Physics 4617/5617: Quantum Physics Course Lecture Notes

Transcription

1 Physics 467/567: Quantum Physics Course Lecture Notes Dr. Donald G. Luttermoser East Tennessee State University Edition 5.

2 Abstract These class notes are designed for use of the instructor and students of the course Physics 467/567: Quantum Physics. This edition was last modified for the Fall 26 semester.

3 V. Quantum Mechanics in Three Dimensions A. Schrödinger Equation in Spherical Coordinates.. In three dimensions, we can write the Hamiltonian operator note that I ll leave the hats off of the operators in this section, but they are there virtually as H = 2 mv2 + V = p 2 2m x + p 2 y + p 2 z + V, V- where or p x h i x, p y h i y, p z h i z, V-2 p h i, V-3 for short. 2. Using this notation in the Schrödinger equation gives i h Ψ t = h2 2m 2 Ψ+V Ψ, where 2 2 x y z 2 is the Laplacian in Cartesian coordinates. V-4 V-5 3. The potential energy V and the wave function Ψ are now functions of r =x,y,z and t. 4. The probability of finding the particle in the infinitesimal volume d 3 r = dx dy dz is Ψr,t 2 d 3 r, and the normalization condition reads Ψ 2 d 3 r =, V-6 V

4 with the integral taken over all space. 5. If the potential is independent of time, there will be a complete set of stationary states, Ψ n r,t=ψ n r e ie nt/ h, V-7 where the spatial wave function ψ n satisfies the time-independent Schrödinger equation: h2 2m 2 ψ n + Vψ n = E n ψ n. V-8 6. The general solution to the time-dependent Schrödinger equation is Ψr,t= c n ψ n r e ient/ h, V-9 with the constants c n determined by the initial wave function, Ψr,, in the usual way. If the potential admits continuum states, then the sum in Eq. V-9 becomes an integral. 7. Separation of Variables. a Typically, the potential is a function only of the distance from the origin. In that case it is natural to adopt spherical coordinates, r, θ, φ. In spherical coordinates, the Laplacian takes the form 2 = r 2 r r 2 + r r 2 sin θ θ sin θ + θ r 2 sin 2 2 θ φ 2 V-. b In spherical coordinates, then, the time-independent Schrödinger equation reads h2 2m r 2 r r 2 ψ r + r 2 sin θ θ +Vψ= Eψ. V 2 sin θ ψ θ + r 2 sin 2 θ V- 2 ψ φ 2

5 c To solve this equation, we will assume that the solution can be represented as the product of separable terms composed of a radial part R and an angular part Y : ψr, θ, φ =RrY θ, φ. V-2 d Putting this into Eq. V- we get h2 Y 2m r 2 d dr r 2dR dr + R r 2 sin θ θ +VRY = ERY. sin θ Y + θ R r 2 sin 2 θ V-3 e Dividing by RY and multiplying by 2mr 2 / h 2 we get + Y R sin θ d dr θ r 2dR 2mr2 dr h 2 [V r E] sin θ Y + θ sin 2 2 Y θ φ 2 =. f The term in the first curly woob-woob-woob bracket depends only upon r, whereas the remainder depends only on θ and φ; accordingly, each must be constant. We will write this separation constant as ll + the reason for choosing this form of the constant will become apparent in VI of the notes, as such Y R d r 2dR 2mr2 dr dr h 2 [V r E] =ll + ; V-4 sin θ Y + sin θ θ θ sin 2 2 Y θ φ 2 = ll +. V-5 2 Y φ 2 Example V. Use separation of variables in Cartesian coordinates to solve the infinite cubical well or particle in a box : V x,y,z=, if x, y, and z are all between and a;, otherwise. V 3

6 a Find the stationary state wave functions and the corresponding energies. Solution a: In the box, we can write the TISE Eq. V-8 as h2 2m 2 ψ x + 2 ψ 2 y + 2 ψ 2 z 2 = Eψ. The separable solution is: ψx,y,z= XxY yzz. Put this in the above equation and divide by XYZ: X d 2 X dx 2 + Y d 2 Y dy 2 + Z d 2 Z dz 2 = 2m h 2 E = k2 x + k2 y + k2 z, where E k2 x + k2 y + k2 z h2, 2m and k x, k y, and k z are three constants. The three terms on the left of this equation are functions of x, y, and z, respectively, so each must be a constant, where kx, 2 ky, 2 and kz 2 are the three separation constants. This leads to three separate differential equations: d 2 X dx 2 = k2 x X; d 2 Y dy 2 = k2 y Y ; d 2 Y dz 2 = k2 z Z. The solution to these three equations are Xx = A x sin k x x + B x cos k x x; Y y = A y sin k y y + B y cos k y y; Zz = A z sin k z z + B z cos k z z. But X =, so B x =;Y =, so B y =;Z =, so B z =. Likewise, Xa = sink x a= k x = n x π/a, n x =, 2, 3,... note that negative values of k are redundant V 4

7 with the positive values and n x since this would give us no wave function. Likewise, k y = n y π/a and k z = n z π/a. So and E = h2 2m π 2 a 2 n 2 x + n 2 y + n2 z nx π ψx,y,z=a x A y A z sin a x ny π sin a y nz π sin a z. We can normalize the three independent solutions separately which was done in Eq. III-32, giving A x = A y = A z = 2/a. So the final solution is ψx,y,z = sin n x π a a x sin n y π a y sin n z π a z ; E = π2 h 2 2ma n 2 2 x + n 2 y + n2 z ; n x,n y,n z =, 2, 3,..., b Call the distinct energies E,E 2,E 3,..., in order of increasing energy. Find E,E 2,E 3,E 4,E 5, and E 6. Determine the degeneracy of each of these energies that is, the number of different states that share the same energy. Note that degenerate bound states do not occur in one dimension, but they are common in three dimensions. Solution b: Set up the following data table: V 5

8 n x n y n z n 2 x + n2 y + n2 z The number of degenerate levels is given by the number of rows available for each energy i.e., n 2 x + n2 y + n2 z. As such, the first 6 energy levels are 4 V 6

9 E = 3 π2 h 2 2ma 2 ; degeneracy d =. E 2 = 6 π2 h 2 2ma 2 ; d =3. E 3 = 9 π2 h 2 2ma 2 ; d =3. E 4 = π2 h 2 2ma 2 ; d =3. E 5 = 2 π2 h 2 2ma 2 ; d =. E 6 = 4 π2 h 2 2ma 2 ; d =6. 8. The Angular Equation. a Taking Eq. V-5 and multiplying by Y sin 2 θ gives sin θ θ sin θ Y + 2 Y θ φ = ll +Y 2 sin2 θ. V-6 b Once again, use separation of variables: Y θ, φ =ΘθΦφ. V-7 Plugging this into Eq. V-6 and dividing by ΘΦ gives sin θ [sin θ ] θ θ ΘΦ + 2 φ 2 ΘΦ = ll + ΘΦ sin2 θ Φ sin θ sin θ Θ +Θ 2 Φ θ θ φ 2 = ll + ΘΦ sin 2 θ Θ sin θ sin θ Θ + 2 Φ = ll + sin 2 θ, θ θ Φ φ 2 V 7

10 or { Θ [ sin θ d sin θ dθ dθ dθ ] } + ll + sin 2 θ + Φ d 2 Φ dφ 2 =. V-8 c The first term of Eq. V-8 is a function only of θ, and the second is a function only of φ, so each must be constant. Let s choose the separation constant m 2 m will later be called the magnetic quantum number, then Θ and [ sin θ d dθ sin θ dθ dθ Φ ] + ll + sin 2 θ = m 2, V-9 d 2 Φ dφ 2 = m2. d The equation for φ is easy with the solution V-2 Φφ =e imφ V-2 actually, there are two solutions: e imφ and e imφ, but we will fold the negative exponents into the positive solution by letting m be negative as well as positive. We also will fold the integration constant into the solution for Θ. e Since Eq. V-2 is nothing more than trigonometric functions in complex space, note that Φφ +2π =Φφ. V-22 In other words, exp[imφ+2π] = expimφ, or exp2πim =. From this it follows that m must be an integer: m =, ±, ±2,... V-23 f The equation for θ becomes sin θ d dθ sin θ dθ dθ +[ll + sin 2 θ m 2 ]Θ=. V-24 V 8

11 The solution to this differential equation is not trivial. It is Θθ =APl m cos θ, V-25 where Pl m is the associated Legendre function, defined by d m Pl m x x2 m /2 P l x, V-26 dx and P l x is the l-th Legendre polynomial. g Legendre polynomials are determined with the Rodrigues formula: P l x d l x 2 l. V-27 2 l l! dx For example, P x =, P x = 2 P 2 x = 4 2 and so on. d dx x2 = x, d 2 x 2 2 = dx 2 3x2, h P l x is a polynomial of degree l inx, and is even or odd according to the parity of l. However, Pl m is not, in general, a polynomial, since if m is odd, it carries a factor x2 : P2 x = 2 3x2, P2 x = x2 /2 d [ dx 2 3x2 ] =3x x 2, P2 2 x = x2 d2 [ dx 2 2 3x2 ] =3 x 2, etc. Since x = cos θ in the associated Legendre functions here, P m l cos θ is always a polynomial in cos θ, multiplied V 9

12 Table V : Some associated Legendre functions, P m l cos θ. P = sin θ P = cos θ P2 2 = 3 sin2 θ P2 = 3 sin θ cos θ P2 = 3 2 cos2 θ P3 3 = 5 sin θ cos 2 θ P3 2 = 5 sin 2 θ cos θ P3 = 3 sin θ 5 2 cos2 θ P3 = 5 2 cos3 θ 3 cos θ if m is odd by sin θ since cos 2 θ = sin θ. Some associated Legendre functions of cos θ are listed in Table V-. i Notice that l must be a non-negative integer for the Rodrigues formula Eq. V-27 to make any sense. Moreover, if m >l, then Eq. V-26 says P m l =. For any given l, then, there are 2l + possible values of m: l =,, 2,...; m = l, l +,...,,,,..., l,l. V-28 j Eq. V-24 is a second-order differential equation: It should have two linearly independent solutions, for some values of l and m. Where are the other solutions? Well, they exist as mathematical solutions to the equation, but they are physically unacceptable because they blow up at θ = and/or θ = π, and do not yield normalizable wave functions. Example V 2. Show that Θθ =A ln[tanθ/2] satisfies the θ equation Eq. V-24 for l = m =. This V

13 is the unacceptable second solution what s wrong with it? Solution: dθ dθ = A θ tanθ/2 2 sec2 = A 2 2 So, d sin θ dθ dθ dθ With l = m =, Eq. V-24 reads sinθ/2 cosθ/2 = = d A =. dθ A sin θ. d sin θ dθ =. dθ dθ So, A ln[tanθ/2] does satisfy Eq. V-24. Note, however, that if θ =, then tanθ/2 = and Θ = A ln = A = Blows up at θ =. Also, Θπ =A ln[tanπ/2] = A ln =A = Blows up at θ = π. k The volume element in spherical coordinates is d 3 r = r 2 sin θ dr dθ dφ, V-29 so the normalizable condition of Equation V-6 becomes ψ 2 r 2 sin θ dr dθ dφ = R 2 r 2 dr Y 2 sin θ dθ dφ =. l It is convenient to normalize R and Y individually: R 2 r 2 dr = and 2π π V Y 2 sin θ dθ dφ =. V-3

14 m The normalized angular wave functions are called spherical harmonics: Y m l θ, φ =ɛ 2l + 4π l m! l + m! eimφ P m l cos θ, V-3 where ɛ = m for m> and ɛ = for m. Note from this equation that Y m l = m Y m l. V-32 Spherical harmonics are orthogonal such that 2π π [Y m l θ, φ] [ Yl m θ, φ] sin θ dθ dφ = δ ll δ mm. V-33 Example V 3. Use Equations V-26, V-27, and V-3 to construct Y and Y 2. Check that they are normalized and orthogonal. Solution: From Eq. V-3, Y =!! P cos θ = P cos θ, 4π 4π where ɛ =. From Eqs. V-26 and V-27, P cos θ =P cos θ =! =, so From Eq. V-3, Y 2 = 5 4π Y = 4π. 3 2 eiφ P2 cos θ = 5 24π eiφ P 2 cos θ, V 2

15 where ɛ = =. From Eq. V-26, Then using Eq. V-27 P 2 x = x 2 d dx P 2x. P 2 x = d 2 x 2 2 = d 4 2 dx 8 dx [2x2 2x] = d 8 dx [4x3 4x] = d 2 dx [x3 x] = 2 3x2 ; so P2 x = x d [ 3 2 dx 2 x2 2] = x 2 3x. Substituting cos θ for x gives P 2 cos θ = cos 2 θ 3 cos θ = 3 cos θ sin θ, and 2 = 5 8π eiφ cos θ sin θ. Y Normalization: Y 2 sin θ dθ dφ = 4π [ π ] [ sin θdθ 2π ] dφ = 4π 22π =. Y2 2 sin θ dθ dφ = 5 8π = 5 4 Orthogonality: Y Y2 = 5 4 = 5 4 π sin2 θ cos 2 θ sin θdθ π 5 sin θ dθ dφ = 4π 8π 2π cos2 θ cos 2 θ sin θdθ cos3 θ + cos5 π θ 3 5 [ 2 3 5] 2 = =. V 3 [ π sin2 θ cos θdθ dφ ] [ 2π ] e iφ dφ.

16 But and So, 2π π e iφ dφ = eiφ i sin2 θ cos θdθ= 2π sin3 θ 3 π = = i cos φ + sin φ 2π = i + i =. Y Y 2 sin θ dθ dφ =. 9. The Radial Equation. a Note that the angular part of the wave function, Y θ, φ is the same for all spherically symmetric potentials, and is independent of this potential = V only affects the radial part of the wave function, Rr, which is determined by the differential equation given in Eq. V-4: d dr r 2dR dr 2mr2 h 2 [V r E] R = ll +R. V-34 b This equation simplifies if we change variables: Let ur rrr, V-35 then using the quotient rule for derivatives we get R = u r dr = rdu/dr u dr r 2 dr r 2 = rdu/dr u dr [ ] d dr r 2 = d r du du dr dr dr dr du = du dr + u rd2 dr 2 du dr = u rd2 dr 2. V 4

17 c Hence making the various substitutions for R in Eq. V- 34 we get r d2 u dr 2mr2 2 h 2 [V E] u r = ll +u r h2 2mr u rd2 dr + h2 2 2mr 2mur h 2 or simplifying gives [V E] = h2 2mr ll +u r, h2 2m d 2 u dr 2 + V + h2 2m ll + r 2 u = Eu. V-36 d This is called the radial equation note that m in this equation represents mass, and not the magnetic quantum number, as was the case in the angular equation. i It is identical in form to the one-dimensional Schrödinger equation Eq. III-8, except that the effective potential, V eff = V + h2 ll + 2m r 2, V-37 contains an extra piece, the so-called centrifugal term, h 2 /2m[ll +/r 2 ]. ii This centrifugal term tends to throw the particle outward away from the origin, just like the centrifugal pseudo- force in classical mechanics. e Meanwhile, the normalization condition becomes u 2 dr =. V-38. We cannot proceed any further without providing a specific potential. As an example, let s consider the infinite spherical well:, if r<a; V r = V-39 if r>a. V 5

18 a Outside the well the wave function is zero; inside the well the radial equation says where d 2 u dr 2 = ll + r 2 k 2 u, k 2mE h V-4. V-4 b For the boundary condition ua = for l =, we get d 2 u dr = 2 k2 u = ur =A sinkr+b coskr. i However, the actual radial wave function is Rr = ur/r, and [coskr/r] blows up as r. So in order for the wave function to be normalizable, we must chose B =. ii This boundary condition then requires sinka = orka = nπ, for some integer n. From this, analogously to the -D infinite square well, the allowed energies that satisfy the boundary condition are E nl = E n = n2 π 2 h 2, n =, 2, 3,.... V-42 2ma2 c Normalizing ur yields A = 2/a. Using the solution to the angular part see Example V-3, Y θ, φ =/ 4π, of the wave equation gives the complete wave equation: ψ n = 2πa sinnπr/a r. V-43 i Notice that the wave function of stationary states are labeled by three quantum numbers, n, l, and m: ψ nlm r, θ, φ. V 6

19 ii The energy, however, depends only upon n and l: E nl = the m quantum states are said to be degenerate with each other since they give the same energy. d The general solution to Equation V-4 for an arbitrary integer l is ur =Arj l kr+brn l kr, V-44 where j l x is the spherical Bessel function of order l, and n l x is the spherical Neumann function of order l. i Bessel functions are defined as j l x x l x For example, j x = sin x x ; j x = x d sin x x dx x d l sin x dx x. V-45 = sin x x 2 cos x x. ii Meanwhile, Neumann functions are defined as n l x x l x For example, n x = cos x ; x n x = x x d dx d l cos x dx x cos x x = cos x x 2. V-46 sin x x. iii Note that the notation d l x dx has the following meaning: if we set l = 3, then 3 { [ ]} d d d d =. x dx x dx x dx x dx V 7

20 iv For small x, note that sin x x x 3 /3!+x 5 /5! and cos x x 2 /2+x 4 /4!, as such: x l j l 2l +!!, n 2l!! l, for x, x l+ V-47 where the double exclamation points mean 2l +!! = l+ [e.g., for l =, 2l+!! =!! = ; l =, 2l +!! = 3!! = 3 = 3; and l =2, 2l +!! = 5!! = 3 5 = 5, and note for the n l equation that!! =! and!! =! ]. So, j x ; n x x ; j x x 3 ; n x x 2. v As can be seen from these trends at small x, the Bessel functions remain finite at the origin, but the Neumann functions blow up there. As such, B l =, so Rr =Aj l kr. V-48 e Now using the boundary condition, Ra =, we must solve the equation j l ka =, V-49 that is, ka is a zero of the l th -order spherical Bessel function. Since spherical Bessel functions are oscillatory see Figure V- and Table V-2, each one has an infinite number of zeros. f The boundary condition requires that k = a β nl, V-5 where β nl is the n th zero of the l th spherical Bessel function. V 8

21 Table V 2: The first four spherical Bessel functions. j = sin x x j = sin x x 2 j 2 = cos x x 3 x sin x x 3 3 x cos x 2 j 3 = 5 x 6 5 sin x 4 x 2 x cos x 3 x Table V 3: The first four spherical Neumann functions. n = cos x x n = cos x x 2 sin x x 3 n 2 = x cos x x 3 3 x sin x 2 5 n 3 = x 6 5 cos x 4 x 2 x sin x 3 x V 9

22 j l x.6 l =.5.4 l =.3.2 l = 2 l = x Figure V : Graphs of the first four spherical Bessel functions. g The wave functions that result from the 3-D Schrödinger equation are ψ nlm r, θ, φ =A nl j l β nl r/a Yl m θ, φ, V-5 with the constant A nl to be determined by normalization. h The allowed energies of these wave functions are E nl = h2 2ma 2 β2 nl. V-52 Each energy level is 2l + -fold degenerate, since there are 2l + different values of m for each value of l see Eq. V-28. V 2

23 B. The Hydrogen Atom.. The Potential Function. a The hydrogen atom consists of a relatively massive, essentially motionless, proton placed at the origin of charge +e, together with a relatively small mass electron of charge e that circles around it, held in orbit by the mutual attraction of opposite charges. b From Coulomb s law, the potential energy in SI units is V r = e2 4πɛ r. V-53 c We can use this potential then in our radial equation Eq. V-36, giving h2 d 2 u 2m dr + e2 2 4πɛ r + h2 ll + 2m r 2 u = Eu. V-54 d We will solve this equation using the analytical solution technique used for the harmonic oscillator. Note that the Coulomb potential admits both continuum states with E >, describing electron-proton scattering and photoionization, and discrete bound states with E<, representing the hydrogen atom. 2. The Radial Wave Function. a We will now determine the bound states for hydrogen. To do this, we will simplify the notation by letting 2mE κ V-55 h note that κ is real since E < for bound states and dividing Eq. V-54 by E which gives u. V-56 κ 2 d 2 u dr 2 = me2 2πɛ h 2 κ κr V 2 + ll + κr 2

24 b Then let so that ρ κr and ρ me2 2πɛ h 2 κ, d 2 u dρ 2 = ρ ρ ll + + ρ 2 u. V-57 V-58 c Next we examine the asymptotic form of the solutions. As ρ, the constant term in the brackets dominates, so approximately d 2 u dρ 2 = u. d The general solution to this equation is uρ =Ae ρ + Be ρ, but e ρ blows up as ρ,sob =. As such uρ Ae ρ for large ρ. V-59 e In the other extreme, as ρ, the centrifugal term dominates when l>, though the result will still apply even when l =, so approximately then d 2 u ll + = u. dρ2 ρ 2 f The general solution to this equation is uρ =Cρ l+ + Dρ l, but ρ l blows up as ρ, so D =. Thus uρ Cρ l+ for small ρ. V-6 g The next step is to peel off the asymptotic behavior, by introducing the new function vρ: uρ =ρ l+ e ρ vρ, V-6 V 22

25 in the hope that vρ will turn out to be simpler than uρ. h Starting out, we get du dρ = ρl e ρ and d 2 u dρ 2 [ l + ρv + ρ dv dρ = ρ l e ρ ll + 2l 2+ρ + v + ρ 2l + ρ dv dρ + v ρd2 dρ 2. i In terms of vρ, the radial equation Eq. V-58 becomes ρ d2 v +2l + ρdv dρ2 dρ +[ρ 2l + ]v =. V-62 j Finally, assume the solution to vρ can be expressed as a power series in ρ: vρ = a j ρ j. j= ] V-63 Differentiating gives dv dρ = ja j ρ j = j +a j+ ρ j. j= j= In the second sum, we replaced j with j +. Note that these two sums are still equivalent, even though they both start from j =, since the j = term equals zero in the second sum. Differentiating again gives d 2 v dρ 2 = j= jj +a j+ ρ j. k Inserting these into Equation V-62 gives j= jj +a j+ ρ j +2l + j +a j+ ρ j V 23 j=

26 2 j= ja j ρ j +[ρ 2l + ] j= a j ρ j =. l Equating the coefficients of like powers gives jj+ a j+ +2l+j+ a j+ 2ja j +[ρ 2l+] a j =, or a j+ = 2j + l + ρ a j. V-64 j + j +2l +2 To determine each coefficient, start with a = A and determine A through normalization. i You might wonder why we didn t just start by using the series solution for uρ in the first place. This is done primarily because a three-term recursion relation a j+2, a j+, and a j would result if the asymptotic e ρ term was not included in the solution. Such a recursion relation is much more difficult to handle than the one derived in Equation V-64. ii For large j this corresponds to large ρ, the recursion relation becomes a j+ 2j = jj + a j = 2 j + a j, so a j 2 j = j! A. iii Suppose for a moment that this were the exact result, then vρ =A 2 j j! ρj = Ae 2ρ, and hence j= uρ =Aρ l+ e ρ, V 24

27 which blows up at large ρ! iv This is precisely the asymptotic behavior we didn t want at large ρ, since the wave functions would no longer be normalizable. There s only one way out of this dilemma: The series must terminate. There must be some maximum integer, j max, such that a jmax + = V-65 beyond which all coefficients vanish automatically. v With this definition, Equation V-64 becomes 2j max + l + ρ =. vi Defining n j max + l + V-66 the so-called principal quantum number, we have ρ =2n. V-67 m The energy E is determined from ρ from Eqs. V-55 and V-56, so E = h2 κ 2 2m = me4 8π 2 ɛ 2 h2 ρ 2. V-68 and the allowed energies are E n = m 2 h 2 e2 4πɛ 2 n = E, n =, 2, 3,... 2 n2 V-69 i This is the famous Bohr formula by any measure the most important result in all of quantum V 25

28 mechanics. This equation written in SI units is equivalent to Equation I-67, which is the Bohr formula given in the cgs unit system. ii Bohr obtained this result in 93 by a serendipitous mixture of inapplicable classical physics and premature quantum theory the Schrödinger equation did not come until 924. n Combining Eqs. V-57 and V-67, we find that where κ = me2 4πɛ h 2 n = an, V-7 a 4πɛ h 2 me 2 =5.29 m =.529 Å V-7 is the so-called Bohr radius the orbital radius of the ground state electron in hydrogen. o It follows from Eq. V-57 that ρ = r an. V-72 p From Eq. V-2, the wave functions that describe the hydrogen atom are labeled by 3 quantum numbers n, l, and m corresponding to the principal, orbital angular momentum, and magnetic also called azimuthal quantum numbers, respectively: ψ nlm r, θ, φ =R nl r Yl m θ, φ, V-73 where referring back to Eqs. V-36 and V-6: R nl r = r ρl+ e ρ vρ. V-74 V 26

29 q To determine vρ, we use Eq. V-67 in Eq. V-64, which gives 2j + l + n a j+ = j + j +2l +2 a j. V-75 r The ground state is defined to be the state that has the lowest energy. This occurs when n =, so E = m 2 h 2 e2 2 = 3.6 ev. V-76 4πɛ This means that one would ionize the atom from the ground state if a photon energy of at least 3.6 ev is imparted on the atom. i From Eq. V-66, the n = state forces both j max and l to as such, m =. ii The wave function for the ground state then becomes ψ r, θ, φ =R r Y θ, φ. V-77 iii The recursion formula of Eq. V-75 truncates after the first term j = yields a =, so vρ is a constant a and R r = a a e r/a. V-78 iv Normalizing this equation gives R 2 r 2 dr = a 2 so a =2/ a. a 2 e 2r/a r 2 dr = a 2a 4 =, v Meanwhile, Y =/ 4π, which gives the final V 27

30 ground state wave function of ψ r, θ, φ = πa 3 e r/a. V-79 s If n = 2, the energy is 3.6 ev E 2 = = 3.4 ev; V-8 4 this is the first excited state or rather, states, since we can have either l = in which case m = or l = with m =,, or +, so there are actually four different states that share this energy. i If l =, the recursion relation gives a = a using j =, and a 2 = using j =, so vρ =a ρ, and hence R 2 r = a 2a r 2a e r/2a. V-8 ii If l =, the recursion formula terminates the series after a single term, so vρ is a constant, and we find R 2 r = a 4a 2 re r/2a. V-82 t For arbitrary n, the possible values of l consistent with Eq. V-66 are l =,, 2,..., n. V-83 u For each l, there are 2l + possible values of m, so the total degeneracy of the energy level E n is dn = n 2l +=n 2. V-84 l= V 28

31 Table V 4: The first few Laguerre polynomials, L q x. L = L = x + L 2 = x 2 4x +2 L 3 = x 3 +9x 2 8x +6 L 4 = x 4 6x 3 +72x 2 96x +24 L 5 = x 5 +25x 4 2x 3 + 6x 2 6x + 2 L 6 = x 6 36x x 4 24x x 2 432x + 72 Table V 5: Some associated Laguerre polynomials, L p q px. L = L 2 = 2 L = x + L 2 = 6x +8 L 2 = x 2 4x +2 L 2 2 = 2x 2 96x + 44 L = L 3 = 6 L = 2x +4 L 3 = 24x +96 L 2 = 3x 2 8x +8 L 3 2 = 6x 2 6x + 2 v The polynomial vρ defined by Equations V-63 and V- 75 is a function well known to applied mathematicians; apart from normalization, it can be written as vρ =L 2l+ n l 2ρ, V-85 where d p L p q p p L q x V-86 dx is an associated Laquerre polynomial see Table V-5, and d q L q x e x e x x q V-87 dx is the q-th Laquerre polynomial see Table V-4. V 29

32 R nl r Legend nl r/a Figure V 2: Graphs of the first few hydrogen radial wave functions, R nl r note that the Bohr radius a in the calculation of R nl r set to unity in this plot. w With these associated Laguerre polynomials, we can now easily determine the radial wave equation see Table V-6 and Figure V-2. x From these polynomial functions, we can now write the final, complete, normalized wave functions for hydrogen: ψ nlm = 2 na 3 n l! 2r l 2r 2n[n + l!] 3 e r/na L 2l+ n l na na Y m l θ, φ. V-88 y Even though this equation is complicated, keep in mind that hydrogen is the only atom for which an analytic solution exists for its wave function. All other atoms require perturbation theory to describe their states. V 3

33 Table V 6: The first few radial wave functions for hydrogen, R nl r. R = 2a 3/2 exp r/a R 2 = a 3/2 r exp r/2a 2 2 a R 2 = a 3/2 r 24 a exp r/2a R 3 = R 3 = R 32 = [ 2 a 3/ a 3/2 6 4 r 8 3 a 3/2 a 2 r 3 a + 2 ] r 2 exp r/3a 27 a r r exp r/3a a a 2 exp r/3a z Of course, these wave functions for hydrogen are orthogonal: ψ nlm ψ n l m r2 sin θ dr dθ dφ = δ nn δ ll δ mm. V With these complete wave functions for hydrogen and their associative eigenvalues, we can derive all of the equations that were presented in I of the notes concerning the spectrum of hydrogen see Eqs. I-4 through I-7 and Eqs. I-67 and I-68. Example V 4. What is the probability that an electron in the ground state of hydrogen will be found inside the nucleus? a First calculate the exact answer, assuming that the wave function Eq. V-79 is correct all the way down to r =. Let b be the radius of the nucleus. Solution a: P = Ψ 2 d 3 r = 2π π b π Ψ 2 r 2 sin θ dr dθ dφ V 3

34 or = 2π dφ = 2π2 = 4 a 3 = π sin θdθ b π b πa 3 e 2r/a r 2 dr a 2 r2 e 2r/a + a3 + 2r a + 2r2 a 2 2 πa 3 e r/a r 2 dr 4 e 2r/a b e 2r/a = 4π πa 3 2r, b a P = + 2b a + 2b2 a 2 e 2b/a. e 2r/a r 2 dr b b Expand the result as a power series in the small number ɛ 2b/a, and show that the lowest order term is the cubic: P 4/3b/a 3. This should be a suitable approximation, provided that b is much less than a which it is. Solution b: P = +ɛ + 2 ɛ2 e ɛ +ɛ + 2 ɛ2 ɛ + ɛ2 2 ɛ3 3! or +ɛ ɛ2 2 + ɛ3 6 ɛ + ɛ2 ɛ3 2 ɛ2 2 + ɛ3 2 ɛ = 2 6 ɛ3 = 2b 3, 6 a P 4 3 b a3. V 32

35 c Alternatively, we might assume that ψr is essentially constant over the tiny volume of the nucleus, so that P 4/3πb 3 ψ 2. Check that you get the same answer this way. Solution c: The ground state wave function is given by Eq. V-79: [ ] ψ 2 2 = πa 3 e = πa, 3 thus 4 4 P πb 3 3 ψ 2 = πb 3 3 πa 3 = 4 b. 3 a3 d Use b 5 m and a.5 m to get a numerical estimate for P. Roughly speaking, this represents the fraction of its time that the electron spends inside the nucleus. Solution d: or P = m.5 5 m 3 = = , P =.7 4. = Example V 5. Use the energy eigenvalue equation i.e., the Bohr formula given by Eq. V-69 to deduce the empirically derived Rydberg formula i.e., Eq. I-7 for the hydrogen atom. Solution: E n = m 2 h 2 e2 4πɛ 2 n 2. V 33

36 Energy is related to a photon s wavelength by E = hc/λ. Setting the photon s energy equal to the energy difference between two bound states: hc = E j E i λ ji = m 2 h 2 e2 4πɛ 2 j 2 i 2 = m 2 h 2 e2 2 4πɛ i, 2 j 2 where j is the principle quantum number of the upper state in the transition and i is the lower state principle quantum number. Solving this for the wavenumber ν ji =/λ ji we get ν ji = = me4 λ ji 8ɛ 2 h3 c i. 2 j 2 To be more precise here, m represents the reduced mass between the electron and proton and not the electron mass alone, so m = m em A m e + M A = = kg m e. m em p = kg 2 m e + m p kg Also, the other constants in this Rydberg formula are ɛ = C 2 s 2 /kg m 3 [farad/meter] and e = C. Using the values for these physics constants, we can define and evaluate the Rydberg constant for hydrogen R to be R = me4 8ɛ 2 h3 c = m = 9, cm, and the Rydberg equation becomes = R λ ji i 2 j 2. Hence proving the empirically derived Eq. I-7 from quantum mechanics for hydrogen where Z =. V 34