Joel Broida
UCSD Fall 2009
Phys 130B
QM II
Homework Set 4
1. Consider the particle-in-a-box problem but with a delta function potential
H ′ (x) = αδ(x − l/2) at the center (with α = const):
∞
∞
H ′ = αδ(x − l/2)
x
0
l/2
l
(1)
(a) Find the first-order correction En to the allowed energies. Why doesn’t
the perturbation affect the states with n even?
(b) Find the first three nonzero terms in the first-order correction to the wave
function (equation (2.10) in the class notes) for the ground state of this
system.
(2)
(c) Find the second-order corrections En , and sum the series to obtain a
closed expression. Hint: Expand the summand in terms of partial fractions. Then look at the cases n = 1, 3 and maybe 5 to try and see the
pattern.
2. The ground-state hydrogen atom wave function is
1 −3/2
ψ100 = √ a0 e−r/a0
π
where using a0 = ~2 /me e2 = 0.529Å is taking the proton to be infinitely
heavy. Assume that the charge of the proton is distributed uniformly throughout the volume of a sphere of radius R = 10−13 cm. Use first-order perturbation theory to find the shift in the ground-state hydrogen-atom energy (in
eV) due to the finite proton size. Hints: (1) Be careful finding the potential
energy of the electron when it has penetrated the nucleus (i.e., the proton).
At a distance r from the nuclear center, let the charge enclosed be Q(r). Find
an expression for this in terms of r and R. Now use the definition of potential
energy F = −∇V along with elementary electrostatics to find V (r) for r ≤ R.
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(2) Justify assuming that the exponential factor in ψ is essentially equal to 1
within the nucleus.
3. In this problem you will work out the perturbation treatment of the (nondegenerate) helium atom ground-state energy. The He atom consists of two
electrons and a nucleus of charge +2e. If we treat the nucleus as infinitely
heavy, we let the origin of our coordinate system be at the nucleus (treated
as a point particle). We may as well let the nucleus have charge +Ze instead
of +2e, because then our treatment will also include heliumlike ions such as
H− , Li+ and Be2+ . The Hamiltonian is
H=−
~2 2
~2 2 Ze2
Ze2
e2
∇1 −
∇2 −
−
+
2me
2me
r1
r2
r12
where the subscripts 1 and 2 on ∇ refer to the coordinates of electrons 1 and
2, r1 and r2 are the distances of the electrons from the nucleus, and r12 is the
distance from electron 1 to electron 2:
r12 = [(x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 ]1/2 .
By using the usual relation between Cartesian and spherical coordinates, it is
possible to write r12 in terms of r1 , θ1 , φ1 , r2 , θ2 , φ2 . It should be clear that
this Hamiltonian represents the separate Hamiltonians of individual electrons
in a hydrogenlike atom, plus an electron-electron interaction potential energy.
The Schrödinger equation is a function of six variables, three for each electron.
But because of the 1/r12 term, the equation is not separable in any coordinate
system, and we must turn to approximation techniques for its solution. To
use perturbation theory, we must write H in the form H = H 0 + H ′ where
H 0 is the Hamiltonian of an exactly soluble problem. Let us write
H0 = −
H′ =
~2 2 Ze2
~2 2 Ze2
∇1 −
−
∇ −
2me
r1
2me 2
r2
e2
.
r12
Then H 0 = H10 +H20 is the sum of two independent hydrogenlike Hamiltonians,
one for each electron. As such, the solution is the product of two unperturbed
wave functions:
(0)
(0)
ψ (0) (r1 , θ1 , φ1 , r2 , θ2 , φ2 ) = ψ1 (r1 , θ1 , φ1 ) ψ2 (r2 , θ2 , φ2 )
with the unperturbed energies
(0)
(0)
E (0) = E1 + E2
defined by
(0)
(0)
(0)
(0)
H1 ψ1 = E1 ψ1
and
2
(0)
(0)
(0)
(0)
H2 ψ2 = E2 ψ2 .
(0)
The energies Ei
are just the usual hydrogenlike-atom energies given by
(0)
Ei
=−
Z 2 e2
;
n2i 2a0
ni = 1, 2, . . .
and hence the zeroth-order energies of our system are given by
2
1
e
1
+
E (0) = −Z 2
n21
n22 2a0
where n1 = 1, 2, 3, . . . and n2 = 1, 2, 3, . . . . The lowest level is non-degenerate
with n1 = n2 = 1, and the zeroth-order wave function is
3/2
3/2
Z
Z
1
1
ψ (0) = 1/2
e−Zr1 /a0 · 1/2
e−Zr2 /a0
a0
a0
π
π
which is simply the product of two independent ground-state wave functions.
Note that what we have is usually written as the 1s(1)1s(2) state, where the
the number in parenthesis denotes which electron is in which state. In this
case, both electrons are in the 1s state (or orbital).
For our unperturbed system, the ground state (i.e., n1 = n2 = 1) has energy
E (0) = −2Z 2
e2
2a0
where −e2 /2a0 = −13.606 eV is the ground-state energy of the hydrogen
atom. The definition of a0 is ~2 /me e2 , but if we replace me by the reduced mass for 4 He (i.e., we no longer regard the nucleus as infinitely heavy),
then −e2 /2a0 becomes −13.604 eV, a somewhat more correct number. Using
this together with Z = 2 for helium, we take the unperturbed (zeroth-order)
ground-state energy to be
E (0) = −8(13.604) = −108.83 eV .
The experimental first ionization energy of He is 24.59 eV, and the second
ionization energy is easily calculated since it is the ionization energy of the
hydrogenlike ion He+ , and is equal to 22 (13.604) = 54.42 eV. Thus the groundstate energy of the helium atom is −(24.59 + 54.42) = −79.01 eV. So we see
that our zeroth-order energy is off by 38%. This shouldn’t be surprising since
we have neglected the term e2 /r12 which is not an insignificant energy.
Now for the perturbation correction.
Evaluate the first-order energy correction E (1) = hψ (0) |H ′ ψ (0) i, and find the
percent error in the ground-state energy. Hints: The integral you must evaluate is over the six variables r1 , θ1 , φ1 , r2 , θ2 , φ2 . Use the fact that 1/r12
3
can be expanded in terms of the spherical harmonics as (see e.g., Jackson’s
Classical Electrodynamics, 3rd ed. Section 3.6)
∞ X
l
l
X
1
4π r<
[Ylm (θ1 , φ1 )]∗ Ylm (θ2 , φ2 )
=
l+1
r12
2l + 1 r>
l=0 m=−l
where r< means the smaller of r1 and r2 , and r> is the larger of these. Use this
expression in your integral, and then multiply and divide by Y00 (Y00 )∗ = 1/4π.
Now use the orthonormality of the spherical harmonics. Next, first integrate
r1 where over the range 0 ≤ r1 ≤ r2 you have r> = r2 , and over the range
r2 ≤ r1 ≤ ∞ you have r> = r1 . This will give you some definite integrals that
you can explicitly do to arrive at a result that depends on r2 . Lastly, you will
have integrals over r2 that you can also explicitly perform. The final result is
a very simple, clean expression in terms of Z, e and a0 .
4. Suppose we want to know how the energy of a system depends on some parameter λ. For example, the energy levels of the one-dimensional harmonic
oscillator are given by
r 1
k
1
En = ~ω n +
=~
n+
2
m
2
and this depends on both k and m. In other words, we want to know ∂En /∂λ.
(a) From the Schrödinger equation H(λ)ψn (λ) = En (λ)ψn (λ) we have En =
hn|H|ni. Assuming that ψn is normalized, i.e., hn|ni = 1, show
D ∂H E
∂En
= n
n .
∂λ
∂λ
(b) We can use the result of part (a) to derive the first-order perturbation
result as follows. As before, write H = H 0 + λV (where 0 ≤ λ ≤ 1) and
(0)
(1)
En = En + λEn + · · · . Now use the result of part (a) to show that
(1)
En = hn|V |ni in agreement with our perturbation theory treatment.
(c) The one-dimensional harmonic oscillator has the Hamiltonian
H =−
1
~2 d2
+ kx2 .
2m dx2
2
Letting λ = k, find hx2 i for any harmonic oscillator stationary-state without evaluating any integrals.
5. Neglecting spin, the interaction energy of a hydrogen atom in a magnetic field
B = Bz ẑ is given by
eBz
H ′ = −µl · B =
Lz .
2me c
(a) Using first-order perturbation theory, find the effect of H ′ on the n = 2
level of hydrogen. Sketch (i.e., diagram) the effect of H ′ on the n = 2
level.
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(b) Now work out what happens if B = Bx x̂. Compare your result with what
you obtained in part (a). Why do you think it came out this way?
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