Electric fields : Stark effect, dipole & quadrupole polarizability. We are often interested in the effect of an external electric field on the energy levels and wavefunction of H and other one-electron atoms so lets consider the atom in a spatially constant electric field. The energy shift due to the electric field is called the Stark effect. If the atom is in an external electrostatic potential ϕ( r ) the Hamiltonian becomes m cm t µ Ze 4πε r + Zeϕ( r n ) eϕ( r e ) Ψ = EΨ where e > and the electric field F = ϕ. Since F is constant ϕ = r i F up to an arbitrary constant, which we ignore. Converting r e & r n to r & R we have where and (Ĥ cm + Ĥ int )Ψ = EΨ Ĥ cm = m cm t +e F i R( Z) Ĥ int = µ Ze 4πε r + e F i r (+ (Z )(m e / m t )) Since the center of mass and the internal motion are independent the wave function Ψ( R, r ) = φ( R)ψ ( r ) factorizes into a product where m cm t +ef i R( Z) φ( R) = E cm φ( R) and
µ Ze 4πε r + q F i r ψ ( r ) = E int ψ ( r ) where q = e(+ (Z )(m e / m t )), an effective charge for Z. The center of mass motion is considered latter and we now focus on the effect of the field on the internal motion of the electron. Before getting into any detail lets consider the order of magnitude of the internal electric field before applying the external field. This Ze field at a distance r from the nucleus is given by and after substituting the 4πε r numerical values of the constants the field strength is (.4399649 ) Z r volts m r in Å. The first Bohr orbit in H has a radius of.5977å so the field in H at that distance from the nucleus is 5.48 volts m but 7 volts m. Laboratory fields vary widely is reasonable and is smaller than the internal field by a factor of approximately 4 suggesting that perturbation theory will be adequate to estimate the change in energy of the one electron atom in typical laboratory fields. with The unperturbed internal Hamiltonian is where Ĥ = µ Ze 4πε r Ĥ ψ nlm = E n ψ nlm and E n = e Z (4πε )a µ n If we measure length in multiples of γ a and the energy in multiples of have Ĥ = Z r with E n = Z n e (4πε )a µ we
Lets choose F to be along the z axis, F = Fẑ Ĥ = Z r + qfr cosθ = Ĥ + qfr cosθ = Ĥ +VF where θ is the angle between the field and the position vector r and ˆV = qr cosθ.in these units F is measured in multiples of with m n in multiples of the electron mass. e (4πε )a and q = + (Z ) / + µ m n Lets first consider the effect of the perturbation on the ground state wave-function ψ ψ with energy E = Z. Since the wave-function and energy in the field are solutions to ( Ĥ + VF ˆ )ψ = Eψ we may expand E &ψ in powers of F ψ = ψ n F n and E = E n F n n= n= The Schrodinger equation becomes ( Ĥ +VF ) ψ n F n = ψ n F n n= n= m= E m F m where ψ n & E n are the n th order corrections to the wave-function and energy. We can write this equation as ( ) ˆVψ p E p k ψ k p F p Ĥ ψ p + δ p k= = p= and since it must be true for any F each coefficient of F p must be zero. Accordingly we recover the usual Rayleigh-Schrodinger perturbation equations Ĥ ψ p + δ p ( ) ˆVψ p p = E p k ψ k k=
The first few equations are p= Ĥ ψ = E ψ, the unperturbed equation where ψ = Z 3 π e Zr and E = Z p=; (Ĥ E )ψ + ( ˆV E )ψ =, the first order equation p=; (Ĥ E )ψ + ( ˆV E )ψ = E ψ, the second order equation p=3; (Ĥ E )ψ 3 + ( ˆV E )ψ = E 3 ψ + E ψ, the third order equation p=4; (Ĥ E )ψ 4 + ˆVψ 3 = E 4 ψ + E ψ Note that since the energy cannot depend on the direction of the electric field all odd order corrections, E,E 3, must be zero. We can verify this explicitly for E as follows. Multiply the first order equation by the ground state wave-function and integrate (Ĥ E ) + ( ˆV E ) = Since Ĥ is Hermitian (Ĥ E ) = (Ĥ E ) = and so the first order correction to the energy is the average of the perturbation over the ground state wave-function which is zero by symmetry. E = ˆV = qz 3 e Zr r 3 dr cos(θ)sin(θ)dθ = Consequently the first order equation becomes (Ĥ E )ψ + ˆVψ = π It happens that one can solve the above perturbation equations exactly to determine ψ n & E n and we will do so before discussing the more common approach of expanding the functions ψ n in terms of the eigenfunctions of Ĥ.
We note that the atom in the field has cylindrical symmetry and look for a solution of the form Since ψ = g(r,θ)ψ (Ĥ E )g(r,θ)ψ =ψ g r + g r (Z r ) + ˆL g r the first order equation becomes g r g r (Z r ) ˆL g = qr cosθ r The most general form of g(r,θ) is g(r,θ) = f (r)p (cosθ) where P (cosθ) is a Legendre polynomial. = Substituting this form into the differential equation for g and noting ˆL P (cosθ) = ( +)P (cosθ) results in = P f '' + f ' ( r Z) ( +) r f = qrp where we note that cosθ = P and consequently only the term with = contributes. Accordingly we have f '' + f ' ( r Z) r f = qr The solution to which is and so f = q r + Zr Z
ψ = q r + Zr Z P (cosθ)ψ To find E we multiply the second order equation by ψ and integrate (Ĥ E ) + ˆV = E Since Ĥ is Hermitian (Ĥ E ) = (Ĥ E ) = we have E = ˆV Doing the integration results in E = 9q 4Z 4 and since the dipole polarizability α is defined by E = αf we have α = 9q Z 4 To find E 4 we use the p=4 equation (Ĥ E )ψ 4 + ˆVψ 3 = E 4 ψ + E ψ which after multiplying by ψ and integrating results in E 4 = ˆV 3 E Note that using the first order equation we can write ˆV 3 = Ĥ E 3 but from the p=3 equation
( Ĥ E )ψ 3 = E ψ ˆVψ which results in and finally ˆV 3 = ˆV E E 4 = ˆV E E So the fourth order correction to the energy depends on the first and second order corrections to the wavefunction and since we have the first order we need the second order correction which we will get from the p= equation (Ĥ E )ψ + ˆVψ = E ψ using ψ = q r + Zr Z P (cosθ)ψ, ˆV = qr cosθ and E = 9q 4Z 4 and writing ψ = g(r,θ)ψ results in using g r g r (Z r ) ˆL g + q r + Zr 3 r Z cos θ = 9q Z 4 g(r,θ) = = f (r)p (cosθ) results in = and since P f '' + f ' ( r Z) ( +) r f + q r + Zr 3 Z cos θ = 9q Z 4 cos θ = P = 3 P + 3
only = & contribute. Letting A = q results in the two differential equations 4 Z f " + f ' ( r Z) 6 r f + 4AZ r + Zr 3 3 = and f " + f ' ( AZ Z) + r + Zr 3 r 3 = 9A The solutions being f = A ( 4 Z r 4 +Zr 3 +5r ) and f = A ( 4 Z r 4 + 6Zr 3 +8r ) so ψ = f ψ + f P (cosθ)ψ As seen above E 4 = ˆV E E and doing the various integrations results in and finally ˆV = 59 3 E 4 = q4 3555 Z 64 q 4 8 q, = Z 6 Z, = 43 q 6 8 Z 6. Since the second hyper-polarizability γ is defined by E = E αf 4 γ F 4 one has γ = 665 q 4 8 Z With these results we can calculate the dipole moment µ of the one electron atom induced by the electric field since by definition µ = de df = αf + 6 γ F 3 In addition to the dipole moment higher moments are induced by the field. For example the zz component of the induced quadrupole moment is given by
Θ zz Θ = ψ r P (cosθ) ψ ψ ψ The numerator is ψ r P (cosθ) ψ = ( r P + r P )F + and r P = 7 q 8 Z & r P 8 = 4 q Z 8 and since ψ ψ = + ( + )F + the induced quadrupole through order F is simply the numerator Θ = 3 4 q F Z 8 What are these moments in the SI system? e Energy and electric field in the SI and au systems are related by E SI = E 4πε a au µ e and F SI = F 4πε a au so if E µ = α F then SI SI SI e E SI = E 4πε a au = µ α F e au au = 4πε a µ α F e SI au 4πε a µ and so α = α πε. The quantity 3 SI au 4 a α SI 4πε polarizability volume. Note it sα is usually reported and is called the 3 3 3 aua 48847. x αaum =. The polarizability ( ) α volume of the H atom in the SI system is then SI =.48847x 3 9 4πε q au Z 4 m3. The third order function is given by ( ) F ( dp ep) Φ = + Φ where 3 3 3 d = ( 6r 6 + 64r 5 + 344r 4 + 85r 3 + 59r + 38r) and 48
e= ( r 6 + 8r 5 + 63r 4 + 8r 3 ) and allows the energy to be calculated through 6 th 4 order or F 6. Series solution for E Usually one cannot solve the differential equations for the various perturbation corrections to the wavefunction and energy as we have done above and one calculates these corrections using the eigenfunctions and eigenvalues of the unperturbed Hamiltonian. For example when the perturbation is ˆV = rf cosθ the second order change in the ground state energy of H is given by the general formula ( E ) = V V ',nlm nlm, where V,nlm = ψ rf cosθ ψ nlm and E nlm = nlm E E nlm n which assumes that the unperturbed functions are complete with respect to the perturbed solutions. There are two classes of eigenfunctions for the H atom, those that describe the bound states, E <, and those that describe the continuum states, E >. Since we have solved the differential equation exactly we know the exact ( ) 9 second order energy, E = F, so lets see what the error is if we use only the 4 bound wavefunctions to calculate the second order correction. Accordingly ( E ) = Fr cosθ nlm E nlm E = F r cosθ nlm n E nlm E = F r cosθ nlm n + nlm n The required matrix element is π r cosθ nlm = dφ sinθ dθ Y cosθ m Yl dr R r r 3 R nl r π and the angular factor can be easily evaluated by observing that ( ) ( ) cos 4π 3 Y θ = so π π π π m dφ sinθd θ cos θ = dφ sinθd θ = δ δ Y Y Y Y m l so l 3 3 l m ( ) r cosθ nlm = 3 δ δ dr R l m r r 3 R n r ( ) = 3 δ l δ m R r R n
E R r R ( ) n = F. Note that only the z 3 n= + summation. n np states contribute to the The radial integral can be evaluated and the series summed (W. Gordon, Annalen der Physik 99, 394, 3) resulting in n 6 = = 85 + 6 9 9 ( ) n ( n ) s n 3 n= ( n + ) E F. F Since the exact E () s = 9 4 F the bound state wave-functions then account for ~8% of the exact result, a significant error! The need to include the continuum solutions in the expansion makes sense physically because in the electric field the electron feels the potential + Fr cos θ and if cosθ < the perturbation will eventually r overwhelm the Coulomb term and instead of the potential going to zero at large r it will eventually become lower than the s energy and the electron will be able to tunnel into the continuum. This is illustrated in the following figure with F = / au potential. -. -. -.3 -.4 -.5 -.6 -.7 s energy H atom in electric field (/ au) -/r -.5r -/r -.5r electron tunnels.76a 7.4a -.8 4 6 8 r(a )