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Quantum Physics 2002 The Hydrogen Atom continued.. Recommended Reading: Harris Chapter 6, Sections 3,4 • Spherical coordinate system •The Coulomb Potential •Angular Momentum • Normalised Wavefunctions •Energy Levels • Degeneracy Put it all together We now have all the components necessary to write down the solutions to the Schrodinger Equation for the Hydrogen Atom. We applied the method of Separation of Variables to wavefunction ψr, θ, φ R(r) θ φ and have found the following solutions for the radial and angular parts. R(r) Rn,l r AssociatedLaguerrePolynomials θ Pl,mL θ AssociatedLegendrePolynomials φ A exp imL φ Azimuthal wavefunction And so the total wavefunction can be written as: ψ n,l ,m r, θ, φ Nn,l ,ml Rn,l r Pl ,ml θ exp iml φ Nn,l ,ml Rn,l r Yl ,ml (, ) l where Nn,l,mL is a normalisation factor which we still have to determine. Note that the wavefunction depends on three quantum numbers, n, the principal quantum number, l = the total angular momentum and ml the z-component of the angular momentum. Example: the wavefunction corresponding to the state n = 1, l = 0, ml = 0, what is the explicit form of this wavefunction ψ 1,0,0 r, θ, φ N1,0,0 R1,0 r P0,0 θ exp i0φ writing the explicit forms of the Laguerre and Legendre polynomials gives 2 r ψ 1,0,0 r, θ, φ N1,0,0 exp a3 2 a0 0 since P0,0() = 1 and exp(i.0.) = 1. Similary for n = 2, l = 2, mL = 1 we find ψ 2,1,1 r, θ, φ N2,1,1 R2,1r P1,1θ exp iφ r 1 r ψ 2,1,1 r, θ, φ N2,1,1 exp sinθ exp iφ 3 2 3a0 2a0 2a0 The angular part of the wavefunction Lets examine the angular part in more detail. We can write the angular part as ψθ, φ θ φ Pl,mL θ exp imL φ as before we can combine these two terms together into a single function Yl,mL(,) the Spherical Harmonics, this function combines the and dependent part of the solution. The Spherical represent the solutions to the Schrodinger equation for a particle confined to move on the surface s a sphere of unit radius. The first few are tabulated on the next page. Note that the Spherical harmonics depends on two quantum numbers l and ml. The Spherical Harmonics Yl,mL(, ) l 0 mL 0 1 0 1 1 2 0 2 1 2 2 Yl,mL(,) 1 2 π 1 3 cos θ 2 π 1 3 sinθe iφ 2 2π 1 5 3 cos2 θ 1 4 π 1 15 sinθ cos θe iφ 2 2π 1 15 sin2 θe 2iφ 4 2π The Spherical Harmonics Yl,mL(, ) continued... l 3 mL 0 3 1 3 2 3 3 Yl,mL(,) 1 7 5 cos3 θ 3 cos θ 4 π 1 21 sinθ 5 cos2 θ 1 e iφ 8 π 1 105 sin2 θ cos θe 2iφ 4 2π 1 35 sin3 θe 3iφ 8 π to see what these wavefunctions look like see the following websites http://www.quantum-physics.polytechnique.fr/en http://www.uniovi.es/~quimica.fisica/qcg/harmonics/harmonics.html http://wwwvis.informatik.uni-stuttgart.de/~kraus/LiveGraphics3D/ java_script/SphericalHarmonics.html Y0,0 Y1,0 Y1,+1 Y1,-1 Y2,+1 Y2,+2 Y2,0 Y2,-1 Y2,-2 z-Component of Angular momentum What happens if we operate on the angular part of the wavefunction with the Lz operator ? L̂ z ψθ, φ L̂ z Pl ,ml θ exp iml φ L̂ z Yl ,m θ, φ l i d d Pl ,ml θ exp iml φ iPl ,ml θ exp iml φ dφ dφ m l Pl ,ml θ exp iml φ m l Yl ,ml θ, φ Thus we see that the Spherical Harmonics are eigenfunctions of the Lzoperator with eigenvalues ml L̂ z Yl ,m θ, φ ml Yl ,m θ, φ l l this shows that the z-component of the angular momentum is quantised in units of What else can we learn from the angular part of the wavefunction ? Total Angular Momentum Since the angular momentum is a vector we can write the total angular momentum as L2 L2x L2y L2z From this (after some math) we find that 2 1 1 L̂2 2 sinθ 2 2 sin θ θ θ sin θ φ Let us operate on the angular part of the wavefunction and see what result it gives us, that is 2 1 1 L̂2 Yl ,ml θ, φ 2 Yl ,ml θ, φ sinθ 2 2 θ sin θ φ sinθ θ 2 l l 1Yl ,ml θ, φ this shows that the the angular part of the wave function is an eigenfunction of the L2 operator, with eigenvalue l l 1 2 Total Angular Momentum Since the square of a vector is equal to the square of its magnitude, this means that the magnitude of the angular momentum can take on only the values L L2 ll 1 2 where l 0, 1, 2, 3... and recall that no physical solution exists unless ml l. So with this restriction we have L Z mL where mL 0, 1, 2, ... l The information we now have is the magnitude of the quantised angular momentum Land the magnitude of the z-component of the angular momentum. For example, if l = 2 then L 22 1 6 and the z-components of the allowed angular momenta are L Z 2,1 , 0, 1, 2 we can represent this graphically as follows Space Quantisation The angular momentum vector can only point in certain directions in space Space Quantisation z LZ mL 2 6 0 θ L ll 1 - - 2 L cosθ z L mL mL θ cos1 l l 1 ll 1 Example What is the minimum angle that the angular momentum vector may make with the z-axis in the case where (a) l = 3 and (b) for l = 1? When l = 3 the magnitude of the angular momentum is L 33 1 12 with 7 possible z-components, mL = -3, -2, -1, 0, 1, 2, 3. The angular momentum vector will make the minimum angle with the z-axis when the z-component is as large as possible, 3 Lz 3 3 1 cosθ θ cos 30 L 12 12 for l = 1: L 11 1 2 and then Lz 1 1 1 cosθ θ cos 45 L 2 2 L Z 3 θ L 12 Notation We can specify the angular momentum states of a single particle as follows l= state Number of states 0 s 1 1 p 3 2 d 5 3 f 7 4 g 9 5 h 11 6 i 13 so for example, if the particle has angular momentum l = 2, then it is called a d-state. In this state the magnitude of the angular momentum is L 22 1 6 and its z-component can have the following values, L Z 2,1 , 0, 1, 2 Note that each angular momentum state has a degeneracy of 2l+1 . Thus for example, a d-state has a degeneracy of 5. this means that there are 5 states with l = 2 which all have the same energy. Degeneracy We have found that the energy levels are given by me 4 1 me 4 1 13.6 En eV 2 2 2 2 2 2 8ε 0h n n 24 πε 0 n note that the energy levels are degenerate n l mL state since they do not depend explicitly on l and 1 0 0 1s ml . Recall that n = 1, 2, 3, 4, … 2 0 0 2s l = 0, 1, 2, 3, … n-1 -1 2p 1 0 2p ml = - l, -(l -1), ….(l -1), l +1 2p so each energy level has a degeneracy of 3 0 0 3s -1 3p n 1 2 2 l 1 n 1 1 3p l 0 +1 3p Example: the n = 2 state has a degeneracy of -2 3d -1 3d 4: 2 0 3d a single 2-s state with l = 0, ml = 0 and a 3+1 3d fold degenerate 2-p state with l = 1, +2 3d ml = -1, 0, 1 Energy -13.6 eV -3.40 eV -3.40 eV -1.51 eV -1.51 eV -1.51 eV Normalisation Just as in the case of one dimensional problems, the total probability of finding the electron somewhere in space must be unity. Using the volume element in polar coordinates, the nomalisation condition becomes all space 2 Ψ n,l,m L dV 1 π 2π 2 2 2 imL φ imL φ R r dr P sin θdθ e e dφ 1 n,l l,mL 0 0 0 The last integral is just 2 so we are left with two integrals. 2 2 Rn,l r dr 1 and 0 π 2 π Pl,mL 2 sinθdθ 1 0 Using these two integrals we can find the correctly normalised Radial and Angular wavefunctions Example Find the angular normalisation constant for P1,+1(). For l = 1, mL = +1 we have P1,+1() = sin (see table). then π π 2 2 2 2 πA1,1 sinθ sinθdθ 2 πA1,1 sin3 θdθ 1 0 0 Carry out the integral, to find 43 1 2 2 π A1 ,1 A1,1 and the correctly normalised wavefunction is 3 P1,1 θ sinθ 8π 3 8π Probability Density What is the probability of finding the electron within a small volume element dV at position r,, ? 2 2 P ψn,l,m dV ψn,l,m r 2 sinθdrdθdφ 2 2 2 2 2 Rn,lYl,m r sinθdrdθdφ Rn,l r dr Yl,m sinθdrdθdφ If we want to find the probability of finding the electron at a distance between r and r + dr (i.e. in a spherical shell of radius r and thickness dr), then we just integrate over and . π 2π 2 2 2 Pdr Rn,l r dr Yl,m sinθdθdφ 0 0 But, since the spherical harmonics are normalised (by definition), so the integral is equal to 1 and we then have 2 Pr dr Rn,l r 2dr or 2 Pr Rn,l r 2 is the probability of finding the electron at a distance r from the nucleus at any angle . What do these distributions look like ? 1s State n = 1, l = 0 R1,0 2 Pr R1,0 r 2 2s State n = 2, l = 0 R2,0 Pr R2,0 2 2 r 2p State n = 2, l = 1 R2,1 2 Pr R2,1 r 2 3s State n = 3, l = 0 R3,0 2 Pr R3,0 r 2 3p State n = 3, l = 1 R3,1 2 Pr R3,1 r 2 3d State n = 3, l = 2 R3,2 2 Pr R3,2 r 2 Summary of Key Points 1) The wavefunctions of the hydrogen atom can be expressed as the product of three one dimensional functions in the variables r, and ψr, θ, φ R(r) θ φ 2) The mathematical form taken by the wavefunction depends upon three quantum numbers n, l, ml. 3) These three quantum numbers can only take integral values subject to the following restrictions: n = 1, 2, 3, … l < n, ml = 0, 1, 2 … l orbitals with l = 0, 1 and 2 are known as s, p and d orbitals respectively. The radial solutions Rn,l(r) are given by the Associated Laguerre Polynomials The angular solutions are given by the Spherical Harmonics Yl,ml( ,) where Yl,ml( ,) =Pl,ml().exp(i ml ), and Pl,ml() are the Associated Legendre Polynomials.