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Chapter 4
The door beckoned, so he pushed through it. Even the street was better lit. He didn’t know how, he just knew—the patron seated at the corner table was his man. Spectacles, unkept hair, old sweater, one bottle, one glass. Elbows on the bar and foot on the rail, he strained for calmness until the barkeep finally appeared. A meek “bourbon, rocks,” left his lips, then he shuffled a bill he thought was a five onto the bar, pocketed the change, tugged on the drink, and let his feet move him to the corner table. “I wanna talk to you....” The stranger slowly greeted him with his eyes, as if he was expected, and replied “What about, son?” Resisting the urge to look away, the words just fell “.. .those Rutherford rays.. .”
Scattering in One Dimension
The free state addressed in the last chapter is the simplest problem because the potential is chosen to be zero. The next simplest problems are those where the potentials are piecewise constant. A potential that is piecewise constant is discontinuous at one or more points. The potential is chosen to be zero in one region and is of non-zero value in other regions without a transition region. A discontinuity in a potential is not completely realistic though these problems do model some realistic systems well. Transistors, semiconductors, and alpha decay (Rutherford rays) are examples for which an abrupt change in potential is a key feature. Though they are not perfect models, piecewise constant potentials do contain some realistic physics and serve to illustrate features of potentials that are very steep.
The primary reason to address these problems is that a discontinuous change in potential is much more mathematically tractable than a continuous change. These problems are usually addressed using the position space, time-independent form of the Schrodinger equation. The strategy is to divide space into regions at the locations where the potential changes and then attain a solution for each region. The points at which the potential changes are boundaries. The wavefunction and its first derivative must be the same on both sides of the boundaries. Equating the wavefunctions and their first derivatives at the boundaries provides a system of equations that yield descriptive information. Also, solving problems in one dimension is usually mathematically attractive and can be realistic. One dimensional situations are often a precursor to multi-dimensional situations.
There is some key quantum mechanical behavior in these problems. You should absorb the idea that a particle has a non-zero probability to appear on the other side of a potential barrier that it does not classically have the energy to surmount. This is known as barrier penetration or tunneling. Also, a particle has a non-zero probability to be reflected at any boundary regardless of energy. For instance, a particle possessing enough energy to surmount a potential barrier can still be reflected. You should learn the terms reflection and transmission coefficients.
A particle is in either a free state (chapter 3), a bound state (chapter 5, 6, and parts of others), or a scattering state. The bound state is described by a potential that holds a particle for a non-zero time period. The scattering state can be described as an interaction of a free particle with a potential that results in a free particle. If a free particle interacts with the potential and does not become bound, the particle is in a scattering state. The hydrogen atom may be ionized. The electron can escape and enter a free state, but it cannot escape without the influence of an external photon. Scattering concerns the situation where there is no external influence.
Finally, you should also appreciate the presence of the postulates of quantum mechanics.
Though the position space, time-independent Schrodinger equation dominates the discussion, re- member that it is simply a convenient form of the sixth postulate.
- Find the reflection and transmission coefficients for a particle of energy E > V 0 incident from the left on the vertical step potential
V (x) =
0 if x < 0 region 1 , V 0 if x > 0 region 2.
Notice that region 2 extends to infinity. This is an approximation to a potential that is very steep but not perfectly vertical, and of significant though not of infinite width. If the step is not vertical, it is difficult to match boundary conditions, and if the step is not of infinite extent, “tunneling” (see problem 3) is possible.
Particles that are incident from the left will either be reflected or transmitted. The reflection coefficient, denoted R , is classically defined as the ratio of intensity reflected to intensity inci- dent. The classical transmission coefficient, denoted T , is the ratio of intensity transmitted to intensity incident. Quantum mechanically, intensity is analogous to probability density. Quantum mechanical transmission and reflection coefficients are based on probability density flux (problem 2). The reflection and transmission coefficients must sum to 1 in either classical or quantum mechanical regimes.
The two physical boundary conditions applicable to this and many other boundary value problems are the wave function and its first derivative are continuous.
A general solution to the Schrodinger equation for a particle approaching from the left is
ψ 1 (x) = A eik^1 x^ + B e−ik^1 x, x < 0 ,
ψ 2 (x) = C eik^2 x, x > 0 ,
where the subscripts on the wavefunctions and wave numbers indicate regions 1 or 2, and
k 1 =
2 mE ¯h
and k 2 =
2 m (E − V 0 ) ¯h
The ki are real because E > V 0. In region 1, the term A eik^1 x^ is the incident wave and B e−ik^1 x describes the reflected wave. C eik^2 x^ models the transmitted wave in region 2. A term like D e−ik^2 x^ , modeling a particle moving to the left in region 2, is not physically meaningful for a particle given to be incident from the left. Equivalently, D = 0 for the same reason. Applying the boundary condition that the wavefunction is continuous,
ψ 1 (0) = ψ 2 (0) ⇒ A e^0 + B e^0 = C e^0 ⇒ A + B = C.
The derivative of the wavefunction is
ψ 1 ′ (x) = A ik 1 eik^1 x^ − B ik 1 e−ik^1 x, x < 0 ,
potential variations. This is a non-trivial calculation that is beyond our scope. The point is that the boundary conditions of continuity of the wavefunction and its first derivative are also justified quantum mechanically.
- (a) Calculate the amplitude C of the transmitted portion of the wavefunction in terms of the coefficient A for the vertical step potential of problem 1.
(b) Show that the ratio | C | 2 / | A | 2 is 6 = T from problem 1.
(c) Rectify the discrepancy between part (b) and the result of problem 1.
Parts (a) and (b) are intended to demonstrate a counter-intuitive fact that is explained in part (c). Use intermediate results from problem 1 to solve for C in terms of A. Form the ratio | C | 2 / | A | 2 which is not the same as T from problem 1. Part (c) requires that you know that flux is velocity times intensity. Since the particle has a different height above the floor of the potential in regions 1 and 2, it has not only a different wavenumber but a different velocity. These velocities are related to the de Broglie relation. The ratio v 2 | C | 2 /v 1 | A | 2 is the transmission coefficient found in problem 1.
(a) Using intermediate results from problem 1, the amplitude of the transmitted portion is
C = A + B = A +
k 1 − k 2 k 1 + k 2
A =
k 1 + k 2 k 1 + k 2
A +
k 1 − k 2 k 1 + k 2
A =
2 k 1 k 1 + k 2
A.
(b) The ratio of the squares of the magnitudes is not the transmission coefficient of problem 1,
∣ ∣ (^) C
∣^2
A
2 k 1 k 1 + k 2
| A | 2
| A | 2 =
4 k^21 ( k 1 + k 2
) 2 6 =^
4 k 1 k 2 ( k 1 + k 2
(c) These differ because the energy, and thus the velocity, of the wave changes as the particle crosses the step. Flux is intensity times velocity. Probability density flux, probability density
times velocity, is appropriate for a quantum mechanical description. Since vi =
pi m
¯hki m
T =
v 2
C
v 1
∣ A
∣^2
¯hk 2 /m
C
¯hk 1 /m
∣ A
∣^2
k 2
C
k 1
∣ A
∣^2
= k 2
2 k 1 k 1 + k 2
| A | 2
k 1 | A | 2 =
4 k 1 k 2 ( k 1 + k 2
consistent with the problem 1.
Postscript: A reflected wave is the same height above the floor of the potential as the incident wave so the reflected wave has the same energy, velocity, and wave number as the incident wave. The velocities cancel in region 1 so are not considered in calculating the reflection coefficient, i.e.,
R =
v 1
B
v 1
∣ A
∣^2
B
∣ A
∣^2
- Determine the transmission coefficient for a particle with E < V 0 incident from the left on the rectangular barrier
V (x) =
V 0 for −a < x < a , 0 for | x | > a.
If the energy of the particle is less than the height of the step, and the “ potential plateau” is of finite length, the particle incident from the left can appear on the right side of a barrier. This is a non–classical phenomena known as barrier penetration or tunneling. Classically, if a ball is rolled up a ramp of height h with kinetic energy K , the ball will roll back down the ramp if K < mgh. A quantum mechanical ball has a non–zero probability that the ball would appear on the other side of the ramp in the case K < mgh.
This is a boundary value problem with three regions because there are two boundaries. The strategy is to require continuity of the wavefunction and its first derivative at all boundaries and solve for the transmission coefficients in terms of the ratios of the squares of the appropriate wavefunction coefficients.
(a) Divide “all space” into three regions at the boundaries of the potential ±a. The wavefunctions consist of a linear combination of waves in both directions in each of the three regions:
ψ 1 (x) = A eikx^ + B e−ikx^ for x < −a , ψ 2 (x) = C eκx^ + D e−κx^ for − a < x < a , ψ 3 (x) = F eikx^ + G e−ikx^ for x > a ,
where k =
2 mE /¯h and κ =
2 m (V 0 − E ) /¯h. Notice that κ is defined so that it is real for E < V 0. This is a technique that leads to minor simplifications. Conclude that G = 0 because it is the coefficient of an oppositely directed incident wave, and therefore, is not physical.
(b) Apply the boundary condition that the wavefunction must be continuous at the boundaries. This yields two equations. Differentiate the wavefunction and then apply the boundary condition that the first derivative of the wavefunction must be continuous at boundaries. This yields two more equations. Eliminating B from the two equations for the left boundary, you should find
2 A e−ika^ =
iκ k
C e−κa^ +
iκ k
D eκx.
Using the two equations at the right boundary to solve for two of the unknown coefficients in terms of the coefficient F , you should find
C eκa^ =
ik κ
F eika^ and D e−κa^ =
ik κ
F eika.
(c) Use the last two equations to eliminate the coefficients C and D from the first equation so
A e−ika^ = F eika
[
cosh (2κa) +
i(κ^2 − k^2 ) 2 kκ
sinh (2κa)
]
⇒ 2 A e−ika^ =
iκ k
C e−κa^ +
iκ k
D eκa. (5)
Multiplying equation (2) by κ and solving for the term with the coefficient D,
D κ e−κa^ = F κ eika^ − C κ eκa
Substituting the right side into equation (4) for D κ e−κa,
C κ eκa^ − F κ eika^ + C κ eκa^ = F ik eika
⇒ 2 C κ eκa^ = F κ eika^ + F ik eika^ = F (κ + ik) eika
⇒ C eκa^ =
ik κ
F eika. (6)
Multiplying equation (2) by κ and solving for the term with the coefficient C,
C κ eκa^ = F κ eika^ − D κ e−κa
Substituting the right side into equation (4) for C κ eκa,
F κ eika^ − D κ e−κa^ − D κ e−κa^ = F ik eika
− 2 D κ e−κa^ = −F κ eika^ + F ik eika^ = F (−κ + ik) eika
⇒ D e−κa^ =
ik κ
F eika. (7)
(c) The signs on the exponentials in equations (6) and (7) are opposite those in equation (5), so
C eκa^ =
ik κ
F eika^ ⇒ C e−κa^ =
ik κ
F eikae−^2 κa,
D e−κa^ =
ik κ
F eika^ ⇒ D eκa^ =
ik κ
F eikae^2 κa,
and now the signs of the exponentials are consistent. Substituting these into equation (5),
⇒ 2 A e−ika^ =
iκ k
ik κ
F eikae−^2 κa^ +
iκ k
ik κ
F eikae^2 κa
F eika 2
iκ k
ik κ
κk κk
e−^2 κa^ +
F eika 2
iκ k
ik κ
κk κk
e^2 κa
F eika 2
2 e−^2 κa^ −
iκ k
e−^2 κa^ +
ik κ
e−^2 κa^ + 2e^2 κa^ +
iκ k
e^2 κa^ −
ik κ
e^2 κa
F eika 2
[
e−^2 κa^ + e^2 κa
iκ^2 kκ
e−^2 κa^ +
ik^2 kκ
e−^2 κa^ +
iκ^2 kκ
e^2 κa^ −
ik^2 kκ
e^2 κa
]
F eika 2
[
e−^2 κa^ + e^2 κa
iκ^2 kκ
e^2 κa^ − e−^2 κa
ik^2 kκ
e^2 κa^ − e−^2 κa
]
= F eika
[
e^2 κa^ + e−^2 κa 2
i (κ^2 − k^2 ) kκ
e^2 κa^ − e−^2 κa 2
)]
⇒ A e−ika^ = F eika
[
cosh (2κa) +
i (κ^2 − k^2 ) 2 kκ
sinh (2κa)
]
(d) The transmission coefficient is the ratio of intensity transmitted, represented by | F | 2 , to the intensity incident, represented by | A | 2. It is conventional to calculate the reciprocal of this ratio to arrive at a compact expression. Both sides are in the polar form of complex numbers, i.e., magnitude and phase. The last result can be arranged as the ratio ∣∣ ∣∣^ A e
−ika F eika
[
cosh (2κa) +
i (κ^2 − k^2 ) 2 kκ
sinh (2κa)
] ∣∣
Now
∣∣^ A e
−ika F eika
[
A e−ika F eika
A∗eika F ∗e−ika
] 1 / 2
[
A A∗
F F ∗
] 1 / 2
[
A A∗^
] 1 / 2
[
F F ∗^
] 1 / 2 =
∣ A
∣ F
| A | 2
| F | 2
[
cosh (2κa) +
i(κ^2 − k^2 ) 2 kκ
sinh (2κa)
] ∣
2 = cosh^2 (2κa) +
(κ^2 − k^2 ) 2 kκ
sinh^2 (2κa),
where we have used the fact the product of complex conjugates is sum of the squares of the real part and the coefficient of the imaginary part. Recalling that cosh^2 (x) = 1 + sinh^2 (x ) ,
| A | 2 | F | 2
= 1 + sinh^2 (2κa) +
κ^4 − 2 k^2 κ^2 + k^4 4 k^2 κ^2
sinh^2 (2κa)
κ^4 − 2 k^2 κ^2 + k^4 4 k^2 κ^2
sinh^2 (2κa)
4 k^2 κ^2 + κ^4 − 2 k^2 κ^2 + k^4 4 k^2 κ^2
sinh^2 (2κa)
κ^4 + 2k^2 κ^2 + k^4 4 k^2 κ^2
sinh^2 (2κa)
κ^2 + k^2
4 k^2 κ^2
sinh^2 (2κa).
Substituting k =
2 mE / ¯h and κ =
2 m (V 0 − E ) / ¯h ,
| A | 2
| F | 2
2 m ¯h^2
(V 0 − E ) +
2 m ¯h^2
E
2 m ¯h^2
E
2 m ¯h^2
(V 0 − E )
) (^) sinh^2
2 a ¯h
2 m (V 0 − E )
4 m^2 ¯h^4
(V 0 − E + E )^2
4 m^2 ¯h^4
4 E (V 0 − E )
sinh^2
2 a ¯h
2 m (V 0 − E )
| A | 2
| F | 2
= T −^1 = 1 +
V 02
4 E (V 0 − E )
sinh^2
2 a ¯h
2 m (V 0 − E )
- Determine the transmission coefficient for a particle with E = V 0 incident from the left on the rectangular barrier defined in the last problem.
Substituting equation (4) into the last equation eliminates the unknown coefficient D ,
2 A ik e−ika^ = ik F eika^ + F ik eika^ (1 − 2 ika) ⇒ 2 A e−ika^ = F eika^ + F eika^ (1 − 2 ika) ⇒ 2 A e−ika^ = F eika^ (2 − 2 ika) ⇒ A e−ika^ = F eika^ (1 − ika)
⇒
A e−ika F eika^
= 1 − ika
and
∣∣^ A e
−ika F eika
∣∣^ A
F
∣∣ = |^ A^ |
| F |
as previously shown,
| A |^2
| F |^2
1 − ika
= 1 + k^2 a^2.
Substituting k =
2 mE / ¯h ,
| A |^2
| F |^2
= T −^1 = 1 +
2 mE ¯h^2
a^2 for E = V 0.
- Show in the limit E → V 0 from below, that the results of problems 3 and 4 are equivalent.
The intent of this problem is to demonstrate self-consistency by examining a limiting case. It is a technique that is difficult to teach and important to learn. Start with the result of problem 3. Expand sinh (x) and ignore the higher order terms so that sinh (x) ≈ x since E ≈ V 0. Of course, in the limit E → V 0 the results of problem 3 and 4 must be the same.
Start with the reciprocal of the transmission coefficient from problem 3. Ignoring higher order terms as negligible in the series expansion of sinh (x) ⇒ sinh (x) ≈ x ,
T −^1 = 1 +
V 02
4 E (V 0 − E )
2 a ¯h
2 m (V 0 − E )
V 02
4 E (V 0 − E )
4 a^2 ¯h^2
2 m (V 0 − E )
V 02
E
a^2 ¯h^2
2 m
2 mV 02 ¯h^2 E
a^2.
Remembering that E ≈ V 0 , T −^1 ≈ 1 +
2 mE ¯h^2
a^2.
- (a) A particle of energy E > V 0 is incident on a rectangular barrier of width 2a. Solve for the value of the transmission coefficient and locate the positions of the maxima and minima in terms of wave number. Plot transmission coefficient versus barrier width for a particle of constant energy and variable barrier width.
(b) Locate the positions of the maxima in terms of particle energy. Plot transmission coefficient versus particle energy where E > V 0 and barrier width is fixed.
This problem introduces a phenomenon known as resonance scattering. At certain values of particle wave number (energy), there is 100% transmission and no reflection. Part (a) should illustrate resonance scattering for an incident particle of fixed energy and a barrier width that is varied. Part (b) illustrates the same phenomenon where particle energy is varied and barrier width is fixed. Start with an intermediate result from problem 10,
T
k 12 − k^22 k 1 k 2
sin^2 (2k 2 a ).
Solve for T instead of its reciprocal. The maxima occur where the sine term is zero at barrier widths of 2 a = nπ/k 2. The minima occur where the sine term is one, where the transmission
coefficient has the value T =
2 k 1 k 2 k^21 + k^22
. Part (b) is likely easiest if started with
T
V 02
4 E (E − V 0 )
sin^2
2 a ¯h
2 m (E − V 0 )
the result of problem 10 in terms of energy. Again, solve for T instead of its reciprocal. This plot is more difficult because the independent variable E appears in five places including the square root of the sine squared term. The important part of this graph is the shape. If you do not have access to a commercial graphing package, try to imagine what the graph should look like, and skip to the solution. You should find the maxima at
E =
π^2 ¯h^2 8 ma^2
n^2 + V 0.
This result is closely related to the infinite square well that is encountered in the next chapter.
(a)
T
k^21 − k 22 k 1 k 2
sin^2 (2k 2 a) =
4 k 12 k 22 4 k 12 k 22
k 14 − 2 k 12 k 22 + k^42 4 k 12 k 22
sin^2 (2k 2 a)
4 k^21 k 22 +
k^41 − 2 k^21 k^22 + k^42
sin^2 (2k 2 a) 4 k^21 k^22
⇒ T =
4 k^21 k^22 4 k^21 k 22 +
k^41 − 2 k^21 k^22 + k^42
sin^2 (2k 2 a)
Maxima of T = 1 occur where sin^2 (2k 2 a) = 0 , i.e., where 2 k 2 a = nπ ⇒ 2 a =
nπ k 2
are
the barrier widths at which maxima occur, noting that the width of the barrier is given to be 2a. The minima occur where the sine squared term is one, so minima occur at
T =
4 k 12 k 22 4 k^21 k^22 +
k 14 − 2 k^21 k^22 + k 24
4 k^21 k^22 k^41 + 2k^21 k^22 + k 24
⇒ T =
2 k 1 k 2 k^21 + k^22
at minima.
If we integrate around the point which is the discontinuity, this is
−
¯h^2 2 m
−≤
d^2 ψ (x) dx^2
dx +
−≤
V (x) ψ (x) dx = E
−≤
ψ (x) dx,
which assumes that the infinite boundary is at x = 0. The quantity ≤ is infinitesimal. If we let ≤ → 0, the right side of the equation is zero. The integral of a finite quantity, ψ (x) , over an arbitrarily small interval is zero. The term with the potential does not vanish because the integral of an infinite quantity, the potential, can be non-zero even for an arbitrarily small interval. So,
lim ≤→ 0
−≤
d^2 ψ (x) dx^2
dx =
2 m ¯h^2
lim ≤→ 0
−≤
V (x) ψ (x) dx
d ψ (x) dx
≤→ 0
d ψ (x) dx
−≤→ 0
2 m ¯h^2
lim ≤→ 0
−≤
V (x) ψ (x) dx
ψ′^ (x)
2 m ¯h^2
lim ≤→ 0
−≤
V (x) ψ (x) dx (1)
is the general value of the discontinuity in the first derivative.
(b) A delta function potential is infinite, so employing equation (1),
∆
ψ′^ (x)
x = 0
2 m ¯h^2
lim ≤→ 0
−≤
− α δ (x)
ψ (x) dx = −
2 m ¯h^2
α ψ (0) ,
because regardless of how closely ≤ approaches zero, −≤ < 0 < ≤ meaning that zero remains within the limits of integration. Since the value of the delta function that makes its argument zero is within the limits of integration, the integral is the wavefunction evaluated at the point that makes the argument of the delta function zero.
Postscript: An infinite potential models a perfectly rigid and perfectly impenetrable wall. The wavefunction must be zero at an infinite potential barrier and in the region of infinite potential because the term V ψ in the Schrodinger equation would be infinite otherwise. There is a non- smooth “corner” in the wavefunction as it goes to zero at any position other than ±∞ , thus there is a discontinuity in the first derivative of the wavefunction. This physically unrealistic model is useful to understanding concepts, is mathematically tractable, and is a first approximation to an abrupt and large potential.
- Calculate the reflection and transmission coefficients for a particle incident on the potential
V (x) = −α δ (x).
This problem is intended to demonstrate how a potential that includes a delta function is treated specifically, and how to treat a discontinu- ity in general. First, write the wavefunction for two regions because the delta function is of “zero” width. The wavenumber is the same in regions 1 and 2 because the particle energy is the same height above the potential “floor” in both regions. A ψ (0) is required. Use either ψ 1 (0) or ψ 2 (0) , because the condition of continuity of the wavefunc- tion at the boundary ensures that they are the same. We use ψ 2 (0) because it has one fewer terms than ψ 1 (0). Classically, the reflection coefficient must be zero. The quantum mechanical result is non-zero.
The solution to the Schrodinger equation in position space is
ψ 1 (x) = A eikx^ + B e−ikx, x < 0 , ψ 2 (x) = C eikx, x > 0 ,
where the infinite potential well is at x = 0. Continuity of the wave function means
A eik(0)^ + B e−ik(0)^ = C eik(0)^ ⇒ A + B = C.
The first derivatives in regions 1 and 2 are
ψ 1 ′ (x) = ik A eikx^ − ik B e−ikx, ψ′ 2 (x) = ik C eikx.
We calculated the general value of the discontinuity of the first derivative in problem 7. This is the difference of the first derivatives in regions 1 and 2, meaning
ψ′ 1 (0) − ψ 2 ′ (0) = −
2 m ¯h^2
α ψ (0)
⇒ ik A eik(0)^ − ik B e−ik(0)^ − ik C eik(0)^ = −
2 m ¯h^2
α ψ (0).
Using ψ 2 (0) = C ,
ik A − ik B − ik C = −
2 m ¯h^2
α C
⇒ ik A − ik B = C
ik −
2 m ¯h^2
α
and we use the continuity condition, A + B = C , to eliminate C , so
ik A − ik B =
A + B
ik −
2 m ¯h^2
α
⇒ ik A − ik B = A
ik −
2 m ¯h^2
α
+ B
ik −
2 m ¯h^2
α
⇒ A
ik
− ik
2 m ¯h^2
α
= B
ik + ik −
2 m ¯h^2
α
⇒ A
2 m ¯h^2
α
= B
2 ik −
2 m ¯h^2
α
⇒ B = A
[(
2 m ¯h^2
α
2 ik −
2 m ¯h^2
α
)]
mαA ik¯h^2 − mα
The reflection coefficient is
R =
∣ B
∣^2
A
mαA∗ −ik¯h^2 − mα
mαA ik¯h^2 − mα
| A | 2
m^2 α^2 k^2 ¯h^4 + m^2 α^2
1 + k^2 ¯h^4
m^2 α^2
The transmission coefficient is
T = 1 − R = 1 −
m^2 α^2 k^2 ¯h^4 + m^2 α^2
k^2 ¯h^4 + m^2 α^2 − m^2 α^2 k^2 ¯h^4 + m^2 α^2
k^2 ¯h^4 k^2 ¯h^4 + m^2 α^2
1 + m^2 α^2
k^2 ¯h^4
- Show in the limit E → V 0 from above, that the result of problems 4 and 10 are equivalent.
This problem has the same intent as problem 5. Start with the result of problem 10. Expand sin (x) and ignore the higher order terms so that sin (x) ≈ x since E ≈ V 0. You must find that in the limit E → V 0 the results of problems 10 and 4 are the same.
- A particle of E > 0 approaches the negative vertical step potential
V (x) =
0 , for x < 0, −V 0 for x > 0.
What are the reflection and transmission coefficients if E = V 0 /3 , and E = V 0 /8?
This problem parallels the discussion of the vertical step potential. It is intended to reinforce the methods of addressing a boundary value problem. It also illustrates another completely non- classical phenomenon. Classically, we expect the reflection coefficients to be zero for any E > 0.
(a) Write the wavefunction. You have only two regions to consider. You should recognize that the general form of the wavefunction is
ψ 1 (x) = Aeik^1 x^ + Be−ik^1 x^ for x < 0 ,
ψ 2 (x) = Ceik^2 x^ + De−ik^2 x^ for x > 0 ,
where k 1 =
2 mE / ¯h and k 2 =
2 m (E − (−V 0 )) / ¯h =
2 m (E + V 0 ) / ¯h. Can you conclude that D = 0 because it is the coefficient of an oppositely directed incident wave?
(b) Apply the continuity conditions to the wavefunction and its first derivative at x = 0. You should have two equations in three unknowns. The square of the coefficient A is the intensity of incidence, and the square of the coefficient B is the intensity of reflection. Since A and B are the two coefficients you want to compare, you should combine your two equations to eliminate C. You should find that A (k 1 − k 2 ) = B (k 1 + k 2 ).
(c) The reflection coefficient is the ratio | B | 2 / | A | 2 , which you can form from your result of part (b). Use the definitions of the wavenumbers to establish R in terms of E and V 0 , then substitute E = V 0 /3 and E = V 0 /8 to get numerical answers. You should find R = 1/9 and 1/ 4 , respectively, and the transmission coefficients follow directly using R + T = 1.
- (a) Find the transmission coefficient for the rectangular well
V (x) =
−V 0 , for −a < x < a , 0 , for | x | > a.
(b) Show that T = 1 in the limit that V 0 = 0.
This problem is provided to give you some practice at solving a boundary value problem. The procedures for this problem strongly parallel problems 3 and 4. You should find
T −^1 = 1 +
V 0 2
4 E (E + V 0 )
sin^2
2 a ¯h
2 m (E + V 0 )
using k 1 =
2 mE / ¯h and k 2 =
2 m (E − (−V 0 )) / ¯h =
2 m (E + V 0 ) / ¯h. Consider the width of the well when V 0 → 0 for part (b).
- The transmission coefficient a particle of E < V 0 approaching a rectangular barrier is
T =
V 02
4 E (V 0 − E )
sinh^2
2 a ¯h
2 m (V 0 − E )
(a) Show that the term
V 02
4 E (V 0 − E )
sinh^2
2 a ¯h
2 m (V 0 − E )
is non–negative and real, and
(b) show that this means R < 1 for the case E < V 0.
The transmission coefficient is non-zero for the case E < V 0 , which is nonsense in a classical regime. Since quantum mechanics “lives” in the world of complex numbers, this problem shows that a non–zero transmission coefficient at a barrier is not a “trick” associated with complex or imaginary numbers, and thus, there is actually a portion transmitted and only a portion reflected when E < V 0. Start with the result of problem 3. There are energies, intrinsically positive and real, two squares, and a product to consider. Can any of those make this term negative? Once you know that the given term is non–negative and real, set it equal to a constant. Part (b) is the algebra of showing that if the constant is positive and real, R is necessarily less than one.
- Plot barrier width versus transmission coefficient for a rectangular barrier of width 2a for the case E = V 0. What widths of the barrier result in 10%, 1%, 0 .1%, and 0 .01% transmission when particle energy is equal to barrier height?
This problem illustrates that even a thin barrier prevents most transmission at E = V 0. A barrier a few wavelengths in width results in miniscule transmission. Start with the reciprocal of the transmission coefficient from problem 4 expressed in terms of energy and half barrier width. Use the de Broglie relationship to express the energy in terms of wavelength, then solve for 2a. The barrier width that results in 10% transmission is slightly less than one particle wavelength.
- Calculate the reflection and transmission coefficients for a particle incident on the potential
V (x) = α δ (x).
Problems 16 and 17 are intended to reinforced the ideas and the procedures in problems 7 and 8. You should find the same reflection and transmission coefficients as found in problem 8.
Chapter 4 Homework Solutions
- For the case E > V 0 , the development is very similar to that of E < V 0. The primary differences are all exponentials have imaginary factors, and we “add zero,” add and subtract the same quantity, to arrive at a form of the transmission coefficient analogous to the case of E < V 0. The wave function is
ψ(x) =
A eik^1 x^ + B e−ik^1 x, for x < −a, C eik^2 x^ + D e−ik^2 x, for −a < x < a, F eik^1 x, for x > a,
where k 1 =
2 mE/¯h and k 2 =
2 m(E − V 0 )/¯h. Note since E > V 0 , we define the wave number in terms of E − V 0 in the region −a < x < a versus V 0 − E in the case E < V 0. The derivative is
ψ′(x) =
A ik 1 eik^1 x^ − B ik 1 e−ik^1 x, for x < −a, C ik 2 eik^2 x^ − D ik 2 e−ik^2 x, for −a < x < a, F ik 1 eik^1 x, for x > a.
The continuity conditions are at x = −a,
A e−ik^1 a^ + B eik^1 a^ = C e−ik^2 a^ + D eik^2 a, (1)
and at x = a, C eik^2 a^ + D e−ik^2 a^ = F eik^1 a. (2)
Applying the boundary condition of continuity of the derivative at x = −a,
A ik 1 e−ik^1 a^ − B ik 1 eik^1 a^ = C ik 2 e−ik^2 a^ − D ik 2 eik^2 a. (3)
At x = a, C ik 2 eik^2 a^ − D ik 2 e−ik^2 a^ = F ik 1 eik^1 a. (4)
Multiplying equation (1) by ik 1 and solving for the term with the coefficient of B,
B ik 1 eik^1 a^ = C ik 1 e−ik^2 a^ + D ik 1 eik^2 a^ − A ik 1 e−ik^1 a.
Substituting the right side into equation (3),
A ik 1 e−ik^1 a^ − C ik 1 e−ik^2 a^ − D ik 1 eik^2 a^ + A ik 1 e−ik^1 a^ = C ik 2 e−ik^2 a^ − D ik 2 eik^2 a
⇒ 2 A k 1 e−ik^1 a^ = C (k 1 + k 2 ) e−ik^2 a^ + D (k 1 − k 2 ) eik^2 a
⇒ 2 A e−ik^1 a^ = C
k 1 + k 2 k 1
e−ik^2 a^ + D
k 1 − k 2 k 1
eik^2 a. (5)
Multiplying equation (2) by ik 2 and solving for the term with the coefficient of D,
D ik 2 e−ik^2 a^ = F ik 2 eik^1 a^ − C ik 2 eik^2 a.
Substituting this into equation (4),
C ik 2 eik^2 a^ − F ik 2 eik^2 a^ + C ik 2 eik^2 a^ = F ik 1 eik^1 a
⇒ 2 C k 2 eik^2 a^ = F (k 1 + k 2 ) eik^1 a
⇒ C =
F
k 1 + k 2 k 2
eik^1 ae−ik^2 a.
Multiplying equation (2) by ik 2 and solving for the term with the coefficient of C,
C ik 2 eik^2 a^ = F ik 2 eik^1 a^ − D ik 2 e−ik^2 a
and using this in equation (4),
F ik 2 eik^1 a^ − D ik 2 e−ik^2 a^ − D ik 2 e−ik^2 a^ = F ik 1 eik^1 a
− 2 D k 2 e−ik^2 a^ = F (k 1 − k 2 ) eik^1 a
D =
F
k 2 − k 1 k 2
eik^1 aeik^2 a.
Substituting the relationships for C and D into equation (5) yields
2 A e−ik^1 a^ =
F
k 1 + k 2 k 2
k 1 + k 2 k 1
eik^1 ae−^2 ik^2 a^ +
F
k 2 − k 1 k 2
k 1 − k 2 k 1
eik^1 ae^2 ik^2 a
F
eik^1 a
k^21 + 2k 1 k 2 + k 22
k 1 k 2
e−^2 ik^2 a^ −
F
eik^1 a
k 12 − 2 k 1 k 2 + k^22
k 1 k 2
e^2 ik^2 a
F
2 k 1 k 2
eik^1 a
[
k^21 e−^2 ik^2 a^ + 2k 1 k 2 e−^2 ik^2 a^ + k^22 e−^2 ik^2 a
− k 12 e^2 ik^2 a^ + 2k 1 k 2 e^2 ik^2 a^ − k 22 e^2 ik^2 a
]
F
2 k 1 k 2
eik^1 a
[
− k 12
e^2 ik^2 a^ − e−^2 ik^2 a
e^2 ik^2 a^ + e−^2 ik^2 a
− k^22
e^2 ik^2 a^ − e−^2 ik^2 a
) ]
F
2 k 1 k 2
eik^1 a^
[
− 2 ik^21 sin(2k 2 a) + 4k 1 k 2 cos(2k 2 a) − 2 ik^22 sin(2k 2 a)
]
= F eik^1 a
[
2 cos(2k 2 a) − i
(k^21 + k^22 ) k 1 k 2
sin(2k 2 a)
]
A e−ik^1 a F eik^1 a^
[
cos(2k 2 a) −
i 2
(k 12 + k^22 ) k 1 k 2
sin(2k 2 a)
]
Since
Ae−ik^1 a F eik^1 a
A
F
∣ =^
| A |
| F |
| A |^2
| F |^2
∣∣cos(2k 2 a) − i 2
(k 12 + k^22 ) k 1 k 2
sin(2k 2 a)
2
= cos^2 (2k 2 a) +
k 12 + k^22 k 1 k 2
sin^2 (2k 2 a).
Adding zero in the form of sin^2 (2k 2 a) − sin^2 (2k 2 a),
| A |^2 | F |^2
= cos^2 (2k 2 a) + sin^2 (2k 2 a) − sin^2 (2k 2 a) +
k^21 + k 22 k 1 k 2
sin^2 (2k 2 a)
[
k 12 + k^22 k 1 k 2
]
sin^2 (2k 2 a). (6)
