Solutions to Problems in Merzbacher, Quantum Mechanics, Third

advertisement
Solutions to Problems in Merzbacher,
Quantum Mechanics, Third Edition
Homer Reid
April 5, 2001
Chapter 7
Before starting on these problems I found it useful to review how the WKB
approximation works in the first place. The Schrödinger equation is
−
~2 d 2
Ψ(x) + V (x)Ψ(x) = EΨ(x)
2m dx2
or
d2
Ψ(x) + k 2 (x)Ψ(x) = 0,
dx2
We postulate for Ψ the functional form
k(x) ≡
r
2m
[E − V (x)].
~2
Ψ(x) = AeiS(x)/~
in which case the Schrödinger equation becomes
i~S 00 (x) = [S 0 (x)]2 − ~2 k 2 (x).
(1)
This equation can’t be solved directly, but we obtain guidance from the observation that, for a constant potential, S(x) = ±kx, so that S 00 vanishes. For a
nonconstant but slowly varying potential we might imagine S 00 (x) will be small,
and we may take S 00 = 0 as the seed of a series of successive approximations
to the exact solution. To be specific, we will construct a series of functions
S0 (x), S1 (x), · · · , where S0 is the solution of (1) with 0 on the left hand side;
S1 is a solution with S000 on the left hand side; and so on. In other words, at
the nth step in the approximation sequence (by which point we have computed
Sn (x)), we compute Sn00 (x) and use that as the source term on the LHS of (1)
to calculate Sn+1 (x). Then we compute the second derivative of Sn+1 (x) and
use this as the source term for calculating Sn+2 , and so on ad infinitum. In
1
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
2
symbols,
0 = [S00 (x)]2 − ~2 k 2 (x)
i~S000
i~S100
=
[S10 (x)]2
[S20 (x)]2
=
····
(2)
2 2
− ~ k (x)
(3)
2 2
(4)
− ~ k (x)
Equation (2) is clearly solved by taking
S00 (x)
= ±~k(x)
⇒
S0 (x) = S00 ± ~
Z
x
k(x0 )dx0
(5)
−∞
for any constant S00 . Then S000 (x) = ±~k 0 (x), so (3) is
p
S10 (x) = ±~ k 2 (x) ± ik 0 (x).
With the two ± signs here, we appear to have four possible choices for S10 . But
let’s think a little about the ± signs in this equation. The ± sign under the
radical comes from the two choices of sign in (5). But if we chose, say, the plus
sign in that equation, so that S00 > 0, we would also expect that S10 > 0. Indeed,
if we choose the plus sign in (5) but the minus sign in (3), then S00 and S10 have
opposite sign, so S10 differs from S00 by an amount at least as large as S00 , in
which case our approximation sequence S0 , S1 , · · · has little hope of converging.
So we choose either both plus signs or both minus signs in (3), whence our two
choices are
p
p
S10 = +~ k 2 (x) + ik 0 (x) or S10 = −~ k 2 (x) − ik 0 (x).
(6)
If V (x) is constant, k(x) is constant, and, as we observed before, the sequence
of approximations terminates at 0th order with S0 being an exact solution. By
extension, if V (x) is not constant but changes little over one particle wavelength,
we have k 0 (x)/k 2 (x) 1, so we may expand the radicals in (6):
ik 0 (x)
ik 0 (x)
S10 ≈ ~k(x) 1 + 2
or S10 ≈ −~k(x) 1 − 2
2k (x)
2k (x)
or
i~k 0 (x)
S10 ≈ ±~k(x) +
.
(7)
2k(x)
Integrating,
Z
i~ x k 0 (u) 0
dx
2 a k(u)
Za x
i~ k(x)
= S1 (a) ± ~
k(u)du +
ln
2
k(a)
a
S1 (x) = S1 (a) ± ~
Z
x
k(u)du +
where a is some point chosen such that the approximation (7) is valid in the
full range a < x0 < x. We could go on to compute S2 , S3 , etc., but in practice
it seems the approximation is always terminated at S1 .
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
3
The wavefunction at this order of approximation is
−1/2
Rx
Ψ(x) = exp(iS1 (x)/~) = eiS1 (a)/~ e±i a k(u)du eln k(x)/k(a)
s
k(a) ±i R x k(u)du
= Ψ(a)
e a
k(x)
= Ψ(a)G± (x; a)
where
(8)
G± (x; a) ≡
s
k(a) ±i R x k(u)du
e a
.
k(x)
(9)
We have written it this way to illustrate that the function G(x, a) is kind of like
a Green’s function or propagator for the wavefunction, in the sense that, if you
know what Ψ is at some point a, you can just multiply it by G± (x; a) to find
out what Ψ is at x. But this doesn’t seem quite right: Schrödinger’s equation
is a second-order differential equation, but (8) seems to be saying that we need
only one initial condition—the value of Ψ at x = a—to find the value of Ψ at
other points. To clarify this subtle point, let’s investigate the equations leading
up to (8). If the approximation (7) makes sense, then there are two solutions
of Schrödinger’s equation at x = a, one whose phase increases with increasing
x (positive derivative), and one whose phase decreases. Equation (8) seems to
be saying that we can use either G+ or G− to get to Ψ(x) from Ψ(a); but the
requirement the dΨ/dx be continuous at x = a means that only one or the
other will do. Indeed, in using (8) to continue Ψ from a to x we must choose
the appropriate propagator—either G+ or G− , according to the derivative of
Ψ at x = a; otherwise the overall wave function will have a discontinuity in its
first derivative at x = a. So to use (8) to obtain values for Ψ at a point x, we
need to know both Ψ and Ψ0 at a nearby point x = a, as should be the case for
a second-order differential equation.
If we want to de-emphasize this nature of the solution with the propagator
we may write
s
1 ±i R x k(u)du
Ψ(x) = C
e a
(10)
k(x)
p
where C = Ψ(a) k(a). In regions where V (x) > E, k(x) is imaginary, so it’s
useful to define
r
2m
κ(x) = −ik(x) =
[V (x) − E]
(11)
~2
and
s
1 ± R x κ(u)du
e a
.
(12)
Ψ(x) = C
κ(x)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
4
If we have a region of space in which the WKB approximation is valid,
knowing the value of Ψ (and its derivative) at one point within the region is
equivalent to knowing it everywhere, because we can use the propagator (9)
to get from that one point to every other point within the region. The WKB
method, however, gives us no way of determining the value of Ψ at that one
starting point. Furthermore, even if we know Ψ at one point within a region of
validity, we can’t use (8) to determine Ψ in other, nonadjacent regions, because
we can’t carry the propagator across regions of invalidity. So basically what we
need is a way of finding one starting value for Ψ(x) in every region of validity
of the WKB approximation.
How do we find such points? Well, one sure-fire way to get starting points
in regions of validity is to identify regions of invalidity, of which there will be
at least one adjacent to each region of validity, and then get values of Ψ at the
boundaries of the regions of invalidity—which will also count as values in the
regions of validity. So we need to identify the regions of invalidity of the WKB
approximation and do a more accurate solution of the Schrödinger equation
there.
The WKB approximation breaks down when k 0 /k 2 1 ceases to hold, which
is true when k ≈ 0 but k 0 6= 0, which happens near a classical turning point of
the motion—i.e., a point x0 at which V (x0 ) = E. But near such a point we may
expand V (x) − E in a Taylor series around the point x0 ; if we keep only the
first (linear in x) term in the series, we arrive at a Schrödinger equation which
we can solve exactly in the vicinity of x0 . To do this, suppose the point x0 is a
classical turning point of the motion, so that V (x0 ) = E. In the neighborhood
of x0 we may expand V (x):
V (x) = E + (x − x0 )V 0 (x0 ) + · · ·
(13)
Then the Schrödinger equation becomes
d2
2m
Ψ(x) − 2 V 0 (x0 )(x − x0 )Ψ(x) = 0.
2
dx
~
The useful substitution here is
1/3
2m 0
u(x) = γ(x − x0 )
γ≡
V (x0 )
~2
so
x(u) =
u
+ x0 .
γ
If we define
Φ(u) = Ψ(x(u))
then
dΦ
dΨ dx
1
=
= Ψ0 (x(u))
du
dx du
γ
1 00
d2 Φ
= 2 Ψ (x(u))
du2
γ
(14)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
5
so (14) becomes
γ2
d2
Φ(u) − γ 3 (x − x0 )Φ(u) = 0
du2
or
d2
Φ(u) − uΦ(u) = 0.
du2
The solution to this differential equation is
Φ(u) = β1 Ai(u) + β2 Bi(u)
(15)
so the solution to the Schrödinger equation (14) is
Ψ(x) = β1 Ai γ(x − x0 ) + β2 Bi γ(x − x0 ) .
For γ(x − x0 ) 1 we have the asymptotic expression
3/2
2
−1/4 β1 − 23 |γ(x−x0 )|3/2
Ψ(x) ≈ π −1/2 [γ(x − x0 )]
e
+ β2 e+ 3 |γ(x−x0 )|
2
(16)
and for γ(x − x0 ) −1 we have
Ψ(x) ≈ π −1/2 |γ(x − x0 )|
−1/4
h
2
π
3/2
|γ(x − x0 )| −
3
4
2
π i
3/2
− β2 sin
.
|γ(x − x0 )| −
3
4
β1 cos
(17)
To simplify these, we need to consider two possible kinds of turning point.
Case 1: V 0 (x0 ) > 0.
In this case the potential is increasing through the turning point at x0 , which
means that V (x) < E for x < x0 , and V (x) > E for x > x0 . Hence the region
to the left of the turning point is the classically accessible region, while the right
of the turning point is classically forbidden. Since V 0 (x0 ) > 0, γ > 0, so for
x < x0 (??) holds. For points close to the turning point on the left side,
r
1/2
2m 0
2m
[E
−
V
(x)]
≈
V
k(x) =
(x0 − x)1/2 = γ 3/2 (x0 − x)1/2
~2
~2
so
|γ(x − x0 )|
−1/4
and
Z
x0
k(u)du = γ
x
3/2
Z
x0
=
r
γ
k(x)
(x0 − x)1/2 du =
x
2
= |γ(x − x0 )|3/2 .
3
(18)
2 3/2
γ (x0 − x)3/2
3
(19)
On the other hand, for points close to the turning point on the left side we
have x > x0 , so γ(x − x0 ) > 0. In this region,
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
κ(x) =
r
1/2
2m 2m 0
= γ 3/2 (x − x0 )1/2
V (x0 )(x − x0 )
V (x) − E ≈
~2
~2
so, for x near x0 ,
Z x
Z x
2
κ(u)du = γ 3/2
(u − x0 )1/2 du = γ 3/2 (x − x0 )3/2
3
x0
x0
and also
|γ(x − x0 )|
−1/4
=
r
γ
.
κ(x)
6
(20)
(21)
(22)
Using (18) and (19) in (17), and (21) and (22) in (16), the solutions to the
Schrödinger equation on either side of a classical turning point x0 at which
V 0 (x0 ) > 0 are
s
Z x0
1 h
π
k(u)du −
2β1 cos
Ψ(x) =
k(x)
4
x
Z x0
π i
k(u)du −
− β2 sin
,
x < x0
(23)
4
x
s
i
R
1 h Rxx κ(u)du
− x κ(u)du
Ψ(x) =
β1 e 0
+ β 2 e x0
,
x < x0
(24)
κ(x)
(we redefined the β constants slightly in going to this equation).
Case 2: V 0 (x0 ) < 0.
In this case the potential is decreasing through the turning point, so the
classically accessible region is to the right of the turning point, and the forbidden
region to the left. Since V 0 (x0 ) < 0, γ < 0. That means that the regions of
applicability of (16) and (17) are on opposite sides of the turning points as they
were in the previous case. The solutions to the Schrödinger equation on either
side of the turning point are
s
i
R
1 h R x0 κ(u)du
− x κ(u)du
β1 e x
+ β 2 e x0
,
x < x0
(25)
Ψ(x) =
κ(x)
s
Z x
π
1 h
k(u)du −
Ψ(x) =
2β1 cos
k(x)
4
x0
Z x
π i
− β2 sin
,
x > x0
(26)
k(u)du −
4
x0
(27)
So, to apply the WKB approximation to a given potential V (x), the first
step is to identify the classical turning points of the motion, and to divide space
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
7
up into regions bounded by turning points, within which regions the WKB
approximation (7) is valid. Then, for each turning point, we write down (23)
and (24) (or (25) and (26)) at nearby points on either side of the turning point,
and then use (10) to evolve the wavefunction from those points to other points
within the separate regions.
We should probably quantify the meaning of “nearby” in that last sentence.
Suppose x0 is a classical turning point of the motion, and we are looking for
points x0 ± at which to make the “handoff” from approximations (16) and
(17) to the WKB approximation These points must satisfy several conditions.
First, the approximate Schrödinger equation (14) is only valid as long as we can
neglect the quadratic and higher-order terms in the expansion (13), so we must
have
0
V (x0 ) .
(28)
|V 00 (x0 )| |V 0 (x0 )| ⇒ 00
V (x0 ) But at the same time, γ must be sufficiently greater than 1 to justify the
approximation (16) (or sufficiently less than -1 to justify (17)); the condition
here is
1/3
−1/3
2m 0
2m 0
1
⇒
.
(29)
V
(x
)
V
(x
)
0 0 ~2
~2
Finally, the points x± must be sufficiently far away from the turning points that
the approximation (7) is valid for the derivative of the phase of the wavefunction;
the condition for this to be the case was
0
2 0
k (x) ~
V
(x
±
)
1
1⇒
(30)
1.
k 2 (x) 2 2m [E − V (x ± )]3/2 If there are no points x0 ± satisfying all three conditions, the WKB approximation cannot be used.
To apply all of this to the problem of bound states in a potential well,
consider a potential like that shown in Figure 1, with two classical turning points
at x = a and x = b. Although there are no discontinuities in the potential
here, the problem may be analyzed in a manner similar to that used in the
consideration of one-dimensional piecewise constant potentials, as in Chapter
6: we divide space into a number of distinct regions, obtain solutions of the
Schrödinger equation in each region, and then match values and derivatives at
the region boundaries.
To divide space into distinct regions in this case, we begin by identifying
narrow regions around the turning points a and b in which the linear approximation (13) is valid. In the narrow region around x = a, we may use (25)
and (26); around x = b we may use (23) and (24). Let the narrow such region
around a be a − 1 < x < a + 1, and that around b be b − 2 < x < b + 2. Then
space divides naturally into five regions: (a) x < a − In this region we are far enough to the left of the turning point that the WKB
approximation is valid, and the wavefunction takes the form (10). However, we
8
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
PSfrag replacements
V (x)
E
a
b
Figure 1: A potential V (x) with two classical turning points for an energy E.
9
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
must throw out the term that grows exponentially as x → −∞, so we are left
with
s
1 − R (a−) κ(u)du
e x
,
x < a − .
(31)
Ψ(x) = A
κ(x)
(b) a − < x < a + In this region we are close enough to the turning point that (13) is valid, so
(25) and (26) may be used.
s
Ra
1 − R a κ(u)du
Ψ(x) =
β1 e x
+ β2 e+ x κ(u)du ,
x < (a − ).
(32)
κ(x)
From (31) and (32) we see that continuity of both the value and first derivative of
Ψ(x) at x = a − requires taking β1 = A, β2 = 0. With this choice of constants,
we achive continuity not only of the value and first derivative of Ψ but also of
all higher derivatives, as must be the case since there is no discontinuity in the
potential.
But now that we know the value of Ψ at x = a − , we also know it at
x = a + , because of course the solution of the Schrödinger equation in the
narrow strip around a (to which (32) is an asymptotic approximation for x < a)
is valid throughout the strip; the same solution that’s valid at x = a − is valid
at x = a + . With β1 = A and β2 = 0, (26) becomes
Ψ(x) = 2A
=A
s
s
1
cos
k(x)
Z
x
k(u)du −
a
π
4
i
Rx
1 h +i(R x k(u)du−π/4)
a
+ e−i( a k(u)du−π/4) ,
e
k(x)
x = (a + )−
(33)
(c) a + < x < b − In this region the WKB approximation (7) is valid, so we may use (8) to find
the wavefunction at any point within the region. Using the expression (33) for
the wavefunction at x = a + , integrating from a + to x in the propagator (9),
and using G+ and G− , respectively, to propagate the first and second terms in
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
10
(33), we obtain for the wavefunction at a point x in this region
s
“R
“R
”
”
Rx
Rx
1
−i a(a+) k(u)du+ (a+)
−π/4
+i a(a+) k(u)du+ (a+)
k(u)du−π/4
+e
Ψ(x) = A
e
k(x)
s
i
Rx
1 h +i(R x k(u)du−π/4)
a
=A
e
+ e−i( a k(u)du−π/4)
k(x)
s
Z x
1
π
= 2A
cos
k(u)du −
k(x)
4
a
s
!
Z b
Z b
1
π
,
a+<x <b−
cos
= 2A
k(u)du −
k(u)du −
k(x)
4
a
x
(34)
Okay, I have now carried this analysis far enough to see for myself exactly
where the Bohr-Sommerfeld quantization condition
Z b
1
k(u)du = n +
π,
n = 1, 2, · · ·
(35)
2
a
comes from, which was my original goal, so I am now going to stop this exercise
and proceed directly to the problems.
Problem 7.1
Apply the WKB method to a particle that falls with acceleration g in a uniform
gravitational field directed along the z axis and that is reflected from a perfectly
elastic plane surface at z = 0. Compare with the rigorous solutions of this problem.
We’ll start with the exact solution to the problem. The requirement of
perfect elastic reflection at z = 0 may be imposed by taking V (x) to jump
suddenly to infinity at z = 0, i.e.
(
mgz, z > 0
V (x) =
∞,
z ≤ 0.
For z > 0, the Schrödinger equation is
d2
Ψ(x) +
dx2
d2
= 2 Ψ(x) +
dx
d2
= 2 Ψ(x) −
dx
0=
2m
[E − mgz] Ψ(x)
~2 2m2 g E
−
z
Ψ(x)
~2
mg
2m2 g
[z − z0 ] Ψ(x)
~2
(36)
11
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
where z0 = E/mg. With the substitution
u = γ(z − z0 )
γ=
2m2 g
~2
1/3
and taking Φ(u) = Ψ(x(u)), we find that (36) is just the Airy equation for Φ(u),
d2
Φ(u) − uΦ(u) = 0
du2
with solutions
Φ(u) = β1 Ai(u) + β2 Bi(u).
Since we require a solution that remains finite as z → ∞, we must take β2 = 0.
The solution to (36) is then
Ψ(x) = β1 Ai γ(z − z0 ) ,
(z > 0).
(37)
For z < 0, I wasn’t quite sure how to account for the infinite potential
jump at z = 0, so instead I supposed the potential for z < 0 to be a constant,
V (z) = V0 , where eventually I’ll take V0 → ∞. Then the Schrödinger equation
for z < 0 is
d2
2m
Ψ(z) − 2 [V0 − E] Ψ(z) = 0
2
dz
~
with solution
r
2m
−kz
[V0 − E].
(38)
Ψ(z) = Ae
,
k=
~2
Matching values and derivatives of (37) and (38) at z = 0, we have
β1 Ai(−γz0 ) = A
γβ1 Ai0 (−γz0 ) = −kA
Dividing, we obtain
1 Ai(−γz0 )
1
=−
γ Ai0 (−γz0 )
k
Now taking V0 → ∞, we also have k → ∞, so the RHS of this goes to zero;
thus the condition is that −γz0 be a zero of the Airy function, which means the
energy eigenvalues En are given by
2m2 g
~2
1/3
En
= xnm ⇒ En =
mg
where xn is the nth root of the equation
Ai(−xn ) = 0.
mg 2 ~2
2
1/3
xn
(39)
Homer Reid’s Solutions to Merzbacher Problems: Chapter 7
12
So that’s the exact solution. In the WKB approximation, the spectrum of
energy eigenvalues is determined by the condition (35). In this case the classical
turning points are at z = 0 and z = z0 , so we have
Z z0
1
n+
π=
k(z) dz
2
0
r
Z
2m z0
=
[E − mgz]1/2 dz
~2 0
r
Z
2m2 g z0
[z0 − z]1/2 du
=
~2
0
r
z0
2m2 g 2
3/2 − (z0 − z) =
2
~
3
0
r
3/2
2
2 2m g E
=
3
~2
mg
so the nth eigenvalue is given by
En =
mg 2 ~2
2
1/3 2/3
1
3
n+
π
.
2
2
Download