Physics 264L: Assignment 8
Made available: Thursday, October 22, 2015
Due:
Wednesday, October 28, 2015, by 7 pm
1. Wave function for diatomic hydrogen H2
Consider a neutral diatomic hydrogen molecule H2 that consists of two electrons labeled 1 and 2 and
two protons labeled 3 and 4 that interact via Coulomb electrical interactions. (Assume for now these
particles are distinguishable so that it makes sense to assign them separate labels.) The corresponding
locations in space of these particles are the three-dimensional position vectors xi = (xi , yi , zi ) and
their corresponding momenta are the three dimensional vector pi = (pix , piy , piz ) where i goes over the
integers 1, 2, 3, and 4.
(a) Given an explicit expression for the total energy E = E(x1 , x2 , x3 , x4 , p1 , p2 , p3 , p4 ) = KE + V
of the H2 molecule.
Note: in classical mechanics, the total energy expressed in terms of momenta vectors and of
position vectors is called the “Hamiltonian” of the system and is often denoted by the symbol H
in honor of the Irish scientist Sir William Hamilton who found a powerful way to rewrite Newtonian
mechanics without having to use vectors or Newton’s second law. So you are writing down the
so-called Hamiltonian of the H2 system, and it is the Hamiltonian, together with some initial
state Ψ0 , that is needed as input to the “black box” machinery of quantum mechanics to calculate
the future state Ψ of the system.
(b) How many variables and on which variables does the wave function Ψ of the entire molecule
depend on?
(c) If the electrons are now considered to be identical and the protons are now considered to be
identical, what are all the symmetries that the H2 wave function Ψ must satisfy, given that
electrons and protons are fermions?
(d) In terms of Ψ, write down an expression for the probability that, at some future time t, the first
electron is found inside a small cube of volume dV1 centered on the vector x1 , the second electron
is found inside a small cube of volume dV2 centered on the vector x2 , the first proton is found
inside a small cube of volume dV3 centered on the vector x3 , and the second proton is found inside
a small cube of volume dV4 centered on the vector x4 .
(e) In practice, one can experimentally measure the position of only one particle at a time in the H2
molecule. Write down an expression involving the wave function Ψ for the probability of observing
at time t an electron of the H2 molecule in a small volume dV that is centered at some location r
in space, independently of where the other three particles are.
Note: this problem ignores the fact that an extra binary label (spin up ↑ or spin down ↓ along some
arbitrary axis) should be included for each electron and proton but the above is enough for you think
through the key issues of what information is needed to set up a multi-particle quantum problem,
and what are the implications of identical particles on the symmetry of the wavefunction of the entire
system. We will discuss these spin labels later in the semester.
2. Application of the infinitely-high square well quantum problem to biomolecules and to
color pigments
In this problem, you apply the energy states of the infinitely-high square well to predict, and so gain
insight about, what kind of molecules are capable of absorbing light from the Sun, which arrives on
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Earth’s surface with the greatest intensity in the visual range of wavelengths, λ ≈ 400 − 750 nm. Such
molecules are essential for organisms like bacteria and plants to extract energy from sunlight, for the
eyes of animals to see light, and to design solar cells that can create electrical energy from sunlight.
In a chemistry course, you have perhaps learned about so-called σ and π chemical bonds that can form
between adjacent carbon atoms, giving cartoons likes this for the carbon double bond in an ethylene
molecule C2 H4 :
The balloon-like shapes represent surfaces of large electron probability density |Ψ|2 where the electron
in that orbital has a large probability of being found inside the balloon; we will learn later in this
course how to calculate these shapes mathematically when we solve the Schrodinger equation for the
hydrogen atom.
It then turns out that a one-dimensional chain (or ring) of carbon atoms can have alternating single
and double bonds, forming a so-called “conjugated system” of carbon bonds. In conjugated systems,
each valence electron in a π orbital delocalizes over the entire length of the conjugated system and so
acts approximately as a free particle in an infinitely-high square well of length L = (N − 1)d, where N
is the number of carbon atoms in the conjugated system and where d is the average bond length.
Here are two biological molecules with conjugated systems:
with the understanding of traditional organic chemistry diagrams that hydrogen atoms bonded to
carbon atoms along a chain are not explicitly shown. The first molecule (a) is chlorophyll a, the
most abundant pigment in plants, while the second molecule (b) is β-carotene, which is the source of
the bright orange color of carrots. (Carotene in your diet makes its way to your eyes in the form of
vitamin A where it absorbs ultraviolet photons that would otherwise damage your retina. And here is
where my biology fails me: why do carrots have this large amount of visual pigment underground in
their roots, where there is no light?) The tan shaded regions in these figures are the conjugated systems
of alternating single and double carbon bonds, along which valence π electrons can move freely.
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From French and Taylor Section 3-5 and from our discussion in class, you know that the discrete energy
values En for a particle of mass m in a one-dimensional tube of length L with infinitely strong walls
capping the tube are given by
π 2 ℏ2 2
n , n = 1, 2, · · · .
(1)
En =
2mL2
From these energy values, you can predict what kind of light a conjugated system can absorb as a
function of its length L with several further pieces of information:
(a) when there are two or more electrons in the same one-dimensional box, they interact weakly with
each other and so the same single-particle energy levels are approximately correct.
Note: this is a remarkable and highly non-obvious result, that the Coulomb potential energy Ke2 /r
between two or more electrons in the same box can be approximately ignored; one would have
expected that if there are two electrons in the same box, all the energy levels change substantially
because of the second electron.
(b) Each energy level can hold at most two electrons, one with spin up and one with spin down
(whatever spin is). You have learned about this assumption if you have taken an introductory
chemistry course.
(c) If En and Em > En are two energy levels of the particle in the box, then a photon of frequency ω
will be absorbed when
hc
ℏω = hf =
= Em − En .
(2)
λ
Note that this assumption lies beyond the Schrodinger equation, which does not take into account
how a photon interacts with a particle.
(d) The “easiest” or lowest-energy photons to absorb in a molecule are those that cause an electron in
the highest occupied energy level to transition to the lowest unoccupied energy level, i.e., between
adjacent energy levels En and En+1 , where En is the highest energy level that contains electrons.
(e) Each carbon atom contributes one π electron in a conjugated chain so a chain of N atoms will
have N electrons in its one-dimensional box. The average carbon-carbon bond length in a conjugate chain is experimentally known to be d ≈ 0.1 nm.
Using this information:
(a) Show that for a conjugated system with an even number N of carbon atoms whose average
separation is d, the molecule will absorb photons of wavelength λ given by:
λ=
4me c 2 (N − 1)2
d
.
πℏ
N +1
(3)
(b) Use Eq. (3) to predict for what range of integers N1 ≤ N ≤ N2 a conjugated system of N carbon
atoms can absorb photons in the visible range, 400 nm ≤ λ ≤ 750 nm. Is your answer consistent
with the values of N for the chlorophyll a and β-carotene molecules above?
(c) How the above two biomolecules absorb light as a function of wavelength has been measured
experimentally and is shown here in terms of light absorption curves:
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(These experimental curves can be obtained by putting the molecules in solution, shining light of
a known initial intensity I0 and known wavelength λ through the solution, and then by measuring
the light intensity I that comes out of the solution. The absorption curve is then proportional to
1 − I/I0 .) A peak at wavelength λ in one of these absorption curves corresponds to the absorption
of a photon of that wavelength by the corresponding molecule.
Is the “infinitely-tall well” model of a conjugated carbon system able to explain the
two peaks of chlorophyll a, with one peak near 400 nm and the second peak near
700 nm?
Note that the two peaks in chlorophyll a’s absorption curve explain why plants are green: chlorophyll a absorbs mainly red and blue photons and reflects the middle range of yellow and green
photons. (But a biology question: why did plants not evolve to absorb light in the central region
of the visual spectrum, for which there are more photons?)
That plants and animals evolved to absorb light mainly in the visual range in turn can be traced
to the surface temperature T ≈ 6000 K of the surface of the Sun since the Sun’s blackbody
equilibrium radiation peaks in the visual range. Other stars have cooler or warmer surfaces and
so the light intensity striking an exoplanet’s surface can peak outside the visual range. So you
can now predict the expected lengths of conjugated molecules that alien creatures might evolve
to use, based on the surface temperature of their star.
3. Symbolic statistical deductions from the wavefunctions of the infinitely-high square well
using Mathematica
The following is a straightforward tutorial to show you how to use Mathematica to evaluate symbolically
quantum-related integrals involving the stationary wavefunctions
√
( nπx )
2
ψn (x) =
sin
,
(4)
L
L
of a particle in an infinitely-tall square well of width L.
For this problem, you are not required to evaluate the integrals analytically although it would
be good practice for you to do so. (I will show you some efficient tricks for calculating moments of
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integrands in the near future.) Instead, make sure that you know how to use Mathematica to evaluate
these kinds of integrals and especially focus on trying to make sense of the values of the moments.
“Making sense” here means making sense physically and mathematically e.g., compare the mean and
standard deviations for position and momentum for a quantum particle in an infinitely tall well (what
you calculate below) with what you would expect for a classical particle moving freely back and forth
with different energies in a one-dimensional tube with capped ends, and also looking at a plot of the
probability density |ψn (x)|2 in the nth state and thinking about what the mean and standard deviation
about the mean correspond to in such a plot.
(a) First show that the average position ⟨x⟩n of a particle in the well that is in the nth energy state
is independent of n and is the middle of the box L/2:
∫
⟨x⟩n =
L
x |ψn |2 dx =
0
L
,
2
(5)
and discuss briefly whether this fact makes sense scientifically and mathematically.
The physical meaning of the average ⟨x⟩n is this: if one prepares many identical experiments
such that the particle is always in the nth energy state ψn , and in each of these experiments you
then measure the location of the particle, you will measure different random values from each
experiment, say the values x1 , x2 , etc. Then the average of these different values is Eq. (5).
The most straightforward way to evaluate this integral using Mathematica’s Integrate command
Integrate[
x ( Sqrt[2/L] Sin[ n Pi x / L ] )^2,
(* integrand *)
{ x, 0, L } (* integration variable and range of integration *)
]
does not lead to the simple answer L/2. (Please verify this by evaluating this code.) The reason is
that Mathematica does not know what you know, that the symbol n represents a positive integer
so that identities such as sin(2nπ) = 0 or cos(2nπ) = 1 can be used to simplify an expression.
To get a more useful answer, use Mathematica’s Simplify command with so-called assertions
about the properties of key variables, like this:
Simplify[
Integrate[
x ( Sqrt[2/L] Sin[ n Pi x / L ] )^2,
{ x, 0, L } (* integration variable and range of integration *)
] ,
Element[ n, Integers ] && n > 0 (* assertions about variable n *)
]
which does yield the desired answer L/2. The Mathematica statement
Element[ n, Integers ]
tells Mathematica that the symbol n is an integer by saying that the variable n is an element of
the set of integers, n ∈ Z. In turn, the Boolean statement
Element[ n, Integers ]
&&
n > 0
states further that n is also a positive symbol hence a natural number. The symbol && is Mathematica’s notation for expressing a Boolean AND between two logical statements that are true or
false, while the notation || represents a Boolean OR.
(b) Use Mathematica to evaluate the second moment ⟨x2 ⟩n of position in the infinite well for the nth
state, giving the answer
(
)
∫ L
L2
3
2
2
2
⟨x ⟩n =
x |ψn | dx =
2− 2 2 ,
(6)
6
π n
0
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and then use this result to calculate the variance σx2 of the particle’s location about its mean ⟨x⟩n
in the nth state:
(
)
⟨
⟩ ⟨ ⟩
6
L2
2
σx2 = (x − ⟨x⟩) = x2 − ⟨x⟩2 =
1− 2 2 ,
(7)
12
π n
and discuss briefly whether this standard deviation makes sense as a function of n: is it reasonable
that σx increases with increasing n, as implied by Eq. (7)? Also show that the standard
deviation σx ≤ L/4 for all states n which tells us that, no matter how big n gets, the particle
does not spread out over the entire width of the box.
(c) Without evaluating any integrals, explain using symmetry (parity) why the average momentum
of a particle in the nth state:
]
]
[
[
∫ L
∫ L
∫ L
d
dψn
∗
∗
∗
⟨p⟩n =
ψn p ψn dx =
ψn −iℏ
ψn dx =
ψn −iℏ
dx
(8)
dx
dx
0
0
0
has the value zero for all n, ⟨p⟩n = 0, and discuss briefly whether this results makes sense
scientifically. (A hint is to think about the physical meaning of Eq. (4) if expressed in terms of
complex exponentials.)
We will discuss why the average momentum involves this strange expression ψ ∗ p ψ in a week or
two. In quantum mechanics, momentum is represented by the operator p = −iℏ∂x and is closely
related to differentation.
Note that when you need to evaluate some integral in quantum mechanics, you should always
check first to see if some symmetry argument allows you to conclude that the integral must
vanish, before spending a lot of time trying to evaluate the integral. This possibility arises when
the potential V of the system is symmetric about some point and if a wavefunction ψ(x) is itself
symmetric or antisymmetric (even or odd) about the point of symmetry.
You can confirm that ⟨p⟩n = 0 by using Mathematica’s symbolic differentation command D, which
can take ordinary derivatives and partial derivatives of symbolic expressions like this
D[ x^n , x ]
gives nx
n−1
(* first derivative of nth power of x *)
and
D[ x^2 Sin[k y] , { y, 2 } ]
(* 2nd-order partial wrt y *)
gives −k x sin(ky) as you should verify directly. (Please look at Mathematicaś documentation
on “Differentation” to see more examples.) Then you would execute the following Mathematica
code to evaluate Eq. (8):
2 2
Simplify[
Integrate[
Sqrt[2/L] Sin[ n Pi x / L ] *
(-I hbar) D[
Sqrt[2/L] Sin[ n Pi x / L ] , (* fn being differentiated *)
x
(* differentiation variable *)
] ,
{ x, 0, L } (* integration variable and range of integration *)
] ,
Element[n, Integers] && n > 0 (* assertions about variable n *)
]
which gives the correct answer 0 when evaluated. In turn, the second moment of the momentum
[
]2
∫ L
∫ L
∫ L
d
d2 ψn
⟨p2 ⟩ =
ψn∗ p2 ψn dx =
ψn∗ −iℏ
ψn dx =
ψn∗ (−iℏ)2
dx,
(9)
dx
dx2
0
0
0
can be evaluated with this Mathematica code:
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Simplify[
Integrate[
Sqrt[2/L] Sin[ n Pi x / L ] *
(-I hbar)^2 D[
Sqrt[2/L] Sin[ n Pi x / L ] , (* fn being differentiated *)
{x, 2}
(* differentiate twice wrt x *)
] ,
{ x, 0, L } (* integration variable and range of integration *)
] ,
Element[n, Integers] && n > 0 (* assertions about variable n *)
]
which gives the value
(
⟨p2 ⟩ =
πℏn
L
)2
.
(10)
Since ⟨p⟩ = 0, we see that the standard deviation σp of the momentum in the nth state is:
σp =
√
πℏ
⟨p2 ⟩ − ⟨p⟩2 =
n,
L
(11)
which increases with n: does it make scientific sense that knowledge of the momentum becomes
more uncertain as the energy of the state increases?
Standard deviations like σx and σp are what appear in the famous uncertainty principles of
quantum mechanics, which we will discuss after the second midterm.
Note that Eq. (10) also gives us the average kinetic energy in the n state, it is
⟨ 2⟩
1 ⟨ 2⟩
π 2 ℏ2
p
=
p n=
n
(12)
2m n
2m
2mL
which are exactly the familiar energy levels Eq. (1) since the potential energy is zero inside the
infinitely tall well and so the particle only has kinetic energy.
⟨ ⟩
⟨ ⟩
(d) Finally, use your improved Mathematica skills to compute the fourth moments x4 n and p4 n
of the nth energy state. (You do not have to discuss whether these values make sense beyond the
fact that they better be positive numbers.) Include in your homework the resulting expressions
and the Mathematica code that you used to evaluate these expressions.
4. Problem 3-6 on page 146 of French and Taylor.
Note that this problem is an important historical insight from the early days of quantum mechanics, it
suggested that when an electron is emitted by radioactive decay of some nucleus, the electron could not
have already existed in the nucleus but had to be created by the decay process. (Does this argument
also apply to emitted positrons?) Note that for part (b)(i), you will need to use the relativistic formula
for the electron’s energy E 2 = (pc)2 + (m0 c2 )2 since since the magnitude of the energy associated with
the energy levels (MeV or bigger) are not small compared to the rest mass of the electron (0.5 MeV).
5. Problem 3-10 on page 148 of French and Taylor.
Comments: you can evaluate the integrals using Mathematica as above, but I would recommend that
you don’t become too dependent on Mathematica and here do the integral by hand. Many integrals
involving trigonometric functions can be evaluated rather easily by expressing the trig functions in
terms of complex exponentials and by then expanding algebraically the integrand to get a sum of
exponentials, which is easy to integrate.
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6. Problem 3-14 on pages 150-151 of French and Taylor.
A comment: solve this problem qualitatively by thinking about the fact that, in order for a quantum
state of energy E to be bound, there cannot be a finite probability of finding the particle as x → ±∞,
i.e, the probability density |Ψ|2 for a bound state must decay to zero at x = ±∞:
lim |Ψ(x, t)|2 = 0.
x→±∞
(13)
By the discussion of French and Taylor related to figures 3-13 and 3-16, this in turn requires the
energy E of any bound state to be less than the asymptotic values V±∞ = limx→±∞ V (x). The
arguments in Section 3-11 further imply that the lowest energy E1 of a bound state must always be
strictly greater than the global minimum of the potential, E1 > minx V (x), else there is no way for the
wave function to decay in the classically forbidden regions. So for 1d potentials V (x), a bound state
can exist if the global minimum of V is strictly less than V± , otherwise there will only be unbound
states such that the particle can be found arbitrarily far in the +x or −x direction.
Although the answer to this problem is given at the back of French and Taylor, your job is to explain
why their answer is the correct.
7. Problem 3-15 on page 151, just answers parts (a) and (f).
8. Problem 3-16 on page 151-152, just answers parts (a) and (e).
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