Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula Chapter 8 Atkins Physical Chemistry (10th Edition) Peter Atkins and Julio de Paula
Trang 1chaPter 8
the quantum theory of motion
The three basic modes of motion—translation (motion
through space), vibration, and rotation—all play an important
role in chemistry because they are ways in which molecules
store energy Gas-phase molecules, for instance, undergo
trans-lational motion and their kinetic energy is a contribution to
the total internal energy of a sample Molecules can also store
energy as rotational kinetic energy and transitions between
their rotational energy states can be observed spectroscopically
Energy is also stored as molecular vibration, and transitions
between vibrational states also give rise to spectroscopic
signa-tures In this Chapter we use the principles of quantum theory
to calculate the properties of microscopic particles in motion
In this Topic we see that, according to quantum theory, a
parti-cle constrained to move in a finite region of space is described
by only certain wavefunctions and their corresponding
ener-gies Hence, quantization emerges as a natural consequence of
solving the Schrödinger equation and the conditions imposed
on it The solutions also bring to light a number of non-classical
features of particles, especially their ability to tunnel into and
through regions where classical physics would forbid them to
be found
This Topic introduces the ‘harmonic oscillator’, a simple
but very important model for the description of molecular
vibrations We see that the energies of oscillator are quantized The acceptable wavefunctions also show that the oscillator may
be found at extensions and compressions that are forbidden by classical physics
The energy of a rotating particle is quantized, but in this Topic
we see that its angular momentum is also restricted to tain values The quantization of angular momentum is a very important aspect of the quantum theory of electrons in atoms and of rotating molecules
cer-What is the impact of this material?
‘Nanoscience’ is the study of atomic and molecular blies with dimensions ranging from 1 nm to about 100 nm and ‘nanotechnology’ is concerned with the incorporation of such assemblies into devices We encounter several concepts
assem-of nanoscience throughout the text In Impact I8.1 we explore
quantum mechanical effects that render the properties of a nanometre-sized assembly dependent on its size
To read more about the impact of this material, scan the QR code, or go to bcs.whfreeman.com/webpub/chemistry/pchem10e/impact/pchem-8-1.html
Trang 2be found.
8A.1 Free motion in one dimension
The Schrödinger equation for a particle of mass m moving
freely in one dimension is (Topic 7B)
−2m2 2 2x =
d
dψ( ) ψ( )and the solutions are (as in eqn 7B.6)
for all values of k Because the energy of the particle is portional to k2, all non-negative values, including zero, of the
pro-energy are permitted It follows that the translational pro-energy of a
free particle is not quantized.
➤
➤ Why do you need to know this material?
The application of quantum theory to translation reveals
the origin of quantization and other non-classical features
of physical and chemical phenomena This material is
important for the discussion of atoms and molecules that
are free to move within a restricted volume, such as a gas
in a container.
➤
➤ What is the key idea?
The translational energy levels of a particle confined to
a finite region of space are quantized, and under certain
conditions particles can pass into and through classically
forbidden regions.
➤
➤ What do you need to know already?
You should know that the wavefunction is the solution of the Schrödinger equation (Topic 7B), and be familiar with the techniques of deriving dynamical properties from the wavefunction by using operators corresponding to the observables (Topic 7C).
Free motion
in one dimension
schrödinger equation (8A.1)
Free motion
in one dimension
functions and energies
wave-(8A.2)
Contents
brief illustration 8a.1: the wavefunction
of a freely-moving particle 318
(a) The acceptable solutions 318
brief illustration 8a.2: the energy of a particle
(b) The properties of the wavefunctions 320
example 8a.1: determining the probability
of finding the particle in a finite region 320
(c) The properties of observables 321
example 8a.2: estimating an absorption
8a.3 Confined motion in two or more dimensions 322
(a) Separation of variables 322
brief illustration 8a.3: the distribution of a
particle in a two-dimensional box 323
brief illustration 8a.4: degeneracies in a
brief illustration 8a.5: transmission probabilities
Trang 3318 8 The quantum theory of motion
The values of the constants A and B depend on how the state
of motion of the particle is prepared:
• If it is shot towards positive x, then its linear
momentum is +k (Topic 7C), and its wavefunction
is proportional to eikx In this case B = 0 and A is a
normalization factor
• If the particle is shot in the opposite direction,
towards negative x, then its linear momentum is −k
and its wavefunction is proportional to e−ikx In this
case, A = 0 and B is the normalization factor.
The probability density |ψ|2 is uniform if the particle is in
either of the pure momentum states eikx or e−ikx According to
the Born interpretation (Topic 7B), nothing further can be said
about the location of the particle That conclusion is
consist-ent with the uncertainty principle, because if the momconsist-entum
is certain, then the position cannot be specified (the operators
corresponding to x and p do not commute and thus correspond
to complementary observables, Topic 8C)
8A.2 Confined motion in one
dimension
Consider a particle in a box in which a particle of mass m is
confined to a finite region of space between two
impenetra-ble walls The potential energy is zero inside the box but rises
abruptly to infinity at the walls at x = 0 and x = L (Fig 8A.1)
When the particle is between the walls, the Schrödinger tion is the same as for a free particle (eqn 8A.1), so the general solutions given in eqn 8A.2 are also the same However, it will prove convenient to use e±ikx = cos kx ± i sin kx (Mathematical
ψ k( )x C= sinkx D+ coskx general solution for 0 ≤ x ≤ L (8A.3)
Outside the box the wavefunctions must be zero as the cle will not be found in a region where its potential energy is infinite:
parti-Forx<0andx L> , ψ k( )x =0 (8A.4)
At this point, there are no restrictions on the value of k and all
solutions appear to be acceptable
(a) The acceptable solutions
The requirement of the continuity of the wavefunction (Topic
7B) implies that ψ k (x) as given by eqn 8A.3 must be zero at the
walls, for it must match the wavefunction inside the material of the walls where the functions meet That is, the wavefunction
must satisfy the following two boundary conditions, or
con-straints on the function at certain locations:
ψ k( )0 0= andψ k( )L =0
As we show in the following Justification, the requirement
that the wavefunction satisfy these boundary conditions implies that only certain wavefunctions are acceptable and
Brief illustration 8A.1 The wavefunction of a
freely-moving particle
An electron at rest that is shot out of an accelerator towards
positive x through a potential difference of 1.0 V acquires a
kinetic energy of 1.0 eV or 0.16 aJ (1.6 × 10−19 J) The
wavefunc-tion for such a particle is given by eqn 8A.3 with B = 0 and
k given by rearranging the expression for the energy in eqn
/
m
or 5.1 nm−1 (with 1 nm = 10−9 m) Therefore the wavefunction is
ψ(x) = Ae 5.1ix/nm
Self-test 8A.1 Write the wavefunction for an electron
travel-ling to the left (negative x) after being accelerated through a
Figure 8A.1 A particle in a one-dimensional region with
impenetrable walls Its potential energy is zero between x = 0
and x = L, and rises abruptly to infinity as soon as it touches the
walls
Trang 4where C is an as yet undetermined constant Note that the
wavefunctions and energy are now labelled with the
dimen-sionless integer n instead of the quantity k.
We conclude that the energy of the particle in a
one-dimen-sional box is quantized and that this quantization arises from
the boundary conditions that ψ must satisfy This is a general
conclusion: the need to satisfy boundary conditions implies that
only certain wavefunctions are acceptable, and hence restricts
observables to discrete values So far, only energy has been
quan-tized; shortly we shall see that other physical observables may
also be quantized
We need to determine the constant C in eqn 8A.6a To do
so, we normalize the wavefunction to 1 by using a standard
integral from the Resource section Because the wavefunction is
zero outside the range 0 ≤ x ≤ L, we use
n n
where the energies and wavefunctions are labelled with the
quantum number n A quantum number is an integer (in some
cases, as we see in Topic 9B, a half-integer) that labels the state
of the system For a particle in a one-dimensional box there is
an infinite number of acceptable solutions, and the quantum
number n specifies the one of interest (Fig 8A.2) As well as
acting as a label, a quantum number can often be used to culate the energy corresponding to the state and to write down the wavefunction explicitly (in the present example, by using the relations in eqn 8A.7)
cal-Justification 8A.1 The energy levels and wavefunctions
of a particle in a one-dimensional box
From the boundary condition ψ k(0) = 0 and the fact that, from
eqn 8A.3, ψ k (0) = D (because sin 0 = 0 and cos 0 = 1), we can
conclude that D = 0 It follows that the wavefunction must be
of the form ψ k (x) = C sin kx From the second boundary
con-dition, ψ k (L) = 0, we know that ψ k (L) = C sin kL = 0 We could
take C = 0, but doing so would give ψ k (x) = 0 for all x, which
would conflict with the Born interpretation (the particle must
be somewhere) The alternative is to require that kL be chosen
so that sin kL = 0 This condition is satisfied if
kL n= π n=1 2, ,…
The value n = 0 is ruled out, because it implies k = 0 and
ψ k (x) = 0 everywhere (because sin 0 = 0), which is
unaccep-table Negative values of n merely change the sign of sin kL
(because sin(−x) = −sin x) and do not result in new solutions
The wavefunctions are therefore
ψ n( )x C= sin (n x Lπ / ) n=1 2, ,…
as in eqn 8A.6a At this stage we have begun to label the
solu-tions with the index n instead of k Because k and E k are related
by eqn 8A.2, and k and n are related by kL = nπ, it follows that
the energy of the particle is limited to E n = n2h2/8mL2, as in eqn
8A.6b
Onedimensional box energy levels (8A.7b)
Brief illustration 8A.2 The energy of a particle in a box
A long carbon nanotube can be modelled as a one-dimensional structure and its electrons described by particle-in-a-box wavefunctions The lowest energy of an electron in a carbon
nanotube of length 100 nm is given by eqn 8A.7b with n = 1:
kg 1100 10× − 9m)2=6 02 10. × −24J
or 0.00602 zJ and its wavefunction is ψ1(x) = (2/L)1/2sin(πx/L).
Self-test 8A.2 What are the energy and wavefunction for the next higher energy electron of the system described in this
functions (8A.7a)
100
Classically allowed energies
12 3
5 4 6 7 8 9 10
n
Figure 8A.2 The allowed energy levels for a particle in a
box Note that the energy levels increase as n2, and that their separation increases as the quantum number increases Classically, the particle is allowed to have any value of the energy in the continuum shown as a shaded area
Trang 5320 8 The quantum theory of motion
(b) The properties of the wavefunctions
Figure 8A.3 shows some of the wavefunctions of a particle in a
one-dimensional box We see that:
• The wavefunctions are all sine functions with the
same amplitude but different wavelengths
• Shortening the wavelength results in a sharper
average curvature of the wavefunction and therefore
an increase in the kinetic energy of the particle (its
only source of energy because V = 0 inside the box).
• The number of nodes also increases as n increase;
the wavefunction ψ n has n − 1 nodes.
• Increasing the number of nodes between walls of a
given separation increases the average curvature of
the wavefunction and hence the kinetic energy of the
• and varies with position The non-uniformity in the
probability density is pronounced when n is small
(Fig. 8A.4) The most probable locations of the particle correspond to the maxima in the probability density
The probability density ψ n2( ) becomes more uniform as n x
increases provided we ignore the fine detail of the increasingly rapid oscillations (Fig 8A.5) The probability density at high quantum numbers reflects the classical result that a particle bouncing between the walls spends, on the average, equal times
at all points That the quantum result corresponds to the cal prediction at high quantum numbers is an illustration of the
classi-correspondence principle, which states that classical
mechan-ics emerges from quantum mechanmechan-ics as high quantum bers are reached
num-Example 8A.1 Determining the probability of finding the particle in a finite region
The wavefunctions of an electron in a conjugated polyene can
be approximated by particle-in-a-box wavefunctions What is
the probability, P, of locating the electron between x = 0 (the left-hand end of a molecule) and x = 0.2 nm in its lowest energy
state in a conjugated molecule of length 1.0 nm?
Method According to the Born interpretation, ψ(x)2dx is the probability of finding the particle in the small region dx located at x; therefore, the total probability of finding the elec- tron in the specified region is the integral of ψ(x)2dx over that
region, as given in eqn 7B.11 The wavefunction of the electron
is given in eqn 8A.7a with n = 1 The integral you need is in the Resource section:
Answer The probability of finding the particle in a region
Now set n = 1, L = 1.0 nm, and l = 0.2 nm, which gives P = 0.05
The result corresponds to a chance of 1 in 20 of finding the
electron in the region As n becomes infinite, the sine term, which is multiplied by 1/n, makes no contribution to P and the classical result for a uniformly distributed particle, P = l/L, is
obtained
Self-test 8A.3 Calculate the probability that an electron in the
state with n = 1 will be found between x = 0.25L and x = 0.75L in
a conjugated molecule of length L (with x = 0 at the left-hand
end of the molecule)
Figure 8A.3 The first five normalized wavefunctions of a
particle in a box Each wavefunction is a standing wave;
successive functions possess one more half wave and a
correspondingly shorter wavelength
Figure 8A.4 (a) The first two wavefunctions, (b) the
corresponding probability densities, and (c) a representation of
the probability density in terms of the darkness of shading
Trang 68A Translation 321
(c) The properties of observables
The linear momentum of a particle in a box is not well defined
because the wavefunction sin kx is not an eigenfunction of the
linear momentum operator However, each wavefunction is a
linear combination of the linear momentum eigenfunctions eikx
and e−ikx Then, because sin x = (e ix − e−ix)/2i, we can write
It follows from the discussion in Topic 7C that half the
meas-urements of the linear momentum will give the value +k and
−k for the other half This detection of opposite directions
of travel with equal probability is the quantum mechanical
version of the classical picture that a particle in a
one-dimen-sional box rattles from wall to wall and in any given period
spends half its time travelling to the left and half travelling to
the right
Because n cannot be zero, the lowest energy that the
parti-cle may possess is not zero (as would be allowed by classical
mechanics, corresponding to a stationary particle) but
= Particle in a box Zero-point energy (8A.10)
This lowest, irremovable energy is called the zero-point energy
The physical origin of the zero-point energy can be explained
in two ways:
• The Heisenberg uncertainty principle requires a
particle to possess kinetic energy if it is confined to a
finite region: the location of the particle is not
completely indefinite (Δx ≠ ∞), so the uncertainty in
its momentum cannot be precisely zero (Δp ≠ 0)
Because Δp = (〈p2〉 − 〈p〉2)1/2 = 〈p2〉1/2 in this case, Δp ≠ 0
implies that 〈p2〉 ≠ 0, which implies that the particle must always have nonzero kinetic energy
• If the wavefunction is to be zero at the walls, but smooth, continuous, and not zero everywhere, then
it must be curved, and curvature in a wavefunction implies the possession of kinetic energy
The separation between adjacent energy levels with quantum
numbers n and n + 1 is
E n+ 1−E n= +n mL h − n h mL = n+ mL h
2 2 2
2 2 2
2 2
Example 8A.2 Estimating an absorption wavelength
β-Carotene (1) is a linear polyene in which 10 single and 11
double bonds alternate along a chain of 22 carbon atoms If
we take each CeC bond length to be about 140 pm, then the
length L of the molecular box in β-carotene is L = 2.94 nm
Estimate the wavelength of the light absorbed by this molecule from its ground state to the next higher excited state
1 β-Carotene
Method For reasons that will be familiar from introductory chemistry, each C atom contributes one p electron to the π-orbitals Use eqn 8A.11 to calculate the energy separation between the highest occupied and the lowest unoccupied levels, and convert that energy to a wavelength by using the Bohr frequency relation (eqn 7A.12)
Answer There are 22 C atoms in the conjugated chain; each contributes one p electron to the levels, so each level up to
n = 11 is occupied by two electrons The separation in energy
between the ground state and the state in which one electron is
mJ
)
Figure 8A.5 The probability density ψ 2(x) for large quantum
number (here n = 50, blue, compared with n = 1, red) Notice
that for high n the probability density is nearly uniform,
provided we ignore the fine detail of the increasingly rapid
oscillations
Trang 7322 8 The quantum theory of motion
8A.3 Confined motion in two or more
dimensions
Now consider a rectangular two-dimensional region of a
sur-face with length L1 in the x-direction and L2 in the y-direction;
the potential energy is zero everywhere except at the walls,
where it is infinite (Fig 8A.6) As a result, the particle is never
found at the walls and its wavefunction is zero there and
every-where outside the two-dimensional region Between the walls,
because the particle has contributions to its kinetic energy from
its motion in both the x and y directions, the Schrödinger
equa-tion has two kinetic energy terms, one for each axis For a
parti-cle of mass m the equation is
This is a partial differential equation (Mathematical background 4), and the resulting wavefunctions are functions of both x and
y, denoted ψ(x,y) This dependence means that the
wavefunc-tion and the corresponding probability density depend on the location in the plane, with each position specified by the values
of the coordinates x and y.
(a) Separation of variables
A partial differential equation of the form of eqn 8A.12
can be simplified by the separation of variables technique
(Mathematical background 4), which divides the equation into
two or more ordinary differential equations, one for each
vari-able We show in the Justification below using this technique
that the wavefunction can be written as a product of functions,
one depending only on x and the other only on y:
ψ ( , ) x y X x Y y= ( ) ( ) (8A.13a)
and that the total energy is given by
where E X is the energy associated with the motion of the
parti-cle parallel to the x-axis, and likewise for E Y and motion parallel
to the y-axis.
or 0.160 aJ It follows from the Bohr frequency condition
(ΔE = hν) that the frequency of radiation required to cause this
λ = 1240 nm The experimental value is 603 THz (λ = 497 nm),
corresponding to radiation in the visible range of the
electro-magnetic spectrum Considering the crudeness of the model
we have adopted here, we should be encouraged that the
com-puted and observed frequencies agree to within a factor of 2.5
Self-test 8A.4 Estimate a typical nuclear excitation energy
in electronvolts (1 eV = 1.602 × 10−19 J; 1 GeV = 109 eV) by
cal-culating the first excitation energy of a proton confined to a
one-dimensional box with a length equal to the diameter of a
nucleus (approximately 1 × 10−15 m, or 1 fm)
Answer: 0.6 GeV
Justification 8A.2 The separation of variables technique applied to the particle in a two-dimensional box
We follow the procedure in Mathematical background 4
and apply it to eqn 8A.12 The first step to confirm that the Schrödinger equation can be separated and the wavefunction
can be factored into the product of two functions X and Y is to note that, because X is independent of y and Y is independent
of x, we can write
∂
2 2
2 2
2 2
2 2
2 2
2 2
Figure 8A.6 A two-dimensional square well The particle is
confined to the plane bounded by impenetrable walls As soon
as it touches the walls, its potential energy rises to infinity
Trang 88A Translation 323
Each of the two ordinary differential equations in eqn
8A.14 is the same as the one-dimensional particle-in-a-box
Schrödinger equation (Section 8A.2) The boundary
condi-tions are also the same, apart from the detail of requiring
X(x) to be zero at x = 0 and L1, and Y(y) to be zero at y = 0
and L2 We can therefore adapt eqn 8A.7a without further
calculation:
X x n1 L2 n x L 0 x L
1
1 2 1
Similarly, because E = E X + E Y, the energy of the particle is
lim-ited to the values
E n n1 2 n L n L h m
1
1
2 2
with the two quantum numbers taking the values n1 = 1, 2, …
and n2 = 1, 2, … independently The state of lowest energy is
(n1 = 1, n2 = 1) and E1,1 is the zero-point energy
Some of the wavefunctions are plotted as contours in Fig 8A.7 They are the two-dimensional versions of the wavefunc-tions shown in Fig 8A.3 Whereas in one dimension the wave-functions resemble states of a vibrating string with ends fixed,
in two dimensions the wavefunctions correspond to vibrations
of a rectangular plate with fixed edges
The first term on the left, (1/X)(d2X/dx2), is independent of y,
so if y is varied only the second term on the left, (1/Y)(d2Y/dy2),
can change But the sum of these two terms is a constant,
2mE/2, given by the right-hand side of the equation Therefore,
if the second term did change, then the right-hand side could
not be constant Consequently, even the second term cannot
change when y is changed In other words, the second term,
(1/Y)(d2Y/dy2), is a constant, which we write −2mE Y/2 By a
similar argument, the first term, (1/X)(d2X/dx2), is a constant
when x changes, and we write it −2mE X/2, with E = E X + E Y
Therefore, we can write
These expressions rearrange into the two ordinary (that is,
single-variable) differential equations
−2m2 dd2x X2 =E X X −2m2 dd2y Y2 =E Y Y (8A.14)
Twodimensional box
energy levels (8A.15b)
Brief illustration 8A.3 The distribution of a particle in a two-dimensional box
Consider an electron confined to a square cavity of side L, and
in the state with quantum numbers n1 = 1, n2 = 2 Because the probability density is
ψ1 22
, (x y, )=L sin πL xsin Lπy
the most probable locations correspond to sin2(πx/L) = 1 and
sin2(2πx/L) = 1, or (x,y) = (L/2, L/4) and (L/2, 3L/4) The least
probable locations (the nodes, where the wavefunction passes through zero) correspond to zeroes in the probability density
within the box, which occur along the line y = L/2.
Self-test 8A.5 Determine the most probable locations of an
electron in a square cavity of side L when it is in the state with quantum numbers n1 = 2, n2 = 3
Answer: points (x = L/4 and 3L/4; y = L/6, L/2, and 5L/6)
(8A.15a)
Two
dimensional box
functions
Trang 9324 8 The quantum theory of motion
We treat a particle in a three-dimensional box in the same
way The wavefunctions have another factor (for the
z-depend-ence), and the energy has an additional term in n L3/ Solution 3
of the Schrödinger equation by the separation of variables
tech-nique then gives
1 2 3
1 2 1 1
2 2 , ,
3 3
πfor
Threedimensional box wavefunctions (8A.16a)
E n n n1 2 3 n L n L n L h m
1
1
2 2
3 3
The quantum numbers n1, n2, and n3 are all positive
inte-gers 1, 2, … that can be varied independently The system has a
zero-point energy (E1,1,1 = 3h2/8mL2 for a cubic box)
(b) Degeneracy
A special feature of the solutions arises when a
two-dimen-sional box is not merely rectangular but square, with L1 = L2 = L
Then the wavefunctions and their energies are
8
, =( + ) Square box energy levels (8.17b)
Consider the cases n1 = 1, n2 = 2 and n1 = 2, n2 = 1:
2 2
Although the wavefunctions are different, they have the same
energy The technical term for different wavefunctions
corres-ponding to the same energy is degeneracy, and in this case we
say that the state with energy 5h2/8mL2 is ‘doubly degenerate’
In general, if N wavefunctions correspond to the same energy,
then we say that the state is ‘N-fold degenerate’.
The occurrence of degeneracy is related to the symmetry
of the system Figure 8A.8 shows contour diagrams of the two
degenerate functions ψ1,2 and ψ2,1 Because the box is square,
one wavefunction can be converted into the other simply by
rotating the plane by 90° Interconversion by rotation through
90° is not possible when the plane is not square, and ψ1,2 and
ψ2,1 are then not degenerate Similar arguments account for the degeneracy of states in a cubic box Other examples of degener-acy occur in quantum mechanical systems (for instance, in the hydrogen atom, Topic 9A), and all of them can be traced to the symmetry properties of the system
8A.4 Tunnelling
If the potential energy of a particle does not rise to infinity
when it is in the wall of the container, and E < V, the
wavefunc-tion does not decay abruptly to zero If the walls are thin (so that the potential energy falls to zero again after a finite dis-tance), then the wavefunction oscillates inside the box, varies
Three
dimensional box
energy levels (8A.16b)
Brief illustration 8A.4 Degeneracies in a dimensional box
two-The energy of a particle in a two-dimensional square box of
side L in the state with n1 = 1, n2 = 7 is
E1 7 2 2 mL h2 mL h
2
2 2
, =( + ) =
This state is degenerate with the state with n1 = 7 and n2 = 1
Thus, at first sight the energy level 50h2/8mL2 is doubly ate However, in certain systems there may be states that are not apparently related by symmetry but are ‘accidentally’
degener-degenerate Such is the case here, for the state with n1 = 5 and
n2 = 5 also has energy 50h2/8mL2 Accidental degeneracy is also encountered in the hydrogen atom (Topic 9A)
Self-test 8A.6 Find a state (n1, n2) for a particle in a rectangular
box with sides of length L1 = L and L2 = 2L that is accidentally
degenerate with the state (4,4)
Answer: (n1 = 2, n2 = 8)
Square box wave-functions (8.17a)
+–
Figure 8A.8 The wavefunctions for a particle confined to a square well Note that one wavefunction can be converted into the other by rotation of the box by 90° The two functions correspond to the same energy True degeneracy is a
consequence of symmetry
Trang 108A Translation 325
smoothly inside the region representing the wall, and oscillates
again on the other side of the wall outside the box (Fig 8A.9)
Hence the particle might be found on the outside of a container
even though according to classical mechanics it has insufficient
energy to escape Such leakage by penetration through a
classi-cally forbidden region is called tunnelling.
The Schrödinger equation is used to calculate the probability
of tunnelling of a particle of mass m incident from the left on
a rectangular potential energy barrier that extends from x = 0
to x = L On the left of the barrier (x < 0) the wavefunctions are
those of a particle with V = 0, so from eqn 8A.2 we can write
ψ = Aeikx+Be− ikx k=(2mE)1 2 /
The Schrödinger equation for the region representing the
bar-rier (0 ≤ x ≤ L), where the potential energy is the constant V, is
−2m2 d2dψ x( )2x +V x E x ψ( )= ψ( ) (8A.19)
We shall consider particles that have E < V (so, according to
classical physics, the particle has insufficient energy to pass
through the barrier), and therefore for which V − E > 0 The
general solutions of this equation are
ψ=Ceκ x+De−κ x κ={ (2m V E− )}1 2 /
as can be verified by differentiating ψ twice with respect to x
The important feature to note is that the two exponentials in
eqn 8A.20 are now real functions, as distinct from the complex,
oscillating functions for the region where V = 0 To the right of
the barrier (x > L), where V = 0 again, the wavefunctions are
ψ = A′eikx k=(2mE)1 2 /
Note that to the right of the barrier, the particle can only be moving to the right and therefore terms of the form e−ikx do not contribute to the wavefunction in eqn 8A.21
The complete wavefunction for a particle incident from the left consists of (Fig 8A.10):
• an incident wave (Ae ikx corresponds to positive momentum);
• a wave reflected from the barrier (Be −ikx corresponds
to negative momentum, motion to the left);
• the exponentially changing amplitudes inside the barrier (eqn 8A.20);
• an oscillating wave (eqn 8A.21) representing the propagation of the particle to the right after tunnelling through the barrier successfully
The probability that a particle is travelling towards positive x (to the right) on the left of the barrier (x < 0) is proportional to
|A|2, and the probability that it is travelling to the right on the
right of the barrier (x > L) is |A′|2 The ratio of these two
prob-abilities, |A′|2/|A|2, which reflects the probability of the particle
tunnelling through the barrier, is called the transmission
prob-ability, T.
To determine the relationship between |A′|2 and |A|2, we need to investigate the relationships between the coefficients
A, B, C, D, and A′ Since the acceptable wavefunctions must
be continuous at the edges of the barrier (at x = 0 and x = L,
function inside barrier
wave-(8A.20)
(8A.18)
Particle in a rectangular barrier
function left of barrier
wave-(8A.21)
Particle in a rectangular barrier
function right of barrier
Incident wave
Transmitted wave Reflected
wave
x
Figure 8A.10 When a particle is incident on a barrier from the left, the wavefunction consists of a wave representing linear momentum to the right, a reflected component representing momentum to the left, a varying but not oscillating component inside the barrier, and a (weak) wave representing motion to the right on the far side of the barrier
Figure 8A.9 A particle incident on a barrier from the left
has an oscillating wave function, but inside the barrier there
are no oscillations (for E < V) If the barrier is not too thick,
the wavefunction is nonzero at its opposite face, and so
oscillates begin again there (Only the real component of the
Trang 11326 8 The quantum theory of motion
After straightforward but lengthy algebraic manipulations of
the above set of equations 8A.22 (see Problem 8A.6), we find
where ε = E/V This function is plotted in Fig 8A.12 The
trans-mission probability for E > V is shown there too The
transmis-sion probability has the following properties:
• T ≈ 0 for E ≪ V;
• T increases as E approaches V: the probability of
tunnelling increases;
• T approaches, but is still less than, 1 for E > V: there
is still a probability of the particle being reflected by the barrier even when classically it can pass over it;
• T ≈ 1 for E ≫ V, as expected classically.
For high, wide barriers (in the sense that κL ≫ 1), eqn 8A.23a simplifies to
T≈16 (1 )eε −ε − 2κ L
The transmission probability decreases exponentially with the
thickness of the barrier and with m1/2 It follows that particles of low mass are more able to tunnel through barriers than heavy ones (Fig 8A.13) Tunnelling is very important for electrons
and muons (mµ ≈ 207 me), and moderately important for
pro-tons (mp ≈ 1840me); for heavier particles it is less important
A number of effects in chemistry (for example, the dependence of some reaction rates) depend on the ability of the proton to tunnel more readily than the deuteron The very rapid equilibration of proton transfer reactions is also a mani-festation of the ability of protons to tunnel through barriers and transfer quickly from an acid to a base Tunnelling of protons between acidic and basic groups is also an important feature of the mechanism of some enzyme-catalysed reactions
isotope-Rectangular potential barrier
transmission probability (8A.23a)
Rectangular potential barrier; κL ≫1
transmission probability (8A.23b)
con-from the acid is computed using 8A.23a, with ε = E/V = 1.995 eV/2.000 eV = 0.9975 and V − E = 0.005 eV (corresponding to
Figure 8A.11 The wavefunction and its slope must be
continuous at the edges of the barrier The conditions for
continuity enable us to connect the wavefunctions in the three
zones and hence to obtain relations between the coefficients
that appear in the solutions of the Schrödinger equation
0 0.2 0.4
Incident energy, E/V
2
10
0.6 0.8 1
Figure 8A.12 The transmission probabilities for passage
through a rectangular potential barrier The horizontal axis
is the energy of the incident particle expressed as a multiple
of the barrier height The curves are labelled with the value
of L(2mV)1/2/ The graph on the left is for E < V and that on
the right for E > V Note that T > 0 for E < V whereas classically
T would be zero However, T < 1 for E > V, whereas classically T
Trang 128A Translation 327
A problem related to tunnelling is that of a particle in a
square-well potential of finite depth (Fig 8A.14) In this kind
of potential, the wavefunction penetrates into the walls, where
it decays exponentially towards zero, and oscillates within the
well The wavefunctions are found by ensuring, as in the
discus-sion of tunnelling, that they and their slopes are continuous at
the edges of the potential Some of the lowest energy solutions
are shown in Fig 8A.15 A further difference from the solutions
for an infinitely deep well is that there is only a finite number
of bound states Regardless of the depth and length of the well,
however, there is always at least one bound state Detailed
con-sideration of the Schrödinger equation for the problem shows
that in general the number of levels is equal to N, with
where V is the depth of the well and L is its length We see that
the deeper and wider the well, the greater the number of bound states As the depth becomes infinite, so the number of bound states also becomes infinite, as we have already seen
Checklist of concepts
quantized
only certain wavefunctions are acceptable and
there-fore restricts observables to discrete values
half-integer) that labels the state of the system
irremovable minimum energy
mechanics emerges from quantum mechanics as high
quantum numbers are reached
three-dimensional box is the product of wavefunctions for the particle in a one-dimensional box
box is the sum of energies for the particle in two or three one-dimensional boxes
two-dimen-sional box corresponds to the state with quantum
num-bers (n1 = 1, n2 = 1); for three dimensions, (n1 = 1, n2 = 1,
n3 = 1)
cor-respond to the same energy
The larger the value of L(2mV)1/2/ (here, 31) the smaller is the
value of T for energies close to, but below, the barrier height.
Self-test 8A.7 Suppose that the junction between two
semi-conductors can be represented by a barrier of height 2.00 eV
and length 100 pm Calculate the probability that an electron
of energy 1.95 eV can tunnel through the barrier
Trang 13328 8 The quantum theory of motion
symmetry of the system
region is called tunnelling.
in the height and width of the potential barrier
than heavy ones
Checklist of equations
Free-particle wavefunctions and
Transmission probability T= +{ (1 eκ L−e−κ L) /2 16 1ε(−ε)}−1 Rectangular potential barrier 8A.23a
T = 16ε(1 − ε)e −2κL High, wide rectangular barrier 8A.23b
Trang 148B Vibrational motion
Atoms in molecules and solids vibrate around their mean tions as bonds stretch, compress, and bend Here we consider one particular type of vibrational motion, that of ‘harmonic motion’ in one dimension
posi-8B.1 The harmonic oscillator
A particle undergoes harmonic motion, and is said to be a harmonic oscillator, if it experiences a restoring force propor-
tional to its displacement:
F= −k xf Harmonic motion restoring force (8B.1)
where kf is the force constant: the stiffer the ‘spring’, the greater
the value of kf Because force is related to potential energy by
F = −dV/dx (see Foundations B), the force in eqn 8B.1
corre-sponds to the particle having a potential energy
V x( )=1k x
2 f 2 Parabolic potential energy (8B.2)
when it is displaced through a distance x from its equilibrium
position This expression, which is the equation of a ola (Fig 8B.1), is the origin of the term ‘parabolic potential energy’ for the potential energy characteristic of a harmonic
parab-oscillator The Schrödinger equation for the particle of mass m
is therefore
Contents
brief illustration 8b.1: the vibration of a
example 8b.1: confirming that a wavefunction
is a solution of the schrödinger equation 332
example 8b.2: normalizing a harmonic oscillator
example 8b.3: calculating properties of a
example 8b.4: calculating the tunnelling
probability for the harmonic oscillator 335
➤
➤ Why do you need to know this material?
The detection and interpretation of vibrational frequencies
is the basis of infrared spectroscopy (Topics 12D and 12E)
Molecular vibration plays a role in the interpretation
of thermodynamic properties, such as heat capacities
(Topics 5E and 15F), and of the rates of chemical reactions
(Topic 21C).
➤
➤ What is the key idea?
The quantum mechanical treatment of the simplest model
of vibrational motion, the harmonic oscillator, reveals
that the energy is quantized and the wavefunctions are
products of a polynomial and a Gaussian (bell-shaped)
function.
➤
➤ What do you need to know already?
You should know how to formulate the Schrödinger
equation given a potential energy function You should
also be familiar with the concepts of tunnelling (Topic 8A)
and the expectation value of an observable (Topic 7B).
a harmonic oscillator, where x is the displacement from
equilibrium The narrowness of the curve depends on the force
constant kf: the larger the value of kf, the narrower the well
Trang 15330 8 The quantum theory of motion
2 2
2mddψ x( )x k xf ψ( )x E x ψ( )
We can anticipate that the energy of an oscillator will be
quan-tized because the wavefunction has to satisfy boundary
con-ditions (as in Topic 8A for a particle in a box): it will not be
found with very large extensions because its potential energy
rises to infinity there That is, when we impose the
bound-ary conditions ψ = 0 at x = ±∞, we can expect to find that only
certain wavefunctions and their corresponding energies are
possible
(a) The energy levels
Equation 8B.3 is a standard equation in the theory of
differen-tial equations and its solutions are well known to
mathemati-cians.1 The permitted energy levels are
where ω (omega) is the frequency of oscillation of a classical
harmonic oscillator of the same mass and force constant Note
that ω is large when the force constant is large and the mass
small It follows that the separation between adjacent levels is
which is the same for all v Therefore, the energy levels form
a uniform ladder of spacing ω (Fig 8B.2) The energy
sepa-ration ω is negligibly small for macroscopic objects (with
large mass) for which classical mechanics is adequate for
describing vibrational motion; however, the energy tion is of great importance for objects with mass similar to that of atoms
separa-Because the smallest permitted value of v is 0, it follows from
eqn 8B.4 that a harmonic oscillator has a zero-point energy
E0 1 2
= ω Harmonic oscillator Zero-point energy (8B.6)
The mathematical reason for the zero-point energy is that
v cannot take negative values, for if it did the wavefunction would not obey the boundary conditions The physical reason
is the same as for the particle in a box (Topic 8A): the cle is confined, its position is not completely uncertain, and therefore its momentum, and hence its kinetic energy, cannot
parti-be exactly zero We can picture this zero-point state as one in which the particle fluctuates incessantly around its equilib-rium position; classical mechanics would allow the particle to
be perfectly still
Atoms vibrate relative to one another in molecules with the bond acting like a spring The question then arises as to what mass to use to predict the frequency of the vibration
In general, the relevant mass is a complicated combination
of the masses of all the atoms that move, with each bution weighted by the amplitude of the atom’s motion That amplitude depends on the mode of motion, such as whether the vibration is a bending motion or a stretching motion, so each mode of vibration has a characteristic ‘effective mass’ For a diatomic molecule AB, however, for which there is only one mode of vibration, corresponding to the stretching and
contri-compression of the bond, the effective mass, μ, has a very
simple form:
μ = m m m mA+ B
A B Diatomic molecule effective mass (8B.7)
When A is much heavier than B, mB can be neglected in the
denominator and the effective mass is μ ≈ mB, the mass of the lighter atom This result is plausible, for in the limit of the heavy atom being like a brick wall, only the lighter atom moves and hence determines the vibrational frequency
1 For details, see our Molecular quantum mechanics, Oxford University
8 v
Potential energy
Figure 8B.2 The energy levels of a harmonic oscillator are
evenly spaced with separation ħω, with ω = (kf /m)1/2 Even in its
lowest energy state, an oscillator has an energy greater than
which is close to the mass of the proton The force constant of
the bond is kf = 516.3 N m−1 It follows from eqn 8B.4, with μ in place of m, that
(8B.4)
Harmonic oscillator energy levels
(8B.3)
Harmonic oscillator schrödinger equation
Trang 168B Vibrational motion 331
The result in Brief illustration 8B.1 implies that excitation
requires radiation of frequency ν = ΔE/h = 90 THz and
wave-length λ = c/ν = 3.3 µm It follows that transitions between
adja-cent vibrational energy levels of molecules are stimulated by or
emit infrared radiation (Topics 12D and 12E)
(b) The wavefunctions
Like the particle in a box (Topic 8A), a particle undergoing
har-monic motion is trapped in a symmetrical well in which the
potential energy rises to large values (and ultimately to infinity)
for sufficiently large displacements (compare Figs 8A.1 and
8B.1) However, there are two important differences:
• Because the potential energy climbs towards infinity
only as x2 and not abruptly, the wavefunction
approaches zero more slowly at large displacements
than for the particle in a box
• As the kinetic energy of the oscillator depends on
the displacement in a more complex way (on account
of the variation of the potential energy), the
curvature of the wavefunction also varies in a more
complex way
The detailed solution of eqn 8B.3 confirms these points
and shows that the wavefunctions for a harmonic oscillator
have the form
where N is a normalization constant A Gaussian function is a
bell-shaped function of the form e−x2
(Fig 8B.3) The precise form of the wavefunctions is
The factor H v (y) is a Hermite polynomial; the form of these
poly-nomials and some of their properties are listed in Table 8B.1 Hermite polynomials, which are members of a class of functions called ‘orthogonal polynomials’, have a wide range of important properties which allow a number of quantum mechanical cal-culations to be done with relative ease Note that the first few
Hermite polynomials are very simple: for instance, H0(y) = 1 and
(8B.9a)
Harmonic oscillator ground-state wavefunction
or 563.4 THz (We have used 1 N = 1 kg m s−2.) Therefore the
separation of adjacent levels is (eqn 8B.5)
E v+1− =E v ( 1 54 57 10 × 0− 34Js) ( ×5 634×1014s) =5 941 1× 0− 2 0J
or 59.41 zJ, about 0.37 eV This energy separation corresponds
to 36 kJ mol−1, which is chemically significant The zero-point
energy, eqn 8B.6, of this molecular oscillator is 29.71 zJ, which
corresponds to 0.19 eV, or 18 kJ mol−1
Self-test 8B.1 Suppose a hydrogen atom is adsorbed on the
surface of a gold nanoparticle by a bond of force constant
855 N m−1 Calculate its zero-point vibrational energy
6 64y6 − 480y4 + 720y2 − 120
* The Hermite polynomials are solutions of the differential equation
H v″− 2yH v′+ 2v H v= 0 where primes denote differentiation They satisfy the recursion relation
Trang 17332 8 The quantum theory of motion
and the corresponding probability density is
( )x N= e−y =N e−x/
The wavefunction and the probability density are shown in Fig
8B.4 Both curves have their largest values at zero displacement
(at x = 0), so they capture the classical picture of the zero-point
energy as arising from the ceaseless fluctuation of the particle
about its equilibrium position
The wavefunction for the first excited state of the oscillator,
the state with v = 1, is
This function has a node at zero displacement (x = 0), and the
probability density has maxima at x = ±α (Fig 8B.5).
The shapes of several wavefunctions are shown in Fig 8B.6 and the corresponding probability densities are shown in Fig 8B.7 At high quantum numbers, harmonic oscillator wave-functions have their largest amplitudes near the turning points
of the classical motion (the locations at which V = E, so the
kinetic energy is zero) We see classical properties emerging in the correspondence principle limit of high quantum numbers (Topic 8A), for a classical particle is most likely to be found at the turning points (where it travels most slowly) and is least likely to be found at zero displacement (where it travels most rapidly)
Figure 8B.4 The normalized wavefunction and probability
density (shown also by shading) for the lowest energy state of a
Figure 8B.5 The normalized wavefunction and probability
density (shown also by shading) for the first excited state of a
Method Substitute the wavefunction given in eqn 8B.9a
into eqn 8B.3 Use the definition of α given in eqn 8B.8
to determine the energy on the right-hand side of eqn 8B.3 and confirm that it matches the zero-point energy given in eqn 8B.6
Answer We need to evaluate the second derivative of the ground-state wavefunction:
2 2
1 2 0
The second and third terms on the left-hand side (in blue)
can-cel and we obtain E=1 k m
ground-state probability density
(8B.10)
Harmonic oscillator
First state wavefunction
Trang 18excited-8B Vibrational motion 333
Note the following features of the wavefunctions:
• The Gaussian function goes very strongly to zero as
the displacement increases (in either direction,
extension or compression), so all the wavefunctions
approach zero at large displacements
• The exponent y2 is proportional to x2 × (mkf)1/2, so
the wavefunctions decay more rapidly for large
masses and stiff springs
• As v increases, the Hermite polynomials become
larger at large displacements (as x v), so the
wavefunctions grow large before the Gaussian
function damps them down to zero: as a result, the
wavefunctions spread over a wider range as v
increases (Fig 8B.7)
8B.2 The properties of oscillators
The average value of a property is calculated by evaluating the expectation value of the corresponding operator (eqn 7C.11,
〈 〉Ω = ∫ψ Ωψ τ* d ) Now that we know the wavefunctions of the harmonic oscillator, we can start to explore its properties by evaluating integrals of the type
Figure 8B.6 The normalized wavefunctions for the first
five states of a harmonic oscillator Note that the number
of nodes is equal to v and that alternate wavefunctions
are symmetrical or asymmetrical about y = 0 (zero
displacement)
v
0 1 2 3 4
18
Probability
density
Displacement
Figure 8B.7 The probability densities for the first five
states of a harmonic oscillator and the state with v = 18
Note how the regions of highest probability density move
towards the turning points of the classical motion as v
Method Normalization is carried out by evaluating the
inte-gral of |ψ|2 over all space and then finding the normalization factor from eqn 7B.3 (N = ∫1/(ψ ψ τ* d ) )1 2 / The normalized
wavefunction is then equal to Nψ In this one-dimensional problem, the volume element is dx and the integration is
from −∞ to +∞ The wavefunctions are expressed in terms of
the dimensionless variable y = x/α, so begin by expressing the integral in terms of y by using dx = αdy The integrals required
are given in Table 8B.1
Answer The unnormalized wavefunction is
a box, for a harmonic oscillator N v is different for each value
of v.
Self-test 8B.3 Confirm, by explicit evaluation of the integral,
that ψ0 and ψ1 are orthogonal
Answer: Show that ∫−∞∞ ψ ψ0 1* dx= 0 by using the information in
Table 8B.1
Trang 19334 8 The quantum theory of motion
(a) Mean values
We show in the following example that the mean displacement,
〈x〉, and the mean square displacement, 〈x2〉, of the oscillator
when it is in the state with quantum number v are
〈 〉x =0 Harmonic oscillator mean displacement (8B.12a)
The result for 〈x〉 shows that the oscillator is equally likely
to be found on either side of x = 0 (like a classical oscillator)
The result for 〈x2〉 shows that the mean square displacement
increases with v This increase is apparent from the probability
densities in Fig 8B.7, and corresponds to the classical
ampli-tude of swing increasing as the oscillator becomes more highly
(v+1)
2 ω, it follows that
〈 〉V =1E
2 v Harmonic oscillator mean potential energy (8B.13b)
The total energy is the sum of the potential and kinetic gies, so it follows at once that the mean kinetic energy of the oscillator is (as could also be shown using the kinetic energy operator)
ener-〈 〉Ek =1E
2 v Harmonic oscillator mean kinetic energy (8B.13c)
The result that the mean potential and kinetic energies of a harmonic oscillator are equal (and therefore that both are
equal to half the total energy) is a special case of the virial theorem:
If the potential energy of a particle has the form
V = ax b, then its mean potential and kinetic energies are related by
Example 8B.3 Calculating properties of a harmonic
oscillator
Consider the harmonic oscillator motion of the HeCl
mol-ecule in Brief illustration 8B.1 Calculate the mean
displace-ment of the oscillator when it is in a state with quantum
number v.
Method Normalized wavefunctions must be used to
calcu-late the expectation value The operator for position along x is
multiplication by the value of x (Topic 7C) The resulting
inte-gral can be evaluated either
• by inspection (the integrand is the product of an odd
and an even function), or
• by explicit evaluation using the formulas in Table 8B.1
The former procedure makes use of the definitions that an
even function is one for which f(−x) = f(x) and an odd
func-tion is one for which f(−x) = −f(x) Therefore, the product of an
odd and even function is itself odd, and the integral of an odd
function over a symmetrical range about x = 0 is zero The
lat-ter procedure using explicit integration is illustrated here to
give practice in the calculation of expectation values We shall
need the relation x = αy, which implies that dx = αdy.
Answer The integral we require is
Both integrals are zero (See Table 8B.1), so 〈x〉 = 0 The mean
displacement is zero because displacements on either side of the equilibrium position occur with equal probability
Self-test 8B.4 Calculate the mean square displacement, 〈x2〉,
of the H − Cl bond distance from its equilibrium position by using the recursion relation in Table 8B.1 twice
Answer: (v+ × 1 )
2 115pm ; eqn 9.12b, with μ in place of m2
(8B.12b)
Harmonic oscillator mean square displacement
(8B.13a)
Harmonic oscillator mean potential energy
Trang 208B Vibrational motion 335
For a harmonic oscillator b = 2, so 〈Ek〉 = 〈V〉, as we have found
The virial theorem is a short cut to the establishment of a
num-ber of useful results, and we use it elsewhere (for example, in
Topic 9A)
(b) Tunnelling
An oscillator may be found at extensions with V > E, which are
forbidden by classical physics, for they correspond to negative
kinetic energy; this is an example of the phenomenon of
tun-nelling (Topic 8A) As shown in Example 8B.4, for the lowest
energy state of the harmonic oscillator, there is about an 8 per
cent chance of finding the oscillator stretched beyond its
clas-sical limit and an 8 per cent chance of finding it with a
classi-cally forbidden compression These tunnelling probabilities are
independent of the force constant and mass of the oscillator
Example 8B.4 Calculating the tunnelling probability for
the harmonic oscillator
Calculate the probability that the ground-state harmonic
oscillator will be found in a classically forbidden region
Method Find the expression for the classical turning point,
xtp, where the kinetic energy vanishes, by equating the
poten-tial energy to the total energy E of the harmonic oscillator The
probability of finding the oscillator stretched beyond a
dis-placement xtp is the sum of the probabilities ψ 2dx of finding
it in any of the intervals dx lying between xtp and infinity, so
evaluate the integral
x
=∫∞ψ v2d
tp
The variable of integration is best expressed in terms of y = x/α
and the integral to be evaluated is a special case of the error
function, erf z, defined as
z
= −1 21 2∫∞ − 2
and evaluated for some values of z in Table 8B.2 (this function
is commonly available in mathematical software packages)
By symmetry, the probability of being found stretched into a
classically forbidden region is the same as that of being found
compressed into a classically forbidden region
Answer According to classical mechanics, the turning point,
xtp, of an oscillator occurs when its kinetic energy is zero,
which is when its potential energy 1
2k xf 2 is equal to its total
energy E This equality occurs when
For the state of lowest energy (v = 0), ytp = 1 and the probability
of being beyond that point is
tp
The normalization constant N0 is calculated from the
expres-sion for N v in Example 8B.2 ( N v=1/(α π1 2 /2v v!)1 2 /):
N0 1 2 0
1 2
1 2
1 21
The integral in the expression for P is written in terms of the
error function erf(1) as
1
1 2 1
In 7.9 per cent of a large number of observations, any
oscil-lator in the state with quantum number v = 0 will be found
stretched to a classically forbidden extent There is the same probability of finding the oscillator with a classically forbid-den compression The total probability of finding the oscilla-tor tunnelled into a classically forbidden region (stretched or compressed) is about 16 per cent
Self-test 8B.5 Calculate the probability that a harmonic
oscil-lator in the state with quantum number v = 1 will be found at
a classically forbidden extension (Follow the argument given
in Example 8B.4 and use the method of integration by parts (Mathematical background 1) to obtain an integral that can be
expressed in terms of the error function.)
Answer: P = 0.056