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L.J. Sham May 8, 2009
Physics 130A Problem Set 5
- Eigenstates in a square well with odd parity Griffiths, Problem 2.29 on p. 82.
- Scatter by a step potential Griffiths, Problem 2.35 on p. 84.
- Esaki diode The tunneling probability for the electrons through a tunnel diode is increased by a factor of 10 when the oxide layer thickness is halved. Find the ratio of the tunneling probabilities when the oxide thickness of the original diode is reduced to one-fourth.
- Scattering by a delta potential The potential of a particle of mass m is given by,
V (x) =
ℏ^2 λ m
δ(x), (1)
where λ may be positive or negative.
(a) In the case of repulsive potential, find the transmission and reflection coefficients inde- pendently for energy E > 0 and show that they add up to unity. (b) in the attractive case, show that the transmission coefficient for the positive incoming particle energy ℏ^2 k^2 / 2 m > 0 has a pole in the continuation to the complex k plane which coincides with the bound state imaginary wave vector.
L.J. Sham May 8, 2009
Physics 130A Problem Set 5
- Eigenstates in a square well with odd parity Griffiths, Problem 2.29 on p. 82.
Solution – Use the potential and notations in Lecture 12. The wave functions are odd in x, i.e.,ψ(−x) = −ψ(x). Hence, the solution may be written as,
for |x| < a, ψ(x) = B sin kx; for |x| > a, ψ(x) = sgn(x)De−κ|x|, (1) where sgn(x) = +1, if x > 0; sgn(x) = − 1 , if x < 0. (2)
We have again excluded the exponentially growing solutions dominating at long distances. By symmetry, the boundary conditions satisfied at x = a will also be satisfied at x = −a. Hence,
B sin ka = De−κa, kB cos ka = −κDe−κa. (3)
Division of the right equation by the left one leads to,
k cot ka = −κ. (4)
Letting ξ = ka, η = κa, we obtain the two relations:
ξ^2 + η^2 = 2 mV a^2 /ℏ^2 , (5) ξ tan
( (^) π 2
= η. (6)
The graphical solutions are also shown as the dashed lines in Fig. 1.
η
0 π/2 π 3π/2 ξ
Figure 1: Square well solutions. The circular segment is given by ξ^2 + η^2 = 16. The dashed lines, ξ tan(ξ + π/2) = η, lead to the odd parity solutions. The roots are circled.
In the limiting case where 2 mV a^2 /ℏ^2 < π/ 2 , there are no odd-parity bound states.
(b) The mix-up can only come from looney thinking from watching too many Looney Tunes. Since the gravitational potential energy is proportional to the height, the figure represents Bugs Bunny’s potential energy as a function of his horizontal distance x perfectly. As he comes off the edge, his height will drop if he knows physics but his total energy (kinetic plus potential) is still a constant E as in the scattering theory. “Thatttt is all, Folks!” (c) Again, E = V 0 / 3. So R = 19 and T = 1 − R = 89.
- Esaki diode The tunneling probability for the electrons through a tunnel diode is increased by a factor of 10 when the oxide layer thickness is halved. Find the ratio of the tunneling probabilities when the oxide thickness of the original diode is reduced to one-fourth.
Solution – The approximate formula for tunneling is [see Lecture 16, Eq. (20)],
Tn ≈ Tpe−^2 κa/n. (17)
The datum given is,
T 2 T 1
e−^2 κa/^2 e−^2 κa^
= eκa^ = 10. (18)
Thus,
T 4 T 1
e−^2 κa/^4 e−^2 κa^
= e
3 2 κa^ = 10^3 /^2 = 32. (19)
- Scattering by a delta potential The potential of a particle of mass m is given by,
V (x) =
ℏ^2 λ m
δ(x), (20)
where λ may be positive or negative.
(a) In the case of repulsive potential, find the transmission and reflection coefficients inde- pendently for energy E > 0 and show that they add up to unity. (b) in the attractive case, show that the transmission coefficient for the positive incoming particle energy ℏ^2 k^2 / 2 m > 0 has a pole in the continuation to the complex k plane which coincides with the bound state imaginary wave vector.
Solution – From Lecture 17, rewrite the Schr¨odinger equation in the form, for the energy eigenstate at energy E > 0 ,
d^2 ψ dx^2
= 2λδ(x)ψ −
2 mE ℏ^2
ψ. (21)
Except at x = 0, the solution for a continuum state, i.e. E > 0 , is
ψ(x) = Aeikx^ + Be−ikx^ if x < 0 , ψ(x) = Ceikx^ + DBe−kx^ if x > 0 , (22)
with the corresponding current density given by,
Jx =
2 mi
Ψ∗(x, t)
∂x
Ψ(x, t) − Ψ(x, t)
∂x
Ψ∗(x, t)
(x) =
ℏk m
(|A|^2 − |B|^2 ) if x < 0 ,
Jx =
ℏk m
(|C|^2 − |D|^2 ) if x > 0. (23)
To find a transmission and reflection coefficient, we inject particles only from the left, say. Then, there are no particles coming in from the right. Hence, D = 0.
To connect the wave functions in the two regions, the boundary conditions are continuity for the wave function,
ψ(x = +0) = ψ(x = −0). (24)
and the change in the first derivative of the wave function by integrating Eq. (21) from a small negative number to a small positive number,
ψ′(+0) − ψ′(−0) = 2λ ψ(0). (25)
(a) Thus, we have
C = A + B, ikC − 2 λC = ik(A − B). (26)
Hence, by multiplying the first equation by ik and adding or subtracting the result to or from the second divided,
C =
k k + iλ
A; B =
λ ik
C =
−iλ k + iλ
A. (27)
T =
C
A
2
k^2 k^2 + λ^2
; R =
B
A
2
λ^2 k^2 + λ^2
Evidently, T + R = 1. (b) Except at x = 0, the solution for a bound state, i.e. E < 0 , is
ψ(x) = Aeκx^ if x < 0 , ψ(x) = Be−κx^ if x > 0 , (29)
where
κ =
− 2 mE/ℏ^2 , (30)
