Joel Broida
UCSD Fall 2009
Phys 130B
QM II
Homework Set 3 Solutions
1. The Hamiltonian for a spherically symmetric potential V (r) is
2
~2
1
∂
2 ∂
H=−
+
+
L2 + V (r) .
2m ∂r2
r ∂r
2mr2
Since V = 0 inside the box and L2 takes derivatives with respect to angles
only, we see that for r ≤ a we have
Hϕ = H(a − r) =
~2
.
mr
Then (since ϕ = 0 for r > a)
Z
Z a
Z
4π~2 a
a−r 2
~2
r dr =
dΩ
(ar − r2 )dr
hϕ|Hϕi =
m
r
m 0
0
=
2π~2 a3
3m
and also
hϕ|ϕi =
Z
dΩ
Z
a
0
(a − r)2 r2 dr = 4π
a5
2πa5
=
.
30
15
Then the variational integral becomes
W =
5~2
hϕ|Hϕi
.
=
hϕ|ϕi
ma2
This is to be compared to the true value π 2 ~2 /2ma2 . The trial function is
then too large by
5 − π 2 /2
= 0.0132118 = 1.32% .
π 2 /2
2. (a) Let us write Hp= T + V where V = V0 for l/4 ≤ x ≤ 3l/4. For the
function ϕ1 = 2/l sin πx/l we have the kinetic energy term
Z
πx
πx d2
~2 2 l
sin
dx
sin
hϕ1 |T ϕ1 i = −
2
2m l 0
l dx
l
Z
π 2 ~2 l 2 πx
=
dx
sin
ml3 0
l
1
π 2 ~2
=
2ml3
=
Z l
2πx
dx
1 − cos
l
0
π 2 ~2
.
2 ml2
The potential energy term is
2
hϕ|V ϕi = V0
l
=
Z
3l/4
sin2
l/4
πx
dx
l
π + 2 ~2
.
2π ml2
Since
2
hϕ|ϕi =
l
Z
l
sin2
0
πx
dx = 1
l
we have
W =
hϕ|Hϕi
= hϕ|T ϕi + hϕ|V ϕi
hϕ|ϕi
2
2
~
1
1
π
+ +
=
2
2 π ml2
= 5.753112
~2
.
ml2
This is only 0.048% larger than the true ground-state energy.
(b) Now repeat this with the function ϕ2 = x(l − x). We have
Z l
~2
d2
hϕ2 |T ϕ2 i = −
x(l − x) 2 x(l − x) dx
2m 0
dx
Z
~2 l
x(l − x) dx
=
m 0
=
~2 l 3
~2
= 0.166667 l5 2
6m
ml
and
hϕ2 |V ϕ2 i = V0
Z
3l/4
l/4
x2 (l − x)2 dx
= 0.026432 l5
2
~2
ml2
and also
hϕ2 |ϕ2 i =
Z
l
0
x2 (l − x)2 dx =
l5 5
l .
30
Therefore
W =
0.166667 + 0.026432 ~2
hϕ2 |Hϕ2 i
=
hϕ2 |ϕ2 i
1/30
ml2
= 5.792969
~2
ml2
which is 0.741% too large.
2
3. We are using the trial function ϕ = e−αx (with α > 0) which goes to zero
as |x| goes to infinity. Therefore, integrating by parts we have the alternative
formulation of the kinetic energy integral hϕ|T ϕi:
Z ∞
Z ∞
d2
d
ϕ 2 ϕ dx =
ϕ ϕ′ dx
dx
dx
−∞
−∞
Z ∞
∞
= ϕϕ′ −
(ϕ′ )2 dx
−∞
=−
Z
−∞
∞
(ϕ′ )2 dx .
−∞
Now we want to evaluate the variational integral
W =
hϕ|Hϕi
hϕ|ϕi
where the potential is attractive, i.e., where H = T − V (x) and V (x) ≥ 0 for
all x. First we have
r
Z ∞
π
−2αx2
e
dx =
hϕ|ϕi =
.
2α
−∞
Next we have
Z ∞
2
d2 −αx2
e
dx
−
e−2αx V (x) dx
2
dx
−∞
−∞
Z
Z
2
∞
∞
2
d −αx2
~2
dx −
e−2αx V (x) dx
e
=
2m −∞ dx
−∞
Z
2~2 α2 ∞ 2 −2αx2
=
x e
dx − I(α)
m
−∞
hϕ|Hϕi = −
~2
2m
Z
∞
e−αx
2
3
where we have defined
I(α) =
Z
∞
2
e−2αx V (x) dx .
−∞
Now
Z
∞
2 −2αx2
x e
−∞
1 ∂
dx = −
2 ∂α
Z
∞
2
e−2αx dx
−∞
r
1 ∂
π
2 ∂α 2α
r
1 π −3/2
=
α
.
4 2
=−
Therefore
~2
2m
r
πα
− I(α)
2
~2 α
W =
−
2m
r
2α
I(α) .
π
hϕ|Hϕi =
and hence
In order that there be a bound state we must have W < 0 or
r
~2 α
2α
<
I(α) .
2m
π
The question is, can we always find an α > 0 such that this relation holds?
Let us define the functions
~2 α
F (α) =
2m
and
r
r
Z
2α
2α ∞ −2αx2
I(α) =
G(α) =
e
V (x) dx .
π
π −∞
Taking the derivatives we have
dF
~2
=
dα
2m
and
dG
=
dα
r
1
2πα
Z
∞
−∞
e
−2αx2
V (x) dx − 2
r
2α
π
Z
∞
2
e−2αx V (x) dx .
−∞
As α → 0, we see that dF/dα remains a positive constant, while the first
integral in dG/dα becomes infinite and the second goes to zero. Graphically,
we have the following general situation:
4
111111
000000
000000
111111
000000
111111
000000
111111
000000
111111
000000
111111
G
F
α
Thus we see that there will always be a region where F < G, and hence there
will always be at least one bound state.
4. (a) In the definition of W we have terms such as
X
hϕ|ϕi =
c∗j ck hfj |fk i .
j,k
But then how do you treat the derivative of this with respect to cl ? What
do you with the terms ∂c∗j /∂cl ? You have to justify setting ∂c∗j /∂cl = 0,
and the way to do this is to show that cj and c∗j may be treated as
independent variables.
(b) From cj = aj + ibj where aj and bj are real, we have
aj =
1
(cj + c∗j )
2
and
bj =
1
(cj − c∗j ) .
2i
Then W (a, b) = W (a(c, c∗ ), b(c, c∗ )) so that
X ∂W ∂ak X ∂W ∂bk
∂W
=
+
=0
∂cj
∂ak ∂cj
∂bk ∂cj
k
k
where we used the conditions ∂W/∂ak = 0 = ∂W/∂bk . Exactly the same
argument shows that
X ∂W ∂ak X ∂W ∂bk
∂W
=
+
= 0.
∗
∂cj
∂ak ∂c∗j
∂bk ∂c∗j
k
k
Conversely, an analogous argument shows that with W = W (c, c∗ ) =
W (c(a, b), c∗ (a, b)) where ∂W/∂ck = 0 = ∂W/∂c∗k , we must also have
∂W/∂aj = 0 = ∂W/∂bj .
P
(c) Now we are able to write ϕ = j cj fj where cj and c∗j are treated as
independent variables. (We have replaced the a’s and b’s by the c’s and
c∗ ’s.) Then
P
∗
j,k cj ck Hjk
W = P
∗
j,k cj ck Sjk
or, equivalently,
W
X
c∗j ck Sjk =
j,k
X
j,k
5
c∗j ck Hjk .
Setting ∂W/∂cl = 0 and letting ∂c∗j /∂cl = 0 we have
W
X
c∗j
j,k
X ∂ck
∂ck
Sjk =
Hjk
c∗j
∂cl
∂cl
j,k
or, since ∂ck /∂cl = δkl ,
X
W
c∗j Sjl =
X
j
c∗j Hjl
j
j
and hence
X
(Hjl − W Sjl ) c∗j = 0 .
(1)
Similarly, setting ∂W/∂c∗l = 0 we have
W
X ∂c∗j
∗ ck Sjk =
j,k
or
∂cl
X
k
X ∂c∗j
j,k
∂c∗l
ck Hjk
(Hlk − W Slk ) ck = 0 .
(2)
Now note that
∗
Hjl
= hfj |Hfl i∗ = hHfl |fj i = hfl |Hfj i = Hlj
and also
∗
Sjl
= hfj |fl i∗ = hfl |fj i = Slj .
Since the summation index j in equation (1) can be replaced by k, we
easily see that taking the complex conjugate of (1) yields (2).
5. We start from
X
k
(α)
(hfi |Hfk i − Wα hfi |fk i) ck
(a) Using ϕα =
(α)
k ck f k
P
= 0;
i = 1, . . . , n .
we have
hfi |H − Wα |ϕα i = hfi |H − Wα |
X
k
(α)
ck f k i
(α)
=
X
hfi |H − Wα |fk ick
=
X
(hfi |Hfk i − Wα hfi |fk i) ck
k
k
= 0.
6
(α)
(3)
(b) Simply evaluate, using part (a):
X (β)∗
hϕβ |H − Wα |ϕα i =
ci hfi |H − Wα |ϕα i = 0 .
i,k
Since this applies to all α and β, taking the complex conjugate and interchanging α and β shows that we also have
hϕα |H − Wβ |ϕβ i∗ = 0 .
(c) First note that the second result in (b) can be written
hϕα |H − Wβ |ϕβ i∗ = h(H − Wβ )ϕβ |ϕα i = hϕβ |H − Wβ |ϕα i = 0
where we used the fact that H is Hermitian and Wβ is real. Equating this
to the first result in (b) we then have
hϕβ |H − Wα |ϕα i = hϕβ |H − Wβ |ϕα i
or
Pm
(Wα − Wβ )hϕβ |ϕα i = 0 .
(d) The function g = α=0 bα ϕα obeys the boundary conditions of the problem because the ϕα ’s do (by assumption). Hence it can also be expanded
in terms of the complete orthonormal set {ψi } as
g=
∞
X
ci ψi .
i=0
where Hψi = Ei ψi and hψi |ψj i = δij . But the coefficients bα were chosen
so that
hg|ψ0 i = hg|ψ1 i = · · · = hg|ψm−1 i = 0
and since ci = hψi |gi, we have
g=
∞
X
ci ψi .
i=m
Furthermore, the bα were also chosen so that g was normalized, and this
then requires that
hg|gi =
∞
X
i,j=m
c∗i cj hψi |ψj i =
∞
X
i=m
2
|ci | = 1 .
Therefore we have
∞
∞
∞
X
X
X
2
|ci | Ei
c∗i cj Ej hψi |ψj i =
c∗i cj hψi |Hψj i =
hg|Hgi =
i,j=m
i,j=m
≥
∞
X
i=m
|ci |2 Em = Em
where we used the fact that Em ≤ Em+1 ≤ · · · .
7
i=m
(e) From parts (b) and (c) we immediately have, for α 6= β,
hϕα |Hϕβ i = Wβ hϕα |ϕβ i = 0
(f) Using part (e) and the original definition of g we have
hg|Hgi =
=
m
X
α,β=0
m
X
α=0
b∗α bβ hϕα |Hϕβ i =
m
X
b∗α bβ Wβ δαβ
α,β=0
2
|bα | Wα .
(g) Since g is normalized, we easily find
hg|gi =
m
X
α,β=0
b∗α bβ hϕα |ϕβ i =
m
X
b∗α bβ δαβ =
m
X
α=0
α,β=0
2
|bα | = 1 .
(h) Since the ϕα ’s were listed in order of increasing values of the variational
integrals, we have W0 ≤ W1 ≤ · · · ≤ Wm , and hence from parts (f) and
(g) we find
hg|gi =
m
X
α=0
|bα |2 Wα ≤
m
X
α=0
|bα |2 Wm = Wm .
(i) From parts (d) and (h) we clearly have
Em ≤ hg|Hgi ≤ Wm
where m = 0, 1, . . . , n − 1.
2
2
6. (a) Our trial function is ϕ = e−cr /a0 so that (letting β = 2c/a20 )
Z ∞
Z ∞
2
2
2
hϕ|ϕi =
e−2cr /a0 r2 dr =
r2 e−βr dr
0
0
∞
2
∂
1 ∂
e−βr dr = −
∂β 0
2 ∂β
√
π −3/2
β
.
=
4
=−
Z
r
π
β
The hydrogen atom Hamiltonian is
2
~2
1
∂
2 ∂
e2
2
H=−
+
+
L
−
2m ∂r2
r ∂r
2mr2
r
8
where the angular momentum operator L2 takes derivatives with respect
to angles only. Thus
2
e2 −cr2 /a20
~2
∂
2 ∂
−cr 2 /a20
−
Hϕ = −
e
+
e
2m ∂r2
r ∂r
r
~2 4c2 r2
6c
e2 −cr2 /a20
= −
−
e
−
2m
a40
a20
r
and therefore
hϕ|Hϕi =
Z
0
∞
−
~2
2m
4c2 r2
6c
− 2
4
a0
a0
−
e2 −2cr2 /a20 2
r dr
e
r
= I1 + I2 + I3
where (remember β = 2c/a20 )
Z
~2 β 2 ∞ 4 −βr2
r e
dr
2m 0
Z
3~2 β ∞ 2 −βr2
r e
dr
I2 =
2m 0
Z ∞
2
I3 = −e2
re−βr dr .
I1 = −
0
Let us first evaluate I3 . Letting u = −βr2 we have du = −2βrdr and
hence
Z ∞
Z −∞
2
1
1
.
re−βr dr = −
eu du =
2β 0
2β
0
So we have
e2 −1
β .
2
We have already evaluated the integral in I2 above when we evaluated
hϕ|ϕi, so we have
√
3 π~2 β −1/2
I2 =
.
8m
And for I1 we take follow the usual procedure
Z ∞
Z ∞
2
2
∂2
e−βr dr
r4 e−βr dr =
2
∂β
0
0
2 r
1 ∂
π
=
2 ∂β 2 β
√
3 π −5/2
β
=
8
I3 = −
9
and hence
I1 = −
√
3 π~2 β −1/2
.
16m
Putting this all together we have
W =
hϕ|Hϕi
3~2 β
2e2
=
− √ β 1/2
hϕ|ϕi
4m
π
r
2e2 2c
3~2 c
−
.
=
2ma20
a0
π
Taking the derivative with respect to c and setting it equal to zero yields
r
2 e2 −1/2
3~2
0=
−
c
2
2ma0
π a0
or
c
1/2
=
r
2 2me2 a0
π 3~2
so that squaring this expression and substituting a0 = ~2 /me2 we obtain
c=
8
.
9π
Substituting this into the above expression for W yields
8 me4
.
W =−
3π 2~2
The exact hydrogen-atom energy levels are given by
En = −
me4
2~2 n2
for n = 1, 2, . . . . In particular, the ground state is for n = 1 so that
E1 = −
me4
.
2~2
Thus our variational integral is too large by a factor of 1 − 8/3π or 15.1%.
(b) The spherical harmonic Y20 is normalized, so we still have
√
π −3/2
β
.
hϕ|ϕi =
4
Since L2 Ylm = l(l + 1)~2 Ylm , we can write the Hamiltonian simply as
2
~2
∂
2 ∂
e2
~2
+
+
l(l
+
1)
−
H=−
2m ∂r2
r ∂r
2mr2
r
10
which for Y20 becomes
H=−
~2
2m
∂2
2 ∂
+
∂r2
r ∂r
+
3~2
e2
−
2
mr
r
The only difference between the integral hϕ|Hϕi in this case and that
in part (a) is the term due to 3~2 /mr2 . Since Y20 is normalized, this is
simply
r
Z
3~2 ∞ 1 −βr2 2
3~2 π
e
r dr =
.
m 0 r2
2m β
As a result, the new W is just the W from part (a) plus the term
r
6~2
12~2
1 3~2 π
c.
=
β=
hϕ|ϕi 2m β
m
ma20
Then taking the derivative of W and setting it equal to zero yields
r
3~2
2 e2 −1/2 12~2
0=
−
c
+
2
2ma0
π a0
ma20
or
c=
8
8
=
.
2
(27) π
729π
Substituting this into
3~2 c
2e2
W =
−
2ma20
a0
r
2c 12~2
c
+
π
ma20
we find, without a lot of difficulty,
W =−
me4
2~2
which is the exact ground-state energy.
7. We consider the linear variation function
r
r
2
2
πx
3πx
sin
+ c2
sin
ϕ = c1 f 1 + c2 f 2 = c1
l
l
l
l
for 0 ≤ x ≤ l, and let E1 ≤ E2 ≤ E2 ≤ · · · be the true energies of the system
given in problem 2. The variational integral is given by (for normalized trial
functions) Wα = hϕα |Hϕα i where Wα is a root of det(Hij −W Sij ) = 0. Given
P
(α)
this Wα , we solve the system j (Hij − Wα Sij )cj = 0 for the coefficients
P (α)
(α)
cj , and then these are used to construct the trial function ϕα = j cj fj .
We assume that the functions ϕα are numbered so that W1 ≤ W2 ≤ W3 ≤ · · · .
And in problem 5 you showed that Em ≤ Wm for each m = 1, 2, . . . , n.
11
(a) Since the potential energy is symmetric, the parity operator commutes
with the Hamiltonian and the exact solutions may be chosen to be parity
eigenstates also. (In other words, the solutions are either even or odd.)
Furthermore, we have shown that in one dimension, the bound states are
non-degenerate, and the states alternate in symmetry, beginning with an
even ground state. The functions f1 and f2 that define ϕ are both even
functions on the interval 0 ≤ x ≤ l. Hence the roots that result from ϕ
will give an upper bound to the two lowest-energy even states, and these
correspond to E1 and E3 .
(b) The functions f1 and f2 are both eigenstates of the Hamiltonian that belong to distinct eigenvalues, and hence they are orthogonal.
(c) The functions f1 and f2 are solutions to the usual particle-in-a-box problem where H = T . These are the solutions to the Schrödinger equation
−
~2 d2 ψ
= Eψ
2m dx2
with the boundary conditions ψ(0) = ψ(l) = 0. The normalized solutions
are
r
nπx
2
sin
;
n = 1, 2, . . .
ψn (x) =
l
l
with the energy eigenvalues
En =
n2 π 2 ~2
.
2ml2
In particular, we have
E1 =
π 2 ~2
= 4.934802201 ~2/ml2
2ml2
and
9π 2 ~2
= 44.41321980 ~2/ml2 .
2ml2
In the notation of this problem, f1 = ψ1 , f2 = ψ3 , ε1 = E1 and ε2 = E3 .
Since the solutions fj are orthonormal, we clearly have
E3 =
Tij = hfi |T fj i = εj hfi |fj i = εj δij .
P
(d) The secular equation is j (Hij − W Sij )cj = 0. The overlap integrals
are simply Sij = hfi |fj i = δij . For the Hij = Tij + Vij we still need to
evaluate the integrals Vij = hfi |V (x)fj i. We have (using the identities
sin2 A = (1/2)(1 − cos 2A) and 2 sin A sin B = cos(A − B) − cos(A + B))
Z
2V0 3l/4 2 πx
1
1
V11 =
dx = V0
+
sin
l l/4
l
2 π
12
V12 = V21
V22
2V0
=
l
V0
=
l
Z
Z
3l/4
l/4
3l/4
2 sin
l/4
3πx
V0
πx
sin
dx = −
l
l
π
3πx
dx = V0
sin
l
2
1
1
−
2 3π
where V0 = ~2 /ml2 . Noting that ε1 = (π 2 /2)V0 and ε2 = (9π 2 /2)V0 , the
secular determinant becomes
ε1 + V0 (1/2 + 1/π) − W
−V0 /π
−V0 /π
ε2 + V0 (1/2 − 1/3π) − W 1
1
1
1
2
= W − W ε1 + V0
+ ε2 + V0
+
−
2 π
2 3π
1
1
1
1
+ε1 ε2 + ε1 V0
+ ε2 V0
−
+
2 3π
2 π
1
V2
1
1
1
2
− 02
+V0
+
−
2 π
2 3π
π
2
= W 2 − W V0 5π 2 + 1 +
3π
4
9π 2 1
9π
π2 1
1
1
+V02
+
+
−
+
4
2 2 3π
2
2 π
1
1
1
1
1
− 2
+
−
+
2 π
2 3π
π
≈ W 2 − 50.5602285962 V0W + 257.679042378656 V02 .
Setting this equal to 0 and solving for W using the quadratic formula
results in the two roots
W1 = 5.750518 ~2/ml2
and
W2 = 44.809711 ~2/ml2 .
The true energies are (for the ground state and the first excited even state)
E1 = 5.750345 ~2/ml2
and
E3 = 44.808373 ~2/ml2
and hence W1 is 0.00300% too high, and W2 is 0.00298% too high. To find
the coefficients c1 and c2 , we only need use one row of the secular equations
because the two rows are linearly dependent. (Their determinant was
zero.) Thus we have the equation
1
V0 (α)
1
(α)
ε1 + V0
− Wα c1 =
+
c .
2 π
π 2
13
Canceling the common factors of V0 this becomes
3
π
π
(α)
(α)
c2 =
+ + 1 − πWα c1 .
2
2
Therefore
(1)
(1)
c2 = 0.00815 c1
and
(α)
(2)
(2)
c2 = −122.7 c1 .
(α)
(1)
Normalizing the function ϕα = c1 f1 + c2 f2 we finally obtain c1 ≈ 1
(2)
and c1 ≈ 0.008 so that
ϕ1 = f1 + 0.008f2
and
ϕ2 = 0.008f1 − 0.9816f2 .
Note that ϕ1 is essentially just f1 . And W1 is 0.045% smaller than the
value found in problem 2 where you used just the function f1 . (This is
as it should be since here we used two functions rather than one, so the
approximation should be better.)
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