Joel Broida
UCSD Fall 2009
Phys 130B
QM II
Homework Set 3
DO ALL INTEGRALS BY HAND, GIVEN THAT
r
Z ∞
π
−ax2
e
dx =
.
a
−∞
1. A particle in a spherical box of radius a has V = 0 for r ≤ a and V = ∞ for
r > a. Use the trial function ϕ = a − r for r ≤ a and ϕ = 0 for r > a to
estimate the ground-state energy. What is the percent error compared to the
true ground-state energy π 2 ~2 /2ma2 ?
2. Consider a one-dimensional potential well as shown in the figure below.
∞
∞
V =0
V = V0
V =0
x
0
l/4
l
3l/4
(a) Use the trial function
ϕ1 =
r
2
πx
sin
;
l
l
0≤x≤l
to estimate the ground-state energy for V0 = ~2 /ml2 , and compare this
with the true ground-state energy E = 5.750345~2/ml2 .
(b) For the same system, use the variation function
ϕ2 = x(l − x) .
2
3. Use the variation method with the trial function ϕ = e−αx (where α > 0)
to show that any one-dimensional attractive potential always has at least one
bound state. Hint: This problem could also be done with the trial function
e−β|x| . However, in this case the first derivative has a discontinuity, and hence
the second derivative blows up. To get around this problem, it is common to
replace the kinetic energy term in the variation integral
Z ∞
d2
~2
ϕ 2 ϕ dx
−
2m −∞ dx
1
with the term
~2
2m
Z
∞
(ϕ′ )2 dx
−∞
where ϕ′ = dϕ/dx. Even if it is not necessary in a particular problem, it
is frequently easier to use this formulation. Explain why this is a valid substitution, and then you may use it to complete this problem. (See Griffiths,
Example 7.3 for another approach.)
4. Let the variation function
ϕ=
n
X
cj f j
i=1
be complex. Then cj = aj + ibj where aj and bj are real numbers. Thus there
are 2n real parameters to be varied.
(a) What is wrong with just immediately letting ∂W/∂cj = 0 just as we did
in the real case?
(b) Use the chain rule to show that the minimization conditions ∂W/∂aj =
0 and ∂W/∂bj = 0 are equivalent to the conditions ∂W/∂cj = 0 and
∂W/∂c∗j = 0.
(c) Show P
that minimization of W leads to equation (1.5) in the class notes
(i.e.,
j (Hkj − W Skj )cj = 0) and its complex conjugate, the latter of
which may then be discarded. This shows that equations (1.5) and (1.6)
in the notes are also valid for complex variation functions.
5. In this problem you will prove that the approximate wave functions obtained in
the linear variation method are orthogonal, and that the approximate energies
obtained are upper bounds to the energies of the n lowest states. Let the
approximate wave function ϕα have the value Wα for the variational integral
P
(α)
and the coefficients cα in the expansion ϕα = nj=1 cj fj . (The label α is
added to distinguish the n different ϕ’s that result from the n roots Wα of
the secular equation.) Let us rewrite equation (1.5) in the notes as
X
(α)
(hfi |Hfk i − Wα hfi |fk i) ck = 0;
i = 1, . . . , n .
(1)
k
(a) Show that hfi |H − Wα |ϕα i = 0 by showing that this integral is equal to
the left side of (1).
(b) Use the result of part (a) to show that hϕβ |H − Wα |ϕα i = 0 and hϕα |H −
Wβ |ϕβ i∗ = 0 for all α and β.
(c) Equate the two integrals in (b), and use the Hermitian property of H
to show that hϕβ |ϕα i(Wβ − Wα ) = 0. This shows that ϕα and ϕβ are
orthogonal if Wα 6= Wβ . (If Wα = Wβ , we can form orthogonal linear
combinations of ϕα and ϕβ that will have the same value for the variational integral.)
(d) Let ϕ0 , ϕ1 , . . . , ϕn−1 be the normalized approximate wave functions
found in the variation method, where the functions are listed in order of
2
increasing value of the variational integral. Let ψ0 , ψ1 , . . . be the true
wave functions of the lowest-energy states. Consider the function
g=
m
X
bα ϕα
α=0
where m ≤ n − 1 and the coefficients bα are chosen so that
hg|ψ0 i = hg|ψ1 i = · · · = hg|ψm−1 i = 0
(e)
(f)
(g)
(h)
(i)
and also so that g is normalized. (Note that there are m + 1 coefficients
bα , and a total of m + 1 conditions imposed on them.) Explain why
hg|Hgi ≥ Em . Hint: See Section 1.2 in the notes.
Use the results of parts (b) and (c) to show that hϕα |Hϕβ i = 0 for α 6= β.
Pm
2
Use the result of part (e) to show that hg|Hgi = α=0 |bα | Wα .
Pm
2
Use the orthonormality of the ϕ’s to show that α=0 |bα | = 1.
Use the results of (f) and (g) to show that hg|Hgi ≤ Wm .
Combine parts (h) and (d) to give the desired result: Wm ≥ Em for
m = 0, 1, 2, . . . , n − 1.
6. For the ground state of the hydrogen atom, consider the Gaussian trial function
2
2
ϕ = e−cr /a0 .
(a) Find the optimum value of c and the percent error in the energy.
(b) Multiply ϕ by the spherical harmonic Y20 , and then minimize the variational integral. This yields an upper bound to the energy of which
hydrogen-atom state?
7. For the system of Problem 2 with V0 = ~2 /ml2 , consider the linear variation
function
r
r
πx
3πx
2
2
ϕ = c1 f 1 + c2 f 2 = c1
sin
+ c2
sin
l
l
l
l
for 0 ≤ x ≤ l. Let the true energies be E0 ≤ E1 ≤ E2 ≤ · · · .
(a) Explain why this function will give upper bounds to the energies E0 and
E2 .
(b) Explain why f1 and f2 are orthonormal. (And don’t just do the integral.)
(c) Note that hfi |Hfj i = hfi |T fj i + hfi |V fj i. Explain why T fj = εj fj (for
0 ≤ x ≤ l), and give the T eigenvalues ε1 and ε2 . Show that hfi |T fj i =
δij εj .
(d) Using the results of parts (b) and (c) to help evaluate the integrals, set up
and solve the secular equation. Then find the variation functions ϕ0 and
ϕ2 that correspond to W0 and W2 . Compare W0 and W2 with the true
energies E0 = 5.750345~2/ml2 and E2 = 44.808373~2/ml2 . Compare W0
with the value W = 5.753112~2/ml2 found in Problem 2 using ϕ = f1 .
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