Problem Set 8
I.
a.
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The potential energy for the finite square well is as follows:
V(x) = -V0
V(x) = 0
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for –a < x < a
otherwise
In this problem, we are investigating bound states (E < 0) that are also odd: (-x)
= - (x).
We need to solve the Schrodinger equation in three regions: x < -a, -a < x < a,
and x > a.
In the region x < -a, V = 0, so the Schrodinger equation is –(ћ2/2m)d2/dx2 = E.
Upon rearrangement: d2/dx2 = -(2mE/ћ2).
This becomes d2/dx2 = -2 by defining  = (-2mE)1/2/ћ. (Observe that  is real
because E < 0.)
A general solution to this differential equation is (x) = Aex + Be-x.
Since e-x   as x  -, B must be 0 for  to be normalizable.
In the region –a < x < a, V = -V0, so the Schrodinger equation is
–(ћ2/2m)d2/dx2 –V0 = E.
Upon rearrangement: d2/dx2 = -[2m(E+V0)/ћ2].
This becomes d2/dx2 = -k2 by defining k = [2m(E+V0)]1/2/ћ.
A general solution to this differential equation is (x) = Csin(kx) + Dcos(kx), but
we're only looking at odd solutions now, so we'll set D = 0.
In the region x > a, V = 0, so the Schrodinger equation is –(ћ2/2m)d2/dx2 = E,
just as it was in the region x < -a.
We rewrite the Schrodinger equation as d2/dx2 = -2, where once again
 = (-2mE)1/2/ћ.
A general solution to this differential equation is (x) = Fex + Ge-x.
Since ex   as x  , F must be 0 for  to be normalizable.
So our wavefunction is
(x) = Aex
(x) = Csin(kx)
(x) = Ge-x
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for x < -a
for –a < x < a
for x > a
To find a formula for allowed energies, we apply the continuity conditions at x =
a.
The continuity of  at x = a gives Csin(ka) = Ge-a.
The continuity of d/dx at x = a gives kCcos(ka) = -G-a.
If we divide the second equation by the first, we obtain kcot(ka) = -, or
–cot(ka) = /k.
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We define z = ka and z0 = (a/ћ)(2mV0)1/2.
Then our equation becomes –cot(z) = /k = (-2mE)1/2/(ћk) through the definition
of .
Next, -cot(z) = [-2mE/(ћ2k2)]1/2 = [-(k2ћ2 – 2mV0)/(ћ2k2)]1/2 by writing E in terms
of k (from the definition of k).
Next, -cot(z) = [(2mV0)/(ћ2k2) – 1]1/2 = [(2mV0)/(ћ2k2) – 1]1/2 = (z02/z2 – 1)1/2.
Check it!
b.
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To determine the number of bound states, we must determine z0.
According to our definition, z0 = (a/ћ)(2mV0)1/2.
We arrive at more convenient units if we multiply the equation by c/c: z0 =
[a/(ћc)](2mc2V0)1/2 = (1 nm/197 eV.nm)(2  511 keV  9 eV)1/2 = 15.4.
It is useful to write z0 as a multiple of /2: z0 = 15.4 = 9.8/2.
We can see from the figure below that whenever z0 increases by /2 (1.57), we
pick up a new intersection with tan(z) or –cot(z).
So in our case, we have nine intersections after the initial intersection, giving us
ten intersections.
Each intersection represents an allowed bound-state energy level and an allowed
wavefunction.
12
sqrt[(z0/z)^2 - 1]
tan(z)
-cot(z)
6
z0
0
0
1.57
3.14
4.71
6.28
7.85
z
9.42
10.99
12.56
14.13
15.7
6-42.
V
0
x
-V0
a.
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In the region x > 0, V = -V0, so the Schrodinger equation is
–(ћ2/2m)d2/dx2 –V0 = E.
Upon rearrangement: d2/dx2 = -[2m(E+V0)/ћ2].
This becomes d2/dx2 = -k22 by defining k2 = [2m(E+V0)]1/2/ћ.
A general solution to this differential equation is
2(x) = Cexp(ik2x) + Dexp(-ik2x), so k2 = [2m(E+V0)]1/2/ћ is the wave number in
the region of positive x.
We're given E = 2V0, so k2 = (6mV0)1/2/ћ.
We're also given (or could derive) ћ2k12/(2m) = 2V0, so m = ћ2k12/(4V0).
So k2 = (3ћ2k12/2)1/2/ћ = (3/2)1/2k1.
b.
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In the region x < 0, V = 0, the Schrodinger equation is –(ћ2/2m)d2/dx2 = E.
Upon rearrangement: d2/dx2 = -(2mE/ћ2) = -k12, where k1 = (2mE)1/2/ћ.
A general solution to this differential equation is
1(x) = Aexp(ik1x) + Bexp(-ik1x).
The first term above represents a beam moving to the right (the incident wave).
The second term represents a beam moving to the left (the reflected wave).
In the region x > 0, 2(x) = Cexp(ik2x). This is the transmitted wave. D = 0
because there is no wave approaching from the right.
We apply the continuity of  at x = 0: A + B = C.
We apply the continuity of d/dx at x = 0: ik1A – ik1B = ik2C.
We use the first equation to eliminate C from the second equation: k1A – k1B =
k2A + k2B.
Solving for B: B = (k1 – k2)A/(k1 + k2).
The reflection coefficient R = |C|2/|A|2 = (k1 - k2)2/(k1 + k2)2.
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Observe that this expression for a downward step is identical to the expression for
an upward step. (For a downward step, k2 > k1, and for an upward step, k2 < k1.)
c.
Since all of the incident beam is either reflected or transmitted, T = 1 – R =
4k1k2/(k1 + k2)2.
d.
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We found in (a) that k2 = (3/2)1/2k1.
So T = 4(3/2)1/2k12/[k1 + (3/2)1/2k1]2 = 4(3/2)1/2/[1 + (3/2)1/2]2 = 0.99.
So if a million particles arrive each second at the potential step, about 990,000 are
transmitted each second.
Classically, all the particles should continue unimpeded since the potential energy
everywhere is less than the particle energy; in fact, the particles gain energy as
they cross the potential step.
6-53.
a.
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Since |(x)|2 is symmetric (even), |(-x)|2 = |(x)|2.
Since (-x) and (x) have the same magnitude, they can only differ by some
factor C whose magnitude is 1: (-x) = C(x). (You might want to say
immediately that C must be 1 or -1. But we need to prove that it's not complex.)
Replacing x with –x yields (x) = C(-x). Substituting this into the equation
above yields (-x) = C2(-x).
So C2 = 1, and C = 1.
b.
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Inside the well, V = 0, and the Schrodinger equation is –(ћ2/2m)d2/dx2 = E.
Upon rearrangement: d2/dx2 = -(2mE/ћ2) = -k2, with k = (2mE)1/2/ћ.
A general solution to this differential equation is (x) = Asin(kx) + Bcos(kx).
(A less useful general solution is Aeikx + Be-ikx. While we prefer the exponential
solutions for reflection/transmission problems, we prefer the sin/cos solutions for
bound states.)
We showed in (a) that (x) must be even or odd. So  will be either a sine or a
cosine.
Let's work with sine first: (x) = Asin(kx). (x) is 0 outside the infinite well, so
continuity requires (x) = 0 at the edge of the well. So 0 = (L/2) = Asin(kL/2).
Sine is 0 when its argument is (integer), so kL/2 = (integer), or k =
2(integer)/L. Thus (x) = Asin[2(integer)x/L] with (integer) = 1, 2, 3…, or
Asin(nx/L) with n = 2, 4, 6….
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Now we can normalize: *dx = A2sin2(nx/L)dx = A2(L/2). (I used the trick
that the integral of a sinusoid over one or more complete periods is half the
distance between the limits of integration.) This equals 1, so A = (2/L)1/2.
Now let's look at cosine: (x) = Bcos(kx). This must go to 0 at the edge of the
well (L/2), so 0 = (L/2) = Bcos(kL/2). Cosine is 0 when its argument is /2 +
(integer), so kL/2 = /2 + (integer), or k = /L + 2(integer)/L. Thus (x) =
Bcos[x/L + 2(integer)x/L] with (integer) = 1, 2, 3…, or Bcos(nx/L) with n =
1, 3, 5….
Normalization gives B = (2/L)1/2.
c.
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We'd defined k = (2mE)1/2/ћ, so E = ћ2k2/(2m).
We can see from our wavefunctions, sin(kx) and cos(kx), that k = n/L, where n
is even for sines and odd for cosines, so all positive integers are allowed.
So E = ћ2n22/(2mL2), which is just what we found when we had a square well
between 0 and L.
6-61.
a.
V
V0
0
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a
x
As always, we must solve the Schrodinger equation in every region.
In the region x < 0, V = 0, so the Schrodinger equation is –(ћ2/2m)d2/dx2 = E.
Upon rearrangement: d2/dx2 = -(2mE/ћ2) = -k2, where k = (2mE)1/2/ћ.
A general solution to this differential equation is Aeikx + Be-ikx.
In the region 0< x < a, V = V0, so the Schrodinger equation is
–(ћ2/2m)d2/dx2 + V0 = E.
Upon rearrangement: d2/dx2 = [2m(V0 – E)/ћ2] = 2, where
 = [2m(V0 – E)]1/2/ћ.
A general solution to this differential equation is Ce-x + Dex.
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In the region x > a, V = 0, so the Schrodinger equation becomes
d2/dx2 = -(2mE/ћ2) = -k2, where again k = (2mE)1/2/ћ.
 A general solution to this differential equation is Feikx + Ge-ikx, but we set G = 0
because there is no wave arriving from the right.
 Now we apply the continuity conditions at x = 0 and x = a.
 The continuity of  at x = 0 yields A + B = C + D.
(1)
 The continuity of d/dx at x = 0 yields ikA – ikB = -C + D.
(2)
-a
a
ika
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 The continuity of  at x = a yields Ce + De = Fe .
(3)
-a
a
ika
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 The continuity of d/dx at x = a yields -Ce + De = ikFe . (4)
 The physics ends here. The rest is tedious algebra. Make sure you master all
the physics before moving on to the algebra!
 We want to eliminate B, C, and D.
 First, let's use Equations (3) and (4) to solve for C and D in terms of F.
 If we multiply Equation (3) by  and add it to Equation (4), we find
2Dea = ( + ik)Feika, or D = ( + ik)Fea(ik-)/(2).
 If we multiply Equation (3) by - and add it to Equation (4), we find
-2Ce-a = (ik - )Feika, or C = ( - ik)Fea(ik+)/(2).
 Now let's plug our equations for C and D into Equation (1):
A + B = ( - ik)Fea(ik+)/(2) + ( + ik)Fea(ik-)/(2).
(5)
 And let's plug our equations for C and D into Equation (2):
ikA – ikB = -( - ik)Fea(ik+)/(2) + ( + ik)Fea(ik-)/(2).
(6)
 We can eliminate B by multiplying Equation (5) by ik and adding it to Equation
(6): 2ikA = (2ik + k2 - 2)Fea(ik+)/(2) + (2ik – k2 + 2)Fea(ik-)/(2).
 So F = 4ikA/[(2ik + k2 - 2)ea(ik+) + (2ik – k2 + 2)ea(ik-)].
 The complex conjugate of the above equation is
F* = -4ikA*/[(-2ik + k2 - 2)ea(-ik+) + (-2ik – k2 + 2)ea(-ik-)].
 Notice that I simply replaced every i with –i.
 So F*F =
162k2A*A/{[42k2 + (k2 - 2)2]e2a - (2ik + k2 - 2)2 - (2ik – k2 + 2)2 +
[42k2 + (2 - k2)2]e-2a}=
162k2A*A/{[42k2 + (k2 - 2)2]e2a – [-42k2 + 4ik(k2 - 2) + (k2 - 2)2] [-42k2 + 4ik(2 - k2) + (2 - k2)2] + [42k2 + (2 - k2)2]e-2a}=
162k2A*A/{[42k2 + (k2 - 2)2]e2a + 82k2 - 2(k2 - 2)2 + [42k2 + (2 - k2)2]e-2a}=
162k2A*A/{[42k2 + (k2 - 2)2](e2a - 2 + e-2a) + 162k2}
Check it!
2
2 2
2
2 2
2 2
= A*A/{sinh (a)[4 k + (k -  ) ]/(4 k ) + 1}
Check it!
2
2
= A*A/{sinh (a)[16E(V0 – E) + (4E – 2V0) ]/[16E(V0 – E)] + 1}
= A*A/{1 + sinh2(a)[16E(V0 – E) + 16E2 – 16EV0 + 4V02]/[16E(V0 – E)]}
= A*A/{1 + sinh2(a)(4V02)/[16E(V0 – E)]}
= A*A/{1 + sinh2(a)/[(4E/V0)(1 – E/V0)]}
 So T = |F|2/|A|2 = 1/{1 + sinh2(a)/[(4E/V0)(1 – E/V0)] }.
b.
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When a >> 1, sinh(a) = (ea + e-a)/2 = ea/2, and sinh2(a) = e2a/4.
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So T = 1/{1 + e2a/[(16E/V0)(1 – E/V0)]}.
The first 1 in the denominator is negligible compared with the very large e2a, so
T = (16E/V0)(1 – E/V0)e-2a.