Šredingerio lygtis yra pagrindinė kvantinės mechanikos lygtis, aprašanti kvantinių dalelių elgesį. Šią lygtį 1925 metais pasiūlė austrų fizikas Ervinas Šriodingeris. Bendru atveju ji užrašoma taip:
H ^ ψ ( r → , t ) = i ℏ ∂ ∂ t ψ ( r → , t ) = − ℏ 2 2 μ ∂ 2 ψ ( r → , t ) ∂ r → 2 + V ( r → ) {\displaystyle {\hat {H}}\psi ({\vec {r}},t)=i\hbar {\partial \over \partial t}\psi ({\vec {r}},t)=-{\hbar ^{2} \over 2\mu }{\partial ^{2}\psi ({\vec {r}},t) \over \partial {\vec {r}}^{2}}+V({\vec {r}})} Čia H ^ {\displaystyle {\hat {H}}} ψ ( r → , t ) {\displaystyle \psi ({\vec {r}},t)} V ( r → ) {\displaystyle V({\vec {r}})} μ {\displaystyle \mu } ℏ {\displaystyle \hbar } i - menamasis vienetas. Lygtis sprendžiama banginės funkcijos atžvilgiu, radus ją galime pilnai aprašyti nagrinėjamą dalelę. Būtent dėl to ši lygtis kartais vadinama antrojo Niutono dėsnio analogu kvantiniame pasaulyje.
Šredingerio lygtis yra vienas iš kvantinės mechanikos postulatų – ją galima užrašyti tik pasirėmus įvairias pasamprotavimais apie banginę dalelių prigimtį, griežto jos išvedimo nėra.
Bendroji Šredingerio lygtis
keisti
Nereliatyvistinėje kvantinėje mechanikoje hamiltonianas yra tiesiog kinetinės ir potencinės energijų suma. Tokiu atveju Šredingerio lygtis atrodo taip:
[ − ℏ 2 2 m ∇ 2 + U ( r → , t ) ] ψ ( r → , t ) = i ℏ ∂ ∂ t ψ ( r → , t ) = E ψ ( r → , t ) {\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+U({\vec {r}},t)\right]\psi ({\vec {r}},t)=i\hbar {\partial \over \partial t}\psi ({\vec {r}},t)=E\psi ({\vec {r}},t)} Čia ∇ 2 {\displaystyle \nabla ^{2}} U ( r → , t ) {\displaystyle U({\vec {r}},t)}
Stacionarioji Šredingerio lygtis
keisti
Jei nagrinėjama sistema yra stacionari, t.y. sistemos hamiltonianas išreikštai nepriklauso nuo laiko t {\displaystyle t}
ψ ( r → , t ) = Φ ( r → ) A ( t ) {\displaystyle \psi ({\vec {r}},t)=\Phi ({\vec {r}})A(t)} Iš čia gauname tikrinių verčių lygtį funkcijai Φ ( r → ) {\displaystyle \Phi ({\vec {r}})}
H ^ Φ ( r → ) = E Φ ( r → ) {\displaystyle {\hat {H}}\Phi ({\vec {r}})=E\Phi ({\vec {r}})} bei A ( t ) {\displaystyle A(t)}
ψ ( r → , t ) = e − i ℏ E t Φ ( r → ) {\displaystyle \psi ({\vec {r}},t)=e^{-{\frac {i}{\hbar }}Et}\Phi ({\vec {r}})} Čia E {\displaystyle E}
Įrodysime, kad Ψ ( x , t ) = e i ( p x − E t ) {\displaystyle \Psi (x,t)=e^{i(px-Et)}} i ∂ Ψ ∂ t = − 1 2 m ∂ 2 Ψ ∂ x 2 {\displaystyle i{\frac {\partial \Psi }{\partial t}}=-{\frac {1}{2m}}{\frac {\partial ^{2}\Psi }{\partial x^{2}}}} p = m v {\displaystyle p=mv} E - energija, m - masė, i - menamasis vienetas.
i ∂ ( e i ( p x − E t ) ) ∂ t = − 1 2 m ∂ 2 ( e i ( p x − E t ) ) ∂ x 2 ; {\displaystyle i{\frac {\partial (e^{i(px-Et)})}{\partial t}}=-{\frac {1}{2m}}{\frac {\partial ^{2}(e^{i(px-Et)})}{\partial x^{2}}};} i ⋅ ( − i E ) e i ( p x − E t ) = − ( i p ) 2 2 m e i ( p x − E t ) ; {\displaystyle i\cdot (-iE)e^{i(px-Et)}=-{\frac {(ip)^{2}}{2m}}e^{i(px-Et)};} E = p 2 2 m = ( m v ) 2 2 m = m v 2 2 . {\displaystyle E={\frac {p^{2}}{2m}}={(mv)^{2} \over 2m}={mv^{2} \over 2}.} Gavome kinetinės energijos formulę, kuri įrodo, kad Šrėdingerio lygtis išspręsta teisingai. x yra koordinate vienmatėje erdvėje (nejudanti), o t yra laikas ir taip kvantinė dalelė aprašoma bangine funkcija.
Šredingerio lygtis sferinėse koordinatėse
keisti
− ℏ 2 2 μ [ 1 r 2 ∂ ∂ r ( r 2 ∂ ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ ∂ θ ) + 1 r 2 sin 2 θ ∂ 2 ∂ ϕ 2 + V ( r ) ] Ψ ( r , θ , ϕ ) = E Ψ ( r , θ , ϕ ) , {\displaystyle -{\hbar ^{2} \over 2\mu }[{1 \over r^{2}}{\partial \over \partial r}(r^{2}{\partial \over \partial r})+{1 \over r^{2}\sin \theta }{\partial \over \partial \theta }(\sin \theta {\partial \over \partial \theta })+{1 \over r^{2}\sin ^{2}\theta }{\partial ^{2} \over \partial \phi ^{2}}+V(r)]\Psi (r,\theta ,\phi )=E\Psi (r,\theta ,\phi ),} − ℏ 2 2 μ r 2 [ ∂ ∂ r ( r 2 ∂ Ψ ∂ r ) + 1 sin θ ∂ ∂ θ ( sin θ ∂ Ψ ∂ θ ) + 1 sin 2 θ ∂ 2 Ψ ∂ ϕ 2 ] + ( V ( r ) − E ) Ψ = 0 , {\displaystyle -{\hbar ^{2} \over 2\mu r^{2}}[{\partial \over \partial r}(r^{2}{\partial \Psi \over \partial r})+{1 \over \sin \theta }{\partial \over \partial \theta }(\sin \theta {\partial \Psi \over \partial \theta })+{1 \over \sin ^{2}\theta }{\partial ^{2}\Psi \over \partial \phi ^{2}}]+(V(r)-E)\Psi =0,}
Ψ ( r , θ , ϕ ) = R ( r ) Y ( θ , ϕ ) , {\displaystyle \;\Psi (r,\theta ,\phi )=R(r)Y(\theta ,\phi ),} Y ∂ ∂ r ( r 2 ∂ R ∂ r ) + R sin θ ∂ ∂ θ ( sin θ ∂ Y ∂ θ ) + R sin 2 θ ∂ 2 Y ∂ ϕ 2 − 2 μ r 2 ℏ 2 ( V ( r ) − E ) R Y = 0 , {\displaystyle Y{\partial \over \partial r}(r^{2}{\partial R \over \partial r})+{R \over \sin \theta }{\partial \over \partial \theta }(\sin \theta {\partial Y \over \partial \theta })+{R \over \sin ^{2}\theta }{\partial ^{2}Y \over \partial \phi ^{2}}-{2\mu r^{2} \over \hbar ^{2}}(V(r)-E)RY=0,}
1 R ∂ ∂ r ( r 2 ∂ R ∂ r ) − 2 μ r 2 ℏ 2 ( V ( r ) − E ) = − 1 Y ( 1 sin θ ∂ ∂ θ ( sin θ ∂ Y ∂ θ ) + 1 sin 2 θ ∂ 2 Y ∂ ϕ 2 ) = l ( l + 1 ) , {\displaystyle {1 \over R}{\partial \over \partial r}(r^{2}{\partial R \over \partial r})-{2\mu r^{2} \over \hbar ^{2}}(V(r)-E)=-{1 \over Y}({1 \over \sin \theta }{\partial \over \partial \theta }(\sin \theta {\partial Y \over \partial \theta })+{1 \over \sin ^{2}\theta }{\partial ^{2}Y \over \partial \phi ^{2}})=l(l+1),}
1 R ( 2 r ⋅ d R d r + r 2 ⋅ d 2 R d r 2 ) − 2 μ r 2 ℏ 2 ( V ( r ) − E ) = − 1 Y [ 1 sin θ ( cos θ ∂ Y ∂ θ + sin θ ⋅ ∂ 2 Y ∂ θ 2 ) + 1 sin 2 θ ⋅ ∂ 2 Y ∂ ϕ 2 ] = l ( l + 1 ) , {\displaystyle {1 \over R}(2r\cdot {dR \over dr}+r^{2}\cdot {d^{2}R \over dr^{2}})-{2\mu r^{2} \over \hbar ^{2}}(V(r)-E)=-{1 \over Y}[{1 \over \sin \theta }(\cos \theta {\partial Y \over \partial \theta }+\sin \theta \cdot {\partial ^{2}Y \over \partial \theta ^{2}})+{1 \over \sin ^{2}\theta }\cdot {\partial ^{2}Y \over \partial \phi ^{2}}]=l(l+1),} l - šalutinis kvantinis skaičius: 0, 1, 2, 3...
Radialinė lygtis:
1 R ( r ) d d r ( r 2 d R ( r ) d r ) − 2 μ r 2 ℏ 2 ( V ( r ) − E ) = l ( l + 1 ) , {\displaystyle {1 \over R(r)}{d \over dr}(r^{2}{dR(r) \over dr})-{2\mu r^{2} \over \hbar ^{2}}(V(r)-E)=l(l+1),} − ℏ 2 2 μ r 2 d d r ( r 2 d R d r ) + [ V ( r ) − E + ℏ 2 2 μ r 2 l ( l + 1 ) ] R ( r ) = 0 ; {\displaystyle -{\hbar ^{2} \over 2\mu r^{2}}{d \over dr}(r^{2}{dR \over dr})+[V(r)-E+{\hbar ^{2} \over 2\mu r^{2}}l(l+1)]R(r)=0;} R ( r ) = u ( r ) r ; d R d r = 1 r d u ( r ) d r − u ( r ) r 2 ; r 2 d R ( r ) d r = r d u ( r ) d r − u ( r ) ; {\displaystyle R(r)={u(r) \over r};\;{dR \over dr}={1 \over r}{du(r) \over dr}-{u(r) \over r^{2}};\;r^{2}{dR(r) \over dr}=r{du(r) \over dr}-u(r);\;} d d r ( r 2 d R d r ) = d d r ( r d u ( r ) d r − u ( r ) ) = d u ( r ) d r + r d 2 u ( r ) d r 2 − d u ( r ) d r = r d 2 u ( r ) d r 2 , {\displaystyle {d \over dr}(r^{2}{dR \over dr})={d \over dr}(r{du(r) \over dr}-u(r))={du(r) \over dr}+r{d^{2}u(r) \over dr^{2}}-{du(r) \over dr}=r{d^{2}u(r) \over dr^{2}},}
− ℏ 2 2 μ r 2 r d 2 u ( r ) d r 2 + [ V ( r ) − E + ℏ 2 2 μ r 2 l ( l + 1 ) ] u ( r ) r = 0 , {\displaystyle -{\hbar ^{2} \over 2\mu r^{2}}r{d^{2}u(r) \over dr^{2}}+[V(r)-E+{\hbar ^{2} \over 2\mu r^{2}}l(l+1)]{u(r) \over r}=0,} − ℏ 2 2 μ d 2 u ( r ) d r 2 + [ V ( r ) + ℏ 2 2 μ r 2 l ( l + 1 ) ] u ( r ) = E u ( r ) ; V e f = V ( r ) + ℏ 2 2 μ r 2 ⋅ l ( l + 1 ) ; E u ( r ) = i ℏ d d r u ( r ) , {\displaystyle -{\hbar ^{2} \over 2\mu }{d^{2}u(r) \over dr^{2}}+[V(r)+{\hbar ^{2} \over 2\mu r^{2}}l(l+1)]u(r)=Eu(r);\;V_{ef}=V(r)+{\hbar ^{2} \over 2\mu r^{2}}\cdot l(l+1);\;Eu(r)=i\hbar {d \over dr}u(r),} − h 2 2 μ d 2 u ( r ) d r 2 + V e f u ( r ) = i ℏ d d r u ( r ) . {\displaystyle -{h^{2} \over 2\mu }{d^{2}u(r) \over dr^{2}}+V_{ef}u(r)=i\hbar {d \over dr}u(r).} Atskyrimas kintamųjų kampinėje lygtyje:
− 1 Y ( 1 sin θ ∂ ∂ θ ( sin θ ∂ Y ∂ θ ) + 1 sin 2 θ ∂ 2 Y ∂ ϕ 2 ) = l ( l + 1 ) , {\displaystyle -{1 \over Y}({1 \over \sin \theta }{\partial \over \partial \theta }(\sin \theta {\partial Y \over \partial \theta })+{1 \over \sin ^{2}\theta }{\partial ^{2}Y \over \partial \phi ^{2}})=l(l+1),} sin θ ∂ ∂ θ ( sin θ ∂ Y ∂ θ ) + l ( l + 1 ) sin 2 θ ⋅ Y ( θ , ϕ ) + ∂ 2 Y ∂ ϕ 2 = 0 , {\displaystyle \sin \theta {\partial \over \partial \theta }(\sin \theta {\partial Y \over \partial \theta })+l(l+1)\sin ^{2}\theta \cdot Y(\theta ,\phi )+{\partial ^{2}Y \over \partial \phi ^{2}}=0,} Y ( θ , ϕ ) = O ( θ ) F ( ϕ ) , {\displaystyle Y(\theta ,\phi )=O(\theta )F(\phi ),} F sin θ ∂ ∂ θ ( sin θ ∂ O ∂ θ ) + l ( l + 1 ) sin 2 θ ⋅ O ( θ ) F ( ϕ ) + O ∂ 2 F ∂ ϕ 2 = 0 , {\displaystyle F\sin \theta {\partial \over \partial \theta }(\sin \theta {\partial O \over \partial \theta })+l(l+1)\sin ^{2}\theta \cdot O(\theta )F(\phi )+O{\partial ^{2}F \over \partial \phi ^{2}}=0,} 1 O sin θ ∂ ∂ θ ( sin θ ∂ O ∂ θ ) + l ( l + 1 ) sin 2 θ = − 1 F ∂ 2 F ∂ ϕ 2 = m 2 , {\displaystyle {1 \over O}\sin \theta {\partial \over \partial \theta }(\sin \theta {\partial O \over \partial \theta })+l(l+1)\sin ^{2}\theta =-{1 \over F}{\partial ^{2}F \over \partial \phi ^{2}}=m^{2},} sin θ ∂ ∂ θ ( sin θ ∂ O ∂ θ ) + [ l ( l + 1 ) sin 2 θ − m 2 ] O = 0 ; {\displaystyle \sin \theta {\partial \over \partial \theta }(\sin \theta {\partial O \over \partial \theta })+[l(l+1)\sin ^{2}\theta -m^{2}]O=0;} ∂ 2 F ∂ ϕ 2 = − F m 2 , F = e i m ϕ , {\displaystyle {\partial ^{2}F \over \partial \phi ^{2}}=-Fm^{2},\;F=e^{im\phi },} kur m - magnetinis kvantinis skaičius.
O ( θ ) {\displaystyle O(\theta )}
sin θ d d θ ( sin θ d O d θ ) + [ l ( l + 1 ) sin 2 θ − m 2 ] O = 0 ; {\displaystyle \sin \theta {d \over d\theta }(\sin \theta {dO \over d\theta })+[l(l+1)\sin ^{2}\theta -m^{2}]O=0;} O ( θ ) = A P l m ( cos θ ) {\displaystyle O(\theta )=AP_{l}^{m}(\cos \theta )} P l m ( x ) = ( 1 − x 2 ) | m | / 2 d | m | P l ( x ) d x | m | ; P l ( x ) = 1 2 l l ! d l d x l ( x 2 − 1 ) l ; {\displaystyle P_{l}^{m}(x)=(1-x^{2})^{|m|/2}{d^{|m|}P_{l}(x) \over dx^{|m|}};\;P_{l}(x)={1 \over 2^{l}l!}{d^{l} \over dx^{l}}(x^{2}-1)^{l};} kadangi, P l ( x ) {\displaystyle P_{l}(x)} l , | m | ≤ l , {\displaystyle |m|\leq l,} P 0 ( x ) = 1 2 0 ⋅ 0 ! d 0 d x 0 ( x 2 − 1 ) 0 = 1 1 ⋅ 1 ⋅ 1 ⋅ 1 = 1 ; P 1 ( x ) = 1 2 1 ⋅ 1 ! d 1 d x 1 ( x 2 − 1 ) 1 = 1 2 ⋅ 2 x = x ; {\displaystyle P_{0}(x)={1 \over 2^{0}\cdot 0!}{d^{0} \over dx^{0}}(x^{2}-1)^{0}={1 \over 1\cdot 1}\cdot 1\cdot 1=1;\;P_{1}(x)={1 \over 2^{1}\cdot 1!}{d^{1} \over dx^{1}}(x^{2}-1)^{1}={1 \over 2}\cdot 2x=x;} P 2 ( x ) = 1 2 2 ⋅ 2 ! d 2 d x 2 ( x 2 − 1 ) 2 = 1 8 d 2 d x 2 ( x 4 − 2 x 2 + 1 ) = 1 8 d d x ( 4 x 3 − 4 x ) = 1 8 ( 12 x 2 − 4 ) = 3 x 2 − 1 2 ; {\displaystyle P_{2}(x)={1 \over 2^{2}\cdot 2!}{d^{2} \over dx^{2}}(x^{2}-1)^{2}={1 \over 8}{d^{2} \over dx^{2}}(x^{4}-2x^{2}+1)={1 \over 8}{d \over dx}(4x^{3}-4x)={1 \over 8}(12x^{2}-4)={3x^{2}-1 \over 2};} P 2 1 ( x ) = ( 1 − x 2 ) | 1 | / 2 d | 1 | ( 3 x 2 − 1 2 ) d x | 1 | = 3 x 1 − x 2 = 3 cos θ sin θ ; {\displaystyle P_{2}^{1}(x)=(1-x^{2})^{|1|/2}{d^{|1|}({3x^{2}-1 \over 2}) \over dx^{|1|}}=3x{\sqrt {1-x^{2}}}=3\cos \theta \sin \theta ;} P 2 2 ( x ) = ( 1 − x 2 ) | 2 | / 2 d | 2 | ( 3 x 2 − 1 2 ) d x | 2 | = ( 1 − x 2 ) d d x ( 3 x ) = 3 ( 1 − x 2 ) = 3 sin 2 θ , {\displaystyle P_{2}^{2}(x)=(1-x^{2})^{|2|/2}{d^{|2|}({3x^{2}-1 \over 2}) \over dx^{|2|}}=(1-x^{2}){d \over dx}(3x)=3(1-x^{2})=3\sin ^{2}\theta ,}
pavyzdžiai Legendarinių funkcijų: P 0 0 = 1 ; P 1 1 = sin θ ; P 0 1 = cos θ ; P 2 2 = 3 sin 2 θ ; P 2 1 = 3 sin θ cos θ ; P 2 0 = 1 2 ( 3 cos 2 θ − 1 ) ; {\displaystyle P_{0}^{0}=1;\;P_{1}^{1}=\sin \theta ;\;P_{0}^{1}=\cos \theta ;\;P_{2}^{2}=3\sin ^{2}\theta ;\;P_{2}^{1}=3\sin \theta \cos \theta ;\;P_{2}^{0}={1 \over 2}(3\cos ^{2}\theta -1);} P 3 3 = 15 sin 3 θ ; P 3 2 = 15 sin 2 θ cos θ ; P 3 1 = 3 2 sin θ ( 5 cos 2 θ − 1 ) ; P 3 0 = 1 2 ( 5 cos 3 θ − 3 cos θ ) . {\displaystyle P_{3}^{3}=15\sin ^{3}\theta ;\;P_{3}^{2}=15\sin ^{2}\theta \cos \theta ;\;P_{3}^{1}={3 \over 2}\sin \theta (5\cos ^{2}\theta -1);\;P_{3}^{0}={1 \over 2}(5\cos ^{3}\theta -3\cos \theta ).}
Tai pavyzdžiai sukamų funkcijų apie z ašį (brėžinis (x, y) plokštumoje), kurios, žinoma yra dvimatės. Kad jos butų apibūdinamos trimatėjėje erdvėje, reikia, kad būtų panaudotas taip pat kampas ϕ {\displaystyle \phi } Y l m ( θ , ϕ ) = ϵ 2 l + 1 ( l − | m | ) ! 4 π ( l + | m | ) ! e i m ϕ P l m ( cos θ ) ; ϵ = ( − 1 ) m , m ≥ 0 ; ϵ = 1 , m < 0. {\displaystyle Y_{l}^{m}(\theta ,\phi )=\epsilon {\sqrt {2l+1(l-|m|)! \over 4\pi (l+|m|)!}}e^{im\phi }P_{l}^{m}(\cos \theta );\;\epsilon =(-1)^{m},m\geq 0;\;\epsilon =1,m<0.}
![{\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+U({\vec {r}},t)\right]\psi ({\vec {r}},t)=i\hbar {\partial \over \partial t}\psi ({\vec {r}},t)=E\psi ({\vec {r}},t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e10eeadc2386bb7c6c434c0e09ea2d1d7a61cfdf)
![{\displaystyle -{\hbar ^{2} \over 2\mu }[{1 \over r^{2}}{\partial \over \partial r}(r^{2}{\partial \over \partial r})+{1 \over r^{2}\sin \theta }{\partial \over \partial \theta }(\sin \theta {\partial \over \partial \theta })+{1 \over r^{2}\sin ^{2}\theta }{\partial ^{2} \over \partial \phi ^{2}}+V(r)]\Psi (r,\theta ,\phi )=E\Psi (r,\theta ,\phi ),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f40d82b3b491d5bd6875b1c2aca37d0f5a905848)
![{\displaystyle -{\hbar ^{2} \over 2\mu r^{2}}[{\partial \over \partial r}(r^{2}{\partial \Psi \over \partial r})+{1 \over \sin \theta }{\partial \over \partial \theta }(\sin \theta {\partial \Psi \over \partial \theta })+{1 \over \sin ^{2}\theta }{\partial ^{2}\Psi \over \partial \phi ^{2}}]+(V(r)-E)\Psi =0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/068201562aa50706d881012e456795ebfb3d2020)
![{\displaystyle {1 \over R}(2r\cdot {dR \over dr}+r^{2}\cdot {d^{2}R \over dr^{2}})-{2\mu r^{2} \over \hbar ^{2}}(V(r)-E)=-{1 \over Y}[{1 \over \sin \theta }(\cos \theta {\partial Y \over \partial \theta }+\sin \theta \cdot {\partial ^{2}Y \over \partial \theta ^{2}})+{1 \over \sin ^{2}\theta }\cdot {\partial ^{2}Y \over \partial \phi ^{2}}]=l(l+1),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdfa2d62caefe3c690b141f9ab5681d0f75598e0)
![{\displaystyle -{\hbar ^{2} \over 2\mu r^{2}}{d \over dr}(r^{2}{dR \over dr})+[V(r)-E+{\hbar ^{2} \over 2\mu r^{2}}l(l+1)]R(r)=0;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75729e6630d2e179707cfc378b93397fe09787ed)
![{\displaystyle -{\hbar ^{2} \over 2\mu r^{2}}r{d^{2}u(r) \over dr^{2}}+[V(r)-E+{\hbar ^{2} \over 2\mu r^{2}}l(l+1)]{u(r) \over r}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2de5c0d410a16753bddfb1da6f6e6266f63c7e4b)
![{\displaystyle -{\hbar ^{2} \over 2\mu }{d^{2}u(r) \over dr^{2}}+[V(r)+{\hbar ^{2} \over 2\mu r^{2}}l(l+1)]u(r)=Eu(r);\;V_{ef}=V(r)+{\hbar ^{2} \over 2\mu r^{2}}\cdot l(l+1);\;Eu(r)=i\hbar {d \over dr}u(r),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/473de65f2a4ddebe416bd110d3d68d4576f3a486)
![{\displaystyle \sin \theta {\partial \over \partial \theta }(\sin \theta {\partial O \over \partial \theta })+[l(l+1)\sin ^{2}\theta -m^{2}]O=0;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b6b7f42c830365bc0685edbfc35283c6de8793d)
![{\displaystyle \sin \theta {d \over d\theta }(\sin \theta {dO \over d\theta })+[l(l+1)\sin ^{2}\theta -m^{2}]O=0;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56185a0c9d19ef06a6ffe30d468835710b381ced)
