7. Variation

Background A course covering modern physics Familiarity with QSHO, H-atom Concepts of primary interest: The Variation Principle Sample calculations: atom ground state The – 1st order perturbation and variation + The Hydrogen Molecular - H2 Tools of the trade:

This handout is keyed to Griffiths Introduction to , 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 EEa11(0)  akk ( EE  1 )  akk ( EE  1 ) [Var.1] kk11 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 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.

22ee 21 2 Hˆ  22    r  2 rr 24mrmrno , 2 4r 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.

  dA 222ee2ru4( rdrA2244 uduA2!) A2 008833  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   1 e EHd ˆ  ererr 22  4 rdr  0 2 rr 24mr  r rr22errr r 22e 2 rer Where: rr  re  22 3   e EHdˆ  eerr 21 err 24  re   r2dr  0  24mr  22 53 e Errdree22rr1211 4 2 4rdr 00  24m   22  53e E41 11eeuu 4 u12 u dr  udu 843200 2m  22 2 22 2  ee Eeeuu u2 4( u du  udu 2)  [Var.2] 00 444mm 4 222 2 1122me 4  Ea22where o mmam24  2 o e Step 3: Vary the parameter(s) to minimize the energy.

11/19/2010 SP425 Notes –Variation Ch7-2 22 2 dE   eme1 0 2 (a0 ) [Var.3] dm44 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  Aebx 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 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 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: HH01 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 cos2.

  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.

  11r 1  (1) Prr (cos22 ) for1 and forr2r1 . [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.)

22sin ddsin 2 Exercise: Show that   where cos = rrˆˆ and r> 00||rr 22 r 12 21 rr122cos 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 ()/12 04422rdrdr [Var.7] 1 a3 4 r 1212  0 00   12 Effective

2 3 2 r1 E(1)Z e e2/Zr10 a e 2/ Zr 20 a1144r 2 dr e  2/ Zr 20 a r 2 dr4 r 2 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 aa00uua 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 aa002/Zr10 a 2/ Zr 10 a a0  2/Zr10 a 2 Ee1 42 3 e 1 2 er11dr   a   22ZZ0   r11  2Zr  0  

2 3 2 2  Z e aa004/Zr10 a 22 2/Zr10 a a0 42 3 e  1r11 dr  e 2r 11 dr  a   22ZZ00r11  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  Z00 0  2 Z  0 0

2 3 2 2 3 3 2 Z eZaa00 uu2 4 e 42 3 ee4uudu udu 4 (2) (3) 16 (2)  a   264ZZ420  0  a0  0   

(1) Ze2255e EZ1  4(1!) (2!) 16(1!) (2) 27.2eV  34eV 64aa00 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. As the experimental value is about – 79 eV, the first order correction has overshot the mark. This situation is rather common. In first order, expectation value of the perturbation is computed in the unperturbed state. In the next order, the ground state wavefunction adjusts to lower the overall energy including that associated with the perturbation.

A variation approach is to be applied instead of further perturbation. The two electron

ZZ33Zra10//( Zra 20 Z 3  Zr 12 r)/ a 0 wavefunction is  = 33ee 3 e. The cloud grows in aa00  a 0 extent to lower the electron-electron repulsion. That is, the effective nuclear charge Z is less than 2. We will treat Z as a variable parameter in , but we will take care to keep the nuclear charge fixed at 2 e, its physical value, in the potential term.

11/19/2010 SP425 Notes –Variation Ch7-6 Ze2 (2Z  )e2 V1    4 r1 4 r1

22222 22 ˆˆ ˆ 22Ze Ze e (2)(2)Z eZe H HH01 1  2   [Var.8] 242mr 1 mrr4 224 1 4r1 4  r2 By this point, nobody wants to launch into another calculation, so let us just assemble

22 3  Z e Z Zra/ 0 ˆ 2 our previous results for  = 3 e , the solution for H0    .  a0 24mr 

222Ze e 22Z 13.6eV ;  Z 227.2 eV ; 5 Z27.2 eV   8  244mrr  12

Ze22e (2)Z e 2 Z 2 27.2eV  ZeV 27.2 ,  (ZZeV 2) 27.2 444 rrr 

222Z ee(2) Z e2 EHˆ 222   2 [Var.9] 2444mrr  12  r

ˆ 225 EH(13.6 eVZ ) 2  4 Z8 Z 2  2( Z 2) Z  2  27 Applying the first derivative test, Z = /16 minimizes the value of E yielding -77.5 eV which is only 1.5 eV high. We will be content with this approximation. Our zero-order estimate was -108.8 eV or about 30 eV low. Our current estimate is about 1.5 eV high 1 or about /3 the error prior to incorporating variation. Improvements to this estimate are usually delayed until graduate school when the Hartree-Fock1 method might be employed.

The Hylleraas Trial Function for the Helium Ground State:

The perturbation term is a function of the electron-electron separation r12. It follows that this parameter could be important in the trial functions. In the presence of this

1 The Hartree–Fock method is also called the self-consistent field method (SCF). One computes the potential for the nucleus plus the charge density of the other electron as a function of r. One then solves the radial equation numerically and inserts the corrected wavefunction into the electron density used to compute the potential. The process is iterated until it converges. http://en.wikipedia.org/wiki/Self_consistent_field_method

11/19/2010 SP425 Notes –Variation Ch7-7  term, the wave function is no longer expected to separate into a function of r1 times  one of r2 .

 ra12//oo r a Hylleraas proposed (,rr12 ) Ae e(1 br 12 ). The function is first normalized to find A(,b). The energy is then evaluated as a function of  and b. That energy is then minimized with respect to variations of the parameters  and b. The

0.364 minimized values are found to be  = 1.849 and b = /ao. The ground state energy estimate for those values is – 78.7 eV.

The Hydrogen Molecular Ion This problem is widely treated, and it can be solved exactly be adopting one of the generalized locally orthonormal coordinate systems, bispherical coordinates. See: http://en.wikipedia.org/wiki/Hydrogen_molecular_ion; The hydrogen molecular ion, dihydrogen cation, or H2+, is the simplest molecular ion. It is composed of two positively-charged and one negatively-charged electron, and can be formed from of a neutral hydrogen . It is of great historical and theoretical interest as, having only one electron, the Schrödinger equation for the system can be solved in a relatively straightforward way due to the lack of electron–electron repulsion (electron correlation). The analytical solutions for the energy eigenvalues [1] are a generalization of the (see Lambert W function and references therein for more details on this function). Consequently, it is included as an example in most textbooks. The first successful quantum mechanical treatment of H2+ was published by the Danish physicist Øjvind Burrau in 1927,[2] just one year after the publication of wave mechanics by Erwin Schrödinger. Earlier attempts using the had been published in 1922 by Karel Niessen[3] and ,[4] and in 1925 by .[5] In 1928, published a review putting together the work of Burrau with the work of and on the hydrogen molecule.[6] Bonding in H2+ can be described as a covalent one-electron bond, which has a formal bond order of one half.[7]

http://en.wikipedia.org/wiki/Bispherical_coordinates, http://www.dartmouth.edu/~chem81/thps/h2plus.html

11/19/2010 SP425 Notes –Variation Ch7-8 http://mathworld.wolfram.com/BisphericalCoordinates.html x = ; y = ;z =



Our model of the hydrogen molecular ion is two fixed () Coulomb force centers separated by R, and an electron moving in the potential field due to the two protons. Actually, the inter-nuclear separation oscillates, and the dumbbell of the nuclei (with the electron) rotates about the center of mass. Our fixed-nuclei model is a first cut.

 r2

R  rr12 rr12 rr1  aoao aaoo  EVEN ee  ODD ee The even combination enhances the electron density between the nuclei leading to bonding; the odd state leads to a decreased density between the nuclei and is anti-bonding.

The goal is to search for a bound state of the hydrogen molecule; that is: a state with energy lower than that of an isolated hydrogen atom plus a proton. Any trial state that

11/19/2010 SP425 Notes –Variation Ch7-9 we choose will provide an upper bound for this energy so any negative energy result could indicate a bound state. The wavefunction will be represented as either the even or odd combination of 1S hydrogenic wavefunctions centered of each of the protons.

rr12 rr12 aoao aaoo  EVEN Ae [ e ]  ODD Ae [ e ] The yet-to-be-normalized wavefunctions are plotted in blue and the probability densities in a contrasting color. Clearly, a state like the even one is required to enhance the probability of the electron being between the nuclei. In such a position, the Coulomb attraction of the electron for the nuclei might overwhelm the Coulomb repulsion of the nuclei. The coordinates are chosen with the origin at the location of  the first proton such that rr1  . The second proton lies a distance R away on the polar

2 2 ½ (positive z) axis. With this choice, r2 = [ R + r – 2 r R cos] . Here, R is in fact the parameter to be varied to find the minimum energy. The electron hamiltonian is: 

222 ˆ  2 ee Hel     244mr12r The energy of the full system includes e2 , the Coulomb energy of the proton pair. 4oR

2222 ˆˆˆ  2 eee HHfull el H pp   24mr124  r 4  R

Note that the protons are treated as classical fixed charges so there is no proton kinetic energy. (The mass of the proton is very large compared to that of rte electron so the proton kinetic energy term is small in any case.) The p-p Coulomb interaction energy is just added to the electron energy.

Normalize the trial functions:1

1 The geometry and change of variable required to solve this problem were treated in the E&M Integration section of the Math Methods notes. Review that handout for additional details and insights.

11/19/2010 SP425 Notes –Variation Ch7-10 2 2  dA  e2/ra12oo e 2 ra /2s e  ( rr 12 )/ a o rdrddAII2 in2 EVEN EVEN   1 1  1 2

2 2  dA  e2/ra12oo e 2 r / a2s e  ( r 1 r2 )/ a o rdrddAII2 in2 ODD ODD    11  1 2 1 A  ; the normalization constant can be assumed real 2I12 I

a  I e2/ra1 o r23 drsin d d 4  (o ) eu u2 du a3 1112 0 o

I  erdrd()/rr12 ao 2 sin d 211

The integrals of the first two terms are equal. The I1 integral is just the standard normalization integral for the hydrogenic 1S state.

(/Ra )22 (/ ra ) 2cos( Rra / 2 ) I  eera/ o oo ordrdd2 sin  2 

First, define some dimensionless variables: s = (r/ao) and S = (R/ao).

22 I  aee32s Ss2cos sS  sdssin dd 2 o 

  22 I  2saese32s Ss2cos sS  indds 2 o 00  The  integration is trivial. Following our practice, the argument of a function in the integrand is chosen as the new variable as we attack the  integral. sSsin d uSssS222cos; du  ; udus Sdsin  Ss222cos sS 

Ss 3 Ss 32suudu 2 ao su   Iaese2 2 o  dseseudu  ds 00SssS S  Ss  The limits on integration are changed when the integration variable is changed. The absolute value is needed as the square root represents a distance.

SsSs Ss Ss euduuu eu edu  uuu eue Ss  Ss Ss   Ss Ss eueuu   eSs(1)( S  s   e () Ss S  s 1)   Ss

Inserting this result into the expression for I2, and breaking the s range into s < S and s > S,

11/19/2010 SP425 Notes –Variation Ch7-11 3  S  2 ao (2)Ss 2 SS2 2s2 I2 e[1] S  s s ds e[1] S  s s ds e [1]  S s s ds S  00s  3 3 22aa332 32 Ieeo SSSSSSS22   e  S 3 o e  S2 SS 66S 2 SS 4234 12 32S 1 I2  aeo 1 S3 S

1 2 1  S 1 3  |A| = 3 22eSS 1 3  [Var.10]  ao   1 2 Ra/ 3 1  o RR1  A| = 3 22e  1aaoo3 ( )   ao  

Apply a sanity check. The normalized form of a 1S state is 1 ra/ o . In the limit that S 3 e  ao 2 ra12//o r ao 3 is large, the cross terms should vanish so ee  d yields  ao from the 3 region around the first proton and  ao from the region around the second proton. Hence in 2 1 the large S limit, |A|  3 . The large S limit works. The result might be valid. 2 ao  Now that the states are normalized, we compute the electron energy. Dirac notation is   adopted as a short hand.  ()rH ˆˆ () rd H rr 222rr 2 12 ee12 ˆ aaoo 2 aaoo HAeeel  ee 244mrr12  rr 222rr 2 12 12 aaoo 2 eeaaoo Ae  e e e 244mrr12  rr r r 2 12  1  2 aaoo a o a o Ae  eEeEe11SS  

rr 22r r 2 121 2 aaooeeao ao Ae  e e e 44rr  21 22e The simplification followed as e-r/ao is an eigenfunction of 2 with the 1S 24mr  energy.

11/19/2010 SP425 Notes –Variation Ch7-12 rrrr 2 1212 ˆ aaaaoooo HEAeeeeel 1S   2 e 2 ra//ra//r a ra ra//ra//r a r a Ae12oo11ee21o e oooe12oo11ee12 e  rr12rr21 4 

The integral in the first set of braces is one as it is the normalization condition. E1S =  e2 = - 13.6 eV. As the two proton positions are equivalent, the first pair of integrals 8 ooa in the second brace set is equal as is the second pair.

2 ra//aara//ra ra 2 HEEAeˆ 2211ooooee21oo 2 e EEA 4II el 11SS  rr12 11SS 34

r2 I  ae()/rr12 ao rdrsin dd I  ae2/ra1 o 1 drsin dd 311o  41o  r2

While tedious, the integrals I3 and I4 can be evaluated using the same methods as we employed to compute the normalization integrals. Their evaluation is left for homework. Following Richard Fitzpatrick, the p-p interaction energy is added and the terms combined to yield E(S) where S = R/ao, the inter-nuclear separation in units of the Bohr radius. .

 2(1)Se22SS (12 S ) e  ES( ) ( 13.6eV ) 1  3  1 2 S  SSSe1(13 )

2(1)Se22SS (12 S ) e E (SeV ) (13.6 ) 3 13.6eVES ( ) binding 1 2 S SSSe1(13 ) As the energy of a hydrogen atom separated by a large distance from a proton would be -13.6 eV, a energy 13.6 eV is added to the total energy before plotting. The energy is plotted in units of 13.6 eV and the separation in units of ao.

11/19/2010 SP425 Notes –Variation Ch7-13 Our model gives Emim =-1.76 eV for R = 2.4928 ao. Fitzpatrick reports the actual bond distance as about 2.5 ao and the energy as -2.8 eV as compared to a hydrogen atom plus a distant proton. Our relatively poor resul t is expected as the trial function has only one adjustable parameter, and the method of variation only finds an upper bound to the ground state energy.

Appendix Computing kinetic energy - a Hermitian approach: The Hermitian conjugate of an operator Qˆ is represented by the symbol Qˆ † and it is defined such that:

 Qqtqtdˆˆ† (,) (,)  (,) qtQqtd(,)  ii ii for any two functions (qi,t) and (qi,t) that are permissible wavefunctions for the problem.

For a 1D problem, the kinetic energy can be computed as:

 11 (,)x tppxtdxˆˆ (,)  pˆ† (,) xt p ˆ (,) xtdx 22mm       1 pxtpxtdxˆˆ(,)  (,) 2 (,)xt  (,) xt dx 22mm xx

  Td 2  (,)xt  (,) xt x [Var.11] 2m   xx  The Hermitian property of the momentum operator was used above. The form of the equation [Var.11] shows that T  is positive.

11/19/2010 SP425 Notes –Variation Ch7-14

Integrals for some variation technique problems: [Var.12]

 sech[wx ] 2 dx  2  w

 tanh[wx ] sech[ wx ] 2 dx  2  ()3w

 2 xwxsech[ dx   2  ()6w3

 2 xwx2 sech[ dx  7 4   ()120w5

 sech[]tanh[]sech[]wx wx wx wx2 dx 1  ()18w

2 In[1]:= Integrate Sech x , x Out[1]= Tanh x     In[3]:= Integrate Sech x D D Sech x , x , x , x   Tanh x 2 2 Out[3]=   Sech x Tanh x 3 3           2 2 In[5]:= Integrate  x Sech x , x, Infinity, Infinity     2 Out[5]= 6      2 In[6]:= Integrate Csch x , x Out[6]= Coth x     In[7]:= Integrate Csch x D D Csch x , x , x , x   Coth x 2 2 Out[7]=  Coth x Csch x 3 3           2 2 In[8]:= Integrate  x Csch x , x, Infinity, Infinity     2 Out[8]= 3      Problems 1.) Derive a result analogous to [Var.11] that is valid for the kinetic energy of a particle in three dimensions.

11/19/2010 SP425 Notes –Variation Ch7-15  2.) Show that TK2 (,)xt  2 (,)xtdx yields [Var.11] after a 2m  x2 simple integration by parts. Show that the 3D result follows from:    []2 ()(    ) and one of the standard integral theorems of vector calculus.

   Td2 ()()    x 2m 

3.) Choose A sech[w x] as a trial function for the variation approach to determining the ground state energy of a quantum harmonic oscillator. The parameter w is a positive real value. Consider using [Var.11]; use the tabulated integrals [Var.12].

4.) Show that B x sech[u x] is orthogonal to A sech[w x]. Treat u as a positive real adjustable parameter that is independent of w in order to estimate the energy of the first excited state of the oscillator. Why do we believe that the state B x sech[u x] is orthogonal to the true ground state of the oscillator as well as to the trial function for the previous problem?

5. You have a complete set n of energy eigenfunctions with energy eigenvalues En:

H |nEn|nn|m = mn. | =  ck |k.

2 2  |= ||ck = 1 and  |H| ||cEkk k k Assume that all the states are non-degenerate. You can approach the ground state by varying the parameters in your solution to minimize the expectation value of the energy. You can find the first excited state by seeking the lowest energy that is orthogonal to the ground state. You must keep the wavefunction normalized at all times. Some suggest that this method does a better job of predicting the ground state energy than it does of predicting the ground state wavefunction. Can you think of any reason why this might be true?

11/19/2010 SP425 Notes –Variation Ch7-16

2 e ? p o s

i t i v r b e e

a l t e d m

o i a

s s

s u i a f i f

t v a l o u e

t i t

s

h e

W h

a 1 . 0 . =

b e

u p p o s S

. E

1 2 1 b | | +

E a | |

e r i g y

s t f

e h e n

v a l o u e

t a t i o n

e e x p

e c h i t c h

h f o w r

b

+

a

=

s e t

a t e

s i d e t r

h o n C

2 2 

22e 6.) For the hydrogen atom, the hamiltonian is: 2 . 24mr 

r2 Assume that that (r) = Ae a2 where a is the parameter to be varied. Normalize the wavefunction. Compute the expectation value of the hamiltonian in this state. Find the value of a that minimizes this energy. Compute the numerical value of a. Compare the a and the minimized energy with the Bohr radius and the ground state energy of the hydrogen atom.

2 2211      1  Recall: r sin  22     222  rr  r  rsin     r sin  

22½½3 2 333 2 2 e 4o First cut answers: TV2½;; amin 3 2 3 ao 2ma 4o a 22 me 22

e2 44e2 8 Emin     27.2eV (  11.5eV 4 oaO 334ooa 3  If you start with a poor model, you usually get poor results!

2  7.) For a QHO, the hamiltonian is: ½kx2 . See equation: [Var.12] 2mx 2 a.) Assume that that (x) = A e- ax2 where a is a constant. Normalize the wavefunction. Compute the expectation value of the hamiltonian in this state. Find the value of a that minimizes this energy. What is that energy? Compare it with ½  c. b.) Assume that that (x) = A sech[cx] where c is a constant. Normalize the wave- function. Compute the expectation value of the hamiltonian in this state. Find the value of c that minimizes this energy. What is that energy? Compare it with ½  c. c.) Repeat with B csch[cx]. This state is orthogonal to A sech[cx]. How can you conclude that the states are orthogonal without computing an integral?

11/19/2010 SP425 Notes –Variation Ch7-17 2 In[1]:= Integrate Sech x , x Out[1]= Tanh x     In[3]:= Integrate Sech x D D Sech x , x , x , x   Tanh x 2 2 Out[3]=   Sech x Tanh x 3 3           2 2 In[5]:= Integrate  x Sech x , x, Infinity, Infinity     2 Out[5]= 6      2 In[6]:= Integrate Csch x , x Out[6]= Coth x     In[7]:= Integrate Csch x D D Csch x , x , x , x   Coth x 2 2 Out[7]=  Coth x Csch x 3 3           2 2 In[8]:= Integrate  x Csch x , x, Infinity, Infinity     2 Out[8]= 3      2  8.) For a QHO, the hamiltonian is: ½kx2 . See equation: [Var.12] 2mx 2 - ax2 a.) Assume that that (x) = A e where a is a parameter. Normalize the wavefunction. Compute the expectation value of the hamiltonian in this state. Find the value of a that minimizes this energy. What is that energy? Compare it with ½  c.

- bx2 b.) Assume that that (x) = B x e where b is a parameter. Normalize the wavefunction. Compute the expectation value of the hamiltonian in this state. Find the

3 value of b that minimizes this energy. What is that energy? Compare it with /2  c. c.) Show that | 2 - x2 d.) Assume that that (x) = C (1 +b x +c x ) e where b, c and  are parameters. Conduct a variation to find the third level of the oscillator. i.) Find the constraints that | = 0 and | = 0 impose. Why are these conditions needed if one wants to use variation to find the third level of the oscillator? ii.) Normalize the constrained form of the trial wavefunction.

11/19/2010 SP425 Notes –Variation Ch7-18 iii.) How many free parameters remain after the constraints are applied? Vary the constrained set of parameters to minimize the energy. e.) It is possible that the variations of part a did not produced the exact ground state. In this case, do the second and third variation attempts necessarily generate energy level estimates that are greater than or equal to the true values?

f.) Based on the general character of energy eigenfunctions and the form of the potential energy, the claim is that the procedure in part b yields an energy that is greater than or equal to the true energy of the second level. Explain why this must be true.

9.) 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 excited state. Estimate the n = 2 energy level for the hydrogen atom problem. Assume the trial wavefunction family: - r 2(r) = B (1 – br) e . Normalize the 2 to find B(,b). Find b() to ensure that 2|1

= 0. Using this b, vary  to minimize the energy of 2.

10.) Evaluate the integrals I3 and I4 that appear in the computation of the energy of the hydrogen mol ecular ion.

11.) It is a good practice to transform integrals to dimensionless forms before they are evaluated. For the hydrogen molecular ion problem, the normalization integrals are rendered dimensionless by defining the lengths in units of the Bohr radius. The process is straightforward because the normalization procedure yields a pure number, one. The integrals needed to evaluate the expectation value of the energy require that a collection of constants with the dimensions of energy be extracted. We might choose e2 E1S =  = - 13.6 eV. Chemists use atomic units in which lengths are measured in 8ooa units of the Bohr radius, but energies are measured in units of the Hartree = e2 = 4ooa

11/19/2010 SP425 Notes –Variation Ch7-19 27.2 eV. Find the dimensionless forms of I3 and I4 that are appropriate when the energy is measured in units of E1S.

12.) The approximate S for maximum bonding in the model is about 2.5. Use the value and compute the probability to find the electron in the region of space ½  <  <  in both EVEN (+) and ODD. Use the results to find the probabilities that the electron is between the protons in each of the two states. The electron is to be considered as between the protons if might be measured to be in the region 0 < z < S = 2.5 ao. Include a figure as part of your solution. 13.) Mega-Problem: Make an estimate of the ground state energy of the H-, the 1 minus hydrogen atom ion (one proton with two electrons). Make your estimate by applying in sequence all the methods applied to the calculation of the ground state energy of helium. Compute the expectation value of r for one of the elections in your final approximate ground state.

Wild guesses: EGS < -0.478 eV and = 16 ao. confidence: Low Additions: 14.) Estimate the energy of the 1s 2s first excited state of helium. Use a Z = 2 hydrogenic wavefunction for the 1s electron and a variable parameter for the effective nuclear charge in the case of the 2s electron. Include the kinetic, electron-nuclear Coulomb and electron-electron Coulomb terms in the energy. The variation 2S state is assumed to be orthogonal to the 1s state (with Z = 2).

 Ae2/raoo and B[1 (r )] e  ra/ 12ssao a.) Find the normalized versions of 1s and 2s. (2 + ) b.) Show that  = /3 is required in order that 1s|2sThat is, the value of  is set once  is known. The only free parameter in the wavefunction is .

11/19/2010 SP425 Notes –Variation Ch7-20 c.) Vary  to find the lowest overall energy state for this family of wavefunctions. What is the energy? Compare it to the 32S and 12S states. Major Point and Warning: This exchange behavior impacts energies if the particles interact via a separation dependent potential. BUT it is wrong to believe that the only energy contribution arises through the direct potential interaction channel. The wavefunctions may be required to bend more or less in the case of odd under spatial exchange which could add a kinetic energy contribution to the net energy shift. Use the full hamiltonian when you evaluate energies.

15.) Estimate the energy splitting between the 32S and 12S metastable states of helium. Use a Z = 2 hydrogenic wavefunction for the 1s electron and Z =1 for the 2s electron. Evaluate the exchange term for the singlet and triplet cases. DO NOT assume that the only contribution is from the e-e Coulomb interaction. Use the full hamiltonian. Integrate[u^2 (2 - 2 u /3) Exp[-u] (Integrate[v2 (2 - 2 v/3) Exp[-v], {v,0,u}]/u 1 + Integrate[v (2 - 2 v/3) Exp[-v], {v,u,Infinity}]),{u,0,Infinity}] = /3

Implies Vint = 1.2 eV for Z = 2 2S state

Integrate[u^2 (1 - 5 u /6) Exp[-u] (Integrate[v2 (1 - 5 v/6) Exp[-v], {v,0,u}]/u 85 + Integrate[v (1 - 5 v/6) Exp[-v], {v,u,Infinity}]),{u,0,Infinity}] = /48.

Implies Vint = 0.9 eV for Z = 1 2S state

16.) Estimate the energy splitting between the 32S and 12S metastable states of helium. Use a Z = 2 hydrogenic wavefunction for the 1s electron and the wavefunction found in 14 for the 2s electron. Evaluate the exchange term for the singlet and triplet cases. Repeat using the effective charge found in the previous problem. DO NOT assume that the only contribution is from the e-e Coulomb interaction. Use the full hamiltonian.

11/19/2010 SP425 Notes –Variation Ch7-21 17.) Use the Hylleraas trial function to estimate the ground state energy of a neutral

  ra12//oo r a helium atom. Hylleraas proposed (,rr12 ) Ae e(1 br 12 ). The function is first normalized to find A(,b). The energy is then evaluated as a function of  and b. That energy is then minimized with respect to variations of the parameters  and b.

0.364 The best values found are:  = 1.849 and b = /ao. The ground state energy for those values is – 78.7 eV.

18.) The H- ion (called the hydride ion) is a bound state on two electrons around a proton. It has recently been adopted in popular culture as a health supplement with powers rivaling those of magnetic bracelets. Hydride exist in liquids and in interstitials in materials. Examine the case of the isolated hydride ion using a trial function of the form proposed by Hylleraas for the ground state of helium. If you find a negative energy, then the hydride ion can exist, and it would have a ground state energy less than that predicted by the ‘Hylleraas’ model. The binding energy of a hydride ion is estimated to be > 0.7 eV. The ion may have a weakly bound excited state. (G. F. W. Drake, Phys. Rev. Lett. 24, 126 (1970).)

19.) Show that small changes in the parameterization do not change the results.

2 22 Suppose that one chose  ()x  Aebx while another chose ()xAe  x /a . Basically,

-2 the exponential length scale is being set in either case. In particular b = b(a) = a . dd Show that if dbEE0then da 0 and hence the same final results will be found.

References: 1. David J. Griffiths, Introduction to Quantum Mechanics, 2nd Edition, Pearson Prentice Hall (2005). 2. Richard Fitzpatrick, Quantum Mechanics Note Set, University of Texas. 2. Ira N. Levine, Quantum Chemistry, 5th Edition, Pearson Prentice Hall (2000).

11/19/2010 SP425 Notes –Variation Ch7-22 Problem Comments: This comment is preliminary. The plots should be reworked to insure that the correct model values were used. Add normalizations at all levels.

CG7.7 Variation using sech[cx] as the trial function for a quantum harmonic oscillator. Plot[{Sqrt[Sqrt[1/Pi]]Exp[-x^2/2], Sech[x]/Sqrt[2], x^2/2},{x,-3,3},PlotRange  {0,1}]

a = 0.155; b = 0.085;Plot[{Sqrt[Sqrt[1/Pi]]Exp[-x^2/2], Sech[x]/Sqrt[2], x^2/2, Sqrt[Sqrt[1/Pi]]Exp[-x^2/2]+(0.155) ((4 x2 - 2)Sqrt[Sqrt[1/Pi]/8]Exp[-x^2/2]) +(0.085) ((12-48 x2+16 x4)Sqrt[Sqrt[1/Pi]/384]Exp[-x^2/2]/Sqrt[1 + a^2+b^2])},{x,- 3,3},PlotRange  {0,1}]

The minimized energy for the sech trial function is 0.5236 . On the left, the sech and the exact ground state are plotted. The energy is correct to within 5%, but the wave variation derived function is not a great approximation to the true ground state wavefunction.

In the second plot, o + 0.155 2 + 0.085 2 is normalized and plotted. That normalized state has an energy of 0.522 .

11/19/2010 SP425 Notes –Variation Ch7-23