Physics 264L: Assignment 9
Made available: Saturday, October 31, 2015
Due:
Thursday, November 5, 2015, by 7 pm
1. Quantum averages behave like classical observables
Consider a particle that is described by a wave function Ψ(t, x) that satisfies the one-dimensional
Schrodinger equation with time-independent potential V (x).
(a) Show that the average value hpi for the momentum in the state Ψ is equal to the particle mass
times the time derivative of the average position hxi:
hpi = m
dhxi
.
dt
(1)
This immediately implies that the average momentum must be zero for any state such that the
average position is constant.
R
R
A suggestion is to start with the expression hxi = x |Ψ|2 dx = x Ψ∗ Ψ dx for the average
position and differentiate both sides with respect to time. Then use the fact that you can pass
the time derivative across the integral sign (please make sure you understand why and when you
can do this), use the fact that Ψ satisfies the Schrodinger equation, somewhere use integration by
parts (maybe more than once), use somewhere the fact that wave functions describing bounded
particles decay to zero at ±∞, and use the definition1 introduced in Assignment 8
·
¸
Z ∞
∂Ψ
∗
Ψ −i~
hpi =
dx.
(2)
∂x
−∞
Your answer in fact helps to explain why one chooses to define hpi in this strange way, in terms
of some derivative.
(b) Use part (a) to show that hpi = 0 for any so-called stationary state, which is a special state that
factors into a function of time times a function of position, Ψ(t, x) = e−iωt ψ(x) = e−i(E/~)t ψ(x),
where ψ is a solution of the time-independent Schrodinger equation with energy E.
(c) Show that2
dhpi
=
dt
À Z ∞ µ
¶
¿
∂V
∂V
−
=
|Ψ|2 dx.
−
∂x
∂x
−∞
(3)
This result is rather interesting since it looks just like Newton’s second law for a point particle of
mass m moving in a potential V (x), when Newton’s law is written in this form:
ma = m
d(mv)
dp
dV
dv
=
=
=F =−
dt
dt
dt
dx
⇒
dp
dV
=−
.
dt
dx
(4)
So Eqs. (1) and Eq. (3) suggest that relations or equations that are familiar from classical physics
have quantum analogs provided one works with the quantum mean values. This helps to explain
why intuition from classical physics is often useful when thinking about quantum mechanical
problems.
1 More generally, the average value h g(p) i of a general function g(p) of the momentum for the state Ψ is defined to be the
R∞
expression −∞
Ψ∗ [g(−i~∂x )Ψ] dx in which the symbol p is replaced by the spatial partial derivative −i~∂x , which then acts
R∞
R∞
˙ ¸ R∞
ˆ
˜
on Ψ. For example, p2 = −∞
Ψ∗ p2 Ψ dx = −∞
Ψ∗ [i~∂x ]2 Ψ dx = −∞
Ψ∗ −~2 ∂x2 Ψ dx.
2 More generally, for a general function f (x), the average value h f (x) i of f (x) in the quantum state Ψ is the weighted average
R∞
of f (x) with the weighting given by the probability density, h f (x) i = ∞
f (x) |Ψ|2 dx.
1
2. The dot product h f | g i of two complex-valued square-integrable functions
In class, we generalized the familiar dot-product of two real-valued vectors a = (a1 , a2 , a3 ) and b =
(b1 , b2 , b3 )
a • b = a1 b1 + a2 b2 + a3 b3 ,
(5)
to the dot product3 h f | g i of two complex-valued square-integrable4 functions f (x) and g(x) via this
integral:
Z ∞
f ∗ (x) g(x) dx.
(6)
h f (x) | g(x) i =
−∞
(a) Show that switching the order of two functions in the dot product gives the complex conjugate of
the original dot product:
h g | f i = h f | g i∗ .
(7)
Although mathematically the order of the functions matters when calculating Eq. (6), the order
does not matter physically because, in any quantum calculation that is compared with experiment,
it is the real-valued magnitude squared |h f | g i|2 that is compared with experiment, and |z|2 =
|z ∗ |2 for any complex number z.
(b) Show that Eq. (6) has properties similar to those of the familiar dot-product:
hf |f i
is always real-valued,
h f | f i ≥ 0,
h f | c g i = c h f | g i,
h f | c1 g1 + c2 g2 i = c1 h f | g1 i + c2 h f | g2 i,
h c1 f1 + c2 f2 | g i = c∗1 h f1 | g i + c∗2 h f2 | g i,
(8)
(9)
(10)
(11)
(12)
for any complex-valued square-integrable functions f , g, f1 , f2 , g1 , and g2 , and for any complexvalued constants c1 and c2 . The only strange features of the dot-product Eq. (6) are that its value
can be a complex number, and linear combinations on the left side lead to complex conjugates of
coefficients.
(c) If Ψ1 (t, x) and Ψ2 (t, x) are two normalized solutions to the one-dimensional Schrodinger equation,
show that the superposition of the two solutions
Ψ = c1 Ψ1 + c2 Ψ2 ,
(13)
is also a solution to the Schrodinger equation for general constant complex coefficients c1 and c2 . If
further these two states are orthogonal so that h Ψ1 | Ψ2 i = 0, then using only the above properties
Eq. (8)-(12) of the dot product (don’t do any integrations!), show that the superposition Eq. (13)
is itself normalized if and only if the coefficients satisfy
2
2
|c1 | + |c2 | = 1.
(14)
Hint: In terms of dot-products, a wave function Ψ is normalized if it is a “unit vector”, which
means that its length-squared is one, which in turn means that that h Ψ | Ψ i = 1.
3. Calculating some quantum amplitudes for quantum states in an infinitely tall well
Use what you learned in the previous problem about the dot product of two complex-valued functions
to gain some new insights about the quantum mechanics of a particle of mass m in a well of width L
with infinitely tall sides.
3 Here we are using a traditional quantum-mechanical notation for the dot product of two quantum states that was introduced
by the theoretical physicist Dirac, the notation is similar to how mathematicians define a dot product via a so-called inner
product (f, g) of two vectors.
4 We need some assumption to make the integrals associated with the dot product Eq. (6) to converge, and an easy physical
assumption is that the states correspond to bound states that decay sufficiently quickly to zero at infinity.
2
It is an axiom of quantum mechanics, confirmed by many experiments, that the probability pn of
measuring an energy value En at time t for a system whose state function is Ψ(t, x), is given by the
magnitude-squared of the dot product of the two functions
2
pn = | h ψn (x) | Ψ(t, x) i | ,
(15)
provided that ψn and Ψ are both normalized wave functions. The complex number obtained by the
dot product
h ψn |Ψ i,
(16)
is called the quantum amplitude to measure the system to be in state ψn (x) with energy En at
time t when the system is currently in the state Ψ. You should think of the quantum amplitude as a
measure of the possibility of transitioning (via some experimental measurement) from the current state
of the system (here Ψ) on the right side of the amplitude to a new state on the left (here ψn ), with
the magnitude-squared of the amplitude Eq. (15) indicating the experimentally measurable probability
(fraction of experiments) that will observe the state ψn when each identical experiment is in the state Ψ.
I gave you a proof in class that, for any two bound energy states ψn (x) and ψn (x) that belong to
different energy states En 6= Em , the quantum amplitude is zero:
h ψn | ψm i = 0,
(17)
i.e., energy states belonging to different energy values are orthogonal (perpendicular). Our goal then
was to understand why different wave functions had different numbers of nodes, but we can now further
interpret Eq. (17) as saying that, if at some time a system is described by the energy state ψm , there
is zero amplitude, and so zero probability |h ψn | ψm i|2 = 0 for an energy measurement of the system
at any time in the future to give a different value for the energy. So solutions ψn (x)e−i(En /~)t of the
time-independent Schrodinger equation are “pure” energy states in that any future measurement of
the energy always gives the same value, En . Problem 4 below gives you a chance to explore a state for
which energy measurements of identical experimental systems in the same state will produce different
energy values,
Much of quantum mechanics consists of calculating the quantum amplitude to start in one state and
to end up in a particular state of energy or to end up in a particular state of momentum or to end
up in a particular state of position. The magnitude-squared of a quantum amplitude then tells you
the probability of an energy experiment recording a corresponding value of E or of a momentum
experiment measuring a momentum value p or of a position experiment measuring a position x for
repeated experiments that are all in the state Ψ. (Because of the so-called uncertainty principles which
you will soon learn about, you cannot do one experiment and measure all quantities of interest, like
you can classically.)
We finally get to
pthe problem of interest: given the fact Eq. (15), consider a particle in the ground
state ψ1,L (x) = 2/L sin(πx/L) of an infinitely tall well of width L. At time t = 0, a quantum wizard
(this problem set is close to Halloween after all) waves his magic wand and instantaneously doubles
the width of the well to 2L by moving the right side of the well at x = L to the position x = 2L,
and then a measurement of the energy of the particle is quickly made. What are the numerical values
of the probabilities p1 , p2 , and p3 that the particle in the new well will be found to have the energy
value E1′ , E2′ , or E3′ where En′ are the energy values of the new wider well with width 2L?
To answer this question, you need to calculate the three quantum amplitudes h ψn,2L | ψ1,L i for n = 1, 2,
and 3 for a particle starting in the ground state ψ1,L of the original well to transition to the state ψ1,2L
or the state ψ2,2L or the state ψ3,2L . (You are welcome to use Mathematica to get the values quickly,
but, if you do, please include your Mathematica code as part of your homework answers.)
4. A time-dependent quantum problem in an infinitely tall well
This is one of the most important homework problems, please make sure you understand it conceptually
and technically.
3
Consider a particle of mass m in an infinitely tall well of width L and consider an initial wave function Ψ0
at time t = 0 that is an equal mixture of the two lowest energy states:
Ψ0 (x) = Ψ(t = 0, x) = A [ψ1 (x) + ψ2 (x)] ,
(18)
where ψn (x) are the usual energy states of the time-independent Schrodinger equation for the well
(Section 3-5 of French and Taylor).
(a) Use
√ the result Eq. (14) above to explain why you can choose the constant A to have the value A√=
1/ 2 in√order for Ψ0 to be normalized. (The normalization constant is not unique, e.g., −1/ 2
or eiφ0 / 2 will work also, but it is arguably an easy choice.)
(b) You showed above that if Ψ1 and Ψ2 are two solutions of the Schrodinger equation, then any
superposition Eq. (13) is also a solution. From this observation, and recalling how we used
separation of variables to find specific solutions of the form ψn (x)e−i(En /~)t for the time-dependent
Schrodinger equation, you should be able to see why the expression
¤
1 £
Ψ(x, t) = √ ψ1 (x)e−iω1 t + ψ2 (x)e−i4ω1 t ,
2
(19)
must be the general solution at time t of the one-dimensional Schrodinger equation that evolves
from the initial state Eq. (18). (If you are not confident of this fact, please verify that Eq. (19) is a
solution by direct substitution of Eq. (19) into the time-dependent Schrodinger equation, followed
by evaluating both sides of the Schrodinger equation.) Here the angular frequency ω1 = E1 /~
associated with the ground state is given by
ω1 =
π2 ~
E1
=
.
~
2mL2
From Eq. (19), show that the probability density |Ψ|2 is the expression
µ
·
¶
µ
¶
¸
³ πx ´
³ πx ´
1
2πx
2πx
|Ψ(t, x)|2 = Ψ∗ Ψ =
sin2
+ sin2
+ 2 sin
sin
cos (3ω1 t) ,
L
L
L
L
L
(20)
(21)
which is time dependent (!) and in fact varies periodically in time with angular frequency 3ω1 .
This is a new, important, and general result, an initial state that is a superposition of two or more
energy states will lead to a time-varying probability density.
Note: you can get some intuition about how Eq. (21) varies with time by running this Mathematica
code to generate a brief animation:
Manipulate[
Plot[
Sin[Pi x]^2 + Sin[2 Pi x]^2 +
{ x, 0, 1 } ,
PlotRange -> { 0, 3.5 }
] ,
{t, 0, 10}
]
2 Sin[Pi x] Sin[2 Pi x] Cos[2 Pi t],
for which I set the box size L = 1 and arbitrarily set the frequency f = 1. You can then see how
the probability density sloshes back and forth periodically over time.
(c) Show that the average position hxi of the particle varies with time according to:
¸
·
L
32
.
hxi = 1 − 2 cos(3ω1 t)
9π
2
(22)
Note that the amplitude of the oscillation is (16/(9π 2 ))L ≈ 0.18L numerically, so the mean does
not move back and forth very much compared to the width L.
4
(d) Use Eq. (1) to efficiently show that the average momentum of the particle at time t is given by
hpi =
8~
sin (3ω1 t) .
3L
(23)
(e) What is the average energy hEi for the system in state Eq. (19)? It is the weighted average
hEi = p1 E1 + p2 E2 ,
(24)
p1 = | h ψ1 | Ψ i |2 ,
(25)
where
is the probability of finding the system in energy state E1 at time t, similarly for p2 . From this
observation, show that the average energy in state Eq. (19) is independent of time and has the
value
E1 + E 2
5π 2 ~2
hEi =
=
.
(26)
2
4mL2
The average value (E1 + E2 )/2 is reasonable since the initial state Eq. (19) is an equal weighting
of the different energy states.
(f) Finally, you can try to make sense of the above expressions by trying to interpret the situation
using classical physics. A classical point particle in this infinitely tall well simply moves back and
forth periodically with a constant kinetic energy and so constant speed v. If you set that kinetic
energy (1/2)mv 2 to the average energy value Eq. (26), show that the classical period Tclassical of
the motion is related to the quantum period Tquantum of hxi by
r
10
Tquantum ≈ 1.1Tquantum
(27)
Tclassical =
9
which means that a classical picture is rather close to the quantum one, at least in terms of the
period.
5. Some properties of the Dirac delta function δ(x)
Two expressions e1 (x) and e2 (x) involving delta functions are said to be equal if the following holds
for any ordinary function f (x):
Z ∞
Z ∞
f (x) e2 (x) dx.
(28)
f (x) e1 (x) dx =
−∞
−∞
From this, show that
δ(x) = δ(−x)
1
δ(cx) =
δ(x),
|c|
dδ(x)
δ(x) = −x
,
dx
(29)
(30)
(31)
where c is some non-zero constant. (A hint for this last relation is to use integration by parts.)
6. Odd-symmetry bound states of the attractive double delta-function potential
By following the discussion given in class, solve for the odd-symmetry bound states ψn (x) with energies En of the attractive double delta-function potential
h
i
V (x) = −α δ(x − L) + δ(x + L) ,
(32)
5
whose unique even solution was discussed in class. A key step is to look for an odd stationary wave
function of the piece-wise form:

Ae−κx
x > L,

κx
B (e − e−κx ) −L < x < L ,
(33)
ψ(x) =

−Aeκx
x < −L
and use continuity of ψ together with integrations over tiny intervals L ± ǫ and −L ± ǫ to deduce this
condition that κ must satisfy:
e−y = 1 − cy,
(34)
where
~2
.
(35)
2mαL
Use the Manipulate command of Mathematica to plot the left and right sides of Eq. (34) simultaneously
as the parameter c is varied to get a visual sense of when solutions of Eq. (34) might exist as the
parameter c varies (which in turn means varying κ for fixed values of m, L, and α). From your
graphical analysis, show that a necessary condition for an odd bound state to exist is that the delta
function potential be strong enough:
~2
α>
.
(36)
2mL
Finally, estimate to one significant digit the numerical value c1 of the coefficient that appears in the
bound state energy Eodd when written in the form Eodd = c1 (~2 /(mL2 )) for the case α = ~2 /(mL).
The corresponding even energy value is E ≈ −0.62(~2 /(mL2 )).
y = 2κL,
and
c=
7. Problem 4-2 on page 185 of French and Taylor. This should be quick and straightforward since it
follows closely what is discussed in the text and in class. For part (e), I recommend that you use
Mathematica to graphically find approximate solutions, as was explained in class.
8. Optional Challenge Problems
(a) Challenge 9.1: Show that Eq. (3) generalizes to
dhpi
= h−∇V i,
dt
(37)
for a particle whose wave function Ψ(t, x, y, z) satisfies the three-dimensional single-particle
Schrodinger equation
−~2 2
i~∂t Ψ =
∇ Ψ + V (x, y, z)Ψ.
(38)
2m
(b) Challenge 9.2: In our various discussions about the Schrodinger equation, we stated the truth of
various statements without proof, so that we could focus on the physics and not get distracted
by the math. For this challenge, prove some of these claims carefully:
i. Prove that the wave functions ψn (x) that satisfy the one-dimensional time-independent
Schrodinger equation can always be chosen to be real functions. (More precisely, a solution ψn can always be chosen to be a linear combination of real functions, in which case we
can just find the real-valued solutions in the first place.)
ii. Prove that the wave functions ψn (x) that satisfy the one-dimensional time-independent
Schrodinger equation can always be chosen to be even or odd functions if the potential V (x) = V (−x) is itself a symmetric function of x.
6
iii. For bound states, show that the separation constant E, that appears as the energy in the onedimensional time-independent Schrodinger equation after applying the technique of separation
of variables Ψ = f (t)ψ(x) to the time-dependent Schrodinger equation, must always be a real
number. (In our discussion in class, we just assumed that E was some constant that was
simultaneously equal to a function of time and to a function of space, but we really need
to know that this E cannot be complex-valued since it corresponds to the observable energy
values in any experiment.)
iv. Show that the lowest possible energy E for a bound state for the one-dimensional Schrodinger
equation is always strictly greater than the global minimum Vmin = minx V (x) of the potential
over all of space.
v. For the one-dimensional time-independent Schrodinger equation with potential V (x), show
that if two functions f (x) and g(x) are both bound-state solutions with the same energy E,
then one function must be a constant times the other. So in one-space dimension, bound
states have a degeneracy of one, there is only one linearly independent wave function for any
given bound-state energy E. (This is not true for freely moving particles since, for a given
energy E, you can have a free particle moving to the right and a free particle moving to the
left with the same energy, so a degeneracy of 2.)
vi. Show that the set of complex-valued square-integrable functions f (x) that live on some interval [a, b] form a vector space in that, if f1 and f2 are in this set, then c1 f1 + c2 f2 are also
in this set.
vii. We discussed in class how the orthonormality h ψm (x) | ψn (x) i = δmn for stationary solutions
of the time-independent Schrodinger equation implied that each wave functions had to have
more and more zeros (nodes) as their energy increased, since a given higher-energy state ψm
had to be orthogonal via Eq. (6) to each of the lower-energy solutions. Show this result more
directly and more insightfully by proving the following result: let ψn be an energy state with
energy En , and let ψm be an energy state with higher energy Em > En and let x0 be a point
where both wave functions vanish. (The values x = ±∞ will always work for this.) Then if
x1 is a second location where ψn vanishes, there has to be at least one point between x0 and
x1 where ψm vanishes, i.e., ψm with higher energy not only has more zeros, the extra zeros
appear in specific ways.
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