6 Wave equation in spherical polar coordinates

Transcription

1 6 Wave equation in spherical polar coorinates We now look at solving problems involving the Laplacian in spherical polar coorinates. The angular epenence of the solutions will be escribe by spherical harmonics. We take the wave equation as a special case: u = 1 c u t The Laplacian given by Eqn. (4.11) can be rewritten as: u = u r + u + 1 ( sin θ u ) 1 u + } {{ r r} r sin θ θ θ r sin θ φ }{{ } raial part angular part. (6.1) 6.1 Separating the variables We consier solutions of separate form u(r, θ, φ, t) =R(r) Θ(θ) Φ(φ) T (t). Substitute this into the wave equation an ivie across by u = RΘΦT : 1 R R r + R rr r + 1 ( 1 sin θ Θ ) 1 1 Φ + r Θ sin θ θ θ r sin θ Φ φ = 1 1 T c T t First separation: r, θ, φ versus t This gives the T equation: which is easy to solve. LHS(r, θ, φ) = RHS(t) = constant = k. 1 1 T c T t = k (6.) 6.1. Secon separation: θ, φ versus r Multiply LHS equation by r an rearrange: 1 ( sin θ Θ ) 1 1 Φ Θ sin θ θ θ sin θ Φ φ = r R R r + r R R r + k r. (6.3) LHS(θ, φ) = RHS(r) = constant = λ We choose the separation constant to be λ. For later convenience, it will turn out that λ = l(l + 1) where l has to be integer. Multiplying the RHS equation by R/r gives the R equation: R r + r R r + This can be turne into Bessel s equation; we ll o this later. [k λr ] R =. (6.4) 44

2 6.1.3 Thir separation: θ versus φ Multiply LHS of Eqn. (6.3) by sin θ an rearrange: ( sin θ sin θ Θ ) + λ sin θ = 1 Φ Θ θ θ Φ φ = m LHS(θ) = RHS(φ) = constant = m. The RHS equation gives the Φ equation without rearrangement: Φ φ = m Φ. (6.5) Multiply the LHS by Θ/ sin θ to get the Θ equation: ( 1 sin θ Θ ) ] + [λ m sin θ θ θ sin Θ =. (6.6) θ 6. Solving the separate equations Now we nee to solve the ODEs that we got from the original PDE by separating variables Solving the T equation Eqn. (6.) is of simple harmonic form an solve as before, giving sinusois as solutions: with ω k = ck. 6.. Solving the Φ equation T t = c k T ω kt, Eqn. (6.5) is easily solve. Rather than using cos an sin, it is more convenient to use complex exponentials: Φ(φ) =e ±imφ Note that we have to inclue both positive an negative values of m. As φ is an angular coorinate, we expect our solutions to be single-value, i.e. unchange as we go right roun the circle φ φ +π: Φ(φ +π) = Φ(φ) e iπm =1 m = integer. This is another example of a BC (perioic in this case) quantising a separation constant. In principle m can take any integer value between an. It turns out in Quantum Mechanics that m is the integer magnetic quantum number an l m l for the z-component of angular momentum. In that context we will see that it is restricte to the range l m l. 45

3 6..3 Solving the Θ equation Starting from Eqn. (6.6), make a change of variables w = cos θ: w = θ w θ = (1 w ) w = 1 cos θ sin θ w (1 w ) w = 1 sin θ θ ( ) 1 w θ θ = 1 sin θ θ = θ sin [ sin θ sin θ ] θ θ, θ = sin θ θ, = 1 [ sin θ ] sin θ θ θ. Eqn. (6.6) becomes ( w (1 w ) w + λ m 1 w ) Θ(w) =. which is known as the Associate Legenre Equation. Solutions of the Associate Legenre Equation are the Associate Legenre Polynomials. Note that the equation epens on m an the equation an solutions are the same for +m an m. It will turn out that there are smart ways to generate solutions for m from the solutions for m = using angular momentum laer operators (see quantum mechanics of hyrogen atom). So it woul be unnecessarily heroic to irectly solve this equation for m. In this course we will only solve this equation for m = Solving the Legenre equation For m = we can write the special case as the Legenre Equation: ((1 w ) w w ) w + λ Θ(w) =. We apply the metho of Froebenius by taking Θ(w) = c i w i Then c i i(i 1)(w i w i ) c i iw i + λc i w i = an rearranging the series to always refer the power w i, [c i+ (i + )(i + 1) + c i (λ i(i 1) i] w i = Since this is true for all w, it is true term by term, an the inicial equation is c i+ ((i + )(i + 1)) = c i (i(i + 1) λ) Start The series switches on when c = amits c an c = Also when c 1 = amits c 1 an c 1 =. 46

4 Termination Note, however c i+ c i for large i. This gives an ill convergent series an for finite solutions the series must terminate at some value of i, which we call l. Thus, for some (quantise) integer value l. λ = l(l + 1) It will turn out in quantum mechanics that l is the orbital angular momentum quantum number Legenre polynomials We enote the solutions the Legenre polynomials P l (w) P l (cos θ) For example: P starts, an terminates with a single term C. P 1 starts, an terminates with a single term C 1. P starts, with C an terminates on C. etc... The first few are P (w) = 1 P 1 (w) = w P (w) = 1 (3w 1) P 3 (w) = 1 (5w3 3w) Exercise: use the recurrence relation c i+ ((i + )(i + 1)) = c i (i(i + 1) l(l + 1)) to verify that these are our series solutions of Legenre s equation Orgthogonality The orthogonality relation is 1 1 P m (w)p n (w)w = π P m (cos θ)p n (cos θ) sin θθ = N m δ mn where N m is a normalisation factor that we o not nee here. In quantum mechanics this is alreay sufficient to cover S, P, D an F orbitals Associate Legenre polynomials As mentione the associate Legenre polynomials can be prouce from Legenre polynomials in quantum mechanics using angular momentum laer operators. Firstly, P l (w) =P l (w) 47

5 Without proof, we can note that it can be shown that if P l (w) satisfies Legenre s equation, then P m l (w) = (1 w m / m ) w P l(w) m will satisfy the associate Legenre polynomial for magnetic quantum number m. As P l is a polynomial of orer l, then the above m-th erivative vanishes for m >lan thus m = l, l +1,...,,... l 1,l General angular solution Putting asie the raial part for the moment, the rest of the general solution is: l Θ(θ)Φ(φ)T (t) = Pl m (cos θ) e imφ (E ml cos ω k t + F ml sin ω k t) l= m= l The angular epenence is given by the combination: P m l (cos θ) e imφ Y l m(θ, φ) These are known as the spherical harmonics (once we inclue a normalisation constant). We ll iscuss these more in Sec What we have not yet establishe is the link between the value of k (an hence ω k ) an the values of m an l. To o this, we woul nee to solve the raial equation for various special cases. 6.3 The spherical harmonics Spherical harmonics {Yl m (θ, φ)} provie a complete, orthonormal basis for expaning the angular epenence of a function. They crop up a lot in physics because they are the normal moe solutions to the angular part of the Laplacian. They are efine as: Yl m (θ, φ) = ( 1)m l +1 (l m)! π (l + m)! P l m (cos θ)e imφ. The extra factor of ( 1) m introuce is just a convention an oes not affect the orthonormality of the functions. The spherical harmonics satisfy an orthogonality relation: π φ π θ sin θ [ Y m 1 l 1 (θ, φ) ] Y m l (θ, φ) =δ l1,l δ m1,m. Note that they are orthonormal, not just orthogonal, as the constant multiplying the prouct of Kronecker eltas is unity Completeness an the Laplace expansion The completeness property means that any function f(θ, φ) evaluate over the surface of the unit sphere can be expane in the ouble series known as the Laplace series: l f(θ, φ) = a lm Yl m (θ, φ), a lm = l= m= l π π θ π φ sin θ [Y m l (θ, φ)] f(θ, φ). Note that the sum over m only runs from l to l, because the associate Laplace polynomials P m l are zero outsie this range. 48

6 6.4 Time inepenent Schroeinger equation in central potential Consier h m ψ(x)+ṽ (r)ψ(x) =Ẽψ(x). We consier solutions of separate form: ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ). Substitute into Schroeinger equation an ivie across by ψ = RΘΦ. m h Multiplying through by r (V (r) E) 1 1 R r r r r R = 1 1 Θ r sin θ θ sin θ θ Θ+ 1 1 Φ r sin θ r m h (V (r) E) 1 R r r r R = 1 1 Θ sin θ θ sin θ θ Θ+ 1 1 Φ sin θ First separation: raial & angular epenence LHS(r) = RHS(θ, φ) = constant = l(l + 1). φ Φ φ Φ Raial equation [ r r r ] m + l(l + 1) + r (V (r) E) R = h The ifferential equation is simplifie by a substitution, u(r) = rr(r) u (r) = R(r)+rR (r) an so [ l(l + 1) + r r u (r) = R (r)+rr (r) = 1 r r r r R + m ] (V (r) E) u(r) = h We take a Coulomb potential an will be consiering boun states, with E<. It is convenient to rewrite in terms of the moulus E an introuce explicit negative sign. We also change variables to ρ = 8m E h r V (r) = e 4πɛ r = e 8m E, 4πɛ hρ an so multiplying by 1 an expressing in terms of u r { ] [ 8m E [ l(l + 1) h + + m ]} ρ ρ h E e 8m E 4πɛ ρ h u(ρ) = We efine λ = e m mc 4πɛ h = α, where α = E E gives us [ ρ 1 l(l + 1) 4 ρ e 1 4πɛ hc λ ρ ] u(ρ) = is the fine structure constant. This 49

7 6.4. Solution by metho of Froebenius We are now (almost!) reay to apply the metho of Froebenius. In principle it coul immeiately be applie an we woul get a an infinite Taylor series that inee solves the equation. However, a close form solution can be obtaine with one extra transformation that removes an over all exponential epenence on ρ. Observe that this equation for large ρ tens to [ ρ 1 ] u(ρ) 4 which has as a normalisable solution u(ρ) e ρ u(ρ) e + ρ which we ignore). We can o rather better by taking a trial solution: (an in aition a non-normalisable solution u(ρ) =e ρ f(ρ). Then, u = e ρ u = e ρ [f (ρ) 1 f(ρ) ]. [f (ρ) f (ρ)+ 14 f(ρ) ]. Now, [ ρ ρ l(l + 1) 4 ρ + λ ρ ] f(ρ) = We now apply the metho of Froebenius for a series substituting f(ρ) = c i ρ i, c i (i)(i 1)ρ i c i (i)ρ i 1 l(l + 1)c i ρ i + λc i ρ i 1 Thus, reexpressing so that all terms are of equal power ρ i 1 c i+1 (i + 1)(i)ρ i 1 c i (i)ρ i 1 l(l + 1)c i+1 ρ i 1 + λc i ρ i 1 i= 1 an we have the inicial equation, c i+1 [i(i + 1) l(l + 1)] = c i [i λ] Series start: The series switches on for c k c i+1 when i(i + 1) = (k 1)k = l(l + 1). The first term c k has k = l + 1. Series termination: If the series oes not terminate, then c i+1 c i i, an f ρ i. This looks i! like the other solution that is a non-normalisable exponential u(ρ) e + ρ which we o not seek. Only if λ = i = n then the series switches off after n l terms. i λ c i+1 = c i i(i + 1) l(l + 1) 5

8 We call n the principal quantum number. Note that for any given l, then n l + 1 as the series commences at k = l + 1. The energy is mc α E = n Thus E = α mc n This energy is consistent with the Hyrogen spectrum (Lymann, Balmer series etc...) Wavefunctions We enote the raial solution for L = l, an principle quantum number n l + 1 as R nl. Using our recurrence relation i n c i+1 = c i i(i + 1) l(l + 1) we have n =1,l = n =,l = n =,l =1 c c c 1 1 c 3 The above energy relation gives us that for each n 8m E ρ = r (6.7) h = αmc n h r (6.8) = r (6.9) n a where a = h is the usual Bohr raius. αmc The solutions are then R 1S 1 ρ e ρ ρ (6.1) = e ρ (6.11) = e r a (6.1) R S 1 ρ ρ e (ρ 1 ) ρ (6.13) = e ρ (1 1 ) ρ (6.14) ( = e r a 1 1 ) r (6.15) a R P 1 ρ e ρ ρ (6.16) = e ρ ρ (6.17) = e r r a (6.18) a 51

9 Note, we have not carefully normalise these raial wavefunctions. For each l the ifferent wavefunctions for the various n>lare orthogonal to each other. The orthogonality relation contains a resiual r factor corresponing to a vestige of the require orthogonality of wavefunctions uner the 3 volume integral. R nl (r)r ml (r)r r = N n N m δ nm A complete orthonormal set can of course be forme as usual but is beyon the scope of the course. This conclues the examinable material of the course 5