� � � � � � � � 8.04 Quantum Physics Lecture XXIIRadial equation for spherically symmetric potential The SE in 3D in spherical coordinates is h¯2 ∂2 2 ∂ L2 − 2m ∂r2 + r∂r ψ(r)+ 2mrψ(r)+ V (r)ψ(r)= Eψ(r) (22-1) 2 using the ansatz ψ(r)= R(r)Y (θ,φ), and inserting for the angular function an eigenfuncctio Y (θ,φ)= Ylm(θ,φ)= �θ,φ|l,m�, (22-2) we have, using L2Ylm(θ,φ)= h¯2l(l + 1)Ylm(θ,φ) after dividing by Ylm for the radial equation, h¯2 ∂2 2 ∂h¯2l(l + 1) −2m ∂r2 + r ∂r +2mr+ V (r) Rnl(r)= EnlRnl(r) . (22-3) 2 Here, we have added two subscripts n, l to the radial wavefunction R(r) and the eigenenergy E because the SE for the radial part of the wavefunction depends on the total angular momentum l of the 3D wavefunction ψ(r). Note. The z-component of angular momentum Lz, and the corresponding magnetic quantum number m, do not appear in the radial equation. We can define an l-dependent effective potential, h¯2l(l + 1) Veff,l = V (r)+ 2 , (22-4) 2mrwhere the additional term is the centrifugal barrier for a particle with angular momenntu �L2� =¯h2l(l + 1). (22-5) The radial equation can be brought into a more familiar-looking form by introducing a new function: u(r)= rR(r) u(r) (22-6) R(r)= r Then, u�r − uu� u R� = = (22-7) 2r2 r − r22u� 2u rR� = r− r(22-8) 23 R�� = u��r − u� u�r2 − u2r = u�� 2u� +2u (22-9) 24 23r− rr − rr∂2 2 ∂ 2 u�� + R(r)= R�� + R� = (22-10) ∂r2 r∂r rr Massachusetts Institute of Technology XXII-1 � � � � 8.04 Quantum Physics Lecture XXIIand the radial equation is h¯2 1 ∂2uh¯2l(l + 1) u(r) u(r)− 2mr ∂r2 +2mr2 + V (r) r = Er (22-11) or h¯2 ∂2 h¯2l(l + 1) unl(r)−2m ∂r2 +2mr2 + V (r) r = Enlunl(r) (22-12) This equation for u(r)= rR(r) has the same form as the 1D SE in the effective potential h¯2l(l + 1) Veff,l(r)= V (r)+ , (22-13) 2mr2 but with slightly different boundary conditions. Therefore, u(r) looks like an anti-Figure I: u(r)= rR(r) has the same form as the 1D SE in the effective potential Veff,l(r), but with slightly different boundary conditions. symmetric solution in all space. Consequences are, e.g., that since an antisymmetric Figure II: u(r) looks like an antisymmetric solution in all space. bound state does not always exist in 1D, that a bound state does not always exist in 3D (in contrast to 1D, where a symmetric bound state always exist in a potential well). 3D wavefunctions u(r) are like antisymmetric 1D wavefunctions in the effective potential h¯2l(l + 1) Vl(r)= V (r)+ . (22-14) 2mr2 Massachusetts Institute of Technology XXII-2 � � � � � � � � � � � 8.04 Quantum Physics Lecture XXIIHydrogen atom Ze2 V (r)= − (and the radial equation is) (22-15) 4π�r → h¯62 ∂2 Ze2 h¯2l(l + 1) − 2m ∂r2 − 4π�r +2mr2 − Eu(r) = 0 (22-16) We introduce a dimensionless position coordinate ρ by ρ2 = 8m¯h|2 E| r2, and define for E< 0 Ze2 8mh¯2 Ze2 8mh¯2 = (22-17) r16π�Eh2 8m hrE8mE|| ¯16π�¯� || || Ze2 m 1 = (22-18) 4π�h¯2|E| ρ λ =: (22-19) ρ The equation can be written as ∂2 λ 1 l(l + 1) ∂ρ2 u + ρ − 4 − ρ2 u = 0 (22-20) Ze2 m mc2 e2 1with ρ = 8m¯|E| r, λ = 4π�¯2|E| = Zα 2|E| , where α = 4π�¯≈ 137.0... is the dimenhh h hc sionless fine structure constant. To solve this equation, we proceed as for the HO: We write a Taylor-expansion solution after having factored out the correct asymptotic behavior. For very large ρ we have d2 1 u = u (22-21) dp2 4 1 2u(ρ) ∝ e− ρ (22-22) For very small ρ, d2 l(l + 1) u = u (22-23) dp2 ρ2 u(ρ) ∝ ρl+1 (22-24) Consequently, we try a solution of the form ρ2u(ρ)= s(ρ)ρl+1 e− 1 (22-25) Massachusetts Institute of Technology XXII-3 ���� � � � � � � � � � � � � � � � � � � � � � � 8.04 Quantum Physics Lecture XXII� 1 � s�(ρ)ρl+1 + s(ρ)(l + 1)ρl − sρl+1 e−1 2ρ (22-26)u�(ρ)= 2u��(ρ)= s��ρl+1 + 2(l + 1)s�ρl + s(l + 1)lρl−1 (22-27) 1�� − 2 s�ρl+1 +(l + 1)sρl (22-28) 1� 1 � s�ρl+1 +(l + 1)sρl − sρl+1 e−1 2(22-29)−22s� (l + 1)ll +1 1ρ = ρl+1 e− s�� + 2(l + 1) s − s� −+s + s2 ρρ2 ρ 4 (22-30) ρλ 1 l(l + 1) λ 1 l(l + 1) u = ρl+1 e− (22-31)ρ − 4 −2 ρ − 4 −sρ2 ρ2 Inserting this into (??) leads to 2(l +1) (l + 1)ll +1 1 λ 1 l(l + 1) s�� + ρ − 1 s� + ρ2 − ρ +4+ ρ − 4 − ρ2 s = 0 (22-32) s�� +2l +2 − 1 s� + λ − l − 1 s = 0 (22-33) ρρ To solve this differential equation, we write a Taylor expansion about ρ = 0: ∞s(ρ)= akρk (22-34) k=0 ∞s�� = akk(k − 1)ρk−2 (22-35) k=0 ∞= ak+2(k + 2)(k + 1)ρk (22-36) k=0 2l +2 2l +2 � − 1 s� = − 1 akkρk−1 (22-37) ρρ ∞= (2l + 2) ak+2(k + 2)ρk − ak+1(k + 1)ρk (22-38) k=0 λ − l − 1 s =(λ − l − 1) ∞ak+1ρk (22-39) ρ k=0 which substituted into (??) results in ρk (k +2)(k +1)ak+2 +2(l +1)(k +2)ak+2 +(λ − l − 1 − k − 1)ak+1 = 0 (22-40) k Massachusetts Institute of Technology XXII-4 = � � � � � � � 8.04 Quantum Physics Lecture XXIIThis must vanish term by term, so we obtain a recursion relation (k + 2)(k +2l + 3)ak+2 =(k + l +2 − λ)ak+1 (22-41) or recursion ak+1 k + l +1 − λ relation for (22-42) ak k +1 k + 2(l + 1) → expansion coefficients If the series does not break off somewhere, we will have for large k, akk 1 ak−1 or∝ ρ 2 . Consequently, we require the series to terminate, which implies λ = k + l + 1 for some ak ∝ k1! , which gives a growth s(ρ) ∝ e+ρ, which is not acceptable for u(ρ)= s(ρ)e− L. Let us call nr = k the integer with that property. It is cutomary to define the principal quantum number as n = nr + l +1 , (22-43) where nr ≥ 0, so n ≥ 0, so n ≥ l + 1, n integer, and Ze2 m λn = (22-44) 4π�h¯2|En| 2mc= Zα (22-45) 2|En| = n (22-46) Consequently, the eigenenergies of the hydrogen atom are 1 2 (Zα)2 eigenenergies of mc (22-47) → hydrogenlike atoms En = −2 n2 This is the same energy eigenspectrum as obtained from the Bohr formula. Note. There are important differences: • The principal quantum number n = nr + l + 1 is really the sum of the radial quantum number nr and the total angular momentum quantum number l. We have obtained the full radial and angular distribution of the electron, which • generalizes the classical concept of an orbit. Massachusetts Institute of Technology XXII-5 � � � � � 8.04 Quantum Physics Lecture XXIIFirst few radial functions ρ22 =8m|En|r (22-48) n h¯2 =8m 1 mc 2 (Zα)2 r 2 (22-49) h¯2 2 n2 (2mcZα)22 =¯r (22-50) hn2 ��22Z 2 1 = r (22-51) 2a0 n2Zr = (22-52) na0 with the Bohr radius h¯2 a0 = (22-53) mcα ρ2 na0Consequently, e− 1 = e− Zr 1. nr = l = 0, n = m = λ, a1 =0 2 a0u(r)= Cρe− 1 ρ = C1 Zr e− Zα (22-54) a0 Zr a0R(r)= u(r)= C2e− r Note. The probability to find the electron between r r2|R(r)|2dr = |u(r)|2dr. 2. (a) nr = 1, l = 0, n =2= λ a1 11 a0 = −1 2= −2· � (22-55) and r + dr is given by (22-56) 22a0u20(r)= Cρe− 1 ρ(1 − 1 ρ)= C� Zr 1 − Zr e− Zr (22-57) 2 a0 2a0 Zr Zr 2a0R20(r)= C�� 1 − 2a0 e− (22-58) Massachusetts Institute of Technology XXII-6 � � 8.04 Quantum Physics Lecture XXII(b) nr = 0, l = 1, n =2= λ a1 a0 =0 → a1 = 0 (22-59) ��2 1 Zr 2u21(r)= Cρ2 e− ρ = C� Zr e− 2a0 (22-60) a0 Zr 2a0R21(r)= C�� Zr e− (22-61) a0 R20 = Rn=2,l=0 and R21 = Rn=2,l=1 are different states that have the same eigenenergy. The occurrence of different eigenstates with the same energy, (or in general quantum number) is called degeneracy. Massachusetts Institute of Technology XXII-7