130_notes.dvi

= −

Ze^2

a 0

[

3

2

2

2

1

2

1

2

+ 2

2

2

1

2

1

2

+ 2

1

2

1

2

− 2

1

1

1

1

]

∆Egs = −

Ze^2

a 0

[

3

8

+

4

8

+

4

8

−

16

8

]

= +

5

8

Ze^2

a 0

=

5

4

Z(13.6 eV) → 34 eV for Z=2

25.8 Homework Problems

1. Calculate the lowest order energy shift for the (0th order degenerate first) excited states of
   Helium ∆E( 2 s,t,l)whereℓ= 0,1. This problem is set up in the discussion of the first excited
   states (See section 25.3). The following formulas will aid you in the computation. First, we
   can expand the formula for the inverse distance between the two electrons as follows.

1

|~r 1 −~r 2 |

=

∑∞

ℓ=0

rℓ<

rℓ>+1

Pℓ(cosθ 12 )

Herer<is the smaller of the two radii andr>is the larger. As in the ground state calculation,

we can use the symmetry of the problem to specify which radius is thelarger. Then we can

use a version of the addition theorem to write the Legendre PolynomialPℓ(cosθ 12 ) in terms

of the spherical hamonics for each electron.

Pℓ(cosθ 12 ) =

4 π

2 ℓ+ 1

∑ℓ

m=−ℓ

(−1)mYℓm(θ 1 ,φ 1 )Yℓ(−m)(θ 2 ,φ 2 )

Using the equationYℓ(−m)= (−1)ℓYℓm∗, this sets us up to do our integrals nicely.

2. Consider the lowest state of ortho-helium. What is the magnetic moment? That is what is
   the interaction with an external magnetic field?


3. A proton and neutron are bound together into a deuteron, thenucleus of an isotope of hydrogen.
   The binding energy is found to be -2.23 MeV for the nuclear ground state, anℓ= 0 state.
   Assuming a potential of the formV(r) =V 0 e

−r/r 0

r/r 0 , withr^0 = 2.8 Fermis, use the variational

principle to estimate the strength of the potential.

4. Use the variational principle with a gaussian trial wave function toprove that a one dimensional
   attractive potential will always have a bound state.


5. Use the variational principle to estimate the ground state energy of the anharmonic oscillator,
   H=p

2

2 m+λx

(^4).

25.9 Sample Test Problems

1. We wish to get a good upper limit on the Helium ground state energy. Use as a trial wave
   function the 1s hydrogen state with the parameter a screened nuclear chargeZ∗to get this
   limit. Determine the value ofZ∗which gives the best limit. The integral〈(1s)^2 | e

2

|~r 1 −~r 2 ||(1s)

(^2) 〉=
5
8 Z
∗α (^2) mc (^2) for a nucleus of chargeZ∗e.