏ π¦ πΏ π₯ + πΏ π₯ πΏ π¦ πΏ 2 , πΏ π§ =0
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Continued π» , πΏ π§ = πΏ 2 2πΌ , πΏ π§ =0 These commutation relations confirm that π ππ are the eigenfunctions for π» , πΏ 2 and πΏ π§ , i.e. π» π ππ = π π+1 β 2 2πΌ π ππ πΏ 2 π ππ =π π+1 β 2 π ππ πΏ π§ π ππ =πβ π ππ
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SchrΓΆdinger equation for hydrogen atom
Some notations Mass: nucleus π π , electron π π , total π CM = π π + π π , reduced π= π π π π π π + π π Position vector: nucleus π π , electron π π , centre of mass (CM) π
= π π π π + π π π π π π + π π , electron relative to nucleus π = π π β π π If π
= π
π₯ π π₯ + π
π¦ π π¦ + π
π§ π π§ , β π
2 = π 2 π π
π₯ π 2 π π
π¦ π 2 π π
π§ 2 (similar for β π 2 ) Classical momentum: nucleus π π = π π π π , electron π π = π π π π , CM π CM = π CM π
, electron relative to nucleus π π =π π Vector without arrow means modulus π= π , etc O positron π π π π electron π π π π π = π π β π π
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Continued Classical energy: πΈ= π π 2 2 π π + π π 2 2 π π β π 2 π = π CM 2 2 π CM + π π 2 2π β π 2 π , CM motion and relative motion are separated Quantum Hamiltonian: π» =β β 2 2 π CM β π
2 β β 2 2π β π 2 β π 2 π S.E. β β 2 2 π CM β π
2 β β 2 2π β π 2 β π 2 π Ξ¦ π , π
=πΈΞ¦ π , π
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Separation of π and π
Let Ξ¦ π , π
=π π
π π β β 2 2 π CM β π
2 π πβ β 2 2π π β π 2 πβ π 2 π ππ=πΈππ divide both sides by ππ: β β 2 2 π CM β π
2 π π β β 2 2π β π 2 π π β π 2 π =πΈ Thus β β 2 2 π CM β π
2 π π = πΈ CM and β β 2 2π β π 2 π π β π 2 π = πΈ π where πΈ CM + πΈ π =πΈ
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Equation of π β β 2 2 π CM β π
2 π= πΈ CM π
This is just a free particle moving in 3D space Plane wave solution π π
=π΄ π π π CM β
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Wave vector: π CM with modulus 2 π CM πΈ CM β and along the direction of π π
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Equation of π β β 2 2π β π 2 πβ π 2 π π= πΈ π π
β β 2 2π β π 2 πβ π 2 π π= πΈ π π β π 2 can be expressed in spherical coordinate system located at nucleus β π 2 = π 2 π π π π ππ + Ξ 2 π 2 From now on, we will omit the subscript of πΈ π for simplicity Let π π =π
π π π,π β β 2 2π π 2 π 2 π
π π π ππ
ππ πβ π 2 ππ
πβπΈ π 2 π
πβ β 2 2π π
Ξ 2 π=0
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Continued divide both sides by π
π: β β 2 π 2 2ππ
π 2 π
π π π ππ
ππ β π 2 πβπΈ π 2 β β 2 2ππ Ξ 2 π=0 Thus β β 2 2ππ Ξ 2 π=π΄ β β 2 π 2 2ππ
π 2 π
π π π ππ
ππ β π 2 πβπΈ π 2 =βπ΄
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Equation of π Rearrange β β 2 2ππ Ξ 2 π=π΄ as Ξ 2 π=β 2ππ΄ β 2 π
π= π π π π π,π , β 2ππ΄ β 2 =βπ π+1 π΄= π π+1 β 2 2π
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Equation of π
β β 2 π 2 2ππ
π 2 π
π π π ππ
ππ β π 2 πβπΈ π 2 =β π π+1 β 2 2π i.e. β β 2 2π π 2 π π π π ππ + π π+1 β 2 2π π 2 β π 2 π π
=πΈπ
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