1/2. e ikx + 1 1/2. e i2kx. 1/2. e i2kx We wish to evaluate Eq. (12.9) using the w(k) function of Eq. (12.10): = e (k k 0 )2 /2(δk) 2 e ikx dk

Transcription

1 CHAPTER 1 Quantum Mechanical Model Systems SECTION Half the particles here have twice the momentum of the other half and therefore also have twice the wavevector of the others. The analog of the wavefunction in Eq. (1.8) is thus ψ 1 1/ e ikx + 1 1/ e ikx. The probability distribution is ψ 1 1/ e ikx + 1 1/ e ikx 1 1/ e ikx + 1 1/ e ikx = e ikx + 1 eikx + 1 = eikx + e ikx = 1 + cos kx. This is an oscillatory function that oscillates between 0 and with a spacing between maxima of π/k. The particles are localized to regions around these maxima (much as a standing waves of water form a regular pattern of maxima and minima), but these regions of maximum probability extend from to. 1. We wish to evaluate Eq. (1.9) using the w(k) function of Eq. (1.10): ψ(x) w(k)e ikx dk = e (k k 0 ) /(δk) e ikx dk. This integral is known. It has the form of the general definite integral 58

2 e p z ± qz dz = π p eq /4p, p > 0, and we can identify p = 1/(δk), q = ix, q = x, and z = k k 0. This gives ψ(x) e (k k 0 ) /(δk) e ikx dk = e ik 0 x e (k k 0 ) /(δk) i(k k 0 )x dk = δk π e ik 0 x e x (δk) /, which, except for the factor δk π, is the function shown in the text on page 39 and plotted (as the real part) in Figure 1.(b). (Whether one has a factor e ik 0 x, as we have here, or a factor e ik 0 x, as in the text, is unimportant. Both have the same real part, cos(k 0 x), and the imaginary part of one is the negative of that of the other.) The definite integral above is a special case of the more general definite integral we have seen before (see Problem 11.): e αy dy = π α. We complete the square in the argument of the exponential: e p z ± qz dz = e p z ± qz q /4p + q /4p dz = e q /4p e p z ± qz q /4p dz = e q /4p e pz + q/p dz then define y = pz + q/p so that dy = p dz. Making these substitutions (with α = 1) completes the derivation., 59

3 1.3 (a) If the source of particles is at x =, then the D coefficient must be zero, since the term in ψ < containing D, e ik < x, represents particles in the region x > 0 moving toward x =. Since there is no source at x = directing particles toward x = and since particles moving toward x = are not reflected back at x =, this coefficient must be zero. (b) The requirement ψ < (0) = ψ > (0) means 1 + B = C, and the requirement that the first derivatives are equal at x = 0 means ik < ik < B = ik > C. Solving these two simultaneous equations for B and C gives (c) The product B*B is B = k < k > k and C = <. k < + k > k < + k > B*B = k < k > k < + k > since the wavevectors k < and k > are both real numbers. As the total energy increases, the potential step becomes an ever-decreasing perturbation; the particle s kinetic energy in the positive x region is almost the same as it was in the negative x region. This means k > is approaching k < as the energy increases, making B*B go to zero. Note as well that C approaches 1 in this high-energy limit, as it should. 1.4 (a) Here C must be zero in order to keep the wavefunction finite as x. (b) Equality of the wavefunctions at x = 0 leads to 1 + B = D, and equality of their first derivatives leads to ik < ik < B = κ > D. Simultaneous solution of these equations gives (c) Here, B*B is B*B = ik < + κ > ik < κ > B = ik < + κ > ik and D = <. ik < κ > ik < κ > * ik< + κ > ik < κ > = ik < + κ > ik < κ > ik < + κ > ik < κ > = k < + κ> k < + κ> = 1. 60

4 (d) The probability of barrier penetration depends on total energy E through the quantity D*D: 4k D*D = <. k < + κ> Since the wavevectors k < and κ > depend on E through k < = me and κ > = m(v E), we can write 4k D*D = < = 4E E + (V E) = 4E V. k < + κ> Since E ranges from 0 to V (because E > V returns us to the energy regime of the previous problem), the probability is maximal when E = V. 1.5 Here, the wavefunctions can be written ψ < = C e ik < x and ψ > = e ik > x + Be ik > x. Note that ψ < has no term corresponding to particles moving towards x = (i.e., no e ik < x term), since particles originate at x = and, while they can be reflected from the potential step at x = 0 and move back to x = in the positive x region, they can only move toward x = in the negative x region. We find expressions for B and C from the requirements that the wavefunctions and their first derivatives are equal at x = 0: 1 + B = C and k > + k > B = k < C. Simultaneous solution of these two equations gives B = k > k < k and C = >. k > + k < k > + k < 61

5 The probability for reflection back to the source at the potential step is B*B, since B is the coefficient of the term representing particles in the positive x region moving towards x =. It is simply B*B = k > k < k > + k <, and since k > approaches k < as E increases, B*B approaches 0 as E. SECTION The zero-point energy of mass m e, the electron mass, in a 1-D box is E 1 = π m e L. Equating this to the electron s mass-equivalent energy, m e c, and solving for L gives L = h m e c = m = m. As an aside, we can repeat this line of reasoning for the zero-point energy of a harmonic oscillator and get another answer of the same magnitude. We write, with X representing the classical turning point of the motion, E 0 = m e X = m e c or X = m e c = m. Thus, the full classical motion of the electron, from X to X, spans 1/π times the Compton wavelength: /m e c = (h/m e c)/π = m. 1.7 Usually, one thinks of classical behavior as the large quantum number limit of a quantum-mechanical system. Here, we will find classical behavior (in the sense defined in this problem the position uncertainty is greater than the distance between successive maxima in the probability distribution) at a surprisingly small quantum number. First, we find x for a particle in a box. The result of Problem 11.3 gives us an easy route, if we recall that <x> = L/: 6

6 ( x) = <x > <x> = L n π L 4 or x L = /. n π (As an aside, note that x/l approaches the constant limit 1/ 1 = 0.89 as n.) We wish to compare x/l to 1/n (the spacing between maxima divided by L) and look for that n = n* for which x/l > 1/n. We make the following table: n x/l /n We find that x > L/n for n 4, which is perhaps a surprisingly small number. 1.8 We write P(x) dx = dt dx = τ 1/ τ 1/ (dx/dt), and with x(t) = X cos ωt, dx/dt = Xω sin ωt. Since τ 1/ = π/ω, we have P(x) dx = dt τ 1/ = dx τ 1/ (dx/dt) = dx ω Xωπ sin ω t = dx πx sin ω t. (We can neglect the minus sign in dx/dt, since probabilities must be positive.) Now we write cos ωt = x/x, and since sin ωt + cos ωt = 1, we have sin ωt = (X x ) 1/ /X so that P(x) dx = dx πx sin ω t = dx π X x 1/ as we wished to show. The normalization integral follows from X P(x) dx = 1 X π X X dx X x 1/. Let x = yx so that dx = X dy. Then the integral above becomes 63

7 1 π X X dx X x 1/ = 1 π 1 1 X dy X y X 1/ = 1 π 1 1 dy 1 y 1/ = π π = At the classical turning point of a harmonic oscillator, x = X v where X v is the classical turning point for state v. Moreover, the energy of state v is E v = kx v / for force constant k. Thus, the Schrödinger equation becomes µ d ψ v dx + kx E v ψ v = d ψ v µ dx + kx kx v ψ v = 0. If x = X v, the terms in parentheses cancel each other, and we are left with d ψ v /dx = 0. The wavefunction has an inflection point at the classical turning point, as Figure 1.8 (and Figure 11. for the ground state) indicate. This must be a general result, since the definition of a classical turning point is that point at which the total energy equals the potential energy and the terms in parentheses in the Schrödinger equation as written above are always (potential energy total energy) The coordinate q used to express the harmonic oscillator wavefunctions in Eq. (1.19) is shown on page 406 in the text to equal v + 1x/X v so that q 1 = 3x/X 1 and q = 5x/X. The Hermite polynomials for these two states are in Table 1.1: H 1 (q) = q and H (q) = 4q. The normalization constants given by Eq. (1.0) are for the wavefunction written in terms of the dimensionless variable q. We want the normalization constants for the wavefunction written in terms of the real distance variable x, and thus we use the expressions listed in the Summary to Chapter 1 on page 440. (See also the comments at the end of Example 1.4 on page 406.) We have N 1 = k π ω 1 1/ and N = k π ω 1 8 1/. Since q can also be written as k/ ω x, we have k/ ω = v + 1/X v. This makes the normalization constants equal to N 1 = 1 3 1/ π X 1 and N = 1 5 1/. 8 π X 64

8 Thus, the entire wavefunctions are ψ 1 = ψ = 1 3 1/ 3x e π X 1 X 3x /X / 5x 4 8 π X e 5x /X. X 1.11 As we used in the previous problem, there is a simple relationship between the classical turning point of any harmonic oscillator state and its parameters m and k. For the ground state, the relationship is X 0 = ω k. Since ω = k/m, this can also be written as X 0 = ω k = k k m = km. Using the values for k and m quoted in the problem, we find X 0 = 0.13 Å for H and X 0 = Å for I. Thus, X 0 /R e = for H and X 0 /R e = for I. Hydrogen vibrates over a total distance (classically) that is about 33% of its bond length while I vibrates over a much smaller 4% or so. 1.1 There is a clear symmetry we can exploit here. We want the ground-state wavefunction to have its maximum at x = 0. If the potential was a particle-in-a-box extending over L/ x L/, we know the ground-state wavefunction would be symmetric about and maximal at x = 0. Likewise, a harmonic potential centered at x = 0 has a ground-state wavefunction symmetric about and maximal at x = 0. Thus, since what we have here is half of each type of potential, we want a continuous wavefunction that is the particle-in-a-box wavefunction for L/ x 0 and a harmonic oscillator wavefunction for x 0. Our usual coordinate system for a particle-in-a-box extends from 0 to L; since the potential we are imagining here has been shifted towards smaller x by L/, the ground-state wavefunction has the sine function of Eq. (1.11) replaced by cosine: ψ(x 0) = L cos πx L 65

9 The harmonic oscillator ground-state wavefunction is ψ(x 0) = 1 π 1/4 e x /X 0 = k 1/4 e k x / ω. X 0 π ω For these to join smoothly at x = 0, they must equal each other and have equal first derivatives. Equal derivatives is assured: both wavefunctions have maxima (and thus zero first derivatives) at x = 0. Continuity of the wavefunctions (along with ω = k/m) gives us the relationship among m, L, and k that we seek: 1/ = k 1/4 = mk 1/4 L π ω π or L = π mk Here, we can exploit the eigenvalue expressions for the creation and annihilation operators given on page 405 in the text: q d dq ψ v = [(v + 1)]1/ ψ v + 1, v = 0, 1,, q + d dq ψ v = (v)1/ ψ v 1, v = 1,, 3, We write the matrix element q nm and expand it as follows: q nm = ψ n qψ m dq = 1 ψ n q d dq + q + d dq ψ m dq = 1 ψ n q d dq ψ m dq + ψ n q + d dq ψ m dq = 1 (m + 1) ψ n ψ m + 1 dq + m ψ n ψ m 1 dq. Since the wavefunctions are orthonormal, the first integral in the last line above vanishes unless n = m + 1 (in which case it equals 1), and the second integral vanishes unless n = m 1 (in which case it, too, equals 1). Thus, if n = m + 1, q nm = q n,n 1 = [ (m + 1)]/ = n/, and if n = m 1, q nm = q n,n + 1 = m/ = [ (n + 1)]/. See also Problem for another representation of these integrals. 66

10 1.14 The matrix representation of the square if the q operator is the square of the matrix representation of the q operator alone: q nm = q ni q im. i = 1 The previous problem gives the pattern of nonzero values for q nm, and armed with this pattern and the equation above, we can deduce that the pattern of nonzero values for q nm must be q = 0 q 01 0 q 10 0 q 1 0 q 1 0 = q 01 q 10 0 q 1 q q 1 q 1 + q 01 q 10 0 q 3 q 1 q 1 q 10 0 q 3 q 3 + q 1 q q 3 q 1 0 q 34 q 43 + q 3 q 3. In particular, we see that q 00 is nonzero and equals q 00 = q01 q 10 = = 1. Problem found <x > = X / where X is the classical turning point. Our result here, q 00 = <q > 00 = 1/, is the dimensionless variant of <x > for the ground state for which q = x/x. Thus, <q > 00 = <x > 00 /X = 1/ Differentiating the generating function gives dh v dq = d dq ( 1)v e q d v e q dq v = ( 1) v qe q d v + 1 e q dq v + 1 = qh v H v + 1. Differentiating this gives 67

11 d H v dq = d(qh v H v + 1 ) = H dq v + q dh v dq dh v + 1 dq = H v + q(qh v H v + 1 ) qh v H v + = ( + 4q )H v 4qH v H v +. If we substitute these expressions for the first and second derivatives of H v into Hermite s differential equation, we find d H v dq q dh v dq + vh v = ( + v)h v + H v + qh v + 1 = 0. To arrive at the recurrence relation shown in the text, define w = v + 1, so that v = w 1. This gives [ + (w 1)]H w 1 + H w + 1 qh w = wh w 1 + H w + 1 qh w = 0. Changing the symbol that indexes the Hermite polynomial in this equation from w to v lets us write the recurrence relation as it is shown in the text, H v + 1 = qh v vh v 1. Finally, if we substitute this expression into our original expression for the first derivative of H v, we find dh v dq = qh v H v + 1 = qh v qh v + vh v 1 = vh v If we expand the Lennard-Jones potential function in a Taylor s series with the aim of finding its effective force constant, k we see that we need the second derivative of V, since k = (d V/dx ) x = x e. We can write V as so that the first derivative is V(x) = D 1 x e 6 x 6 + x e 1 x 1 dv(x) dx = D 1x e 6 x 7 1x e 1 x 13 and the second derivative is 68

12 d V(x) 6 dx = D 1 7x e x x e x 14. Evaluating this at x = x e gives the answer we seek: k = d V(x) dx x = x e = D 1 7x e 6 xe xe 1 xe 14 = 7D x e. SECTION A particle moving freely in 3-D space has the same Hamiltonian whether the space is confined to a finite region (the 3-D particle-in-a-box potential) or extends to infinity. The Hamiltonian is given by Eq. (1.1): H = m x + y + z = m. Since the three directions in space x, y, and z are equivalent and independent of each other, the most general 3-D free particle wavefunction has the form ψ(x,y, z) = Ae i(k x x + k y y + k z z) + Be i(k x x + k y y + k z z) = Ae ik r + Be ik r where k x, k y, and k z are the wavevectors for motion along the x, y, and z directions, respectively, and where k is the vector with components k x, k y, and k z, r is the vector with components x, y, and z, and k r represents the vector dot product of k and r. The three quantum numbers needed to describe the state of the system (including its energy) are the wavevector components since k x = p x = mv x, k y = p y = mv y, k z = p z = mv z, and E = k x + k y + k z m. Each wavevector can vary continuously from to (although ± implies an infinite velocity, which relativity does not allow) The -D particle-in-a-box quantum numbers are n x and n y, and each can range from 1 to in integral steps. Since the box is a square (all sides of length L), the total energy expression is 69

13 E = E x + E y = π ml n x + n y. To find all the states with energy E 9 π /ml, we can define the dimensionless energy ε = E/( π /ml ) = (n x + ny )/, which we can tabulate: n x \n y / 5 17/ 5/ 4 13/ / 9 5/ 4 17/ 10 5/ 3 Entries in italics have ε > 9 and thus correspond to states we are not interested in here. This shows us that the state with ε = 1 (or E = π /ml ) is nondegenerate, as are those with ε = 4 (E = 4 π /ml ) and those with ε = 9 (E = 9 π /ml ). The states with ε = 5/ (E = 5 π /ml ), with ε = 13/ (E = 13 π /ml ), and with ε = 17/ (E = 17 π /ml ) are doubly degenerate. To locate the most probable position of the particle in any one state, we look at the square of the wavefunction. If we define the x and y coordinates so that the edges of the box extend over the ranges 0 x, y L, the square of the wavefunction is ψ 1/ (x,y) = ψ nx(x)ψny(y) = sin n x πx L L = L sin n x πx L sin n y πy L. 1/ sin n y πy L L Wherever this function is a maximum for any (n x, n y ) choice locates the most probable position or positions. An easy way to see this probability is as density plot (an analog of a contour plot) that plots ψ in two dimensions as a gradation in shade that is proportional to the value of ψ. Such plots are shown at the top of the next page for the states (n x, n y ) = (1, 1), (3, 1), (1, 3), and (, ), drawn so that the darkest shade locates the maximum or maxima in ψ. 70

14 L (n x, n y ) = (1, 1) (n x, n y ) = (1, 3) L y y 0 0 x L 0 0 x L L (n x, n y ) = (3, 1) L (n x, n y ) = (, ) y y 0 0 x L 0 0 x L The probability peaks in the center of the box (at x = y = L/) for the (1, 1) state. The states (1, 3) and (3, 1) are related by simple symmetry: rotate the picture of one by 90 to get the picture of the other. State (1, 3) peaks at x = L/ and y = L/6, L/, and 5L/6, while state (3, 1) peaks at Y = L/ and x = L/6, L/, and 5L/6. State (, ) peaks at the four positions (x, y) = (L/4, L/4), (3L/4, L/4), (L/4, 3L/4), and (3L/4, 3L/4) For an isotropic 3-D harmonic oscillator, ω = k/m is the only parameter needed to characterize the energy of the system. There are three independent quantum numbers that specify a particular state and energy of such a system: v x, v y, and v z, and each can range from 0 to. The energy expression is E = E x + E y + E z = ω v x v y v z + 1 = ω v x + v y + v z + 3, 71

15 and we can use it to tabulate ε = E/ ω for various combinations of quantum numbers: v x v y v z ε v x v y v z ε v x v y v z ε v x v y v z ε / / / 0 1 9/ / / 0 1 9/ 1 0 9/ / 0 0 7/ 1 0 9/ / / 0 0 7/ 0 1 9/ / / 0 0 7/ 1 0 9/ / We see that the lowest energy state is nondegenerate, at E = 3 ω/, the next highest state is three-fold degenerate at E = 5 ω/, the next is six-fold degenerate at E = 7 ω/, and the fourth state is ten-fold degenerate at E = 9 ω/. 1.0 We first convert the C C bond energy, 00 kj mol 1, to joule per molecule units: 00 kj mol molecules mol 1 = J molecule 1. Now we seek a quantum number n that, when substituted in the energy expression in Example 1.5 with m = mass of one He atom = kg and r m = radial distance the He can move inside the C 60 cage = 0.45 Å = m, equals or exceed this energy. Thus, we solve the following expression for n: E n = J = π mr m n = ( J)n, and find n = 9.0, so that, since n must be an integer, any state with n 9 or so gives the He atom sufficient energy to escape the C 60 cage. (Of course, the 00 kj mol 1 bond energy is not a very accurately known number here, and since our answer n = 9.0 is so close to 9, it may well be that the state with n = 9 is not really sufficiently energetic. On the other hand, the model itself is quite crude, and we should expect the motion of a He atom inside C 60 to be more complex than what we have indicated here. As mentioned in Example 1.5, there are other states for this system in which the He atom has angular momentum about the center of the cage. These should also be considered.) 7

16 SECTION We start with the Cartesian representation of L z given by Eq. (1.31): L z = i x y y x and follow the steps outlined in the problem to transform to plane polar coordinates. Using the relations r = (x + y ) 1/ and θ = tan 1 (y/x), we can derive r x y = r y x = θ x y = θ y x = x x + y 1/ = x r y x + y 1/ = y r y x 1 + (y/x) = y x + y = y r 1 x 1 + (y/x) = x x + y = x r. We use these expressions to build the operator through the chain rule: so that x y = x r y x r + θ y x θ = x y r y x = y r x y r + θ x y 73 r + x r θ θ = y x r r y r θ x y y x = x y r r + x r θ y x r r y r θ = xy r r + x r θ xy r r + y r θ = x + y r θ = θ. This proves that L z in plane polar coordinates equals ( /i)(/θ).

17 1. We note first that the four first derivatives among the two coordinate systems variables that we derived in the previous problem can also be written r x y = x r = cos θ r y x = y r = sin θ θ = y x y r = sin θ r θ = x y x r = cos θ. r Using these expressions in the chain rule lets us write F = cos θ F sin θ x y r θ r F = sin θ F + cos θ y x r θ r F θ r F θ r. We substitute from these six expressions as called for into the general expressions for the second derivatives: F x = r x y r = cos θ F y = r y x r = sin θ F + θ x y x y θ r cos θ r sin θ r F + θ y x y x θ r sin θ r + cos θ r F x y θ F y x θ sin θ r + cos θ r θ cos θ r sin θ r θ sin θ r + cos θ r θ θ and expand each expression: 74

18 F x = cos θ F r + sin θ F cos θ sin θ r r r + sin θ r F cos θ sin θ + θ r F θ F y = sin θ F r + cos θ F cos θ sin θ + r r r + cos θ r F cos θ sin θ θ r θ. F rθ Adding these and recalling that sin θ + cos θ = 1 gives us = F x + F y = F r + 1 F r r + 1 F r θ = 1 r r 1 F r r + 1 F r θ. F rθ 1.3 We start with the expressions for the components of the angular momentum operator written in spherical polar coordinates as given in Eq. (1.3): L x = i L y = i L z = i sin φ cot θ cos φ θ φ cos φ cot θ sin φ θ φ φ, then we square each operator: L x L x = sin φ θ cot θ cos φ φ sin φ cot θ cos φ θ φ 75

19 L y L y = cos φ cot θ sin φ θ φ L z L z = φ φ, cos φ cot θ sin φ θ φ and expand the products: L x L x = cos φ sin φ cot θ + csc θ φ + cos φ cot θ φ + cos φ sin φ cot θ θφ + sin φ θ + cos φ cot θ θ L y L y = cos φ sin φ cot θ + csc θ φ + sin φ cot θ φ cos φ sin φ cot θ θφ + cos φ θ + sin φ cot θ θ L z L z = φ. We add these to form the operator we seek, L = L x L x + L y L y + L z L z. Fortunately, several terms cancel, and the identity sin φ + cos φ = 1 simplifies several others: L = = = cot θ + 1 φ + θ + cot θ θ θ + cot θ θ + 1 sin θ 1 sin θ 76 φ θ sin θ θ + 1 sin θ φ. To express L in Cartesian coordinates, we use Eq. (1.31) for each component operator:

20 L x = i L y = i L z = i y z z y z x x z x y y x, then we square and expand each component. Since all three operators are so similar, once we square and expand one in detail, we can immediately write the other two by analogy: L x L x = y z z y y z z y = y z y z y z z y z y y z + z y z y = y z y yz y zy z zy + z z yz y = y z yz + z zy y y y z z. Thus, the squares of the other two components are and L y L y = z x zx + x xz z z z x x L z L z = x y xy + y yx x x x y y. We add these three expressions to form the operator for the square of the total angular momentum, L = L x L x + L y L y + L z L z, shown at the top of the next page. Note that this version of the operator is considerably more complicated than the version written in spherical polar coordinates earlier in this problem. Consequently, the Cartesian representation is rarely used. 77

21 L = y z yz + z zy y y y z z + z x zx + x xz z z z x x + x y xy + y yx x x x y y = x y + + y z x + + z z x + y yz + zx + xy zy xz yx x x + y y + z z. 1.4 The expression in part (a), [L x,l y ] = i L z, is easiest to prove using Cartesian coordinates: [L x,l y ] = L x L y L y L x = i y z z y i z x x z i z x x z = y + yz yx x zx z z + zx yx zy + zy xz z xy xy z + x + xz y zy = x y y x = i x i y y x = i L z. i y z z y The expression in part (b), [L,L z ] = 0, is easiest to prove in spherical polar coordinates, and the calculus is perhaps a bit easier and more transparent if we write the operator L in the following expanded way, which we first encountered in the previous problem: L = 1 sin θ = θ sin θ θ + 1 sin θ θ + cot θ θ + 1 sin θ φ φ. 78

22 This lets us write L L z as and L z L as L L z = = 3 L z L = i i φ = 3 i θ + cot θ θ + 1 sin θ φ 3 θ + cot θ φ θφ + 1 sin θ i 3 φ 3 φ θ + cot θ θ + 1 sin θ φ 3 θ + cot θ φ θφ + 1 sin θ 3 φ 3. Subtracting these to form the commutator clearly gives zero. For part (c), we substitute the definition L + = L x + il y into the expanded commutator of interest: [L z,l + ] = L z L x + il y L x + il y L z = L z L x L x L z + i L z L y L y L z = [L z,l x ] + i[l z,l y ] = i L y + i( i L x ) = L x + il y = L +. For part (d), we again substitute the definition L + = L x + il y in the commutator of interest here and write [L +,L ] = [ L x + il y, L ] = [L x,l ] + i[l y,l ]. (Note how commutator algebra is distributive. This lets us jump directly and confidently from the second to the third step above.) Since L commutes with every angular momentum component operator, both of the final two commuta- 79

23 tors equal zero, as, therefore, does the operator [L +,L ]. Thus, these two operators commute. To see if L + and L commute, we write [L +,L ] = [ L x + il y, L x il y ] = L x + il y L x il y L x il y L x + il y = L x ilx L y + il y L x + L y Lx ilx L y + il y L x L y = i L y L x L x L y = i[l y,l x ] = i i L z = L z. Since the final result is not identically zero, these operators do not commute. 1.5 Let ψ m be an eigenfunction such that L z ψ m = m ψ m. Then we argue as follows: L z L ψ m = L z L ψ m = L z L L L z + L L z ψ m = [L z,l ] + L L z ψ m = L + L L z ψ m = L L z ψ m L ψ m = L (m )ψ m L ψ m = (m 1) L ψ m. Thus, the function L ψ m has an eigenvalue for the L z operator that is one unit lower than ψ m itself. 1.6 If l = 1 and m = 1, the eigenfunction ψ lm = ψ 1,1 must satisfy the following equation: L L z ψ1,1 = L ψ 1,1 L z ψ1,1 = ψ 1,1 ψ 1,1 = ψ 1,1 so that L x +Ly =. This, along with Lz = and L =, lets us construct the diagram at the top of the next page for the classical vector L. The L 80

24 vector lies somewhere on the surface of the shaded cone in this figure; the Uncertainty Principle does not allow us to say exactly where. 45 z L = y x For the general case of l = m, we note that the length of L is l (l + 1) and the length of its z component is m = l. Thus, we can make the following diagram of the general classical vector geometry: z l (l +1) θ l We see that cos θ = l/ l (l + 1), and thus as l, cos θ 1, or θ To write L + in spherical polar coordinates, we substitute the spherical polar forms of L x and L y into the definition of L +. Since we will need L later on in this problem, we can derive both at once. We also take advantage of the identity e iφ = cos φ + i sin φ and find 81

25 L ± = L x ± il y = i sin φ cot θ cos φ θ φ ± i cos φ cot θ sin φ θ φ = (±i cos φ sin φ) cot θ(cos φ ± i sin φ) i θ φ = ie i ±iφ cot θ e±iφ θ φ = e ±iφ θ + i cot θ φ. Next, we take l = m, Θ l,l (θ) = sin l θ, and Φ l (φ) = e imφ = e ilφ and operate on ΘΦ with L +. Since L + raises the z component one unit, but for m = l, this component is already as large as it can be, the operator should yield zero: L + Θ l,l (θ)φ l (φ) = L + sin l θ e ilφ = e iφ θ + i cot θ φ sinl θ e ilφ = e iφ le ilφ cos θ sin l 1 θ l e ilφ cot θ sin l θ = e iφ le ilφ cos θ sin l 1 θ l eilφ cos θ sin θ sinl θ = e iφ le ilφ cos θ sin l 1 θ l e ilφ cos θ sin l 1 θ = 0. Finally, we set l = and operate on (sin θ)φ (φ) = (sin θ)e iφ with L : L Θ, (θ)φ (φ) = L sin θ e iφ = e iφ i cot θ φ θ sin θ e iφ = e iφ e iφ cot θ sin θ e iφ cos θ sin θ = e iφ e iφ cos θ sin θ sin θ e iφ cos θ sin θ = e iφ e iφ cos θ sin θ e iφ cos θ sin θ = 4 e iφ cos θ sin θ. 8

26 Since we have not included normalization constants in our wavefunctions, this result is only proportional to ψ,1, but we can see that it has the correct functional dependence on θ and φ for this wavefunction: for m = 1, we should find a factor e iφ, and we do; for l = and m = 1, we should find the factors (cos θ sin θ) that represent the θ dependence of the Y,1 (θ,φ) spherical harmonic function (see Table 1.), and we do. 1.8 With the help of Table 1., which lists the spherical harmonic functions we need and shows us that the φ part of each term in the Theorem s sum vanishes, since φ appears in the spherical harmonic functions as e imφ and this term times its complex conjugate gives simply 1, we write Unsöld s Theorem for the particular case of l = 1 as 1 m = 1 Y 1m (θ,φ) = Y 1, 1 + Y 1,0 + Y 1,1 = 1 3 π 4 sin θ + 3 cos θ sin θ = 3 4π sin θ + cos θ = 3 4π = constant. For l =, we follow the same argument, using two trigonometric identities to simplify our result, sin θ = 1 cos θ and sin 4 θ cos 4 θ = sin θ cos θ: m = Y m (θ,φ) = Y, + Y, 1 + Y,0 + Y,1 + Y, = 1 π sin4 θ sin θ cos θ cos θ 1 5 = 16π 9cos4 θ 6cos θ + 3sin 4 θ + 1cos θ sin θ = 16π 9cos4 θ 6cos θ + 3sin 4 θ + 1cos θ 1 cos θ = 16π 6cos θ + 3 sin 4 θ cos 4 θ = 16π 6cos θ + 3 sin θ cos θ = 16π 3sin θ + 3cos θ + 1 = 5 4π = constant. 83

27 One can show that, in general, the constant is (l + 1)/4π. 1.9 We follow the logic used in Example 1.7 here, using the energy expression in Eq. (1.4) with R = a 0, the Bohr radius of Eq. (1.45) with µ = m e. We find l(l + 1) l(l + 1) E = = m e a 0 4πε = l(l + 1) e 4 m e m 0 (4πε 0 ) e m e e = l(l + 1) ( J) = l(l + 1) (13.6 ev). Note that the energy constant e 4 m e /(4πε 0 ) = e /(4πε 0 )a 0 = E 1, which is the negative of the usual H atom ground state energy (see Eq. (1.46b)), appears here Equating the attractive force between the proton and electron to the centrifugal force of an orbiting electron in the Bohr model gives us e 4πε 0 a = m e aω. The Bohr quantization condition, m e a ω = n, gives us an expression for the orbiting frequency ω: ω = n m e a, that, when substituted in the first equality, gives e 4πε 0 a = m e a n m e a, or a = n 4πε 0 m e e = n a 0. If E = T = m e a ω /, we can use the expressions above for ω to write an expression for a ω that, when substituted in the energy expression, gives something we can arrange in the form of Eq. (1.5): 84

28 E = m e a ω = n m e e 4 m e n = m e m e a = n m e a n 4 4(4πε 0 ) = e 4 m e (4πε 0 ) 1 n = e (4πε 0 )a 0 1 n, where, in the last step, we have collected those constants that define the Bohr radius, a 0 = (4πε 0 ) /m e e If we scale a m proton radius up to 1 mm (a magnification factor of ), and if we define the proton distance from the center of mass to be r p and the electron distance r e, then we can write two expressions relating r p and r e. The first is that they add to give our measure of a 1s H atom radius, which we are taking to be <r> so that r p + r e = <r>. The second is the definition of center of mass: m p r p = m e r e. The 1s state value for <r> is 3a 0 / = m, which, when scaled by our magnification factor, becomes about 66 m. Solving the two expressions for r p and r e gives r p = m e m <r> and r m p + m e = p <r>. e m p + m e The magnified numerical values are r p = 36 mm and r e = 66 m. (All current theories and experiments on electrons, by the way, indicate that the electron has no radius; it is a true point.) In terms familiar to American readers, if we place the center of mass on the goal line of an American football field, the proton is about 1.4 in. into the end zone and the electron is on the opponent s 8 yard line. 1.3 From the general expression E = hc/λ for the emission wavelength λ accompanying an energy change E, we can write λ = hc E = hc E n E m = hc E 1 n E 1 m = hcn m E 1 m n = hc E 1 n m n m. With E 1 = e /(4πε 0 )a 0, we identify the constant in the general expression, Eq. (11.3): nm = (4πε 0 )a 0 hc/e. Setting m = gives the constant in Eq. (11.), the Balmer formula. 85

29 1.33 The differences in size and ionization energy for the three isotopic forms of hydrogen are due to the different reduced masses µ of each isotope. The reduced mass enters the Bohr radius expression and the ground-state energy expression (the negative of which is the ionization potential): a 0 = 4πε 0 µe and IP = E 1 = e (4πε 0 )a 0. Using the full number of significant figures in the mass of the electron along with the accurate nuclear masses given in the problem and the conversion factor between atomic mass units and kg, we can calculate reduced masses: m µ H = e m H = m e + m 31 kg H m µ D = e m D = m e + m 31 kg D m µ T = e m T = m e + m 31 kg. T From these, the following radii and ionization potentials follow: a 0 (H) = 4πε 0 e = m µ H IP H = e (4πε 0 )a 0 (H) = J a 0 (D) = 4πε 0 e = m µ D IP D = e (4πε 0 )a 0 (D) = J 86

30 a 0 (T) = 4πε 0 e = m µ T IP T = e (4πε 0 )a 0 (T) = J. The size differences are inconsequential and unobservable, but the ionization energy differences are experimentally observable The general rules that guide us here are: the total number of nodes = n 1 and the total number of spherical nodes = n l 1. For wavefunction (a), spherically symmetric tells us l = 0 (an s state), and three spherical nodes tells us 3 = n 0 1 or n = 4. It also follows that m = 0. For wavefunction (b), the three planar nodes (and thus no spherical nodes) again tell us 3 = n 1 or n = 4. Since there are no spherical nodes, 0 = n l 1 = 4 l 1 or l = 3 (an f state). This is the 4f xyz wavefunction shown in Figure 1.16(e), and we cannot uniquely identify m. Wavefunction (c) has one spherical and one planar node so that = n 1 or n = 3. The one spherical node tells us 1 = n l 1 = 3 l 1 or l = 1 (a p state). Since the nodal plane is the xy plane, the wavefunction is cylindrically symmetric about the z axis. This is the 3p z wavefunction (m = 0). Wavefunction (d) has two spherical nodes and two nonspherical (conical) nodes: 4 = n 1 or n = 5 and = n l 1 = 5 l 1 or l = (a d state). Again, m = 0 because conical nodes are cylindrically symmetric about z. This is the 5d z wavefunction. For (e), the symmetry tells us m = 0, and the single nodal plane (which must be the xy plane) tells us l = 1 so that n =. This is the p z wavefunction of in Figure 1.14(d) From Table 1.4, we see that the angular portion of the p z wavefunction is simply cos θ. For the 3d z wavefunction, it is 3 cos θ 1. Polar plots of r = cos θ and r = 3 cos θ 1 are shown below. 87

31 p z 3d z These plots show the symmetry and rough spatial extent of the electron probability density plots shown in Figures 1.14(d) and 1.15(a) For the 1s state, <r> = 3a 0 / = m, while a 0 = m. The radius r* that encloses 90% of the electron probability in this state is given implicitly by the equation π 0.90 = dφ 0 π sin θ dθ 0 0 r* ψ 1s r dr = 4 a o 3 r* e r/a 0r dr. 0 If we let x = r/a 0, this integral becomes x* 0.90 = 4 e x x dx 0, 88

32 which cannot be evaluated in closed form. Numerical integration gives x* =.661, or r* =.661a 0 = m, a larger measure of the 1s atomic size than either a 0 or <r>. It is interesting to look at a graph of probability versus r for this state just to see how rapidly the probability is approaching 1: Probability r/a 0 At the <r> value, which is 1.5 on the scale of this figure, the probability is only 0.6 or so From Example 1.8, the radial distribution function, rdf, is defined as 4πr R nl where R nl is the radial part of the wavefunction for the state with quantum numbers n and l. These radial factors are listed in Table 1.3, and thus we can write rdf 1s = 4πr 4 3 e r/a 0 = 16π 3 r e r/a 0 a 0 a 0 rdf pz = 4πr 1 r 3 e r/a 0 = π 5 r 3 e r/a 0. 4a 0 a 0 6a 0 We find the maxima in these functions by differentiation: drdf 1s dr drdf pz dr = 16π a 0 3 = π 6a 0 5 re r/a 0 r a 0 e r/a 0 3r e r/a 0 r3 3 e r/a 0. a 0 89

33 When these derivatives equal zero, we are at the maximum. For 1s, this happens when (r r /a 0 ) = 0 or when r = a 0. For p z, we want 3r r 3 /a 0 = 0 or r = 3a 0. Finally, we find the s radial distribution function: rdf s = 4πr 1 3 r e r/a 0 = πr 8a a 0 3 r e r/a 0. 0 a a 0 0 This function is zero at r = 0 and (neither of which counts as a node) and at r = a 0, which is the node location. The graph in Example 1.8 on page 43 in the text shows this node as well as the maximum in the 1s rdf at r = a The general expression for the ionization potential of a one-electron atom of nuclear charge Z is IP(Z) = Z E 1 where E 1 is the H atom ground-state energy. For U 91+, Z = 9 and IP = 8464E 1 = ev. The general expression for <r> is <r> = a 0 Z 3n l(l + 1). If Z = 9, n = 1, and l = 0, we find <r> = 3a 0 /184 = a 0 = Å. If we consider only l = 0 states, for some n = n* the U 91+ size will exceed the H atom 1s size, given by <r> = 3a 0 /. Thus, we write 3a 0 = 3a 0 9 n* or n* = 9 so that (since n* must be integral) n* 10. We can make a plot of <r> 1s = 3a 0 /Z versus IP(Z ) = Z E 1 for Z = 1 through 9 if we solve each expression for Z and equate the results. We find <r> = (3a 0 /) ( E 1 /IP) 1/, which is graphed below in a log log plot. 90

34 10 <r>/a H U IP/E The Earth s speed v in its orbit around the sun is v = πr τ = π( m) 10π m s 1 s so that its angular momentum is L = mvr = ( kg) ( m s 1 ) ( m) kg m s The huge number represents an angular momentum quantum number l. (The correct quantum expression L = l(l + 1) reduces to L = l in the large l limit.) An H atom must have n at least this large in order to have such an angular momentum. Thus, if we take n = l = and substitute into the general expression for <r>, we find <r> = a 0 3n l(l + 1) = a 0 3n n n a 0 n or <r> a 0 = m. This huge number, larger than the radius of the known Universe, indicates that while n can increase without limit in the simplest theory of atomic hydrogen, states with very large n do not have physical significance The L z operator is ( /i)(/φ), and the p x and p y hybrid functions depend on φ as shown in the text on page 434: p x cos φ and p y sin φ. Since we 91

35 know cos φ/φ = sin φ and sin φ/φ = cos φ, these hybrid functions are not eigenfunctions of L z. In fact, the L z operator turns one function into something at least proportional to the other Table 1.5 lists the general form of the hybrid H-atom wavefunctions, and for the 3p x and 3p y functions, we find 3p x = 1 Ψ Ψ 31 1 and 3p y = i Ψ 311 Ψ Table 1.4 in turn gives the hydrogenic wavefunctions Ψ 31±1 as Ψ 31±1 = 1 π 1 3/ e r/3a 0 6r r a a 0 0 a 0 sin θe±iφ. Substitution and simplification (using cos φ = e iφ + e iφ and i sin φ = e iφ e iφ ) gives 3p x = 3p x = π a 0 3/ π a 0 3/ 6r a 0 r a 0 cos φ sin θ e r/3a 0 6r a 0 r a 0 sin φ sin θ e r/3a 0. The one radial node in each of these hybrids occurs when 6r/a 0 r /a 0 = 0, or at r = 6a 0. The 3p x hybrid has a nodal plane wherever cos φ = 0, which is the y-z plane with x = 0, exactly like the p x hybrid, and the 3p y hybrid has a nodal plane wherever sin φ = 0, which is the x-z plane with y = 0, exactly like the p y hybrid. 1.4 We follow the logic of the previous problem again here, starting with the general forms of the hybrids in Table 1.5: 9

36 3d xz = 1 Ψ 31 + Ψ 3 1 3d yz = i Ψ 31 Ψ 3 1 3d xy = i Ψ 3 Ψ 3. The hydrogenic wavefunctions we need are in Table 1.4: Ψ 3±1 = Ψ 3± = 1 3/ e r/3a 0 r 81 π a 0 a π a 0 3/ e r/3a 0 r sin θ cos θ e±iφ a 0 sin θ e ±iφ. Substitution and simplification again gives the hybrids: r 3d xz = 3/ e r/3a 0 cos φ cos θ sin θ 81 π a 0 a 0 r 3d yz = 3/ e r/3a 0 sin φ cos θ sin θ 81 π a 0 a 0 r 3d xy = 3/ e r/3a 0 cos φ sin φ sin θ. 81 π a 0 a 0 Using the relationships between Cartesian and spherical polar coordinates x = r sin θ cos φ, y = r sin θ sin φ, and z = r cos θ, we can see that the 3d xz hybrid contains xz/r = cos φ cos θ sin θ, the 3d yz hybrid contains yx/r = sin φ cos θ sin θ, and the 3d xy hybrid contains xy/r = cos φ sin φ sin θ. Thus, 3d xz is zero wherever xz = 0 and similarly for the other two hybrids. SECTION

37 1.43 Equation (1.53b) gives us the eigenvalues for the S z operator: S z α = ( /)α and S z β = ( /)β. The eigenvalue expressions for the square of S z are therefore S z α = Sz S z α = S z α = S z α = 4 α S z β = Sz S z β = S z β = S z β = 4 β. Since the expectation value of an operator that gives an exact eigenvalue is just the eigenvalue, we can write, for either the α or β spin state, <S z > = /4. Thus, <S > = 3<S z > = 3 /4 = (1/)(1/ + 1), in accord with Eq. (1.53a). GENERAL PROBLEMS 1.44 For scenario (a), the H O bending angle, we note first that the H O H bending angle θ varies between 0 (the two H atoms overlap) and 180 (the molecule is linear). Moreover, we know that water has an equilibrium (lowest energy) structure with θ 105. As θ 0, the potential energy should rise rapidly as the two H atoms bump into and repel each other; the θ = 0 limit should not be reachable. As θ 180, the potential should reach a maximum, since, by symmetry, the linear configuration divides configurations that are bent to one side or the other of linear. This leads to the qualitative sketch shown below. V(θ) θ 94

38 We can graph the potential for scenario (b) exactly: it is the Coulombic repulsive potential energy between two identical charges, V(r 1 ) = e /(4πε 0 )r 1, with the zero of energy taken at r 1 =. It looks like this: 4 V(r 1 )/10 18 J r 1 /10 10 m The methyl cyanide, CH 3 CN, to methyl isocyanide, CH 3 NC, isomerization in part (c) is characterized by an isomerization angle θ that we can take to be 0 in the cyanide configuration and 180 in the isocyanide configuration. Since the cyanide configuration is the more stable of the two, V(θ) must have an absolute minimum at θ = 0. But since the isocyanide is also stable, there must be a secondary minimum in V(θ) at 180 that is higher in energy than the absolute minimum. For all angles in between, the energy must rise smoothly to a maximum and then fall again, leading to the following qualitative diagram: V(θ) θ As the O H group is rotated around the C O bond, the alcoholic H alternately eclipses and staggers hydrogens on the methyl group. Thus, every 10, there 95

39 is a maximum in V(θ) at each eclipse, and 60 past each maximum, half way between successive maxima, we encounter a minimum representing each staggered conformation. All the maxima have the same height, all the minima have the same height, and θ ranges from 0 to 360. These are the characteristics of a function we can write as V(θ) = V 0 [1 + cos(3θ)] where the constant V 0 measures the potential difference between the minima and maxima. This function is shown below. V 0 V(θ) θ Note that there is no sound theoretical argument that says the cosine (or sine our choice is arbitrary) function accurately represents the true V(θ). Any function with the right number of wiggles of uniform height could perhaps do as well or better, but the cosine is the simplest function with a chance at being correct, and experiments have shown that it can do a good job of reproducing some of the phenomena associated with this sort of internal motion If the new variables are x = (x + y)/ and y = (y x)/, we can solve these equations for the original variables: x = (x y )/ and y = (x + y )/, and substitute these expressions into the potential energy function V(x, y): V(x, y) = k 0 (5x + 5y + 6xy) = k 0 5 x y + 5 x + y + 6 x y x + y = k 0 (8x + y ) = 1 (16k 0 ) x + 1 (4k 0 ) y. 96

40 Not only does this variable change remove the cross term that precluded a separation of variables solution, the final form of V shows at a glance that the potential is simply a sum of harmonic oscillator potentials in the two independent variables x and y. Written this way, we can easily identify the force constants k x and k y for these two directions: k x = 16k 0 and k y = 4k 0. The energy expression for a mass m bound to this -D potential is therefore E = ω x v x ω y v y + 1, ω x = 4 k 0 m and ω y = k 0 m Classically, the electron and positron in positronium move toward each other, moving symmetrically about a center of mass that, since e and e + have the same mass, is always half way between them. For states of zero orbital angular momentum, these two particles move along a line. The only difference between the Ps atom quantum mechanical problem and the ordinary H atom problem is the reduced mass: for H, µ H = m e m p /(m e + m p ) m e, but for Ps, µ Ps = m e m e /(m e + m e ) = m e / = µ H /. Since the ionization energy of H is the negative of the ground-state energy: IP = E 1, and since E 1 µ (see Eq. (1.46a)), the Ps ionization energy is half that of H. Since the Bohr radius is proportional to 1/µ (see Eq. (1.45)), the Ps Bohr radius is twice that of H. The energy level expression for Ps is identical to that for H with the exception of the change in the energy constant: E n (H) = J n, E n (Ps) = J n = J n. For H, the longest wavelength a ground-state atom can absorb is that which takes the atom from n = 1 to n =. It is 11.5 nm, the longest wavelength of the Lyman series mentioned on page 351 and shown in Figure 11.1 in the text. The same calculation holds for Ps: λ = hc E = = hc E (Ps) E 1 (Ps) hc J J 1 = 43 nm, 97

41 exactly twice the H atom n = 1 wavelength. In a classical picture of the Ps ground state, the positron electron motion is along a line connecting both particles through the center of mass. The classical turning point of this motion for Ps is 4a 0, found by writing E 1 (Ps) = the total energy = V(r = classical turning point) = e /(4πε 0 )r and solving for r. The time τ it takes the electron (say) to go from its turning point, in to r = 0, out to the opposite turning point, then back to r = 0 and on to the original turning point must be four times the time to move from the first turning point to r = 0. (This time is the same for the positron to make its mirror-image motion.) It can be shown that this total time is four times as long as the time for the electron in the Bohr model of H to make one Bohr orbit around the proton, which for Ps is a total time τ s. Thus, in 0.1 ns = s, the Ps mean lifetime, the electron and positron make about classical orbits around their mutual center of mass The density of states for a particle in a 1-D box is worked out as Example 0.1 on page 754 in the text. We can follow the logic shown there to derive the density of states for a particle-on-a-ring, introduced in Example 1.6 on page 419 in the text. The energy expression is E = m /MR for a particle of mass M with quantum number m = 0, ±1, ±,. One difference between this system and the particle-in-a-box is that every state here except m = 0 is doubly degenerate. It introduces little error to derive the general expression assuming all states are doubly degenerate, and thus we write the density of states ρ(e) as ρ(e) = dm de = dm de 1 m = MR 1 = MR 1/ MR 1 = R M E If V(r) = (constant)r n, the general form of the Virial Theorem becomes <T> = <r V r > = <(constant)rnrn 1 > = n<v>. Since E = <T> + <V>, we can write E = <T> + <V> = n <V> + <V> = n + <V> = <T> + n <T> = n + n <T>. If n = (harmonic), we see that <T> = <V>. For n = 1 (Coulombic), <T> = <V>/. 98

42 1.49 If we equate the total energy of a state of the H atom, E n, to the effective potential energy (the sum of the centrifugal potential and the Coulombic potential), we can write E n = e 1 l(l + 1) = (4πε 0 )a 0 n µr e (4πε 0 )r, a 0 = (4πε 0 ) µe. With the definition suggested in the problem, ρ = r/a 0, this can be written 0 = = e (4πε 0 )r e 1 (4πε 0 )a 0 n l(l + 1) µr e (4πε 0 )r e 1 (4πε 0 )a 0 n l(l + 1)a 0 e (4πε 0 )r = 1 r 1 n l(l + 1)a 0 r = ρ 1 1 n l(l + 1) ρ or n ρ ρ l(l + 1)n = 0. A quadratic solution for ρ gives ρ = n ± n n l(l + 1). The two solutions represent the two classical turning points. Note that for l = 0 (i.e., s states), the expression simplifies to ρ = n and ρ = 0, which simply means that r = 0 can be reached in s states. In units of a 0, the outer classical turning points (the values for ρ with the + sign chosen) are 3 for 4s, for 4p, for 4d, and 4 for 4f. These can be compared to the <r> values for the same states: 4, 3, 1, and 18, respectively, using the expression for <r> given on page 431 in the text If we take p = /R, so that T = /m e R and thus F out = dt/dr = /m e R 3, we can write the force balance equation F out + F in = m e R 3 e (4πε 0 )R = 0. Solving this expression for R gives R = (4πε 0 )/m e e = a 0, the usual Bohr radius. The Virial Theorem says E = T = /m e a 0 = E 1, the exact H atom ground-state energy. Thus, this simple model gives the same energy and size parameter as the Bohr theory, which, in turn, gives the same energy and size as the complete quantum mechanical theory. 99