= e2. = 2e2. = 3e2. V = Ze2. where Z is the atomic numnber. Thus, we take as the Hamiltonian for a hydrogenic. H = p2 r. (19.4)

Transcription

1 Chapte 9 Hydogen Atom I What is H int? That depends on the physical system and the accuacy with which it is descibed. A natual stating point is the fom H int = p + V, (9.) µ which descibes a two-paticle system with educed mass µ = mm m + M m, if m M. (9.) What is V? It seems easonable to adopt the known electostatic potential between electic chages. Does this continue to apply in the atomic ealm? Only expeiment can answe. In the hydogen atom, H, thee is an electon, of chage e, and a poton of chage +e. The electostatic potential is e( e) Fo ionized helium, He +, the potential is e( e) = e. = e, and fo doubly ionized lithium, Li ++, we have 3e( e) In geneal, fo a single-electon atoms, = 3e. (9.3a) (9.3b) (9.3c) V = Ze, (9.3d) whee Z is the atomic numnbe. Thus, we take as the Hamiltonian fo a hydogenic atom H = p µ Ze. (9.4) 67 Vesion of Octobe, 0

2 68 Vesion of Octobe, 0 CHAPTER 9. HYDROGEN ATOM I Hee, we ignoe the spin of the electon and the poton. The equations of motion fo the elative position and momentum, which ae conjugate vaiables, ae [, p] = i h, (9.5) d dt = H [, H] = i h p = p µ, dp dt = [p, H] = H i h = Ze 3. (9.6a) (9.6b) Is angula momentum conseved? (It must be, since H is a spin-independent scala.) dl dt = d dt ( p) = p µ p 3 Ze = 0. (9.7) Is thee anything else, besides H itself, of couse, that is conseved? An affimative suggestion comes fom the known dynamic situation with the Newtonian potential, Ze GmM, (9.8) descibing a planet about the sun, fo example. It is familia that the obit is an ellipse which holds steady in space. This fact is unique to the / potential, to the invese squae law of foce. The clue to finding what is the associated conseved quaity is to conside d dt. In the following, we will ignoe the noncommutativity of opeatos, and only state the necessay modifications at the end: d dt = v d, (9.9) dt whee so d dt = v If we define the axial vecto A by d dt = d dt = v, (9.0) ( v) ( v) 3 = 3 = 3 L µ = Ze d dt p L µ. (9.) A = µzep L, (9.) we see that A is constant, d A = 0. (9.3) dt

3 69 Vesion of Octobe, 0 cosθ + d d θ Figue 9.: Conside a point a distance d fom a fixed line, called the diectix. Let and θ be pola coodinates fom the point as oigin, with the angle measued fom the nomal to the diectix. The ellipse is the locus of points which have a fixed atio e between the distance fom the point and the distance cosθ + d fom the diectix. Obseve that A L = L µzep L L = 0, (9.4) because both tems vanish, if we ignoe noncommutativity. Leaving the symbols unchanged, but thinking of astonomy, we ecognize the elliptical obit: A = µze p L = L = e cosθ, (9.5) µze whee e = A, and θ is the angle between A and, so = L /µze e cosθ. (9.6) which is the pola equation of an ellipse, if e <, with eccenticity e. [The definition of an ellipse such that the sum of the distances fom the two foci is constant, + = constant, (9.7) is moe familia, but an altenate definition is the locus of points that make a constant atio e between the adial distance, and the hoizontal distance fom a fixed line, the diectix, = e( cosθ + d), = ed e cosθ. (9.8) See Fig. 9..] Back to quantum mechanics. All that is needed is to make A Hemitian by symmetic multiplication: p L (p L L p). (9.9)

4 70 Vesion of Octobe, 0 CHAPTER 9. HYDROGEN ATOM I Fo example, the z component of this is (p xl y p y L x L x p y + L y p x ) = (p xl y + L y p x ) (p yl x + L x p y ). (9.0) Recall that if A and B ae two Hemitian opeatos, the symmetic poduct is Hemitian: (AB + BA) = (BA + AB). (9.) The measue of non-commutativity hee is (p L + L p) z = p x L y p y L x + L x P y L y p x = [p x, L y ] [p y, L x ] = i hp z, (9.) o in geneal, p L + L p = i hp. (9.3) Theefoe the Hemitian combination is (p L L p) = p L i hp = L p + i hp. (9.4) It is still tue in quantum mechanics that L and A ae othogonal, A L = L (p L i hp) L = 0 (p L L i hp L) = 0. µze µze (9.5) So, A L = L A = 0. (9.6) Since thee ae 6 vaiables,,p, and 6 constants L,A, the constant H cannot be independent of L and A. Initially, ignoing the non-commutativity, we obseve A = µze p L + (µze (p L) (p L). (9.7) ) The tiple scala poduct hee is actually L, and the squae of p L is just p L (p L) = p L. Thus, A = + ( ) p µz e 4 µ Ze L = + µz e 4 HL, (9.8) o solving fo H, H = µz e 4 A L. (9.9) As an aside, note fo Boh cicula obits, whee A = 0 and L = n h, H = µz e 4. (9.30) n h This looks ight, except fo the question about why n = 0 is excluded.

5 7 Vesion of Octobe, 0 The necessay quantum mechanical modification is indicated by the (p L) calculation. The latte should be [ ] (p L L p) = ( L p + i hp) (p L i hp) because = (L p) (p L) + i hl p p + i hp p L + h p = p (L + h ), (9.3) (L p) (p L) = L p (p L) = L [p(p L) p L] = p L, (9.3) since p is a scala, and hence commutes with L. But the stuctue of H must emege in Eq. (9.8), so the coect expession must be A = + µz e 4 H(L + h ). (9.33) Now we tun to the commutation elation of L and A. Of couse, and A is a vecto, so, fo example, L L = i hl. (9.34) [A x, L y ] = i ha z. (9.35) How about A A? It must be a constant vecto, and theefoe a multiple of L and A. A is not possible, since it has an odd numbe of s and ps, when we ecognize that =. On the othe hand, L has an even numbe. The commutato i h [ k, p l ] emoves an even numbe, so A A must have an even numbe of s and ps. Theefoe we conclude A A = i hcl, o [A x, A y ] = i hcl z, (9.36) whee C is a scala constant. Find C fom [A, A ] = µz e 4H[A, L ], (9.37) since [A, H] = 0 because da/dt = 0. Then, fo example, [A z, A ] = [A z, A x + A y ] = i hc(a xl y + L y A x A y L x L x A y ) on the one hand, and = i hc(a L L A) z, (9.38) [A z, L ] = [A z, L x + L y] = i h(l x A y + A y L x L y A x A x L y ) = i h(a L L A) z, (9.39)

6 7 Vesion of Octobe, 0 CHAPTER 9. HYDROGEN ATOM I so compaing these two we find We have theefoe, [L x, L y ] = i hl z, [A x.l y ] = i ha z, [A x, A y ] = i h C = µz H. (9.40) e4 ( ) H µz e 4 L z, (9.4) etc. Now define J (±) = ( L ± ) µz e 4 H A, (9.4) and we find (denote tempoaily s = µz e 4 /( H)) [J (+) x, J (+) y ] = 4 [L x + sa x, L y + sa y ] = i h 4 (L z + L z + sa z + sa z ) = i hj (+) z, (9.43a) [J x ( ), J y ( ) ] = 4 [L x sa x, L y sa y ] = i h 4 (L z + L z sa z sa z ) = i hj ( ) z, (9.43b) [J x (+), J y ( ) ] = 4 [L x + sa x, L y sa y ] = i h 4 (L z L z + sa z sa z ) = 0, (9.43c) etc. We see that J (±) ae two independent angula momenta. These momenta have equal magnitude: (J (±) ) = 4 ( L ± s(a L + L A) + s A ), (9.44) because the coss tem vanishes accoding to Eq. (9.6). Now substituting in Eq. (9.33), and ecalling the definition of s, the L cancels, and we have (J (±) ) = ( µz e 4 ) 4 H h, (9.45) and denoting the common value of (J (±) ) by j(j+) h, j = 0, /,, 3/,,..., we find ( µz h [4j(j + ) + ] = h (j + ) e 4 ) = = µz e 4 H E, (9.46) whee E is the enegy eigenvalue. Denote j + = n, whee n =,, 3,.... We obtain the Boh enegy levels, E n = µz e 4, (9.47) n h

7 73 Vesion of Octobe, 0 without encounteing n = 0. The Boh enegy levels follow fom an exact quantum-mechanical teatment. The fact that this agees with expeiment validates ou use of the Coulomb potential. It is not meely good ageement with the spectum of hydogen that is obtained, but though the nuclea mass M dependence in µ = mm/(m +m). This is in ageement with the small shift between the H and He + specta, the latte coected fo the facto of Z = 4. Rydbeg atoms, whee an oute electon is in a highly excited state, can also be descibed appoximately by this enegy fomula. The numbe of states having a given enegy, that is, the degeneacy of the enegy levels, is the numbe of states fo which n = j + has a definite value. Since thee ae j + values of m fo each of J (+) z and J ( ) z, this numbe is (j + ) (j + ) = n =, 4, 9,.... (9.48) The state of lowest enegy, with n =, is unique. Since L = J (+) + J ( ), and this state is one fo which J (+) = J ( ) = 0, this state is also chaacteized by the angula momentum quantum numbes l = 0, m = 0. In geneal, we can denote the states of the hydogen atom by the values of these thee quantum numbes, n, l, m. (9.49) The lowest enegy state obeys L, 0, 0 = 0, A, 0, 0 = 0. (9.50) The latte equation eads ( ) (p L i hp), 0, 0. (9.5) µze Because this is a zeo angula momentum state, this equation eads ( + i hp ) µze, 0, 0 = 0. (9.5) We constuct a diffeential equation fo the wavefunction by multiplying on the left by a position eigenstate, ( 0 = + i hp ) ( µze, 0, 0 = + i h ) h µze i ψ 00 ( ), (9.53) whee the gound-state wavefunction is ψ 00 ( ) = 00. (9.54) Define the Boh adius by a 0 = h µe. (9.55)

8 74 Vesion of Octobe, 0 CHAPTER 9. HYDROGEN ATOM I Then this diffeential equation eads ( + a ) 0 Z ψ 00 ( ) = 0. (9.56) The adial component of this vecto equation is ( + a ) 0 ψ 00 () (9.57) Z whee we ve simplified the notation by dopping the pimes, and ecognizing that the gound-state wavefunction depends only on the distance fom the nucleus, =, which follows fom the othe components of the gadient. This is as expected fo a state of zeo angula momentum. The solution to this equation is immediate: ψ 00 () = Ae Z/a0. (9.58) To detemine A, we apply the nomalization condition, (d) ψ 00 () =. (9.59) Hee the volume element is and so we have (choosing the phase to be zeo) 4πA 0 (d) = 4π d, (9.60) ( d e Z/a0 = 4πA a ) 3 0 Z Since the last integal is, the nomalization constant is 0 dxx e x. (9.6) ( ) Z 3 / A =, (9.6) and so the nomalized gound-state wavefunction of the hydogenic atom is πa 3 0 ( ) Z 3 / ψ 00 () = e Z/a0. (9.63) πa 3 0 The pobability amplitude is concentated in a egion of chaacteistic size a 0 /Z, and it, and the coesponding pobability, ae sketched in Fig. 9.. The next-to-lowest enegy levels ae n = o j = /. These fo states coespond to the combination of two spins of /. This is inteesting, because these spin-/ s cetainly have no eal existence. The state with m (+) = m ( ) = +, i.e., n =, m =, and theefoe l =, is chaacteized by (J x + ij y ) (+) = 0, (J x + ij y ) ( ) = 0, (9.64)

9 75 Vesion of Octobe, 0 Ψ 4Π Ψ Figue 9.: Gound-state wavefunction fo the hydogen atom (Z = ), and the squae of the wavefunction (multiplied by 4π ), plotted vesus in units of the Boh adius a 0 because it is both the highest m (+) state and the highest m ( ) state. Equivalently, it is the highest m state, and satisfies as well The coesponding wavefunction satisfies (L x + il y ) = 0, (9.65) (A x + ia y ). (9.66) (L x + il y ) = (L x + il y )ψ ( ) = L + ψ ( ) = 0. (9.67) In the latte equation, L + is a diffeential opeato. Late, we will discuss the geneal appoach to solving this poblem in spheical pola coodinates, using spheical hamonics. Fo the pesent, howeve, it will suffice to note that Now we note L + = (yp z zp y + izp x ixp z ) h i [(y ix) z z( y i x )] = h[ (x + iy) z + z( x + i y )]. (9.68) and that if we wite the wavefunction as ( x + i y )(x + iy) = = 0, (9.69) ψ = (x + iy)f(), = x + y + z, (9.70)

10 76 Vesion of Octobe, 0 CHAPTER 9. HYDROGEN ATOM I we find L + ψ = h(x + iy)[ (x + iy) z + z( x + i y )]f() [ = h(x + iy)f () (x + iy) ( )] z + z x + i y [ = h(x + iy)f () (x + iy) z ( x )] + z + iy = 0. (9.7) Thus the angula momentum condition (9.67) is satisfied. To detemine the adial function f, we use the second equation (9.66). Recall that A = µze(p L i hp), (9.7) and then (p L) x + i(p L) y = p y L z p z L y + i(p z L x p x L z ) = ip z L + i(p x + ip y )L z. (9.73) Theefoe, since L + = 0, (L z h) = 0, (9.74) we have [ x + iy (A x + ia y ) = On the wavefunction, this means ( x + iy + a 0 Z ] µze ( i h)(p x + ip y ) = 0. (9.75) ( x + i )) ψ () = 0, (9.76) y again dopping the pimes on eigenvalues. Now using the constuction (9.70) and the identity (9.69), we see that ( x + i y )f() = x + iy so the diffeential equation (9.76) eads which has solution a 0 Z So the wavefunction is d f(). (9.77) d d f() + f() = 0, (9.78) d f() = Ce Z/(a0). (9.79) ψ () = C(x + iy)e Z/a0 = C sin θe iφ e Z/a0, (9.80) witing this in spheical pola coodinates, and ecalling that cosφ + i sinφ = e iφ. (9.8)

11 77 Vesion of Octobe, 0 The constant C is detemined, up to a phase, by the nomalization condition, that the pobability of finding the electon somewhee is unity, (d) ψ () =, (9.8) o π π = d dθ sin θ dφ C sin θe Z/a = C d 4 e Z/a0 π d cosθ( cos θ) 0 = 8π 3 C ( a 0 Z ) 5 which gives the magnitude of the nomalization constant: 0 ( Z C = ( dxx 4 e x = 64π C a ) 5 0, (9.83) Z a 0 ) 5/ 8 π, (9.84) so, witing the wavefunction in a fom compaable to that fo the gound state, ψ () = ( ) 3/ Z Z (x + iy)e Z/a0 π a 0 a 0 = ( ) 5/ Z 8 sinθe iφ e Z/a0. (9.85) π a 0 The pobability of finding the electon in a given elements of volume is ψ (, θ) = ( ) 5 Z sin θ e Z/a0. (9.86) 64π a 0 In the homewok, you will wok out the thee othe wavefunctions fo n =, those coesponding to l =, m =, 0, and the state with l = m = 0. It is clea that to poceed futhe systematically, we need to lean moe about combining angula momenta, and descibing the angula pat of the wavefunctions systematically. So we will now etun to that subject.