Research: Science and Education
Quantum Alchemy: Transmutation of Atomic Orbitals S. M. Blinder Department of Chemistry, University of Michigan, Ann Arbor, MI 48109-1055;
[email protected] The modern theory of molecular structure and reactivity is based to a major extent on the energies and geometries of atomic orbitals. These originated as electron wave functions in the exact solution of the Schrödinger equation for the hydrogen atom. A comprehensive review of atomic orbitals was published in this Journal by Berry (1). Definitive pictorial representations for both atomic and molecular orbitals were computed by Wahl (2). I will show in this paper how the atomic orbitals are, in principle, all related to one another, making use of both geometrical symmetries and more arcane mathematical transformations. The box shows normalized real Cartesian forms for hydrogenic orbitals of principal quantum numbers n = 1, 2, 3. For compactness, these are expressed in atomic units, whereby ⁄ = m = e = 1. We assume infinite nuclear mass or alternah tively interpret m as the reduced mass mM/(m + M). The unit of length is the Bohr radius a0 = h⁄ 2/me 2 and the unit of energy is the hartree, E0 = e 2/a 0. To convert these entries to conventional units with arbitrary nuclear charge Ze, multiply each x,y,z and r by Z/a 0 and multiply the wave function by (Z/a 0)3/2. For example, the 3pz orbital written in full is
ψ3pz =
5/2
2 Z 81 π a 0
6 – Zr ze r/3a 0 a0
These atomic orbitals are illustrated in Figure 1. Dark gray and light gray indicate, respectively, positive and negative regions of the wave functions (the radial nodes of the 2s and 3s orbitals Real Cartesian Forms of Hydrogenic Orbitals in Atomic Units
ψ1s = 1 e r π 1 ψ2s = 1 – r e r/2 2 2 2π ψ2p = 1 ze r/2 z 4 2π ψ2p , ψ2p analogous x
are obscured). These pictures are stylized representations of atomic orbitals and should not be interpreted as quantitatively accurate. The functions in the box are solutions of the Schrödinger equation expressed in atomic units: 1 1 1 ∇ 2 – r ψn l m = E n ψn l m , E n = 2 2 2n
(1)
and Cartesian coordinates so that
∇2 =
∂2 ∂2 ∂2 + + and r = ∂x 2 ∂y 2 ∂z 2
x 2 + y2 + z2
Geometrical Symmetry Atomic orbitals occur in degenerate sets related by rotational transformations, some more obvious than others. We require a fundamental quantum-mechanical theorem on degenerate eigenfunctions. Suppose an operator M commutes with the Hamiltonian Ᏼ for the system; that is, [M,Ᏼ] ≡ M Ᏼ – ᏴM = 0
(2)
The solutions for a d-fold degenerate energy level of the Schrödinger equation can be written Now
Ᏼψnk = Enψnk,
k = 1, 2, …, d
(3)
MᏴ ψnk = Ᏼ(Mψnk) = En(M ψnk)
(4)
showing that M ψnk is also an eigenfunction of the same energy. It must therefore be equal to a linear combination of the known degenerate eigenfunctions of energy En: M ψnk = c1ψn + c2ψn + … + cd ψn 1
2
d
(5)
where c1, c2, … are constants, some or possibly all being equal to zero.
y
ψ3s = 1 27 – 18r + 2r 2 e r/3 81 3π ψ3p = 2 6 – r ze r/3 z 81 π ψ3p , ψ3p analogous x
y
1 3z 2 – r 2 e r/3 81 6π ψ3dzx = 2 zxe r/3 81 π ψ3dyz, ψ3dxy analogous
ψ3dz 2 =
ψ3d 2 2 = x –y
1 x 2 – y 2 e r/3 81 π
Figure 1. Hydrogenic atomic orbitals.
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The operators representing the three components of orbital angular momentum
L z = ih x ∂ – y ∂ ∂y ∂x
et cyc.
(6)
all commute with the Hamiltonian. The notation “et cyc.” means that the same relation holds with the Cartesian variables (x,y,z) replaced in cyclic permutation by (y,z,x), and again by (z,x,y). Noting that
∂r = x , ∂x r
∂r = y , ∂y r
∂r = z ∂z r
(7)
it follows that the operators Lx, Ly, Lz acting on a function of r alone give zero. Thus Lx,y,z ψns = 0
(8)
For operations with L on a p or d orbital it is sufficient to consider just their Cartesian factors in the box. For example
L x z = ih y ∂ – z ∂ z = ihy ∂z ∂y showing that
Figure 2. Rotational transformations of orbitals.
⁄ Lxψ 2p = ihψ 2p et cyc.
(10)
⁄ Ly ψ 2p = +ihψ 2p et cyc.
(11)
z
analogously,
(9)
z
y
x
However, Lxψ 2p = 0, and likewise for y and z. The geometrical transformations of the three 2p orbitals are summarized in Figure 2, with constant numerical factors suppressed. Clearly the 3p orbitals follow an analogous pattern. For the 3d orbitals, the following relationships are easily obtained: Lx(xy) = const xz, Ly(xy) = const yz x
Lz(xy) = const (x 2 – y 2), Lz(x 2 – y 2) = const xy (12)
L R
Figure 3. Planetary orbit showing angular momentum L and aphelion R.
Lx(3z2 – r2) = const yz, Ly(3z2 – r2) = const xz with the obvious permutations. These transformations among d orbitals are also represented in Figure 2 (omitting nonstandard forms such as dy -z ). It becomes quite apparent from the diagram how the 3dz orbital can have the same energy as the “four-leaf clover” varieties. It must be confessed that we have so far done nothing more than reexpress the well-known ladder (raising and lowering) operators for angular momentum, whereby 2
2
2
(Lx ± iLy)Yl,m = const Yl,m±1
(13)
However this has produced explicit relationships among the real Cartesian forms of the orbitals, which are more often used in chemistry.
attracting bodies, the orbital angular momentum L = r × p is a constant of the motion. In general, an orbit will precess about the center of force. The two special cases associated with nonprecessing orbits are the harmonic oscillator and the inversesquare attraction. Specifically, the vector R in Figure 3, which connects the sun to the aphelion (the most distant point of the orbit), is a constant of the motion, over and above the angular momentum L. As is always the case in physics, a constant of the motion is associated with a symmetry of the system, which, in the quantum version of the problem, leads to degeneracy. To deduce the additional constant of the motion, we begin with the vector equation for Newton’s second law applied to a Coulomb attraction:
Hidden Symmetry A more challenging task is to relate the sets of degenerate orbitals with the same principal quantum number but different angular momentum: (2s, 2p), (3s, 3p, 3d), etc. This has its roots in Kepler’s laws of planetary motion. The nonrelativistic two-body gravitational problem has exact solutions in which the planet describes an elliptical orbit around the sun, which sits at one of the foci, as shown in Figure 3. For any central force, which depends only on the scalar distance between the 392
F=
dp Ze 2r = dt r3
(14)
where the linear momentum is given by p = m dr/dt. Now take the vector product of eq 14 with the angular momentum L = r × p:
L×
dp Ze 2 = L×r dt r3
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Research: Science and Education
Remembering that dL/dt = 0 because L is a constant of the motion, and going through the vector manipulations
L × r = m r × dr × r = mr 2 dr – mr r ⋅ dr dt dt dt
(16)
eq 15 reduces to
d L × p + mZe 2 r = 0 r dt
(17)
The dimensionless quantity 1
A = mZe 2 L × p +
r r
(18)
known as the Runge–Lenz vector (3), is the additional constant of the motion for the Kepler–Coulomb problem. The equation for the orbit can be obtained from A r = A r cos θ = (mZe 2)1L 2 + r
(19)
Now that we have obtained another set of operators that commute with the Hamiltonian, we can play the same games we did with angular momentum. The following results were derived by Lohr (7):
(20)
Azψ1s = 0, Azψ2s = const ψ2pz, Azψ2pz = const ψ2s (22) Azψ2px = Azψ2py = 0
This can be solved to give 1
m Ze 2 L 2 r= 1 – A cos θ
which is the equation for an ellipse in polar coordinates (4). Therefore, the Runge–Lenz vector A is parallel to the aphelion R with magnitude A equal to the eccentricity of the orbit. The quantum-mechanical solution of the Kepler–Coulomb problem was first worked out by Pauli (5). The quantum analog of the Runge–Lenz vector contains the symmetrized operator product 1 L × p –p × L 2 to take account of the noncommutativity of components of L and p. Reverting to atomic units and Z = 1, the z component of the Runge–Lenz operator can be written
A z = 1 Lx py – Ly px – px Ly + p y Lx + z = r 2 2 2 2 ∂ ∂ ∂ ∂2 z +z –y –x – ∂ +z 2 2 ∂z r ∂y∂z ∂z∂x ∂x ∂y
Figure 4. Runge–Lenz transformations.
(21)
with analogous forms for Ax and Ay. To do the complicated algebra and calculus, it is helpful to make use of one of the symbol-manipulating computer programs such as Mathematica or Maple. After somewhat lengthy algebra, it can be shown that Ax, Ay, Az all commute with the hydrogen atom Hamiltonian. A system that commutes with the orbital angular momentum is said to transform under the continuous symmetry group of rotations in 3 dimensions, designated SO(3) by mathematicians. The hydrogen atom, by virtue of its additional symmetry, can be shown to transform under SO(4), the symmetry group for rotations in 4-dimensional space (6 ). This accounts for the additional degeneracy of its eigenvalues. (In some references, these groups are designated SO3 and SO4, which should provide some amusement for chemists.)
For principal quantum number n = 3: Azψ3s = const ψ3pz, Azψ3pz = const ψ3s + const′ ψ3dz
2
Azψ3px = const ψ3dzx, Azψ3py = const ψ3dyz Azψ3dz = const ψ3pz, Azψ3dxy = Azψ3dx –y = 0 2
2
2
(23)
Azψ3dyz = const ψ3py , Azψ3dzx = const ψ3px Axψ3dxy = const ψ3py , Axψ3dx –y = const ψ3px 2
2
Some of the Runge–Lenz transformations among the 3s, 3p, and 3d orbitals are pictured in Figure 4. It has thus been demonstrated that hydrogenic orbitals of differing angular momentum but the same principal quantum number are indeed degenerate despite their dissimilar geometry. Supersymmetry: Radial Ladder Operators To complete our atomic orbital “alchemy”, it remains to transform orbitals to different values of the principal quantum number. Schrödinger (8) in 1940 developed a technique for solving second-order differential equations by a factorization method. There is extensive subsequent literature on this approach but we will limit ourselves to one simple result. We need to express the hydrogenic eigenfunctions in spherical polar coordinates: ψ nlm(r, θ , φ) = R nl (r)Ylm(θ, φ)
(24)
and represent the radial function by R nl (r) = const F nl ( ρ )eρ /2
(25)
where ρ ≡ 2Zr/na0. We state without proof that the quantum number n is increased by 1 in the operation
F n+1, l ρ = const ρ d + n + 1 – ρ Fnl ρ dρ
(26)
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Thus the square bracket represents a ladder operator for n, analogous to eq 13 for m. Note that this operator is neither Hermitian nor does it commute with the Hamiltonian. As an illustration, we operate on the 2s orbital, which has n = 2, l = 0, ρ = r, F20 = const (1 – ρ/2). We find
ρ d + 2+1–ρ dρ
ρ ρ2 1– = 3 – 3ρ + 2 2
(27)
The last expression reduces to const × (27 – 18r + 2r ) = F30 after setting ρ = 2r/3. 2
In many-electron atoms, the degeneracy among orbitals with different (n, l ) is removed. When electrons no longer experience the pure Coulomb field of a bare nucleus, the orbital energies rearrange themselves into the familiar sequence: 1s < 2s < 2p < 3s < 3p < 4s ~ 3d, … . Let us define a modified Runge–Lenz operator for an electron moving in a spherically symmetrical potential V (r): Ã = L × p – rV (r)
(28)
which reduces to the conventional operator when V (r) = Z/r. Let each orbital be, in principle, an eigenfunction of a oneelectron Schrödinger equation ˜ φ (r) = φ (r) Ᏼ (29) n
n n
with an effective Hamiltonian ˜ = 1⁄ ∇ 2 + V (r) Ᏼ
(30)
2
This might, for example, be a Hartree–Fock-type computation. ~ Now à no longer commutes with Ᏼ. In fact, ˜ Ã] = 1⁄ r ∇ 2V (r) (31) [Ᏼ, 2
Let us now evaluate the integral of this commutator between the 2s and 2pz solutions to eq 29: ˜ Ã |2s – 2p | Ã Ᏼ|2s ˜ = 1⁄ 2p |z∇ 2V (r)|2s (32) 2p |Ᏼ z
z
z
2
z
This can be solved for
2p – 2s =
2pz|z 2V(r)|2s 22pz|Ã z |2s
(33)
which reduces to zero, as it should, for a Coulomb potential. Let us do a very rough estimation of ∆ for the helium atom, approximating φ1s, φ2s, φ2p by the corresponding hydrogenic orbitals for Z = 2. Considering the 1s2s and 1s2p configurations of He (neglecting singlet–triplet differences), the
394
Z –1 – Z + 1 e 2Zr r R
(34)
2V(r) = 4 Z 3e 2Zr + 4 π Zδ r
(35)
V(r) = and
(which is equivalent to Poisson’s equation, ∇ 2Φ = 4πρ). Evaluating the integrals in eq 33, we obtain 2p – 2s ≈ +4/181 hartree ≈ +0.60 eV
Broken Symmetry
z
potential V (r) on the outer electron is produced by the nucleus plus the 1s electron. For φ1s(r) = (Z 3/π)1/2 e-Zr, a simple calculation gives
The experimental energy differences between the 1s2p and 1s2s configurations in helium are 0.6 eV and 1.1 eV for the singlet and triplet states, respectively. We have thus reproduced the correct ordering of orbital energies (2s < 2p) with the approximately correct magnitude. This is however not a recommended way to calculate orbital energy differences— my purpose was to show how removal of the degeneracy is related to the Runge–Lenz vector. Conclusion It has been demonstrated that all the hydrogenic atomic orbitals can, in concept, be transmuted into one another by an appropriate sequence of transformations. This would have delighted the medieval alchemists! Literature Cited 1. Berry, R. S. J. Chem. Educ. 1966, 43, 283–299. 2. Wahl, A. C. Science 1966, 151, 961; Sci. Am. 1970, 222 (April), 54–70. Also, J. Chem. Educ. 1985, 62 (3), entire issue. 3. Runge, C. Vector Analysis; Dutton: New York, 1919; p 79 ff. Lenz, W. Z. Phys. 1924, 24, 197. An essentially equivalent approach was used by Laplace in Mécanique Céleste (ca 1800). 4. See, for example, Goldstein, H. Classical Mechanics; AddisonWesley: Cambridge, MA, 1953; p 76 ff. 5. Pauli, W. Z. Phys. 1926, 36, 336; English translation in van der Waerden, B. L. Sources of Quantum Mechanics; Dover: New York, 1968; p 387 ff. See also Baym, G. Lectures on Quantum Mechanics; Benjamin: Reading, MA, 1974; p 175 ff. 6. Englefield, M. J. Group Theory and the Coulomb Problem; Wiley: New York, 1972. 7. Lohr, L. L. Int. J. Quantum Chem. 1976, 10, 799–809. 8. Schrödinger, E. Proc. R. Irish Acad. 1940, A46, 9–16. See also David, C. W. Am. J. Phys. 1966, 34, 984–985.
Journal of Chemical Education • Vol. 78 No. 3 March 2001 • JChemEd.chem.wisc.edu
