Física

Ejercicios Física

Physics 125c Course Notes Approximate Methods Solutions to Problems 040415 F. Porter

1

Exercises
1. Prove the theorem quoted in section ??: Theorem: If we have a normalized function |ψ such that E0 ≤ ψ|H|ψ ≤ E1 , then E0 ≥ ψ|H|ψ − Hψ|Hψ − ψ|H|ψ 2 . E1 − ψ|H|ψ (1) (2)

Solution: The theorem is equivalent to the statement ψ|H 2 ψ − ψ|H|ψ
2

≥ ( ψ|H|ψ − E0 )(E1 −ψ|H|ψ ).

(3)

Notice that if we add a constant A to H, obtaining H = H + A (hence also En → En = En + A), both sides of this inequality are unaltered. The left hand side is a measure of the width of the energy distribution and is not altered by shifting the energy scale. Likewise, the right hand side only depends on energy differences. Thus, as long as the spectrum of H is bounded below, theproblem is equivalent to a problem where the spectrum is non-negative. In particular, we may simplify by taking E0 = 0. Hence, consider: ψ|H 2 ψ − ψ|H|ψ
2

− ψ|H|ψ (E1 − ψ|H|ψ ) = ψ|H 2 ψ − E1 ψ|H|ψ 2 |cn |2 En − E1 |cn |2 En =
n=0 n=0

(4) (5) (6)

=
n=0

|cn | En (En − E1 ) |cn |2 En (En − E1 ),
n=1

2

= ≥ 0,

since E0 = 0,

(7)

since each term in the sum is non-negative. 12. Let us pursue our variational approach to the estimation of ground state energy levels of atoms to the “general” case. We consider an atom with nuclear charge Z, and N electrons. The Hamiltonian of interest is: H(Z, N ) = Hkin − ZVc + Ve
N

(8) (9) (10)

Hkin =
n=1

p2 n 2m

,

where Vc = α Ve = α 1 n=1 |xn | 1 N ≥j>k≥1|xk − xj |
N

(11) (12) (13) (14)

m = electron mass α =fine structure constant.

Denote the ground state energy of H(Z, N ) by −B(Z, N ), with B(Z, 0) = 0. (a) Generalize the variational calculation we performed for the ground state of helium to the general Hamiltonian H(Z, N ). Thus, select your “trial function” to be a product of N identical “hydrogen atom ground state” functions. Determine the resulting lower ˆ bound B(Z, N ) on B(Z, N ) (i.e., anupper limit on the ground state energies). Solution: Let Ze be the fixed nuclear charge in the Hamiltonian, and let z be the effective Z variational parameter. The trial wave function we are told to use is thus:
N

ψZN (x1 , . . . , xN ) =
n=1

z 3 − az (rn ) e 0 , πa3 0

a0 =

1 . mα

(15)

The expectation value of the total kinetic energy for this trial function is: p2 1 Hkin = N ψ| 1|ψ = Nz 2 mα2 . (16) 2m 2 The expectation value of the potential energy of the electrons in the nuclear electric field is: −ZVc = −NZzmα2 . 2 (17)

The expectation value of the potential energy of the electrons in the fields of the other electrons is: N (N − 1) 1 5 z mα2 . 2 24 Putting these terms together, we have Ve = H(Z, N )
z

(18)

1 5 = mα2 Nz z − 2Z + (N − 1) 2 8

(19)

Weminimize with respect to z: 0= d 2 5 5 z − 2Zz + (N − 1)z = 2z − 2Z + (N − 1). dz 8 8 5 (N − 1) 16 (20)

Thus, the minimum occurs at z=Z− (21)

The variational bound on the (negative of the) ground state energies is then: ˆ B(Z, N ) = − H(Z, N )
min 2 1 5 2 = mα N Z − (N − 1) . (22) 2 16

(b) Make a simple table comparing your variational bounds with the observed ground state energies forlithium, beryllium, and nitrogen. Note that a simple web search for “ionization potentials” will get you a multitude of tables of observed values, or you can look at a reference such as the CRC Press’s Handbook of Chemistry and Physics. The table entries are typically of the form: B(Z, N ) − B(Z, N − 1). Solution: A Google search on “ionization potentials” results in many suitable hits, including:http://www.chemistrycoach.com/ionization potentials f.htm The ionization potentials for lithium, beryllium, and nitrogen are reproduced in Table 2b. The predicted bounds on the energies, according to Eqn. 22, are compared with the observed values in Table 2b.

3

Ionization Potentials for Lithium, Beryllium, and Nitrogen (from http://www.chemistrycoach.com/ionization potentials f.htm) 1st...