7. Variation Background A course covering modern physics Familiarity with QSHO, H-atom Concepts of primary interest: The Variation Principle Sample calculations: Hydrogen atom ground state The Helium Atom – 1st order perturbation and variation + The Hydrogen Molecular Ion - H2 Tools of the trade: This handout is keyed to Griffiths Introduction to Quantum Mechanics, 2nd Ed. It is not designed to be used independently. The Variation Principle: The ground state of a quantum system has a very special property; it is the state with the lowest possible energy. The hamiltonian is an operator for the energy, and it has a complete set of eigenfunctions n each with its energy eigenvalue En. The completeness property ensures that any allowed wavefunction can be represented as a linear combination of those eigenstates. akk with k 2 2 ak 1. It follows that the energy expectation value for the state is Ea kkE. k k We will label the ground state as n = 1, and all other states by indices greater than 1. 22 2 EEa11(0) akk ( EE 1 ) akk ( EE 1 ) [Var.1] kk11 Our goal is to find the ground state, the lowest energy state. Any change that reduces |ak| for k > 1 lowers E and adjusts to be closer to the desired ground state form. (To make life simpler, we assume that the ground state is non-degenerate.) Sample Calculation: The Hydrogen Atom Ground State for use with Griffiths QM Contact: [email protected] Start by choosing (r) to be a function with any number of adjustable parameters. Any adjustment of the parameter that lowers the energy expectation value moves closer to - the real ground state. Guided by hindsight, the trial wavefunction chosen is (r) = A e r where is the loan adjustable parameter. The parameter A is set by normalization, and A is expected to be a function of the adjustable parameters. 22 2 2 22ee1 Hˆ r 2 rr 24mrmrno , 2 4r The simpler (independent of angles) form of the hamiltonian is applicable because the trial wave function has only radial dependence. Step 1: Choose an allowable wavefunction. It must be normalized. 2222ru22442 dA ee4( rdrA uduA2!) A 008833 3 3 2 As the wavefunction can be chosen to be real, a possible A = . Step 2: Evaluate the energy for arbitrary values of the parameter(s). 22 3 rr 1 22e EHdˆ ere 4 rdr 0 2 rr 24mr r rr22errr r 22e 2 rer Where: rr re 22 3 ˆ rr 21rre 2 EHd ee e 24 re rdr 0 24mr 22 5322rr 11 2 e Errdree12 4 4rdr 00 24m 22 5311uu12 e E41 ee 4 u u dr udu 843200 2m 22 2 22 2 ee Eeeuu u2 4( u du udu 2) [Var.2] 00 444mm 4 222 2 1122me 4 Ea22where o mmam24 2 o e Step 3: Vary the parameter(s) to minimize the energy. 11/19/2010 SP425 Notes –Variation Ch7-2 22 2 dE eme1 0 2 (a0 ) [Var.3] dm44 The magic value for is the inverse of the Bohr radius. Our trial function becomes 1 ra/ o 1()r 1/2 3/2 e which is the true ground state wavefunction. (We found the ‘exact’ ao answer because the initial trial wavefunction family contained the target function.) One of the problems for this section is to repeat the process above using a Gaussian trial wave function. Extension to first excited states: One can attempt to find the first excited state by constructing states that are orthogonal to the ground state and that are normalized. Varying the parameters of such functions you might find an approximation to the first - r excited state. Trial functions: 2(r) = B (1 – br) e . Normalize the 2 to find B(,b). Find b() to ensure that 2|1 = 0. Using b(), vary to minimize the energy of 2. What is you use a different representation for essentially the same parameter? 2 22 Suppose that one chose ()x Aebx while another chose ()xAe x /a . The results dd would be unchanged. The minimization conditions db EE 0and da 0 would lead to the same wavefunction and the same minimum energy. The Helium Atom: Helium is an atom with one electron too many. As a first step, we will model helium using the single electron hydrogenic wave functions, then include first order perturbation theory and finally add parameter variation to generate an approximate wavefunction and ground state energy. 2Z e2 The one electron (He+) problem has the hamiltonian 2 where Z = 2. The 24mr previous solution of the hydrogen atom problem is adequate to provide the answers. The doubled nuclear charge acts on the electron to half the linear dimension parameter 11/19/2010 SP425 Notes –Variation Ch7-3 and halving the electron-nuclear separation while doubling the nuclear charge 22 quadruples the potential energy. Using the particle in the box model, ( E n ) n 2ma2 suggests that halving the length scale quadruples the kinetic energy as well. If you 1 want something more formal, the virial theorem result is that K = -½ V for a /r potential. The two energies must scale in the same manner ( [linear extent]-2). 54.4 eV The single electron energies are – /n2, and the wavefunctions have half the spatial extent of the hydrogenic functions. The two electrons fill the n = 1 level so the ground state energy is – 108.8 eV. The only problem is that the experimental result is nearer to -79.0 eV. The model needs to be improved. The most obvious omission is the electron-electron repulsion. The total hamiltonian 22222 ˆˆ ˆ 22Ze Ze e should be: HH01 H 2 . The first order 224mr4 1 m r2 4 r12 perturbation correction is the expectation value of the perturbation in the unperturbed state. 2 ˆˆe (1) H ; E11 gs H gs ; gs (, rrt 12111 ,) SS (,) rt (,) rt2 [Var.4] 4 r12 The evaluation is not as challenging as it first appears. While studying electrostatic boundary value problems, we derived the relation: 1 r ˆˆ (1) Pr (21 r) where r> is the larger of r1 and r2. ||rr21 0 r Choosing the coordinate 2 polar axis to lie along rrr12,ˆˆ1 becomes cos2. 1 r P (cos ) [Var.5] r (1) 2 12 0 r 11/19/2010 SP425 Notes –Variation Ch7-4 The wavefunction has no angular dependence so the orthogonality of the Legendre polynomials kills all but the = 0 contribution. Our effective relation for spherically symmetric states becomes. 11r 1 (1) Prr (cos22 ) for1 and forr2r1 . [Var.6] rr12 0 r 1 r2 (In the case, that the wavefunctions have angular dependence, the orthogonality -1 relations often can be used to reduce the effective expression for (r12) to a few terms.) 22sin ddsin 2 Exercise: Show that where cos = rrˆˆ and r> 00||rr 22 r 12 21 rr122cos rr 12 2 is the larger of r1 and r2. As sin d = 2, equation [Var.6] gives the correct effective 0 1 2 2 value for |rr21 |. Recommended change of variable: u = r1 + r2 – 2 r1r2 cos. 3 3 Z Zra/ 0 Z Z ()/rr12 a 0 The wavefunction 1S = 3 e so the time independent form of = 3 e . a0 a0 2 3 2 Ee(1) Z Zrra()/12 0 e er Zrra ()/12 04422rdrdr [Var.7] 1 a3 4 r 1212 0 00 12 Effective 2 3 2 r1 (1)Z e 2/Zr10 a 2/ Zr 20 a11 2 2/ Zr 20 a 2 2 E e e 44r dr e r dr4 r dr 122 a3 4 rr 2211 0 00 12 r1 It’s a tedious evaluation of several simple integrals using integration by parts. 2 3 2 2/Zr a r1 2/ Zr a 2/ Zr a E(1) 4 Z e eer101 2022drer 20 drrdr 122 a3 r 2211 0 00 1 r1 Reducing the inner integrals: rZ1 222/r10a 2/Zr 10a 1 2/Zr20 a 22 2/Zr 10 a aa00uua 0 r e r22 dr e r 22 dr e u du e u du 1 00r 22Zr1 Z 2Z 0 1 Integrate[Exp[-u] u^2 ,{u,0,c}]= 2-(2+c (2+c)) -c Integrate[Exp[-u] u ,{u,c,Infinity}]= (1+c) -c FullSimplify[c^(-1) (2-(2+c (2+c)) -c)+(1+c) -c] = (-c (-2-c+2 c))/c 11/19/2010 SP425 Notes –Variation Ch7-5 2 3 2 2 (1) Z e 2/Zr10 a aaa00 2/Zr10 a 0 2/Zr10 a 2 Eeee1 42 3 222ZZrZr 1 2r11dr a0 0 11 2 3 2 2 (1) Z e aa002/Zr10 a 2/ Zr 10 a a0 2/Zr10 a 2 Ee1 42 3 e 1 2 er11dr a 22ZZ0 r11 2Zr 0 2 3 2 2 Z e aa004/Zr10 a 22 2/Zr10 a a0 42 3 e 1r11 dr e 2r 11 dr a 22ZZ00r11 2Zr 0 Time Out: Sanity check. Is there any chance that the expression above is correct? First, everything inside the braces is a pure number. Both terms have the same dimensions. () 2 3 222 2 2 Z eea0 2 e rdr11 Tracking the dimensions only: 4 3 2Z rdr11 4 4 ength Energy. () a 4 ao 0 There is a chance that the expression is not pure nonsense. Return to the calculation. 2 3 2 2 (1) Z e a0 E 4 3 1 a 2Z 0 332 4/Zr10 a aa004Zr1 42ZZ 2/Zr10 a a0 2Z edre41aa11 2 ar adr1 0 44Zr1 Z00 0 2 Z 0 0 2 3 2 2 3 3 2 Z eZaa00 uu2 4 e 42 3 ee4uudu udu 4 (2) (3) 16 (2) a 264ZZ420 0 a0 0 (1) Ze2255e EZ1 4(1!) (2!) 16(1!) (2) 27.2eV 34eV 64aa00 8 4 8 The energy estimate including the correction from first order perturbation for the electron-electron repulsion is -108.8 eV + 34.0 eV = - 74.8 eV.