Lecture 4 Quantum mechanics in more than one-dimension

Transcription

1 Lecture 4 Quantum mechanics in more than one-dimension

2 Background Previously, we have addressed quantum mechanics of 1d systems and explored bound and unbound (scattering) states. Although general concepts carry over to higher dimension, without symmetry, states of the Schrödinger operator are often inaccessible. In such situations, we must exploit approximation methods to address properties of the states perturbation theory. However, when degree of symmetry is high, the quantum mechanics can often be reduced to a tractable low-dimensional theory. Here we address three-dimensional problems involving a central potential (e.g. an atom) where the system has full rotational symmetry.

3 Background Previously, we have addressed quantum mechanics of 1d systems and explored bound and unbound (scattering) states. Although general concepts carry over to higher dimension, without symmetry, states of the Schrödinger operator are often inaccessible. In such situations, we must exploit approximation methods to address properties of the states perturbation theory. However, when degree of symmetry is high, the quantum mechanics can often be reduced to a tractable low-dimensional theory. Here we address three-dimensional problems involving a central potential (e.g. an atom) where the system has full rotational symmetry.

4 Rigid diatomic molecule Consider quantum mechanics of a rigid diatomic molecule with nuclear masses m 1 and m 2, and fixed bond length, r. Since molecule is rigid, coordinates specified by centre of mass R = m 1r 1 + m 2 r 2 m 1 + m 2 and orientation, r = r 2 r 1 (with r = r). With total mass M = m 1 + m 2, and moment of inertia, I = µr 2, where µ = m 1m 2 m 1 + m 2 is reduced mass, Ĥ = ˆP 2 2M + ˆL 2 2I with ˆP = i R and ˆL = r ˆp is internal angular momentum.

5 Rigid diatomic molecule Ĥ = ˆP 2 2M + ˆL 2 2I Since internal and centre of mass degrees of freedom separate, wavefunction can be factorized as ψ(r, R) =e ik R Y (r). Internal component of wavefunction, Y (r), describes quantum particle moving on a sphere with r = r constant a rigid rotor, Ĥ rot = ˆL 2 2I Eigenstates of rotor are states of angular momentum operator, ˆL 2. Indeed, in any quantum mechanical system involving a radial potential, angular momentum is conserved, i.e. [Ĥ, ˆL] = 0 and angular component of wavefunction indexed by states of ˆL 2.

6 Angular momentum: commutation relations To explore quantum rotor model, Ĥrot, we must therefore address properties of the angular momentum operator. Following the usual quantization procedure, the angular momentum operator defined by ˆL = r ˆp where [ˆp i, r j ]= i δ ij. Using this relation, one may show that components of angular momentum operators obey commutation relations, [ˆL i, ˆL j ]=i ɛ ijk ˆL k e.g. [ˆL x, ˆL y ]=i ˆL z ɛ ijk is antisymmetric tensor (Levi-Civita symbol) ɛ 123 =1= ɛ 213 (together with all permutations) while other components are zero.

7 Angular momentum: eigenvalues Since angular momentum, ˆL is a vector quantity, it may be defined by magnitude, ˆL 2, and direction. As components of ˆL are mutually non-commuting, a common set of eigenstates for any two can not be constructed. They do, however, commute with ˆL 2 (exercise) therefore, we will seek eigenbasis of ˆL 2 and one direction, say ˆL z, ˆL 2 a, b = a a, b, ˆLz a, b = b a, b To find states a, b, we could turn to coordinate basis and express ˆL 2 and ˆL z as differential operators however, before doing so, we can learn much using operator formalism (cf. harmonic oscillator).

8 Angular momentum: raising and lowering operators ˆL 2 a, b = a a, b, ˆLz a, b = b a, b Let us then define operators ˆL ± = ˆL x ± iˆl y Since [ˆL 2, ˆL i ] = 0, ˆL 2 (ˆL ± a, b ) =ˆL ±ˆL2 a, b = a(ˆl ± a, b ), i.e. ˆL ± a, b is also eigenstate of ˆL 2 with eigenvalue a. From commutation relations, [ˆL i, ˆL j ]=i ɛ ijk ˆLk, we have [ˆL z, ˆL ± ]=[ˆL z, ˆL x ± iˆl y ]=i (ˆL y iˆl x )=± (ˆL x ± iˆl y )=± ˆL ± Therefore, while ˆL ± conserve eigenvalue a, they do effect projection, ˆL z ˆL± a, b = ˆL ±ˆLz a, b +[ˆL z, ˆL ± ] a, b =(b ± )ˆL ± a, b if ˆL z a, b = b a, b, ˆL ± a, b is either zero, or an eigenstate of ˆL z with eigenvalue b ±, i.e. ˆL ± a, b = C ± (a, b) a, b ±

9 Angular momentum: raising and lowering operators ˆL 2 a, b = a a, b, ˆLz a, b = b a, b Let us then define operators ˆL ± = ˆL x ± iˆl y Since [ˆL 2, ˆL i ] = 0, ˆL 2 (ˆL ± a, b ) =ˆL ±ˆL2 a, b = a(ˆl ± a, b ), i.e. ˆL ± a, b is also eigenstate of ˆL 2 with eigenvalue a. From commutation relations, [ˆL i, ˆL j ]=i ɛ ijk ˆLk, we have [ˆL z, ˆL ± ]=[ˆL z, ˆL x ± iˆl y ]=i (ˆL y iˆl x )=± (ˆL x ± iˆl y )=± ˆL ± Therefore, while ˆL ± conserve eigenvalue a, they do effect projection, ˆL z ˆL± a, b = ˆL ±ˆLz a, b +[ˆL z, ˆL ± ] a, b =(b ± )ˆL ± a, b if ˆL z a, b = b a, b, ˆL ± a, b is either zero, or an eigenstate of ˆL z with eigenvalue b ±, i.e. ˆL ± a, b = C ± (a, b) a, b ±

10 Angular momentum: raising and lowering operators ˆL 2 a, b = a a, b, ˆLz a, b = b a, b Let us then define operators ˆL ± = ˆL x ± iˆl y Since [ˆL 2, ˆL i ] = 0, ˆL 2 (ˆL ± a, b ) =ˆL ±ˆL2 a, b = a(ˆl ± a, b ), i.e. ˆL ± a, b is also eigenstate of ˆL 2 with eigenvalue a. From commutation relations, [ˆL i, ˆL j ]=i ɛ ijk ˆLk, we have [ˆL z, ˆL ± ]= ± ˆL ± Therefore, while ˆL ± conserve eigenvalue a, they do effect projection, ˆL z ˆL± a, b = ˆL ±ˆLz a, b +[ˆL z, ˆL ± ] a, b =(b ± )ˆL ± a, b if ˆL z a, b = b a, b, ˆL ± a, b is either zero, or an eigenstate of ˆL z with eigenvalue b ±, i.e. ˆL ± a, b = C ± (a, b) a, b ±

11 Angular momentum: raising and lowering operators ˆL ± a, b = C ± (a, b) a, b ± To fix normalization, a, b a, b = 1, noting that ˆL ± = ˆL, ˆL ± a, b 2 a, b ˆL ±ˆL ± a, b = a, b ˆL ˆL± a, b Then, since ˆL ˆL ± = ˆL 2 x + ˆL 2 y ± i[ˆl x, ˆL y ]=ˆL 2 ˆL 2 z ˆL z, ˆL ± a, b 2 = a, b (ˆL 2 ˆL 2 z ˆL z ) a, b = a b 2 b 0 Since a 0 and b is real, must have b min b b max, a, b max ˆL +ˆL + a, b max = a bmax 2 b max =0 a, b min ˆL ˆL a, b min = a bmin 2 + b min =0 i.e. a = b max (b max + ) and b min = b max.

12 Angular momentum: raising and lowering operators ˆL ± a, b = C ± (a, b) a, b ± To fix normalization, a, b a, b = 1, noting that ˆL ± = ˆL, ˆL ± a, b 2 a, b ˆL ±ˆL ± a, b = a, b ˆL ˆL± a, b Then, since ˆL ˆL ± = ˆL 2 x + ˆL 2 y ± i[ˆl x, ˆL y ]=ˆL 2 ˆL 2 z ˆL z, ˆL ± a, b 2 = a, b (ˆL 2 ˆL 2 z ˆL z ) a, b = a b 2 b 0 Since a 0 and b is real, must have b min b b max, a, b max ˆL +ˆL + a, b max = a bmax 2 b max =0 a, b min ˆL ˆL a, b min = a bmin 2 + b min =0 i.e. a = b max (b max + ) and b min = b max.

13 Angular momentum: raising and lowering operators ˆL ± a, b = C ± (a, b) a, b ± To fix normalization, a, b a, b = 1, noting that ˆL ± = ˆL, ˆL ± a, b 2 a, b ˆL ±ˆL ± a, b = a, b ˆL ˆL± a, b Then, since ˆL ˆL ± = ˆL 2 x + ˆL 2 y ± i[ˆl x, ˆL y ]=ˆL 2 ˆL 2 z ˆL z, ˆL ± a, b 2 = a, b (ˆL 2 ˆL 2 z ˆL z ) a, b = a b 2 b 0 Since a 0 and b is real, must have b min b b max, a, b max ˆL +ˆL + a, b max = a bmax 2 b max =0 a, b min ˆL ˆL a, b min = a bmin 2 + b min =0 i.e. a = b max (b max + ) and b min = b max.

14 Angular momentum: raising and lowering operators a = b max (b max + ) and b min = b max For given a, b max and b min determined uniquely cannot be two states with the same a but different b annihilated by ˆL +. If we keep operating on a, b min with ˆL +, we generate a sequence of states with ˆL z eigenvalues b min +, b min +2, b min +3,. Only way for series to terminate is for b max = b min + n with n integer, i.e. b max is either integer or half odd integer.

15 Angular momentum: eigenvalues Eigenvalues of ˆL z form ladder, with eigenvalue b = m and m max = l = m min. m known as magnetic quantum number. Eigenvalues of ˆL 2 are a = l(l + 1) 2. ˆL 2 l, m = l(l + 1) 2 l, m ˆL z l, m = m l, m Both l and m are integer or half odd integers, but spacing of ladder of m always unity.

16 Angular momentum: raising and lowering operators ˆL 2 l, m = l(l + 1) 2 l, m, ˆLz l, m = m l, m Finally, making use of identity, ˆL ± l, m 2 (ˆL2 = l, m ˆL ) 2 z ± ˆL z l, m we find that ˆL + l, m = l(l + 1) m(m + 1) l, m +1 ˆL l, m = l(l + 1) m(m 1) l, m 1

17 Representation of the angular momentum states Although we can use an operator-based formalism to construct eigenvalues of ˆL 2 and ˆL z it is sometimes useful to have coordinate representation of states, Y lm (θ, φ) = θ, φ l, m. Using the expression for the gradient operator in spherical polars, = ê r r + ê θ 1 r θ + ê φ 1 r sin θ φ with ˆL = i r, a little algebra shows, ˆL z = i φ, ˆL± = e ±iφ (± θ + i cot θ φ ) [ 1 ˆL 2 = 2 sin θ θ(sin θ θ )+ 1 ] sin 2 θ 2 φ

18 Representation of the angular momentum states ˆL z = i φ, ˆL ± = e ±iφ (± θ + i cot θ φ ) Beginning with ˆL z = i φ, i φ Y lm (θ, φ) =m Y lm (θ, φ) since equation is separable, we have the solution with l m l. Y lm (θ, φ) =F (θ)e imφ N.B. if l (and therefore m) integer, continuity of wavefunction, Y lm (θ, φ +2π) =Y lm (θ, φ), is assured. [Not so if l is half-integer.]

19 Representation of the angular momentum states Y lm (θ, φ) =F (θ)e imφ, ˆL± = e ±iφ (± θ + i cot θ φ ) (Drawing analogy with procedure to find HO states) to find F (θ), consider state of maximal m, l, l, for which ˆL + l, l = 0. Making use of coordinate representation of raising operator 0= θ, φ ˆL + l, l = e iφ ( θ + i cot θ φ ) Y ll (θ, φ) = e i(l+1)φ ( θ l cot θ) F (θ) i.e. θ F (θ) =l cot θf (θ) with the solution F (θ) =C sin l θ. The 2l states with values of m lower than l generated by repeated application of ˆL on l, l. [ l m Y lm (θ, φ) =C( θ + i cot θ φ ) sin l θe ilφ] }{{} ˆL

20 Representation of the angular momentum states Y lm (θ, φ) =F (θ)e imφ, ˆL± = e ±iφ (± θ + i cot θ φ ) (Drawing analogy with procedure to find HO states) to find F (θ), consider state of maximal m, l, l, for which ˆL + l, l = 0. Making use of coordinate representation of raising operator 0= θ, φ ˆL + l, l = e iφ ( θ + i cot θ φ ) e ilφ F (θ) = e i(l+1)φ ( θ l cot θ) F (θ) i.e. θ F (θ) =l cot θf (θ) with the solution F (θ) =C sin l θ. The 2l states with values of m lower than l generated by repeated application of ˆL on l, l. [ l m Y lm (θ, φ) =C( θ + i cot θ φ ) sin l θe ilφ] }{{} ˆL

21 Representation of the angular momentum states Y lm (θ, φ) =F (θ)e imφ, ˆL± = e ±iφ (± θ + i cot θ φ ) (Drawing analogy with procedure to find HO states) to find F (θ), consider state of maximal m, l, l, for which ˆL + l, l = 0. Making use of coordinate representation of raising operator 0= θ, φ ˆL + l, l = e iφ ( θ + i cot θ φ ) e ilφ F (θ) = e i(l+1)φ ( θ l cot θ) F (θ) i.e. θ F (θ) =l cot θf (θ) with the solution F (θ) =C sin l θ. The 2l states with values of m lower than l generated by repeated application of ˆL on l, l. [ l m Y lm (θ, φ) =C( θ + i cot θ φ ) sin l θe ilφ] }{{} ˆL

22 Representation of the angular momentum states Eigenfunctions of ˆL 2 are known as spherical harmonics, Y lm (θ, φ) = ( 1) m+ m [ 2l +1 4π where the functions Pl m(ξ) = (1 ξ2 ) m/2 2 l l! associated Legendre polynomials. ] 1/2 (l m )! P m l (cos θ)e imφ (l + m )! d m+l dξ m+l (ξ 2 1) l are known as I see no reason why you should ever memorize these functions! Y 00 =1 Y 10 = cos θ, Y 11 = e iφ sin θ Y 20 = 3 cos 2 θ 1, Y 21 = e iφ sin θ cos θ, Y 22 = e 2iφ sin 2 θ States with l =0, 1, 2, 3,... are known as s, p, d, f,...-orbitals. Note symmetries: Y l, m =( 1) m Y lm and ˆPY lm =( 1) l Y lm.

23 Representation of the angular momentum states Eigenfunctions of ˆL 2 are known as spherical harmonics, Y lm (θ, φ) = ( 1) m+ m [ 2l +1 4π where the functions Pl m(ξ) = (1 ξ2 ) m/2 2 l l! associated Legendre polynomials. ] 1/2 (l m )! P m l (cos θ)e imφ (l + m )! d m+l dξ m+l (ξ 2 1) l are known as I see no reason why you should ever memorize these functions! Y 00 =1 Y 10 = cos θ, Y 11 = e iφ sin θ Y 20 = 3 cos 2 θ 1, Y 21 = e iφ sin θ cos θ, Y 22 = e 2iφ sin 2 θ States with l =0, 1, 2, 3,... are known as s, p, d, f,...-orbitals. Note symmetries: Y l, m =( 1) m Y lm and ˆPY lm =( 1) l Y lm.

24 Representation of the angular momentum states Eigenfunctions of ˆL 2 are known as spherical harmonics, Y lm (θ, φ) = ( 1) m+ m [ 2l +1 4π where the functions Pl m(ξ) = (1 ξ2 ) m/2 2 l l! associated Legendre polynomials. ] 1/2 (l m )! P m l (cos θ)e imφ (l + m )! d m+l dξ m+l (ξ 2 1) l are known as As an example of the first few (unnormalized) spherical harmonics: Y 00 =1 Y 10 = cos θ, Y 11 = e iφ sin θ Y 20 = 3 cos 2 θ 1, Y 21 = e iφ sin θ cos θ, Y 22 = e 2iφ sin 2 θ States with l =0, 1, 2, 3,... are known as s, p, d, f,...-orbitals. Note symmetries: Y l, m =( 1) m Y lm and ˆPY lm =( 1) l Y lm.

25 Representation of the angular momentum states radial coordinate fixed by Re Y lm (θ, φ) and colours indicate relative sign of real part.

26 Rigid rotor model After this lengthy digression, we return to problem of quantum mechanical rotor Hamiltonian and the rigid diatomic molecule. Eigenstates of the Hamiltonian, Ĥ = ˆP 2 2M + ˆL 2 2I given by ψ(r, r) =e ik R Y l,m (θ, φ) with eigenvalues E K,l = 2 K 2 2M + 2 2I l(l + 1) where, for each set of quantum numbers (K,l), there is a 2l + 1-fold degeneracy. With this background, we now turn to general problem of 3d system with centrally symmetric potential, V (r) (e.g. atomic hydrogen).

27 The central potential When central force field is entirely radial, the Hamiltonian for the relative coordinate is given by Using the identity, Ĥ = ˆp2 2m + V (r) ˆL 2 =(r ˆp) 2 = r i ˆp j r i ˆp j r i ˆp j r j ˆp i = r 2ˆp 2 (r ˆp) 2 + i (r ˆp) with r ˆp = i r = i r r, find ˆp 2 = ˆL 2 r 2 2 r 2 [ (r r ) 2 + r r ] Noting that (r r ) 2 + r r = r 2 2 r +2r r, we obtain the Schrödinger equation, [ 2 2m ( r ) r r + V (r)+ ˆL ] 2 2mr 2 ψ(r) =Eψ(r)

28 The central potential When central force field is entirely radial, the Hamiltonian for the relative coordinate is given by Using the identity, Ĥ = ˆp2 2m + V (r) ˆL 2 =(r ˆp) 2 = r i ˆp j r i ˆp j r i ˆp j r j ˆp i = r 2ˆp 2 (r ˆp) 2 + i (r ˆp) with r ˆp = i r = i r r, find ˆp 2 = ˆL 2 r 2 2 r 2 [ (r r ) 2 + r r ] Noting that (r r ) 2 + r r = r 2 2 r +2r r, we obtain the Schrödinger equation, [ 2 2m ( r ) r r + V (r)+ ˆL ] 2 2mr 2 ψ(r) =Eψ(r)

29 The central potential When central force field is entirely radial, the Hamiltonian for the relative coordinate is given by Using the identity, Ĥ = ˆp2 2m + V (r) ˆL 2 =(r ˆp) 2 = r i ˆp j r i ˆp j r i ˆp j r j ˆp i = r 2ˆp 2 (r ˆp) 2 + i (r ˆp) with r ˆp = i r = i r r, find ˆp 2 = ˆL 2 r 2 2 r 2 [ (r r ) 2 + r r ] Noting that (r r ) 2 + r r = r 2 2 r +2r r, we obtain the Schrödinger equation, [ 2 2m ( r ) r r + V (r)+ ˆL ] 2 2mr 2 ψ(r) =Eψ(r)

30 The central potential [ 2 2m ( r ) r r + ˆL ] 2 2mr 2 + V (r) ψ(r) =Eψ(r) From separability, ψ(r) =R(r)Y l,m (θ, φ), where ( [ 2 r ) ] 2m r r + 2 l(l + 1) + V (r) R(r) =ER(r) 2mr 2 Finally, setting R(r) = u(r)/r, obtain one-dimensional equation [ 2 r 2 ] 2m + V eff(r) u(r) =Eu(r), V eff (r) = 2 l(l + 1) + V (r) 2mr 2 with boundary condition u(0) = 0, and normalization, d 3 r ψ(r) 2 = r 2 dr R(r) 2 = dr u(r) 2 =1 So, for bound state, lim r u(r) 0 0 a r 1/2+ɛ with ɛ> 0.

31 The central potential [ 2 2m ( r ) r r + ˆL ] 2 2mr 2 + V (r) ψ(r) =Eψ(r) From separability, ψ(r) =R(r)Y l,m (θ, φ), where ( [ 2 r ) ] 2m r r + 2 l(l + 1) + V (r) R(r) =ER(r) 2mr 2 Finally, setting R(r) = u(r)/r, obtain one-dimensional equation [ 2 r 2 ] 2m + V eff(r) u(r) =Eu(r), V eff (r) = 2 l(l + 1) + V (r) 2mr 2 with boundary condition u(0) = 0, and normalization, d 3 r ψ(r) 2 = r 2 dr R(r) 2 = dr u(r) 2 =1 So, for bound state, lim r u(r) 0 0 a r 1/2+ɛ with ɛ> 0.

32 The central potential [ 2 2m ( r ) r r + ˆL ] 2 2mr 2 + V (r) ψ(r) =Eψ(r) From separability, ψ(r) =R(r)Y l,m (θ, φ), where ( [ 2 r ) ] 2m r r + 2 l(l + 1) + V (r) R(r) =ER(r) 2mr 2 Finally, setting R(r) = u(r)/r, obtain one-dimensional equation [ 2 r 2 ] 2m + V eff(r) u(r) =Eu(r), V eff (r) = 2 l(l + 1) + V (r) 2mr 2 with boundary condition u(0) = 0, and normalization, d 3 r ψ(r) 2 = r 2 dr R(r) 2 = dr u(r) 2 =1 So, for bound state, lim r u(r) 0 0 a r 1/2+ɛ with ɛ> 0.

33 The central potential: bound states [ 2 r 2 ] 2m + V eff(r) u(r) =Eu(r), V eff (r) = 2 l(l + 1) + V (r) 2mr 2 Since u(0) = 0, we may map Hamiltonian from half-line to full with the condition that we admit only antisymmetric wavefunctions. Existence of bound states can then be related back to the one-dimensional case: Previously, we have seen that a (symmetric) attractive potential always leads to a bound state in one-dimension. However, odd parity states become bound only at a critical strength of interaction. So, for a general attractive potential V (r), the existence of a bound state is not guaranteed even for l = 0.

34 Atomic hydrogen [ 2 r 2 ] 2m + V eff(r) u(r) =u(r), V eff (r) = 2 l(l + 1) + V (r) 2mr 2 The hydrogen atom consists of an electron bound to a proton by the Coulomb potential, V (r) = e2 1 4πɛ 0 r and, strictly speaking, m denotes the reduced mass (generalization to nuclear charge Ze follows straightforwardly). Since we are interested in finding bound states of proton-electron system, we are looking for solutions with E < 0. Here we sketch the methodology in outline for details, refer back to IB.

35 Atomic hydrogen [ 2 r 2 ] 2m + V eff(r) u(r) =Eu(r), V eff (r) = 2 l(l + 1) + V (r) 2mr 2 To simplify equation, set ρ = κr, where κ = 2mE ( ρu(ρ) 2 = 1 2ν ) l(l + 1) + ρ ρ 2 u(ρ), 2ν = e2 κ 4πɛ E At large separations, 2 ρu(ρ) u(ρ) and u(ρ) e ρ. Near origin, dominant term for small ρ is centrifugal component, 2 ρu(ρ) l(l + 1) ρ 2 u(ρ) for which u(ρ) ρ l+1.

36 Atomic hydrogen [ 2 r 2 ] 2m + V eff(r) u(r) =Eu(r), V eff (r) = 2 l(l + 1) + V (r) 2mr 2 To simplify equation, set ρ = κr, where κ = 2mE ( ρu(ρ) 2 = 1 2ν ) l(l + 1) + ρ ρ 2 u(ρ), 2ν = e2 κ 4πɛ E At large separations, 2 ρu(ρ) u(ρ) and u(ρ) e ρ. Near origin, dominant term for small ρ is centrifugal component, 2 ρu(ρ) l(l + 1) ρ 2 u(ρ) for which u(ρ) ρ l+1.

37 Atomic hydrogen Finally, defining u(ρ) =e ρ ρ l+1 w(ρ), equation for w(ρ) reveals that ν = e2 κ 4πɛ 0 2E must take integer values, n principal quantum number, i.e. ( e 2 E n = 4πɛ 0 ) 2 m n 2 1 n 2 Ry Therefore, ρ = κ n r, where κ n = 2mE n = e2 m 1 4πɛ 0 2 n = 1 a 0 n, with the atomic Bohr radius. a 0 = 4πɛ 0 e 2 2 m = m Formally, the set of functions w nl (ρ) =L 2l+1 n l 1 (2ρ) are known as associated Laguerre polynomials L k p(z).

38 Atomic hydrogen Finally, defining u(ρ) =e ρ ρ l+1 w(ρ), equation for w(ρ) reveals that ν = e2 κ 4πɛ 0 2E must take integer values, n principal quantum number, i.e. ( e 2 E n = 4πɛ 0 ) 2 m n 2 1 n 2 Ry Therefore, ρ = κ n r, where κ n = 2mE n = e2 m 1 4πɛ 0 2 n = 1 a 0 n, with the atomic Bohr radius. a 0 = 4πɛ 0 e 2 2 m = m Formally, the set of functions w nl (ρ) =L 2l+1 n l 1 (2ρ) are known as associated Laguerre polynomials L k p(z).

39 Atomic hydrogen Translating back from ρ to r, R nl (r) =Ne Zr/na 0 ( Zr na 0 ) l L 2l+1 n l 1 (2Zr/na 0) For principal quantum number n, and l = n 1, R n,n 1 r n 1 e Zr/na 0. ( Z R 10 =2 a 0 R 21 = ) 3 2 e Zr/a 0 ( Z a 0 ) 3/2 ( Zr a 0 ) R 20 = 1 2 ( Z a 0 ) 3/2 ( e Zr/2a 0 Zr a 0 ) e Zr/2a 0

40 Atomic hydrogen But why the high degeneracy? Since [Ĥ, ˆL] = 0, we expect that states of given l have a 2l + 1-fold degeneracy. Instead, we find that each principal quantum number n has an n 2 -fold degeneracy, i.e. for given n, all allowed l-states degenerate. As a rule, degeneracies are never accidental but always reflect some symmetry which we must have missed(!) In fact, one may show that the (Runge-Lenz) vector operator ˆR = 1 2m (ˆp ˆL ˆL ˆp) e2 r 4πɛ 0 r is also conserved by the Hamiltonian dynamics, [Ĥ, ˆR] = 0. From this operator, we can identify generators for the complete degenerate subspace (cf. ˆL ± ) a piece of mathematical physics (happily) beyond the scope of these lectures.

41 Atomic hydrogen But why the high degeneracy? Since [Ĥ, ˆL] = 0, we expect that states of given l have a 2l + 1-fold degeneracy. Instead, we find that each principal quantum number n has an n 2 -fold degeneracy, i.e. for given n, all allowed l-states degenerate. As a rule, degeneracies are never accidental but always reflect some symmetry which we must have missed(!) In fact, one may show that the (Runge-Lenz) vector operator ˆR = 1 2m (ˆp ˆL ˆL ˆp) e2 r 4πɛ 0 r is also conserved by the Hamiltonian dynamics, [Ĥ, ˆR] = 0. From this operator, we can identify generators for the complete degenerate subspace (cf. ˆL ± ) a piece of mathematical physics (happily) beyond the scope of these lectures.

42 Atomic hydrogen But why the high degeneracy? Since [Ĥ, ˆL] = 0, we expect that states of given l have a 2l + 1-fold degeneracy. Instead, we find that each principal quantum number n has an n 2 -fold degeneracy, i.e. for given n, all allowed l-states degenerate. As a rule, degeneracies are never accidental but always reflect some symmetry which we must have missed(!) In fact, one may show that the (Runge-Lenz) vector operator ˆR = 1 2m (ˆp ˆL ˆL ˆp) e2 r 4πɛ 0 r is also conserved by the Hamiltonian dynamics, [Ĥ, ˆR] = 0. From this operator, we can identify generators for the complete degenerate subspace (cf. ˆL ± ) a piece of mathematical physics (happily) beyond the scope of these lectures.

43 Summary For problems involving a central potential, Ĥ = ˆp2 2m + V (r) Hamiltonian is invariant under spatial rotations, Û = e i θe n ˆL. This invariance implies that the states separate into degenerate multiplets l, m with fixed by angular momentum l. ˆL 2 l, m = l(l + 1) 2 l, m, ˆLz l, m = m l, m The 2l + 1 states within each multiplet are generated by the action of the angular momentum raising and lowering operators, ˆL ± l, m = l(l + 1) m(m ± 1) l, m ± 1

44 Summary In the case of atomic hydrogen, Ĥ = ˆp2 2m e2 4πɛ 0 r an additional symmetry leads to degeneracy of states of given principal quantum number, n, ( e 2 E n = 4πɛ 0 ) 2 m n 2 1 n 2 Ry The extent of the wavefunction is characterized by the Bohr radius, a 0 = 4πɛ 0 e 2 2 m = m