Physics 264L: Assignment 11
Made available: Saturday, November 28, 2015
Due:
Friday, December 4, 2015, by 6 pm
1. A wave packet that initially decays exponentially
A free particle of mass m in one space dimension has a quantum state Ψ at time t = 0 described by a
function Ψ0 (x)
Ψ(t = 0, x) = Ψ0 (x) = Ae−a|x| ,
(1)
where a > 0 is some positive real-valued decay constant and A is a normalization constant. Eq. (1)
represents a wave packet since Ψ0 (x) is spatially localized, i.e., the function decays to zero on both
sides, in the limit x → ∞ and in the limit x → −∞.
(a) Draw a sketch of Ψ0 (x) and determine (with a brief explanation) the physical units of the parameters A and a.
(b) What value should the coefficient A have so that Ψ0 (x) is normalized?
(c) Show that, when the normalized initial state Eq. (1) is written as a superposition of free-particle
momentum states
[
]
∫ ∞
1
Ψ0 (x) =
c(k) √ eikx dk,
(2)
2π
−∞
the coefficients c(k) (called the Fourier coefficients of the function Ψ0 ) are given by the expression
√
a
2a
c(k) =
.
(3)
2π k 2 + a2
Also draw a sketch of the function c(k) for the case a = 1 and −5a ≤ k ≤ 5a so you understand
the shape of the curve c(k).
Hint: Equation [2.103] on page 62 of the assigned reading in Griffiths (available on Sakai). Please
evaluate the integral by hand, without Mathematica, to maintain or strengthen your mathematical
skills.
(d) Show by direct integration (again by hand, without using Mathematica) that, for c(k) given by
Eq. (3), the area under the curve |c(k)|2 is 1, that is
∫ ∞
|c(k)|2 dk = 1.
(4)
−∞
This implies that |c(k)|2 can be interpreted as a wavenumber probability density in that the
expression |c(k)|2 dk can be interpreted as the probability for an experimental measurement of
the wavenumber of the particle in the state Ψ0 (x) to yield a wavenumber value in the infinitesimal
interval [k, k + dk].
Note: You have established a particular result of a general theorem of Fourier analysis known as
“Parseval’s theorem” which states that, for any mathematically reasonable function f (x) (“reasonable” means that the function doesn’t have too many discontinuities too close together),
∫ ∞
∫ ∞
|f (x)|2 dx =
|c(k)|2 dk,
(5)
∞
−∞
1
√
∫∞
where c(k) = (1/ 2π) ∞ f (x)e−ikx dx is the Fourier coefficient of f (x). So if f (x) is normalized
in “position space”, then the function c(k) that gives the Fourier coefficients of f as a function
of k (this is called the Fourier transform of f (x)), is normalized in “wavenumber space” and vice
versa.
(e) Based on the previous problem, show that the probability prob[0 ≤ p ≤ ℏa] for a momentum
measurement at time t = 0 of the particle’s momentum p to yield a value in the interval [0, ℏa]
approximately has the value 0.41:
prob[0 ≤ p ≤ ℏa] =
2+π
≈ 0.41 .
4π
(6)
Some comments: you need to do the integration by hand (which involves little work given what
you have already done for part (d)), but it is fine if you then use Mathematica to verify your
answer. If you do so, you should use the Assumptions part of the Integrate command to tell
Mathematica what are the properties of various symbols in the integrand or integration bounds,
which will help Mathematica to get its answer. For example, one could type this Mathematica
∫L
code to evaluate the definite integral −L Ψ0 (x) dx for some symbolic length L:
Integrate[
A Exp[- a Abs[x] ] ,
(* integrand Eq. (1) *)
{ x, -L, L } ,
(* integrate between -L and L *)
Assumptions ->
( Elements[a, Reals] && a > 0 ) &&
( Elements[L, Reals] && L > 0 )
]
The last two lines tell Mathematica that the symbol a is a positive real number in the form of
the Boolean expression “a is an element of the set of real numbers R” (a ∈ R) and that “a > 0”.
The second Boolean statement indicates that the symbol L is also a positive real number.
(f) For the initial state Eq. (1), use symmetry or Mathematica
to calculate the √
averages ⟨x⟩, ⟨p⟩,
√
⟨x2 ⟩, ⟨p2 ⟩, and so deduce the standard deviations σx = ⟨x2 ⟩ − ⟨x⟩2 and σp = ⟨p2 ⟩ − ⟨p⟩2 , and
so verify that a position-momentum uncertainty principle holds for Eq. (1), namely
σx σp ≥
ℏ
.
2
(7)
Explain briefly why this makes sense visually by discussing the shapes of the function Ψ(t = 0, x)
and of the function c(q) in Eq. (3) for the regime a quite large and for the regime a quite small.
Hint: since |c(k)|2 is the probability density for the wavenumber k, it is perhaps quickest to
calculate averages of powers pn of momentum via
∫ ∞
n
n
n
k n |c(k)|2 dk,
(8)
⟨p ⟩ = ⟨(ℏk) ⟩ = ℏ
∞
rather than use the fact that p = −iℏ∂x .
(g) From what you have learned about how to construct a time-dependent solution Ψ(t, x) of the
Schrodinger equation by expressing an initial state Ψ(0, x) in terms of a sum of energy states:
∑
∑
cn e−i(En /ℏ)t ψn (x),
(9)
cn ψn (x) ⇒ Ψ(t, x) =
Ψ(0, x) =
n
n
2
you can see that the time-dependent state Ψ(t, x) that evolves from the initial state Eq. (1) is
given by:
(
)
∫ ∞
1
Ψ(t, x) =
c(k)e−i(E(p)/ℏ)t √ eikx dk
(10)
2π
∞
[ (
)]
∫
a3/2 ∞
1
ℏk 2
=
(11)
exp
i
kx
−
t
dk.
π ∞ k 2 + a2
2m
The integral in Eq. (11) does not have an antiderivative that can be obtained in terms of familiar
functions, but it is straightforward to approximate the integral’s value numerically at successive
values of time t and so discover qualitatively how the probability density |Ψ(t, x)|2 changes over
time.
Run the following Mathematica code and summarize the qualitative features of
how the probability density |Ψ(t, x)|2 changes over time. (A warning: this code
will take a few minutes to complete on most computers.)
fn[ t_ ] := Plot[
(* plot |Psi(t,x)|^2 at time t *)
( 1. / Pi^2 ) Abs[
NIntegrate[
1/(k^2 + 1) Exp[ I (k x - k^2 t) ] ,
{ k, -Infinity, Infinity}
]
]^2 ,
{ x, -4, 4 } ,
(* plot for -4 <= x <= 4 *)
PlotRange -> { 0, 1 }
(* plot vertical axis in range [0,1] *)
]
g1 = fn[ 0. ] ;
(* plots at some representative times *)
g2 = fn[ 0.1 ] ;
g3 = fn[ 0.4 ] ;
g4 = fn[ 1.0 ] ;
Show[ g1, g2, g3, g4 ]
Here I have arbitrarily chosen the decay constant a = 1 so that a numerical value can be obtained,
and I have chosen the mass to have a value so that the coefficient of k 2 is one. A key point is that,
by using a modest amount of Mathematica, it is straightforward for you to get some scientific
intuition about rather complicated mathematical expressions. Mathematica has no trouble dealing
with complex exponentials or integrals over infinite ranges.
2. Transmission coefficient T (E) for a potential with two delta-function wells
Consider a particle of mass m moving in a one-dimensional potential V (x) given by two identical delta
function wells of equal strength α, and at locations x = ±L:
V (x) = −αδ(x − L) − αδ(x + L).
(12)
We used this potential previously as a simple way to model the attractive binding of two identical
atoms in a diatomic molecule.
(a) For a scattering state with energy E > 0 arriving from x = −∞, show that the transmission
coefficient T , which is the fraction of particles in the right-going beam that continues on to x =
+∞, is given by the expression
T =
8g 4
,
(8g 4 + 4g 2 + 1) + (4g 2 − 1) cos(ϕ) − 4g sin(ϕ)
3
(13)
where the parameters g and ϕ are defined by
g=
ℏ2 k
,
2mα
ϕ = 4kL,
k=
√
2mE
.
ℏ
(14)
Some suggestions:
i. The wave function ψ(x) has a piece-wise representation with different forms in the three
different spatial regions
ψI (x)
= Aeikx + Be−ikx ,
x ≤ −L
ψII (x) = Ceikx + De−ikx , −L ≤ x ≤ L
ψIII (x) = F eikx ,
x ≥ L,
(15)
since V = 0 in all three regions. Note that we do not want to use the symmetry V (x) = V (−x)
of the potential here since the problem is physically asymmetric with a beam of particles
arriving from the left and being transmitted to the right. Also, ψIII has only one term since,
if particles arrive from the left and scatter off the potential, there can only be particles moving
to the right of x = L.
ii. Obtain four equations involving the unknown coefficients of the wave functions ψI , ψII , and
ψIII by using continuity of the wave function ψ(x) at the point x = −L and separately at
the point x = L, and by integrating the time-independent Schrodinger equation first over the
interval [−L − ϵ, −L + ϵ] and then integrating a second time over the interval [L − ϵ, L + ϵ],
while taking the limit ϵ → 0 in both cases.
iii. You want to eliminate the coefficients C and D associated with ψII and obtain a relation
between A and F since T = |F/A|2 . From your four equations, you should be able to add and
subtract one pair of equations to express C in terms of F and D in terms of F . By adding
and subtracting the other two equations, you can then express C in terms of A and B, and D
in terms of A and B. Equating the two different expressions for C and the two different
expressions for D will then give two linear equations in the unknowns A, B, and F , from
which you can eliminate B to finally get a relation between A and F .
(b) Show that for tiny positive values of the energy 0 < E ≪ 1, the transmission coefficient has the
mathematical form
T ∝ E.
(16)
√
Some hints: since g ∝ k ∝ E, small E implies that the parameter g is tiny so all higher powers
of g that are in some sum can be dropped as negligible compared to other terms in the sum.
(However, you cannot drop the√g 4 in the numerator of Eq. (13) since that does not appear in a
sum.) Also note that ϕ ∝ k ∝ E also becomes tiny when E is tiny so you can use small angle
approximations of the trig functions such as cos(x) ≈ 1 − (1/2)x2 and sin(x) ≈ x − (1/6)x3 .
(c) Show that for large energies E ≫ 1, the transmission coefficient has the approximate functional
form:
c(E)
,
(17)
T ≈1−
E
where c is a quantity that varies with E but is bounded in magnitude so never gets large, i.e., T
“decays” to 1 roughly as 1/E for large energies, and has the value 1 in the limit E → ∞.
(d) Check your analytical results Eqs. (16) and (17) by using Mathematica to plot Eq. (13) as a
function of energy E for the parameter values L = 1 and α = (1/1.3)ℏ2 /(2m).
Suggestions: it is easier
to plot T (k) versus k than T (E) versus E so plot T (k) vs k and keep
√
in mind that k ∝ E so the horizontal k-axis in your plot is like a distorted energy axis. The
following Mathematica code
4
Clear[ k ] ;
g = 1.3 k ;
phi = 4 k ;
(* have set L = 1 *)
tt = 8 g^4 / (
(8 g^4 + 4 g^2 + 1) + (4 g^2 - 1) Cos[phi] - 4 g Sin[phi]
) ;
Plot[ {1., tt} , { k, 0, 10 }, PlotRange -> {0, 1} ]
is one way to generate a plot for some representative parameter values. Is this plot consistent with
your above analytical results? Are there any transmission resonances, which are special values
of k or E such that you get perfect transmission with T = 1?
(e) Recall from our discussion in class that, for a potential V (x) = −αδ(x) with a single deltafunction well, the transmission coefficient T remains unchanged if the well is turned into a barrier
by changing the sign of the potential strength, α → −α.
Discuss briefly whether this observation remains true for the two-delta function potential Eq. (12):
does anything change qualitatively or substantially if the two delta functions are now made repulsive by making the substitution α → −α in Eq. (13), where α is considered to be a positive
number? As one example, does the wave function in Eq. (15) change its qualitative form? Does
the transmission coefficient T change qualitatively in its mathematical form, e.g., do the resonances go away or change in some way? (Try repeating the plot of part (d) above with the sign
of α reversed.)
3. French and Taylor problem 9-11 on page 418, but write your answer in the form:
)
( √
1
1
=1+
sin2 β 1 + ϵ ,
T
4ϵ(1 + ϵ)
where
ϵ=
E
V0
and β =
√
L 2mV0
,
ℏ
(18)
(19)
are dimensionless parameters.
For part (c) of problem 9-11, you can use some Mathematica code like this:
beta = 10.5 Pi
tt = 1 / (
1 + ( 1 / (
Plot[
tt,
{ ee, 0, 0.5
PlotRange ->
]
;
4 ee (1 + ee) ) ) Sin[ beta Sqrt[1 + ee] ]^2 ) ;
},
{0, 1}
where you might want to try some bigger horizontal plotting ranges (say E/V0 up to 1 or 2) to get a
full sense of how the transmission coefficient behaves as a function of energy. You can also try using
the Manipulate command to see how the transmission coefficient depends on the parameter β. (Note
that you do not want to use the symbol E in the above piece of code to represent energy since E in
Mathematica has the preassigned meaning of e ≈ 2.718, the base of the natural logarithm.)
4. Transmission coefficient T for a rectangular barrier: three cases
Once you have obtained the transmission coefficient T for a rectangular well, Eq. (18), it turns out
you can quickly and easily obtain general formulas for T for the case of a monoenergetic beam of
5
particles with energy E striking a rectangular barrier of height V0 if you are willing to assume that
various limits should exist mathematically. There are three cases to work out for a rectangular barrier:
E > V0 (beam energy higher than the top of the barrier), E = V0 (beam energy equal to the top of
the barrier), and 0 < E < V0 , the case of quantum tunneling, when the particle is not bound (E > 0)
but is below the height of the barrier.
Please obtain the corresponding three formulas for the tunneling coefficient by carrying out the following
calculations:
(a) For the case E > V0 , it turns out that the transmission coefficient T (ϵ) for a rectangular barrier of
height V0 can be obtained from the expression Eq. (18) for a rectangular well by simply reversing
the sign of the symbol V0 → −V0 (changing the well into a barrier). Show that reversing the sign
of V0 gives the first formula for the transmission coefficient:
Case E > V0 > 0 :
( √
)
1
1
=1+
sin2 β ϵ − 1
T
4ϵ(ϵ − 1)
(20)
You could alternatively obtain this expression by working through the equations obtained by
imposing continuity of ψ and continuity of dψ/dx at x = 0 and at x = L and you would find that
you get the same result.
(b) A next case is when the energy E of the beam has exactly the same value E = V0 as the top of
the rectangular barrier. One could guess (or hope) that this case could be obtained by taking the
limit E → V0 of Eq. (20). Carry out this limit to show that
Case E = V0 > 0:
1
β2
=1+
.
T
4ϵ
(21)
But also explain what is the new mathematical form of the piece of the wave function ψII (x) in
the region 0 ≤ x ≤ L when E = V0 . (And one could again derive Eq. (21) by using continuity of ψ
and dψ/dx at x = 0 and at x = L to discover that you obtain the same result as taking the limit,
but you do lose track of the fact that the wave function has changed its form mathematically in
the middle region.)
(c) Finally, show that the transmission coefficient for the case 0 < E < V0 , which is also the case of
quantum tunneling since all classical particles with energy less than the maximum of some barrier
would reflect from the barrier, is given by
Case 0 < E < V0 :
)
( √
1
1
=1+
sinh2 β 1 − ϵ ,
T
4ϵ(1 − ϵ)
(22)
where the hyperbolic function sinh (pronounced as “sinch” by many people but also just pronounced as “hyperbolic sine”) and other hyperbolic trig functions are defined by
sinh(x) =
ex − e−x
,
2
cosh(x) =
ex + e−x
,
2
tanh(x) =
sinh(x)
.
cosh(x)
(23)
Two suggestions:
i. From the definition of sinh and from Euler’s formula, show that the sin and sinh functions
are related by:
sin(ix) = i sinh(x),
(24)
where x is a real variable.
ii. In Eq. (18), reverse the sign of V0 (and take into account the fact that β depends on V0 ) to
convert the well to a barrier and then use Eq. (24) to deduce Eq. (22).
6
I hope you are impressed with how much can be deduced with little effort by this trick of “analytic
continuation”, in which some expression involving real variables is assumed to remain correct when
parts of an expression become complex valued as certain variables change their values. (But this
does not always work, there are cases where the analytic continuation of some expression may not
be meaningful physically.)
5. French and Taylor problem 9-14 on page 419. A short simple but useful numerical evaluation of Eq. (22)
above.
6. French and Taylor problem 9-19 on page 420.
7. French and Taylor problem 9-22 on pages 421.
7