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Lecture Note on Solid State Physics
Lattice waves
Masatsugu Suzuki and Itsuko S. Suzuki
Physics, Department, State University of New York at Binghamton
(December 6, 2007)
Abstract
A lecture note on the lattice waves in the solid is presented. In a crystal each atom are
coupled with the neighboring atoms by spring constants. The collective motion of atoms
leads to a well-defined traveling wave over the whole system, leading to the collective
motion, so called phonon. Here the equation of motion of atoms around thermal
equilibrium position will be discussed in terms of several methods, which include
numerical calculation on the eigenvalue problem (based on the Mathematica) and the
translation operators in the quantum mechanics. We show that all these methods lead to
the same conclusion, the existence of lattice wave, phonon in the quantum mechanics.
For students who just start to study the solid state physics, it may be difficult to
understand the validity of the assumption that the deviation of the displacement of the
atoms from the thermal equilibrium is well described by a traveling wave. We
numerically solve the eigenvalue-problem of the motion of atoms in the linear chain
(typically 50 - 100 atoms) using Mathematica. We will give a bit of evidence for the
existence of the normal modes propagating along the linear chain.
Content
1. Introduction
2. Lattice waves
2.1 Overview
2.2 One dimensional case: longitudinal mode
2.3 Born-von Karman boundary condition
2.4 First Brillouin zone
2.5 Normal modes
2.6 Continuum wave equation
3. Eigenvalue-problem; solution using Mathematica
4. First Brillouin zone and group velocity
4.1 Definition of the group velocity
4.2 The physical meaning of the first Brillouin zone
4.3 Standing wave
4.4 General property of the group velocity
5. Determination of force constants
5.1
The system with the nearest neighbor interaction
5.2
System with long-ranged interactions
6. Vibration of square lattice
7. Two atoms per primitive basis
8. The number of modes; degree of freedom
8.1 One-dimensional case
8.2 Three-dimensional case
9. Classical Model
1
9.1 Theory of the transverse wave in a string
9.2 Energy density of the elastic wave
10. Quantum mechanical approach: phonon
10.1 Annihilation and creation operators
10.2 Symmetry of lattice and translation operator
11. Crystal momentum
12. Semi-classical approach
12.1 Simple case
12.2 General case
13. Inelastic neutron scattering by crystal with lattice vibration
13.1 Scattering cross section
13.2 Energy and momentum conservation
Conclusion
References
Appendix
A.
Simple harmonics (1D) in quantum mechanics
B.
Translation operator in quantum mechanics
1.
Introduction
Phonons are a quantum mechanical version of a special type of vibrational motion,
known as normal modes in classical mechanics, in which each part of a lattice oscillates
with the same frequency. These normal modes are important because, according to a
well-known result in classical mechanics, any arbitrary vibrational motion of a lattice can
be considered as a superposition of normal modes with various frequencies; in this sense,
the normal modes are the elementary vibrations of the lattice. Although normal modes are
wave-like phenomena in classical mechanics, they acquire certain particle-like properties
when the lattice is analyzed using quantum mechanics (see wave-particle duality.) They
are then known as phonons.
There have been many excellent textbooks on the physics of lattice waves. Typical
books1-14 which we read during the preparation of writing this lecture note, are presented
in References. These books are very useful for our understanding physics.
2
2.1
Lattice waves
Overview
Consider the elastic vibrations of a crystal with one atom in the primitive cell. We
want to find the frequency of an elastic wave in terms of the wavevector k and the elastic
constants. When a wave propagates along the x-direction, entire planes of atoms move in
phase with displacements either parallel or perpendicular to the direction of k.
We can describe with a single co-ordinate us the displacement of the plane s from its
equilibrium position.
2
Fig,.1 (Left figure) (Dashed lines) Planes of atoms when in e quilibrium. (Solid lines)
Planes of atoms when displaced as for a longitudinal wave. The coordinate u
measures the displacement of the planes. (Right figure) Plane of atoms as
displaced during passage of transverse wave.
For each wavevector there are three modes; one of longitudinal polarization, two of
transverse polarization. We assume that the elastic response of the crystal is a linear
function of the forces. Or the elastic energy is a quadratic function of the relative
displacement of any two points in the crystal. The forces on the plane s caused by the
displacement of the plane s+p is proportional to the difference us+p - us of their
displacements.
For brevity, we consider only nearest-neighbor interactions, so that p = ±1. The total
force on s comes from planes s ± 1.
Fig.2 The displacements of atoms with mass M are denoted by us-1, us, and us+1. The
repeat distance is a in the direction of the wavevector k. The direction of us is
parallel to the direction of the wavevector k for the longitudinal wave and is
perpendicular to the direction of the wavevector k for the transverse wave
3
2.2
One dimensional case: longitudinal mode
We start our discussion with the Lagrangian for the displacement of the s-th plane,
given by
Ls = Ts − Vs
1
1
mu s2 − C[(u s +1 + a − u s − a ) 2 + (u s + a − u s −1 − a) 2 ] .
2
2
1
1
= mu s2 − C[(u s +1 − u s ) 2 + (u s − u s −1 ) 2 ]
2
2
The Lagrange’s equation for this system is derived as
d ⎛ ∂L ⎞ ∂L
⎟=
⎜
,
dt ⎜⎝ ∂u s ⎟⎠ ∂u s
or
1
mus = − C[2(us +1 − us )(−1) + 2(u s − us −1 )]
2
,
= −C (−us +1 + us + us − u s −1 )
=
(2.1)
(2.2)
(2.3)
= C (u s +1 − 2us + us −1 ) = Fs
where Fs = C (us +1 − us ) − C (us − us −1 ) is the effective force on the s-th plane (Hooke’s
law), C is the force constant between nearest-neighbor planes, C L ≠ CT (CL: force
constant for longitudinal wave, CT: force constant for transverse wave). It is convenient
hereafter to regard C as defined for one atom of the plane, so that Fs is the force on one
atom in the plane s.
The equation of motion of the plane s is
d 2u s
M
(2.4)
= C (u s +1 − 2u s + u s −1 ) ,
dt 2
where M is the mass of an atom in the s-th plane. Suppose that this equation has the
traveling wave solutions of the form
u s = ue i ( ksa −ωt ) .
(2.5)
Note that the validity of this assumption will be verified by solving directly the
eigenvalue problem (see Sec. 2.5). The boundary condition is illustrated below.
Fig.3 One-dimensional array of equal masses and springs. This is the simplest model of
a vibrational band..
Alternative representation of the Born-von Karman boundary condition. The object
connecting the ion on the extreme left with the spring on the extreme right is a massless
regid rod of length L = Na.
⇓
4
Fig.4 The Born-von Karman periodic boundary condition for the linear chain.
2.3
Born-von Karman boundary condition
The Born-von Karman or periodic boundary condition
u s = u s+ N ,
u s = e i ( ksa −ωt ) .
The periodic boundary condition requires that
2π
or
,
k=
e ikNa = 1 ,
a N
(2.6)
(2.7)
(2.8)
Fig.5 N modes for − π / a ≤ k ≤ π / a
where
N N
N
N
,− + 1, , − 1, .
2
2
2
2
N
For each k, there are one longitudinal mode and two transverse modes.
∴(1+2)N modes = 3N modes
((very important))
N atoms: each atom has 3 freedoms 3N
u s = e i ( ksa −ωt ) ,
2π
k′ = k +
n
(n: integer).
a
The displacement of the n-th atom for the wave k′
=−
i[( k _
2π
(2.8)
(2.9)
n ) sa −ωt ]
a
(2.10)
u s = e i ( k ′sa −ωt ) = e
= e i ( ksa −ωt ) .
So the displacement us is the same as for the wavevector k, for any atom whatsoever.
Thus the wave k′ cannot be physically differentiated from the wave k.
2.4
First Brilouin zone
What range of k is physically significant for elastic waves?
⇒ Only those in the first Brilloiun.
u s +1
(2.11)
= e ika .
us
The range −π to π for the phase ka covers all independent values of the exponential.
5
− π ≤ ka ≤ π
−
or
π
a
≤k≤
π
a
.
(2.12)
In the continuum limit (a = 0),
k max =
π
a
→∞.
(2.13)
Fig.6 The relation between k’ and k = k’+2π/a.
Suppose that
k′ = k +
2π
n
a
(n: integer)
(2.14)
Then
2π
i ( k ′+ n ) a
u s +1
a
= e ika = e
= e ik ′a .
(2.15)
us
Thus the displacement can always be described by a wavevector within the first Brillouin
2π
zone. Note that G =
n is a reciprocal lattice vector. Thus by subtraction of an
a
appropriate reciprocal lattice vector from k, we always obtain an equivalent wavevector
in the 1st zone: k = k′ + G, where k′ is the wavevector in the first Brillouin zone.
2.5
Normal modes
We solve the equations of motion
d 2u s
M
= C (u s +1 − 2u s + u s −1 ) ,
dt 2
by assuming that
u s = e i ( ksa −ωt ) .
Then we have
d 2u s
= −ω 2 u s ,
2
dt
u s ±1 = e ± ika u s ,
[
(2.16)
(2.17)
(2.18)
(2.19)
]
− ω 2 Mu s = C e ika + e − ika − 2 u s ,
leading to the dispersion relation
2C
4C 2 ka
C
ω 2 = (2 − 2 cos ka) =
(1 − cos ka) =
sin ( ) ,
2
M
M
M
or
6
(2.20)
(2.21)
ω=
4C
ka
| sin( ) | .
M
2
(2.22)
The boundary of the first Brillouin zone lies at k = ±
π
a
.
dω 2
dω 2Ca
= 2ω
=
sin ka ,
dk
dk
M
dω
π
⇒
= 0 at k = ± .
dk
a
(2.23)
(2.24)
2
C
M
Fig.7 The dispersion curve (ω vs k) for a monatomic linear lattice with nearest neighbor
interactions only. The first Brillouin zone is the segment between –π/a and π/a. w
is linear for small k and that dω/dk vanishes at the zone boundaries (k = ±π/a.
ω2 =
4C
⎛ ka ⎞
sin 2 ⎜ ⎟
M
⎝ 2⎠
or
ω=
4C
⎛ ka ⎞
sin ⎜ ⎟
M
⎝ 2⎠
2.6
dispersion relation
(2.25)
Continuum wave equation
We consider an original wave equation given by
d 2u s
M
= C (u s +1 − 2u s + u s −1 ) .
(2.26)
dt 2
We assume that us(x) = u(x, t). In the small limit of a = Δx, sa = x and s is continuous
variable. Under this assumption, u(x, t) can be expanded using a Taylor expansion,
Δx ∂u ( x, t ) (Δx) 2 ∂ 2u ( x, t )
us +1 (t ) = u ( x + Δx, t ) = u ( x, t ) +
+
+ O(Δx)3 , (2.27)
∂x 2
1! ∂x
2!
Then the original wave equation can be rewritten as
7
∂ 2u ( x, t )
M
∂t 2
= C[u ( x, t ) +
Δx ∂u ( x, t ) (Δx) 2 ∂ 2u ( x, t )
Δx ∂u ( x, t ) (Δx) 2 ∂ 2u ( x, t )
+
−
u
x
t
+
u
x
t
−
+
2
(
,
)
(
,
)
]
1! ∂x
2!
1! ∂x
2!
∂x 2
∂x 2
(2.28)
or
M
∂ 2 u ( x, t )
∂ 2 u ( x, t ) ,
]
= C (Δx) 2
2
∂t
∂x 2
(2.29)
or
2
2
∂ 2 u ( x, t ) C
C 2 ∂ 2 u ( x, t )
2 ∂ u ( x, t )
2 ∂ u ( x, t )
,
(
x
)
a
=
v
=
Δ
=
∂t 2
M
∂x 2
M
∂x 2
∂x 2
(2.30)
which is the continuum elastic wave equation with the velocity of sound given by
C
v=
a.
(2.31)
M
3.
Eigenvalue-problem; solution using Mathematica
We solve the eigenvalue-problem given by
(3.1)
ω 2us = K (−us +1 + 2us − us −1 ) ,
where s = 1, 2, …, N, K = C/M. These equations can be expressed using a matrix M (for
example, N x N) and column matrix (1xN);
MU = ω2U.
(3.2)
2
Here ω is the eigenvalue. The number of eigenvalues is N. For example, the matrix M
and the column matrix U for N = 12 (for example), are given by
⎛ 2K
⎜
⎜− K
⎜ 0
⎜
⎜ 0
⎜ 0
⎜
⎜ 0
M =⎜
⎜ 0
⎜ 0
⎜
⎜ 0
⎜ 0
⎜
⎜ 0
⎜ 0
⎝
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
and
8
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0
0
0
0
0
0
0
0
0
0
−K
2K
−K
0 ⎞
⎟
0 ⎟
0 ⎟
⎟
0 ⎟
0 ⎟⎟
0 ⎟
⎟ ,(3.3)
0 ⎟
0 ⎟
⎟
0 ⎟
0 ⎟
⎟
−K⎟
2 K ⎟⎠
⎛ u1 ⎞
⎜ ⎟
⎜ u2 ⎟
⎜u ⎟
⎜ 3⎟
⎜ u4 ⎟
⎜u ⎟
⎜ 5⎟
⎜u ⎟
U = ⎜ 6 ⎟ , for each eigenvalue.
(3.4)
u
⎜ 7⎟
⎜ u8 ⎟
⎜ ⎟
⎜ u9 ⎟
⎜u ⎟
⎜ 10 ⎟
⎜ u11 ⎟
⎜u ⎟
⎝ 12 ⎠
Using the Mathematica (program called Eigenvalues), we can calculate the dispersion
relation of ωk vs k = (π/a) (n/N) with N = 1, 2, 3,…, N-1, N. For simplicity, here we
choose N = 100. For comparison we also make a plot of the solution given
by ω = 4C / M | sin(ka / 2) | . We find that there is a good agreement between these two
curves. K = C/M.
((Mathematica-1))
The mathematica program is attached to this lecture note.
Phonon dispersion relation for one atom in the unit cell. The Mathematica program is
presented in the Appendix.
w
2 K
1.0
0.8
0.6
0.4
0.2
p n
20
40
60
80
100
a 100
Fig.8 The dispersion relation with red (from Mathematica calculation) and blue
( ω = 4 K sin(ka / 2) with K = C/M). The x axis is k = (π/a) (n/N) with n = 1,
2, …, N (= 100). The y xais is ω /(2 K ) . The agreement between these two
results are excellent.
9
((Mathematica-2)) The Mathematica program used is attached to this lecture note.
Here we use the Mathematica program (Eigensystem) to obtain the eigenvalues and
the corresponding eigenfunction U. For each eigenvalue, we can calculate the deviation
(us) of the position of the atom from the equilibrium position, as a function of the
equilibrium position xs = s a, where s = 1, 2, 3,….. ,N. Here we consider the eigenvalue
problem (N = 50). We show the plot of Un=(u1, u2, u3, …, uN) as a function of the location
(xs = s a), where n = 1, 2, …, N. u = 1 . We find that the wavelength λn is given by λn =
2Na/n. The corresponding wavenumber qn is equal to kn = (2π/λn) = (π/a)(n/N). The
deviation us is well described by the form sin[πs(n/N)] = sin[kn(sa)] ( ≈ eik n sa ).
u
0.2
0.1
x
10
20
30
40
50
a
-0.1
-0.2
Fig.9 Plot of u = (u1, u2, u3, …., u50) as a function of xs = s a for the eigenvalues ωi (i =
1, 2, 3, …, 10). ω1<ω2<…<ω10. Note that u is normalized ( u = 1 ). N = 50.
10
81 <
82 <
u
u
0.20
0.2
0.1
0.0
-0.1
-0.2
0.15
0.10
x
10 20 30 40 50 a
83 <
u
0.2
0.1
0.0
- 0.1
- 0.2
85 <
u
x
10 20 30 40 50 a
86 <
u
0.2
0.1
0.0
- 0.1
- 0.2
x
10 20 30 40 50 a
87 <
u
x
10 20 30 40 50 a
88 <
u
0.2
0.1
0.0
-0.1
-0.2
x
10 20 30 40 50 a
89 <
u
0.2
0.1
0.0
- 0.1
- 0.2
84 <
u
10 20 30 40 50 a
0.2
0.1
0.0
- 0.1
- 0.2
10 20 30 40 50 a
0.2
0.1
0.0
-0.1
-0.2
x
0.2
0.1
0.0
- 0.1
- 0.2
x
x
10 20 30 40 50 a
810<
u
0.2
0.1
0.0
-0.1
-0.2
x
10 20 30 40 50 a
x
10 20 30 40 50 a
Fig.10 Plot of u = (u1, u2, u3, …., u50) as a function of xs = s a for the eigenvalues ωi (i =
1, 2, 3, …, 10). ω1<ω2<…<ω10. The wavenumber kn = (π/Na)n. The value of n is
denoted in each figure.
11
85 <
810<
u
u
0.2
x
0.1
0.0
- 0.1 1020 304050 a
- 0.2
815<
0.2
x
0.1
0.0
-0.1 10 203040 50 a
-0.2
820<
u
u
0.2
x
0.1
0.0
- 0.1 1020 304050 a
- 0.2
825<
0.2
x
0.1
0.0
-0.1 10 203040 50 a
-0.2
830<
u
u
0.2
x
0.1
0.0
- 0.1 1020 304050 a
- 0.2
835<
0.2
x
0.1
0.0
-0.1 10 203040 50 a
-0.2
840<
u
u
0.2
x
0.1
0.0
- 0.1 1020 304050 a
- 0.2
845<
0.2
x
0.1
0.0
-0.1 10 203040 50 a
-0.2
850<
u
0.2
x
0.1
0.0
- 0.1 1020 304050 a
- 0.2
u
0.2
x
0.1
0.0
-0.1 10 203040 50 a
-0.2
Fig.11 Plot of u = (u1, u2, u3, …., u50) as a function of xs = s a for the eigenvalues ωn (n =
5, 10, 15, 20, 25, 30, 35, 40, 45, 50). ω5<ω10<…<ω50. u = 1 . N = 50. The
wavenumber kn = (π/Na)n. The value of n is denoted in each figure.
12
4.
4.1
First Brillouin zone and group velocity
Definition of the group velocity
The transmission velocity of a wave pocket is the group velocity
∂ω
vg =
= ∇ k ω (k ) .
∂k
This is the velocity of energy propagation in the medium.
vg =
∂ω
Ca 2
⎛1 ⎞
=
cos⎜ ka ⎟ ,
∂k
M
⎝2 ⎠
v g = 0 at k =
(4.1)
(4.2)
π
a
The wave is a standing wave: zero net transmission velocity for a standing wave. Note
that the phase velocity is defined by v p = ω / k .
Long wavelength limit
When ka « 1,
⎡ 1
⎤ 1
1 − cos ka = 1 − ⎢1 − (ka) 2 ⎥ ≅ (ka) 2 .
⎣ 2!
⎦ 2
(4.3)
Then
ω2 =
2C 1 2 2 C 2 2
k a =
k a
M 2
M
or
ω=
Ca 2
k,
M
(4.4)
∂ω
Ca 2
=
= v0
at
ka = 0.
(4.5)
∂k
M
The velocity of sound is independent of frequency in this limit. Thus ω = vk , exactly as
in the continuum theory of elastic waves --- in the continuum limit a = 0 and thus ka = 0.
vg =
4.2
The physical meaning of the first Brillouin zone
We discuss the physical meaning of the first Brillouin zone. To this end, we consider
π
π 2π 7π
and k = k0 + G =
+
=
.
the case of us with k = k0 =
3a
3a a
3a
(i)
k = k0 =
π
.
3a
At t = 0, we have u s = ue i ( ksa −ωt ) at t = 0. We make a plot of
π
⎛π ⎞
] = cos⎜ s ⎟ ,
⎝3 ⎠
as a function of s, where s = 0, 1, 2, 3, …, where u = 1 in Fig.17.
π 2π 7
+
= π
(ii) k = k0 + G =
3a a 3a
Re[us ] = Re[e
i ( s − ωt )
3
We also make a plot of Re[us ] = Re[e
⎛ 7π ⎞
i⎜
s⎟
⎝ 3 ⎠
⎛ 7π ⎞
s ⎟ at t = 0 in Fig.17.
] = cos⎜
⎝ 3 ⎠
13
0.5
s
Re[u /u]
1
0
-0.5
-1
0
1
2
3
4
5
6
7
s
Fig.12 Plot of Re[ei ( ksa −ωt ) ] at t = 0 for k = k0 + G = 7π/3 (the red solid line) and k = k0 =
π/3a (the dotted blue line).
As shown in Fig.12, the wave represents by the solid curve (k = 7π/3a) conveys no
information not give by the dashed curve (k = π/3a). Because the wave displacement is
defined only at lattice points, the propagation of a large wavenumber (k0 + G) lying
outside the first Brillouin zone is identical to a short wavenumber k0 lying inside the first
Brillouin zone. In a continuum, the amplitudes would have a value everywhere, as
represented by the continuous lines, so that both wavenumbers would be distinguishable.
4.3
Standing wave
At the zone boundary k = ± π / a , the group velocity is equal to zero, implying that no
energy is propagated. The amplitude Re[us] at time t is described
[
]
[
]
Re[us ] = Re uei ( ksa −ωt ) = Re u (−1) s e − iωt = u (−1) s cos ωt
(4.1)
s
The alternate atoms oscillate in opposite phases, because (-1) = -1 for odd integers s and
(-1)s = 1 for even integers s. Note that the wave is a standing wave. A standing wave is an
example of wave motion with zero group velocity. The wave moves neither to the right
nor to the left. This situation is equivalent to Bragg reflections of x-ray.
Fig.13
The condition for the Bragg reflection in the reciprocal lattice. k − k ' = 2π / a (or
k = π / a and k ′ = −π / a )
The Bragg reflection arises when k = ± π / a . Even if we excite only the state k = π/a. we
should obtain k = -π/a through the Bragg reflection. The superposition of these two wave
leads to a standing wave,
1 i ( ksa −ωt )
1 −iωt isπ
e
e
e ± e −isπ
± e i ( k ′sa −ωt ) =
2
2
(4.6)
1 −iωt ⎧ 2 cos sπ ⎫
−iωt ⎧cos sπ
e ⎨
=
⎬ = 2e ⎨
2
⎩2i sin sπ ⎭
⎩i sin sπ
[
]
[
]
14
This implies that the wave cannot propagate in a lattice, but through successive
reflections back and forth, a standing wave is set up.
4.4
General property of the group velocity
There are two main features of the phonon dispersion relation;
(i)
E = ħωk is an even function of k; ω− k = ωk or E− k = Ek , where ħ is the Dirac’s
constant (ħ = h/2π) and h is the Planck’s constant.
(ii)
Periodicity in Ek + G = Ek with G [= (2π/a) x integer] is the reciprocal lattice.
Fig.14 Values of k in the first Brillouin zone related by the symmetrical relation (Ek+G =
E k and E-k = E k). k = ±π/a is the boundary of the first Brillouin zone (|k|≤π/a).
We now consider the value of the group velocity at the Brillouin zone boundary.
From the condition Ek = E− k , we have E(3) = E(4). From the condition Ek = Ek + G , we
have E(3) = E(5). Therefore, we have E(4) = E(5). On taking δ → 0, the group velocity at
the boundary of Brillouin zone is defined as [ E (5) − E (4)] / 2δ , which reduces to zero
(dΕk/dk→0).
((Note))
It follows that from the condition ( Ek = E− k ), we have E(1) = E(2). On taking δ→0,
the group velocity defined by [ E (2) − E (1)] / 2δ reduces to zero (dΕk/dk→0). On applying
the periodicity condition Ek = Ek + G this result can immediately be extended as follows.
dΕk/dk→0 at k = 0, ±2π/a, ±4π/a,….. This prediction (only from the symmetry
consideration) that the group velocity is equal to zero at k = 0, may be inconsistent with
the result derived from the linear chain model above described. In the linear chain model,
it is predicted that there is a discontinuous jump in the group velocity at k = 0, from -v0
(v0 = a C / M ) at k = 0- to –v0 at k = 0+. We note that this motion at k = 0 is a translation
of the crystal as a whole and it is therefore not property of a vibration.
5.
5.1
Determination of force constants
The system with the nearest neighbor interaction
We can make a statement about the range of the forces from the observed dispersion
relation for ω. The generalization of the dispersion relation to p nearest planes is found to
be
15
ω k2 =
2
M
∑ [1 − cos( pka )]C
p >0
π /a
M
∫ dkω
π
2
k
− /a
p
,
cos(rka) = 2∑ C p
p >0
(5.1)
π /a
∫π dk [1 − cos( pka)]cos(rka) .
(5.2)
− /a
Fig.15 Linear chain having the nearest-neighbor coupling (C1), the second nearestneighbor coupling (C2), and the third nearest-neighbor coupling (C3), and the
Here
π /a
π /a
⎡ sin(rka) ⎤
dk cos(rka) = ⎢
= 0,
∫
⎣ ra ⎥⎦ −π / a
−π / a
(5.3)
and
π /a
π /a
− /a
− /a
1
∫ dk cos( pka) cos(rka) = π∫ dk 2 [cos{( p + r )ka} + cos{( p − r )ka}]
π
π /a
⎧ 1 2π
1
⎪
=
dk cos[( p − r )ka] = ⎨ 2 a
∫
2 −π / a
⎪⎩ 0
for p = r
,
otherwise
(5.4)
Thus
π /a
M
∫π dkω
2
k
cos(rka) = −2π
− /a
Cr
,
a
(5.5)
or
π /a
Ma
(5.6)
dkω k2 cos( pka) ,
∫
2π −π / a
gives the force constant at range pa, for a structure with a monatomic basis.
Cp = −
5.2
System with long-ranged interactions
Fig.16 Linear chain with the nearest-neighbor and the next nearest-neighbor interactions.
Here we assume that there are the second-, third,…nearest neighbor interactions in the
system in addition to the nearest neighbor-interaction. Then the Lagrangian L is given by
16
L = T −V
∞
1
1
= ∑ Mu s2 − ∑∑ C p (u s − u s + p ) 2
s 2
s p =1 2
.
{
(5.6)
}
∞
1
1
= ∑ Mu s2 − ∑ C p (u s − u s + p ) 2 + (u s − u s − p ) 2 −
s 2
p =1 2
The Lagrange equation is
d ∂L
∂L
=
,
dt ∂u s ∂u s
or
∞
∞
p =1
p =1
(5.7)
Mu s = −∑ C p (u s − u s + p ) − ∑ C p (u s − u s − p )
(5.8)
∞
= ∑ C p (u s + p − 2u s + u s − p )
p =1
Here we assume a traveling wave given by u s = ue i ( ksa −ωt ) . Then we have
(
∞
)
− Mω 2 u s = ∑ C p e ipka − 2 + e −ipka u s ,
(5.9)
p =1
or
(
∞
)
− Mω 2 = ∑ C p e ipka − 2 + e −ipka ,
(5.10)
p =1
or
ω2 =
2
M
∞
C
C
∑ C (1 − cos pka ) = M [1 − cos(ka)] + M [1 − cos(2ka)] + ... ,
p =1
p
1
2
(5.11)
which includes the higher harmonics terms through long-range spring constants. The
value of Cp can be determined experimentally from the phonon dispersion relation.
6.
Vibration of square lattice4
Fig.17 Square array of lattice constant a. The displacements are normal to the plane of
the lattice.
17
We consider transverse vibrations of a planar lattice of rows and columns of identical
atoms, and let u(l, m) denote the displacement normal to the plane of the lattice of the
atom in the l-th column and m-th row. The mass of each atom is M, and C is the force
constant for nearest neighbor atoms.
The equation of motion is expressed by
d 2u (l , m)
M
= C{[u (l + 1, m) + u (l − 1, m) − 2u (l , m)] + [u (l , m + 1) + u (l , m − 1) − 2u (l , m)]
dt 2
(6.1)
We assume that the solution of u(l, m) is given by
u (l , m) = u (0) exp[i (lk x a + mk y a − ωt ) ,
(6.2)
where a is the spacing between nearest-neighbor atoms. The equation of motion is
satisfied only if
2C
ω=
2 − cos(k x a ) − cos(k y a ) ,
(6.3)
M
where the first Brillouin zone is
−
π
a
≤ kx ≤
π
a
and −
π
a
≤ kx ≤
π
a
.
(6.4)
Fig.18 Brillouin zone of the two-dimensional square lattice with a lattice constant a.
Bragg reflections occur at the zone boundary of the first Brillouin zone.
18
Fig.19 Three-dimensional plot of the dispersion relation (of the 2D square lattice. The
height (ω) is plotted as a function kxa vs kya (in the 2D wavevector-plane).
7.
Two atoms per primitive basis
We consider a cubic crystal where atoms of mass M1 lie on one set of planes and
atoms of mass M2 lie on planes interleaved between those of the first set.
Fig.20 (a) Diatomic linear chain, There are two atoms with masses M1 and M2 per unit
cell (lattice constant a). (b) The plane of atoms stacked alternatively (lattice
constant a).
The Lagrangian of the system is given by
1
1
L = M 1u s2 + M 2 v s2 +
2
2
1
− C (u s +1 − v s ) 2 + (v s − u s ) 2 + (u s − v s −1 ) 2 +
2
The Lagrange’s equations are as follows.
d ∂L
∂L
=
,
dt ∂u s ∂u s
or
[
19
.
]
(7.1)
(7.2)
1
M 1u s = − C [− 2(v s − u s ) + 2(u s − v s −1 )],
2
(7.3)
or
M 1u s = −C (− v s + 2u s − v s −1 ) = C (v s + v s −1 − 2u s ) .
(7.4)
Similarly
d ∂L
∂L
=
,
(7.5)
dt ∂v s ∂v s
1
M 2 v s = − C [− 2(u s +1 − v s ) + 2(v s − u s )] = C (u s +1 − 2v s + u s ) .
(7.6)
2
Let a denote the repeat distance of the lattice in the direction normal to the lattice plane.
Equation of motion is given by
⎧
d 2u s
= C (v s + v s −1 − 2u s )
⎪⎪M 1
dt 2
,
(7.7)
⎨
2
⎪M d v s = C (u + u − 2v )
s +1
s
s
⎪⎩ 2 dt 2
Here we assume that each plane interacts only with its nearest-neighbor and that the force
constants are identical between all pairs of nearest-neighbor planes. Suppose that us and
vs have the forms of
⎧⎪u s = ue i ( ska −ωt )
.
(7.8)
⎨
⎪⎩v s = ve i ( ska −ωt )
Then
− ω 2 M 1u = Cv (1 + e − ika ) − 2Cu
.
(7.9)
− ω 2 M 1v = Cu (e ika + 1) − 2Cv
When the determinant of the coefficients of u and v vanishes, the homogeneous linear
equations have a non-trivial solution.
2C − M 1ω 2 − C 1 + e − ika
(7.10)
= 0,
− C 1 + e ika 2C − M 2ω 2
or
M 1 M 2ω 4 − 2C ( M 1 + M 2 )ω 2 + 2C 2 (1 − cos ka ) = 0 .
(7.11)
We examine this equation in the limiting cases
(i) ka « 1
(ii) ka = ±π at the zone boundary
(
)
(
)
(i) ka « 1
1 2 2
k a
2
M 1 M 2ω 4 − 2C ( M 1 + M 2 )ω 2 + C 2 k 2 a 2 = 0 ,
1 − cos ka ≈
Or
20
(7.12)
ω =
2
C (M 1 + M 2 ) ± C 2 (M 1 + M 2 ) 2 − M 1M 2C 2 k 2 a 2
M 1M 2
C (M 1 + M 2 ) ⎡
M 1M 2C 2 k 2 a 2
=
⎢1 ± 1 − 2
M 1M 2
C (M 1 + M 2 ) 2
⎢⎣
≈
⎤
⎥
⎥⎦
.
C (M 1 + M 2 ) ⎡ ⎧
M 1 M 2 C 2 k 2 a 2 ⎫⎤
1
1
±
−
⎢ ⎨
2
2 ⎬⎥
M 1M 2
⎢⎣ ⎩ 2C ( M 1 + M 2 ) ⎭⎥⎦
Then we have the two roots for ka « 1
⎛ 1
1 ⎞
⎟⎟ (optical branch),
+
ω 2 = 2C ⎜⎜
⎝ M1 M 2 ⎠
1
C
2
2
ω =
k 2 a 2 (acoustic branch).
M1 + M 2
(*) The extent of the first Brillouin zone is −
π
a
(7.14)
(7.15)
≤k≤
π
(ii) ka = ±π
M 1 M 2ω 4 − 2C ( M 1 + M 2 )ω 2 + 4C = 0
M 1ω 2 − 2C M 2ω 2 − 2C = 0
or
2C
2C
ω2 =
;
ω2 =
.
M1
M2
(
(7.13)
)(
)
a
.
(7.16)
(7.17)
ω
optical
phonon
branch
acoustic
phonon
branch
-π/a
π/a
0
k
for M1 > M2
Fig.21 The dispersion curve (ω vs k) for a diatomic linear chain with nearest neighbor
atoms interacting with interaction C only. The masses of the atoms are M1 and M2;
The first Brillouin zone is the segment between –π/a and π/a. There is an acoustic
mode (lower branch) and an optical mode (upper branch).
21
We consider the particle displacement in TA and TO branches. For the optical branch at k
= 0, we have
− ω 2 M 1u = 2Cv − 2Cu
M
2C
2C
u
=− 2
: out-of-phase
(7.18)
=
=
2
v 2C − M 1ω
M1
⎛ 1
1 ⎞
⎟⎟
2C − M 1 2C ⎜⎜
+
⎝ M1 M 2 ⎠
For the acoustical branch at k = 0, we have
− ω 2 M 1u = 2Cv − 2Cu .
(7.19)
2
Since ω = 0 , then we have u = v . : in-phase
Fig.22 Nature of the vibration in the acoustical and optical branch of a vibrational
spectrum.
u s = ue i ( ska −ωt ) , v s = ve i ( ska −ωt ) .
(7.20)
Acoustical waves in a diatomic linear lattice ( k =
π
2a
)
Fig.23 Transverse acoustic waves in a diatomic linear lattice.
u cos( ska) = v cos( ska) .
(7.21)
22
Optical waves in a diatomic linear lattice ( k =
π
2a
)
Fig.24 Transverse optical waves in a diatomic linear lattice. Figure represents a snapshot
of the wave motion.
M 1u + M 2 v
= 0 ⇒ The velocity of center of mass does not move.
M1 + M 2
The atoms vibrate against each other, but their center of mass is fixed. If the two atoms
carry opposite charges, we may excite a motion of this type with the electric field of a
light wave.
vc ≡
Fig.25 Origin of the name of optical mode
So that the branch is called the optical branch. The atoms (and their center of mass) move
together, as in long wavelength acoustical vibrations, whence the term acoustical branch.
Fig.26 Origin of the name of acoustic mode
This is a frequency gap at k max = ±
π
a
of the first Brillouin zone.
((Note))
If we look for solutions in the gap with ω real, then k will be complex, so that the
wave is damped in space.
((Mathematica-3)) phonon dispersion for the system with two atoms in unit cell
23
wêw0
1.5
1.0
0.5
-2
-1
1
2
kaêp
Fig.27 The dispersion curve of the diatomic linear chain with two atoms in a unit cell.
The ratio M2/M1 is varied as a parameter between 2, 3, 4, … and 10.
8.
8.1
The number of modes; degree of freedom
One-dimensional case
We consider the degree of freedom for N atoms in the linear lattice chain. There are N
atoms (each unit cell has one atom). Each atom has three degrees of freedom. One for
each of the x, y, z directions. Then we have 3N degree of freedoms. This indicates that
there are 3N modes in the system. This implies that the number of allowed k values in a
single branch is just N for each Brillouin zone; 2N transverse acoustic (TA) modes and N
longitudinal acoustic (LA) mode.
8.2
by
Three-dimensional case
We consider the lattice waves in the 3D system. The displacement vector us is given
u s ~ e ik ⋅ R n ,
(7.1)
where Rn is the position vector of the atom located in the equilibrium positions of the
lattice.
R n = n1a1 + n2 a 2 + n3 a 3 ,
(7.2)
a1, a2, and a3 are the primitive lattice vectors along the x, y, and z directions. From the
boundary condition, we have
(7.3)
eik ⋅( R n + N1a1 ) = eik ⋅R n , or
e ik x N1a1 = 1 ,
leading to the selected values of wavenumber kx
2π
2π
kx =
l1 =
l1 for a1 = a2 = a3 = a.
(7.4)
N1a1
N1a
Similarly, we have
2π
2π
2π
2π
ky =
l2 =
l2 ,
and
kz =
l3 =
l3
(7.5)
N 2 a2
N 2a
N 3a3
N 3a
Here l1, l2, and l3 are integers given by
24
⎧
⎪
N
N
N
⎪ l1 = − 1 ,− 1 + 1, , 1
2
2
2
⎪
⎪
N1
⎪
N2 N2
N
(7.6)
,−
+ 1, , 2 .
⎨l2 = −
2
2
2
⎪
N2
⎪
⎪
N3 N3
N3
⎪ l3 = − 2 ,− 2 + 1, , 2
⎪
N3
⎩
This means that there are N (=N1N2N3 ) modes, where N is the number of unit cells in the
system.
Each modes has 3 degrees of freedom (1 longitudinal mode and 2 transverse modes).
Then the total number of modes is 3 × N = 3N.
(a)
In the case of one atom per unit cell, we have 3N degree of freedom, with N
longitudinal acoustic mode and transverse acoustic mode 2N
(b)
We consider the number of degrees of freedom of the atoms. With p atoms in the
primitive cell and N primitive cells, there are pN atoms. Each atom has three degrees of
freedom, one for each of the x, y, z directions, making a total of 3pN degrees of freedom
for the crystal. The number of allowed k values in a single branch is just N for one
Brillouin zone. Thus, the one LA and two TA branches have a total of 3N modes. The
remaining (3p - 3) x N degrees of freedom are accommodated by the optical branches.
3 acoustical branches
1 longitudinal acoustical (LA) mode
2 transverse acoustical (TA) mode
3p - 3 optical branches
(p - 1) longitudinal optical (LO) mode
2(p - 1) transverse optical (TO) mode
For p = 2, for example, we have 1 LA, 1 LO modes, and 2 TA and 2 TO modes.
9.
9.1
Classical Model
Theory of the transverse wave in a string
25
Fig.28 Oscillation of one-dimensional continuum
Suppose that a traveling wave is propagating along a string that is under a tension Ts.
Let us consider one small element of length Δx. The ends of the element make small
angle θA and θB with the x axis. The net force acting on the element along the y-axis is
∑ Fy = T sin θ B − T sin θ A
,
(9.1)
= T (sin θ B − sin θ A ) ≈ T (tan θ B − tan θ A )
or
⎛ ∂y ⎞ ⎛ ∂y ⎞
(9.2)
Fy = T [⎜ ⎟ − ⎜ ⎟ ] .
⎝ ∂x ⎠ B ⎝ ∂x ⎠ A
We now apply the Newton’s second law to the element, with the mass of the element
given by m = μΔx ,
Fy = ma y = μΔx
∂2 y
.
∂t 2
(9.3)
Then we have
∂2 y
⎛ ∂y ⎞ ⎛ ∂y ⎞
μΔx 2 = Ts [⎜ ⎟ − ⎜ ⎟ ]
∂t
⎝ ∂x ⎠ B ⎝ ∂x ⎠ A
,
⎛ ∂y ⎞ ⎛ ∂y ⎞
⎜ ⎟ −⎜ ⎟
2
2
μ ∂ y ⎝ ∂x ⎠ B ⎝ ∂x ⎠ A ∂ y
=
= 2
Δx
∂x
Ts ∂t 2
which leads to a wave equation given by
∂2 y μ ∂2 y 1 ∂2 y
,
=
=
∂x 2 Ts ∂t 2 v 2 ∂t 2
where v is the velocity of the sound,
T
v= s .
(9.4)
(9.5)
(9.6)
μ
Here we use the Taylor expansion,
26
∂2 y
∂ ⎛ ∂y ⎞
⎛ ∂y ⎞
⎛ ∂y ⎞
⎛ ∂y ⎞ ⎛ ∂y ⎞
− ⎜ ⎟ = Δx ⎜ ⎟ = Δx 2 .
⎜ ⎟ −⎜ ⎟ =⎜ ⎟
∂x
∂x ⎝ ∂x ⎠
⎝ ∂x ⎠ B ⎝ ∂x ⎠ A ⎝ ∂x ⎠ x + Δx ⎝ ∂x ⎠ x
9.2
(9.6)
Energy density of the elastic wave
2
⎛ ∂y ⎞
When the length of element changes to Δx to Δx 1 + ⎜ ⎟ , the work done in the
⎝ ∂x ⎠
system (= Wc) of the conserved system is given by
2
2
2
1 ⎛ ∂y ⎞
T
⎛ ∂y ⎞
⎛ ∂y ⎞
− Ts Δx 1 + ⎜ ⎟ + Ts dΔ = −Ts dΔ[1 + ⎜ ⎟ − 1] = − s Δx⎜ ⎟ . (9.7)
2 ⎝ ∂x ⎠
2 ⎝ ∂x ⎠
⎝ ∂x ⎠
Since the potential energy ΔU is related to Wc by ΔU = −Wc , the potential energy ΔU is
given by
2
T
⎛ ∂y ⎞
ΔU = s Δx⎜ ⎟ .
2 ⎝ ∂x ⎠
The kinetic energy contribution ΔK is given by
(9.8)
2
1 ⎛ ∂y ⎞
ΔK = μΔx ⎜ ⎟ .
2 ⎝ ∂t ⎠
Then the energy density is given by
(9.9)
2
ΔE = ΔK + ΔU = Δx[
2
Ts ⎛ ∂y ⎞ μ ⎛ ∂y ⎞
⎜ ⎟ + ⎜ ⎟ ].
2 ⎝ ∂x ⎠
2 ⎝ ∂t ⎠
(9.10)
10.
Quantum mechanical approach: phonon
10.1 Annihilation and creation operators1,3
We assume U = (u1, u2,
,uN) for the eigenvalue ω = ωk. The displacement u(x) is
expresses using a Dirac delta function,
u ( x) = ∑ usδ ( x − sa ) ,
(10.1)
s
with L = Na . The Fourier transform of u(x) is given by
1
1
Uk =
dxu ( x)e − ikx =
us e − iksa .
∑
∫
N
N s
The inverse Fourier transform of Uk is
1
1
1
eiksaU k =
∑
∑
∑ us 'e−iks 'aeiksa
N k
N k N s'
1
= ∑ us ' ∑ eik ' a ( s − s ')
N s'
k'
N
= ∑ usδ s , s '
N s
= us
or
27
(10.2)
(10.3)
1
e iksaU k .
∑
N k
The Lagrangian of the system is given by
L = T −V
us =
(10.4)
1
⎡M
2⎤.
= ∑ ⎢ u s2 − C (u s +1 − u s ) ⎥
2
⎦
s ⎣ 2
We use the notation
1
us =
U k eiksa .
∑
N k
Since us is real, it is required that
*
U −k = U k .
Using these relations, we have
1
U kU k 'ei ( k + k ') sa = ∑∑U kU − k .
∑s us2 = N ∑∑∑
k k' s
k
s
Here we use
N
∑e
i ( k + k ') sa
s =1
= Nδ k ', − k .
(10.5)
(10.6)
(10.7)
(10.8)
(10.9)
Similarly
∑ (u
s +1
s
1
2
− us ) = 2∑U kU − k (1 − cos ka) = ∑ Mωk2U kU − k ,
k
k k
(10.10)
where
2C
(10.11)
(1 − cos ka) .
M
Then L can be rewritten as
⎛M
⎞
Mωk2
(10.12)
L = ∑ ⎜⎜ U kU − k −
U kU − k ⎟⎟ .
2
k ⎝ 2
⎠
The linear momentum Pk conjugate to Uk is defined by
∂L
Pk =
= MU − k .
(10.13)
∂U k
Then Hamiltonian H is obtained as
1 ⎛ 1
1
⎞
H = ∑ PkU k − L = ∑ ⎜ Pk P− k + Mωk2U kU − k ⎟ .
(10.14)
2 k ⎝M
2
⎠
k
Note that we define the Fourier transform of the linear momentum by
1
ps =
e −iksa Pk
∑
N k
(10.15)
1
iksa
Pk =
∑ e ps
N s
+
with P− k = Pk .
(10.16)
Let us work it out by the operator technique, starting from the commutation relations,
[us , ps ' ] = i δ s , s ' .
(10.17)
ω k2 =
28
Then the commutation relation is preserved in [U k , Pk ' ] = i δ k , k ' , since
[U k , Pk ' ] = [
1
N
∑e
−iksa
us ,
s
1
N
∑p
−ik 'sa
s
]
s
1
e −iksa [u s , ps ' ]e ik 's 'a
∑∑
N s s'
.
(10.18)
i
−iksa ik 's 'a
= ∑∑ e e δ s ,s '
N s s'
i
= ∑ e −i ( k −k ') sa = i δ k ,k '
N s
Thus our new displacements and momenta are canonically conjugate, and noncommuting, if they are of the same wavenumber; otherwise they are dynamically
independent operators. The Hamiltonian is rewritten as
1 ⎛ 1 +
1
⎞
+
H = ∑ ⎜ Pk Pk + Mωk2U k U k ⎟ .
(10.19)
2 k ⎝M
2
⎠
The final step is the introduction of annihilation and creation operators defined by
1
Mωk +
1
ak+ =
(
Uk −
iPk )
2
M ωk
,
(10.20)
1
Mωk
1
+
ak =
(
Uk +
iPk )
2
M ωk
=
where ak and ak+ act to destroy, and create a phonon of waveuumber, k and energy ωk ,
respectively. One can get the expressions for Pk and Uk from the above equations for ak
and ak
+
Pk = i
M ωk
+
( ak − a− k ) ,
2
Uk =
( a k + a− k ) ,
(10.21)
+
2 Mωk
The annihilation and creation operators satisfy the commutation
[a , a ]
k
+
k'
= δ k ,k ' ,
(10.22)
(10.23)
[a k , a k ' ] = [a k+ , a k+' ]
= 0,
(10.24)
The transformed Hamiltonian is
1⎞
1⎞
⎛
⎛
H = ∑ ωk ⎜ ak+ ak + ⎟ = ∑ ωk ⎜ N k + ⎟ ,
(10.25)
2⎠ k
2⎠
⎝
⎝
k
where N k = ak+ ak . So that each phonon may be regarded as possessing an energy ωk .
The total Hamiltonian is the sum of the Hamiltonian of independent linear oscillators of
angular frequency ωk. The various properties of the operators and eigenstates of the
Hamiltonian are seen in the Appendix A.
10.2
Symmetry of lattice and translation operator13,15
29
We now consider the displacement defined by
u ( x) = x u = ∑ us x s .
(10.26)
s
When we use the Dirac notation x s = δ ( x − sa ) , the ket vector is described by
u = ∑ us s .
(10.27)
s
We introduce the translation operator given by T(a). The Hamiltonian H is invariant
under the translation of the system by the lattice constant a. In other words,
T (a) commutes with the Hamiltonian H. The eigenstate of H should be simultaneously
the eigenstate of T (a) . T(a) is a unitary operator, but not a Hermite operator. Then the
eigenvalue of T(a) is a complex number (see the Appendix B for more detail).
Since T (a ) s = s + 1 , the ket s is not an eigenstate of T(a). Suppose that us has the
form of exp(isak), where k is a real parameter. u is a linear combinationof s with s = 0,
1, …, N-1.
u = ∑ eisak s .
(10.28)
s
When T(a) is applied to the eigen ket u , we get
T (a ) u = ∑ eisak T (a ) s = ∑ eisak s + 1
s
=e
−ika
∑e
s
i ( s +1) ak
s + 1 = e −ika u
,
(10.29)
s
which means that u is the eigenket of T(a). Similarly, we repeat this process N times,
[T (a )]N u = e − ikNa u = u ,
(10.30)
since [T (a )]N =1 (we use the periodic boundary condition). Then we have e − iNka =1, or
2π
2π
k=
n where n =0, 1, 2,…, N-1 ( 0 ≤ k <
). Alternatively we chose the value of k
Na
a
as the N states for −
π
≤k<
π
(the first Brillouin zone).
a
a
It follows from this discussion that when u = uk are introduced, the N harmonic
oscillators (N being the number of atoms in the crystal) becomes uncoupled and that to
each uk corresponds to one separate oscillator with the angular frequency ωk. This
motion is called a normal mode of vibration or a normal mode. This motion does not
describe the motion of a single atom in the crystal but, rather, of all atoms in it.
A few examples will show how this works. We take a normal mode for which k
equals the reciprocal lattice G (=2π/a). In this case, the coefficient us equals unity. This
motion is a translation of the crystal as a whole and it is therefore not property a vibration.
The case just considered corresponds of course to the center (k = 0) of the Brillouin zone.
In order to have another example, we consider the case where k equal to a point at the
zone boundary of the Brillouin zone. The displacement vector is out of phase in going
from one cell to the next, whether the successive lines show the evolution of the motion
in time, the harmonic function depicted corresponding to the particular frequency of this
normal mode.
30
11.
Crystal momentum
A phonon of k will interact with particles such as photons, neutrons, and electrons as
if it had a momentum k . However, a phonon does not carry physical momentum. The
physical momentum of a crystal is given by
∂L
Pk =
= MU − k
∂U k
,
(11.1)
1
iksa
=M
∑ us e
N s
where
1
1
Uk =
dxu ( x)e − ikx =
(11.2)
∑ use−iksa .
∫
N
N s
We assume that u s is independent of s; us = u . Then the momentum Pk is evaluated as
M
Pk =
u ∑ eiksa
N s
M
=
u[1 + eika + e 2ika + .... + eika ( N −1) ] ,
(11.3)
N
= M N uδ k ,0
2πr
where k =
(r = 0, ±1, ±2,…, ±N/2). The mode k = 0 represents a uniform translation
Na
of the crystal as a whole. Such a translation does carry momentum. For most practical
purpose, a phonon acts as if its momentum were ħk, sometimes called the crystal
momentum.
12
Semiclassical approach
12.1 Simple case
The energy of a lattice vibration is quantized. The quantum of energy is called a
phonon. Elastic waves in crystals are made up of phonons. Thermal vibrations in crystals
are thermally excited phonons. The energy of an elastic mode of angular frequency ωk is
1⎞
⎛
(12.1)
ε k = ⎜ nk + ⎟ ω k ,
2⎠
⎝
when the mode is excited to quantum number n k , where the mode is occupied by n
1
ω is the zero pint energy of the mode. We consider the wave of the
2
mode k with the amplitude.
u = u k cos(ω k t − kx) ,
(12.2)
where u is the displacement of a volume element from its equilibrium positions at x in the
crystal.
uk = u0 cos(ωt − kx) = u0 (cos ωt cos kx + sin ωt sin kx) .
(12.3)
The energy in the mode is
εk = [ K + P ] = 2 K
(∵ K = P )
(12.4)
photons. The term
31
when it is averaged over time.
2
1 ⎛ ∂u ⎞
K = ρ⎜ ⎟ ,
(12.5)
2 ⎝ ∂t ⎠
∂u
(12.6)
= u 0ω k [− sin ω k t cos kx + cos ω k t sin kx ] ,
∂t
where ρ is the mass density.
1 2 2
2
2
2
2
∫ KdV = 2 ρu 0 ω k ∫VdV [− sin ω k t cos kx + cos ω k t sin kx .
(12.7)
− 2 sin ω k t cos ω k t sin kx cos kx]
Here we have
⎧ x = 0 ~ Na (= L)
⎪
⎪
2π
N⎞
π
⎛
(12.8)
=
⎜ = 0,±1, ,± ⎟ ,
⎨k =
N
aN
2
⎛
⎞
⎝
⎠
⎪
a⎜ ⎟
⎪⎩
⎝2⎠
from the boundary condition: cos(kNA) = 1 ⇒ kNa = 2π ). Then we obtain
Na
Na
0
0
2
∫ dx cos kx = ∫ dx
1
(1 + cos 2kx ) = 1 Na + [k sin 2kx]0Na = 1 Na .
2
2
2
⎛ 2π ⎞
(∵ 2kNa = 2⎜
⎟ Na = 4π ).
⎝ aN ⎠
Similarly, we have
Na
Na
1
2
∫0 dx sin kx = 2 Na ,
∫0 dx sin kx cos kx = 0 .
Then we have
V 2 2
V 2 2
2
2
∫ KdV = 4 ρu0ω (sin ωt + cos ωt ) = 4 ρu0ω .
The time average kinetic energy is
T
V
V
2 2 1
K = ρu 0 ω
dt = ρu 02ω 2 .
∫
4
4
T 0
1
ε , we have
2
1
1⎛
1⎞
V
ρu 02ω k2 = ε = ⎜ nk + ⎟ ω k ,
4
2
2⎝
2⎠
(12.9)
(12.10)
(12.11)
(12.12)
Since K =
(12.13)
or
1⎞
⎛
2 ⎜ nk + ⎟
1
2⎠
u0 .
⇒ u eff =
u 02 = ⎝
ρVω k
2
Since ρV = NM ,
32
(12.15)
1⎞
⎛
2 ⎜ nk + ⎟
2⎠
.
(12.16)
u 02 = ⎝
NMω k
This relates the displacement in a given mode to the phonon occupancy n of the mode.
An optical mode with ω close to zero is called a soft mode.
12.2 General case
The Lagrangian L is given by
L = T −V
1
⎡M
2⎤
= ∑ ⎢ us2 − C (us +1 − us ) ⎥ .
2
⎦
s ⎣ 2
= L(u1 , u2 ,.., u N −1 , u1 , u2 ,.., u N −1 )
The linear momentum conjugate to us, is given by
∂L
ps =
= Mu s .
∂u s
Then the Hamiltonian H can be derived as
H = ∑ ps us − L
(12.17)
(12.18)
s
1
M 2
2
u s + ∑ C (u s +1 − u s ) .
2
s
s
s 2
1
M
2
= ∑ u s2 + ∑ C (u s +1 − u s )
2
s
s 2
Now we calculate the total energy for the case of u s = u cos( ska − ωt ) .
= ∑ Mu s2 − ∑
u s = u (−1)(−ω ) sin( ska − ωt ) ,
(12.19)
(12.20)
or
E=
1
Mu 2ω 2 ∑ [1 − cos 2(ωt − ska)]
4
s
,
(12.21)
1 2
1
⎡
⎤
+ Cu (1 − cos ka)∑ ⎢1 − cos 2(ωt − ska − ka)⎥
2
2
⎦
s ⎣
NM
(mass density), V = Na.
with ρ =
V
The dispersion relation is given by
2C
ω2 =
(1 − cos ka) .
(12.22)
M
Then the total energy is
⎧
1
1
⎡
⎤⎫
E = Mu 2ω 2 ⎨∑ [1 − cos 2(ωt − ska)] + ∑ ⎢1 − cos 2(ωt − ska − ka)⎥ ⎬ .
4
2
⎦⎭
s ⎣
⎩ s
(12.23)
The time-average is
33
T
E =
1
1
1
Edt = Mu 2ω 2 ∑1 = Mu 2ω 2 N
∫
T0
2
2
s
1
= ρVu 2ω 2
2
where
T=
2π
ω
, ρ=
,
(12.24)
NM NM
, NM = ρV
=
V
Na
In general, we use
u s = ∑ u k cos( ska − ωt )
(12.25)
k
Then we have
1
ρV ∑ u k2ω k2
(12.26)
2
k
This energy is compared with the result derived from the quantum mechanics.
1
1⎞
⎛
ρV ∑ u k2ω k2 = ∑ ω k ⎜ nˆ k + ⎟ .
2
2⎠
⎝
k
k
or
(12.27)
1
1⎞
⎛
ρVu k2ω k2 = ω k ⎜ nˆ k + ⎟ ,
(12.28)
2
2⎠
⎝
or
1⎞
1⎞
⎛
⎛
2 ⎜ nˆ k + ⎟ 2 ⎜ nˆ k + ⎟
2⎠
2⎠
,
(12.29)
u k2 = ⎝
= ⎝
ρVω k
NMω k
where ρV = NM . Here we define the effective amplitude as
1
(12.30)
ukeff =
uk ,
2
where
1⎞
⎛
⎜ nˆk + ⎟
(ukeff )2 = ⎝ NMω 2 ⎠ .
(12.31)
k
E =
eff
If the root-mean square of the average displacement (= < (uk ) 2 > ) is comparable to
the lattice constant a, the system may melt (Lindeman criterion).8
13.
13.1
Inelastic neutron scattering by crystal with lattice vibration
Scattering cross section
34
Fig.29 The neutron scattering by phonons in the system. ki is the wavevector of the
incident neutron and kf is the wavevector of the outgoing neutron. Q is the
scattering vector. Q = kf – ki. |ki| = |kf| = 2π/λ, λ is the wavelength of neutron.
We consider the neutron diffraction by a crystal with lattice vibration. The scattering
form factor is given by
− iQ ⋅ R j
,
(13.1)
F (Q) = ∑ be
j
where Q (=kf – ki) is the scattering vector, kf and ki are the wave vectors of incident and
outgoing neutrons, and b is scattering amplitude (which is independent of Q). The vector
Rj is the actual position of the atom that ought to have been at the lattice Rj0. For Rj = Rj0
(periodic configuration), we have a Bragg condition
− iQ ⋅ R
(13.2)
∑ e j = Nδ (Q − G ) .
j
When Rj = Rj(0) + uj, we have
(0)
F (Q) = b∑ exp[−iQ ⋅ (R j + u j )]
j
= b∑ exp[−iQ ⋅ (R j )] − ibQ ⋅ ∑ u j exp(−iQ ⋅ R j )
(0)
j
(0)
.
(13.3)
j
The first term is a Bragg reflection. We assume that the displacement vector uj is given
by
u j = e(q)[exp(iq ⋅ R (j0 ) ) + exp(−iq ⋅ R (j0) )]u (q) ,
(13.4)
where q is the wavevector of phonon (k is not used in this section to avoid confusion),
e(q) is the polarization vector and u(q) is the displacement amplitude. Then the second
term is rewritten as
F2 (Q) = −ib[Q ⋅ e(q)]∑{exp[i (q − Q) ⋅ R (j0 ) ] + exp[−i (q + Q) ⋅ R (j0 ) ]}u (q)
j
. (13.5)
= −iNb[Q ⋅ e(q)]u (q){δ (Q + q − G ) + δ (Q − q − G )}
Then the scattering intensity S(Q) is proportional to
S (Q) = N 2b 2 [Q ⋅ e(q)]2 | u (q) |2 ,
(13.6)
where Q = G ± q .
35
Fig.30 The relation between ki, kf, Q (= kf – ki), q and G. ki and kf are the wavevectors of
incident and outgoing neutrons. ki and kf are on the Ewald sphere with radius (|ki|
= |kf| = 2π/λ, λ is the wavelength of neutron). Q is the scattering vector. q is the
wavevector of phonon. G is the reciprocal lattice vector. Q = G + q.
When we use
1⎞
⎛
⎜ nq + ⎟
2⎠
| u (q) |2 = ⎝
,
(13.7)
NMωq
the Intensity S(Q) can be rewritten as
1⎞
⎛
⎜ nq + ⎟
2⎠
.
(13.8)
S (Q) = Nb 2 [Q ⋅ e(q)]2 ⎝
Mωq
Here we need to take into account of the energy conservation. The kinetic energy of the
incident neutron and the outgoing neutrons is given by ε and ε’. The energy conservation
law is valid; ε ' = ε + ωq for the absorption of phonon with ωq and ε ' = ε − ωq for the
emission of phonon with ωq. Finally we get the dynamic structure factor given by
[Q ⋅ e(q)]2 ωq
2
S (Q, ω ) = Nb
[( nq + 1)πδ (ω − ωq ) + nq πδ (ω + ωq )] ,
(13.9)
2
Mωq
where ω = (ε '−ε ) / ,
+
aˆq nq = nq + 1 nq + 1
for the creation of phonon,
(13.10)
aˆq nq = nq nq − 1
for the destruction of phonon.
(13.11)
(see the Appendix A). Using the factor [Q ⋅ e(q)]2 , we can select the branch; if Q ⊥ e(q) ,
this branch does not contribute to the inelastic neutron scattering. We note that S(Q) is
related to S(Q, ω) through
1
S (Q) =
S (Q, ω )dω .
(13.12)
2π ∫
13.2 Energy and momentum conservation
We consider the matrix element given by the integral over the entire space of the
system,
∫ exp(−ik f ⋅ r) exp(iq ⋅ r) exp(ik i ⋅ r)dr = ∫ exp[−i(k f − k i − q) ⋅ r]dr .
36
(13.13)
This integral is not equal to zero only when
k f − ki − q = G .
(13.14)
The integral over the time is given by
∫ exp(−iωk f t ) exp(±iωqt ) exp(iωk i t )dt = ∫ exp[i(−ωk f ± ωq + ωk i )t ]dt . (13.15)
is not equal to zero only when
ω = ωk f − ωk i = ±ωq
(energy conservation).
(13.16)
Conclusion
We have shown that the well-defined lattice waves propagate over the crystal,
forming a so-called phonon as a quantization of the lattice waves. Phonon has the dual
characters of wave and particle, which is essential to the quantum mechanics. Phonon is
one of bosons, obeying the Bose-Einstein statistics. Phonons will be seen to play an
important role in any phenomena for which the energy of importance is comparable to
ω , the energy of the phonon in question. In the BCS (Bardeen-Cooper-Schrieffer)
model for the superconductivity, a specific interaction between electrons can lead to an
energy gap separated from excited states by a energy gap. The formation of Cooper pairs
is due to the electron-phonon interaction. The first electron interacts with the lattice and
deforms it; the second electron sees the deformed lattice. Thus the second electron
interacts with the first electron through the lattice deformation.
REFERENCES
1.
J.M. Ziman, Electrons and Phonons (Oxford at the Clarendon Press, 1960).
2.
J.M. Ziman, Principles of the Theory of Solids (Cambridge University Press.1964).
3.
J.M. Ziman, Elements of Advanced Quantum Theory (Cambridge University Press,
Cambridge, 1969).
4.
C. Kittel, Introduction to Solid State Physics, sixth edition (John Wiley & Sons,
New York, 1986).
5.
C. Kittel, Quantum Theory of Solids (John Wiley & Sons, New York, 1963).
6.
G.H. Wannier, Elements of Solid State Theory (Cambridge at the University Press,
1960).
7.
R.E. Peierls, Quantum Theory of Solids (Oxford at the Clarendon Press, 1964).
8.
D. Pines, Elementary Excitation in Solids (W.A. Benjamin, Inc New York 1964).
9.
W. Cochran, The dynamics of atoms in crystals (Edward Arnold (Publishers) Ltd.,
London,1973).
10.
N.W. Ashcroft and N.D. Mermin, Solid State Physics (Holt, Rinehart and
Winston, New York, 1976).
11.
R.A. Levy, Principles of Solid State Physics (Academic Press, New York, 1968).
12.
A.A. Maradudin, I.M. Lifshitz, A.M. Kosevich, W. Cochran, and M.J.P.
Musgrave, Lattice Dynamics (W.A. Benjamin, Inc. New York, 1969).
13.
S.L. Altman, Band Theory of Solids. An Introduction from the Point of View of
Symmetry (Oxford University Press, New York 1991).
14.
R. Huebener, Electrons in Action, Roads to Modern Computers and Electronics.
(Wiley-VCH Verlag GmbH & Co. KGaA, 2005).
15.
J.J. Sakurai, Modern Quantum Mechanics (Addison Wesley, New York, 1994).
37
APPENDIX
Simple harmonics (1D) in quantum mechanics15
A
(a) Annihilation and creation operator
The commutation relation
[xˆ, pˆ ] = i .
The Hamiltonian of the simple harmonics is
1 2 mω02 2
ˆ
H=
pˆ +
xˆ .
2m
2
The eigenvalue-problem of the simple harmonics
Hˆ n = ε n n ,
(A.1)
(A.2)
(A.3)
with
1⎞
⎛
(A.4)
2⎠
⎝
where n = 0, 1, 2, 3,....
In the { x } representation, the wave function of the simple harmonics can be described as
ε n = ⎜ n + ⎟ ω0 ,
2
⎛
d 2 mω02 2 ⎞
⎜⎜ −
(A.5)
+
x ⎟⎟ x n = ε n x n .
2
2
⎝ 2m dx
⎠
Here we introduce the creation operator and annihilation operators given by
ipˆ ⎞ 1 ⎛⎜ mω0
ipˆ ⎞⎟
⎜⎜ xˆ +
⎟⎟ =
xˆ +
,
mω0 ⎠
m ω0 ⎟⎠
2 ⎜⎝
2⎝
β ⎛
ipˆ ⎞ 1 ⎛⎜ mω0
ipˆ ⎞⎟
⎜⎜ xˆ −
⎟⎟ =
aˆ + =
xˆ −
,
mω0 ⎠
m ω0 ⎟⎠
2 ⎜⎝
2⎝
aˆ =
β ⎛
(A.6)
(A.7)
with
mω0
β=
(A.8)
(
)
1
aˆ + aˆ + =
2β
xˆ =
(aˆ + aˆ ),
+
2mω0
(A.9)
1 mω0
1 m ω0
(
(aˆ − aˆ + ),
aˆ − aˆ + ) =
i
2
2β i
[xˆ, pˆ ] = 1 2 mω0 aˆ + aˆ + , aˆ − aˆ + = − aˆ, aˆ + ,
i
i
2β
pˆ =
(
or
)
[
]
[
]
[aˆ, aˆ ] = 1̂
+
aˆ + aˆ =
(A.11)
(A.12)
⎞
ipˆ ⎞⎛
1
ipˆ ⎞ β ⎛ 2
pˆ
⎜⎜ xˆ + 2 2 − i
⎟⎟ =
⎟⎟⎜⎜ xˆ +
⎜⎜ xˆ −
[ pˆ , xˆ ]⎟⎟
2 ⎝
mω0 ⎠⎝
mω0 ⎠ 2 ⎝
m ω0
mω0
⎠
β ⎛
2
(A.10)
2
38
2
(A.13)
or
aˆ + aˆ =
1 ⎛ ˆ 1
⎞
⎜ H − ω0 ⎟
2
ω0 ⎝
⎠
(A.14)
or
1⎞
⎛
Hˆ = ω0 ⎜ Nˆ + ⎟ ,
2⎠
⎝
(A.15)
where
Nˆ = aˆ + aˆ .
The operator N̂ is Hermitian since
(A.16)
+
Nˆ + = (aˆ + aˆ ) = aˆ + aˆ = Nˆ .
(A.17)
The eigenvectors of Ĥ are those of N̂ , and vice versa since [ Hˆ , Nˆ ] = 0 ,
[ Nˆ , aˆ ] = aˆ + aˆ , aˆ = aˆ + aˆaˆ − aˆaˆ + aˆ = aˆ + , aˆ aˆ = −aˆ ,
[ Nˆ , aˆ + ] = aˆ + aˆ , aˆ + = aˆ + aˆaˆ + − aˆ + aˆ + aˆ = aˆ + aˆ , aˆ + = aˆ + .
Thus we have the relations
[ Nˆ , aˆ ] = −aˆ ,
and
[ Nˆ , aˆ + ] = aˆ + ,
1⎞
⎛
Nˆ n = ω0 ⎜ n + ⎟ n .
2⎠
⎝
[ Nˆ , aˆ ] n = −aˆ n ,
From the relation
[
[
]
]
[
]
[
]
(A.18)
(A.19)
(A20)
(A.21)
(A.22)
( Nˆ aˆ − aˆNˆ ) n = −aˆ n ,
(A.23)
Nˆ (aˆ n ) = (n − 1)aˆ n .
(A.24)
or
â n is the eigenket of N̂ with the eigenvalue (n-1).
aˆ n ≈ n − 1 .
(A.25)
From the relation
[ Nˆ , aˆ + ] n = aˆ + n ,
( Nˆ aˆ − aˆ Nˆ ) n = − aˆ n ,
or
Nˆ (aˆ + n ) = (n + 1)aˆ + n .
(A.26)
(A.27)
(A.28)
aˆ + n is the eigenket of N̂ with the eigenvalue (n+1).
aˆ + n ≈ n + 1
(A.29)
Now we need to show that n should be either zero or positive integers: n = 0, 1, 2, 3,….
We note that
n aˆ + aˆ n = n n n ≥ 0 ,
(A.30)
39
n aˆaˆ + n = n aˆ + aˆ + 1 n = (n + 1) n n ≥ 0 .
(A.31)
The norm of a ket vector is non-negative and the vanishing of the norm is a necessary and
sufficient condition for the vanishing of the ket vector. In other words, n ≥ 0. If n = 0,
aˆ n = 0. If n ≠ 0, â n is a nonzero ket vector of norm n n n .
If n>0, one successively forms the set of eigenkets,
aˆ n , aˆ 2 n , aˆ 3 n , …. aˆ p n , , belonging to the eigenvalues, n-1, n-2, n-3,….., n-p,
This set is certainly limited since the eigenvalues of N̂ have a lower limit of zero. In
other words, the eigenket aˆ p n ≈ n − p , or n-p = 0. Thus n should be a positive integer.
Similarly, one successively forms the set of eigenkets,
2
3
p
aˆ + n , aˆ + n , aˆ + n , …. aˆ + n , belonging to the eigenvalues, n+1, n+2, n+3,….., n+p,
Thus the eigenvalues are either zero or positive integers: n = 0, 1, 2, 3, 4,
.
+
The properties of â and â
(A.32)
1.
aˆ 0 = 0 ,
since 0 aˆ + aˆ 0 = 0 .
2.
aˆ + n = n + 1 n + 1 ,
[ Nˆ , aˆ + ] n = aˆ + n ,
Nˆ aˆ + n = aˆ + Nˆ n + aˆ + n = (n + 1)aˆ + n .
(A.33)
(A.34)
(A.35)
aˆ + n is an eigenket of N̂ with the eigenvalue (n + 1).
Then
aˆ + n = c n + 1 .
(A.36)
Since
2
2
n aˆaˆ + n = c n + 1 n + 1 = c ,
(A.37)
n aˆ + aˆ + 1 n = n + 1 = c ,
we obtain
c = n +1 .
2
(A.39)
aˆ n = n n − 1 ,
[ Nˆ , aˆ ] n = −aˆ n ,
(A.41)
Nˆ aˆ n = aˆNˆ n − aˆ n = (n − 1)aˆ n ,
(A.43)
or
(A.40)
3.
(A.42)
â n is an eigenket of N̂ with the eigenvalue (n - 1)
Then
aˆ n = c n − 1 .
(A.44)
Since
40
2
(b)
2
n aˆ + aˆ n = c n − 1 n − 1 = c = n ,
(A.45)
c= n
(A.46)
Basis vectors in terms of 0
We use the relation
1 = aˆ + 0 ,
1 +
1 +2
(aˆ ) 0 ,
aˆ 1 =
2
2!
1 +
1 +3
(aˆ ) 0 ,
3 =
aˆ 2 =
3
3!
----------------------------------------------------1 +n
1 +
(aˆ ) 0 ,
n =
aˆ n − 1 =
n
n!
The expression for x̂ n and p̂ n
2 =
xˆ n =
(aˆ + aˆ ) n
+
2 mω 0
=
2mω0
(A.48)
(
)
n +1 n +1 + n n −1 ,
(
(A.49)
)
m ω0 +
m ω0
i aˆ − aˆ n =
i n +1 n +1 − n n −1 .
(A.50)
2
2
Therefore the matrix elements of â , â + , x̂ , and p̂ operators in the { n } representation
(
pˆ n =
)
are as follows.
n' aˆ n = nδ n ', n −1 ,
(A.51)
n' aˆ + n = n + 1δ n ', n +1 ,
n' xˆ n =
2 mω 0
(
(A.52)
)
n + 1δ n ', n +1 + nδ n ', n −1 ,
(
(A.53)
)
m ω0
n + 1δ n ', n +1 − nδ n ', n −1 .
(A.54)
2
Mean values and root-mean-square deviations of x̂ and p̂ in the state n .
n' pˆ n = i
n xˆ n = 0 ,
(A.55)
n pˆ n = 0 ,
(A.56)
1⎞
⎛
,
(A.57)
n xˆ 2 n = ⎜ n + ⎟
2 ⎠ mω 0
⎝
(Δp )2 = n pˆ 2 n = ⎛⎜ n + 1 ⎞⎟m ω0 .
(A.58)
2⎠
⎝
The product ΔxΔp is
1⎞
1
⎛
ΔxΔp = ⎜ n + ⎟ ≥
(Heisenberg’s principle of uncertainty),
(A.59)
2⎠
2
⎝
Note that
(Δx )2 =
41
(aˆ
xˆ 2 =
+
+ aˆ )(aˆ + + aˆ ) =
(aˆ aˆ
+
+
+ aˆaˆ + aˆ + aˆ + aˆaˆ + ) ,
2mω0
2mω0
m ω0 +
m ω0 + +
(
(
pˆ 2 =
aˆ − aˆ )(aˆ + − aˆ ) =
aˆ aˆ + aˆaˆ − aˆ + aˆ − aˆaˆ + ) ,
2
2
and
( )
n aˆ +
2
n =0,
(A.61)
(A.62)
n aˆ 2 n = 0 ,
(A.63)
n aˆ + aˆ + aˆaˆ + n = n 2aˆ + aˆ + 1 n = 2n + 1 ,
(A.64)
Mean potential energy
1
1
1
2
V = mω 2 n xˆ 2 n = mω 2 (Δx ) = ε n .
2
2
2
Mean kinetic energy
1
1
(Δp )2 = 1 ε n .
K =
n pˆ 2 n =
2m
2m
2
Thus we have
V = K .
(Virial theorem)
B.
(A.60)
Translation operator in quantum mechanics15
Here we discuss the translation operator Tˆ (a ) in quantum mechanics,
ψ ' = Tˆ (a ) ψ ,
(A.65)
(A.66)
(A.67)
(B.1)
or
ψ ' = ψ ' Tˆ + (a) .
(B.2)
In an analogy from the classical mechanics, it is predicted that the average value of x̂
in the new state ψ ' is equal to that of x̂ in the old state ψ plus the x-displacement a
under the translation of the system
ψ ' xˆ ψ ' = ψ xˆ + a ψ ,
or
ψ Tˆ + (a) xˆTˆ (a) ψ = ψ xˆ + a ψ ,
or
Tˆ + (a) xˆTˆ (a) = xˆ + a1̂ .
Normalization condition:
ψ ' ψ ' = ψ Tˆ + (a)Tˆ (a) ψ = ψ ψ ,
or
Tˆ + (a)Tˆ (a ) = 1 .
[ Tˆ (a) is an unitary operator].
From Eqs.(B3) and (B4), we have
xˆTˆ (a) = Tˆ (a)( xˆ + a) = Tˆ (a) xˆ + aTˆ (a) ,
42
(B.3)
(B.4)
or the commutation relation:
[ xˆ, Tˆ (a)] = aTˆ (a) .
From this, we have
xˆTˆ (a ) x = Tˆ (a) xˆ x + aTˆ (a) x = ( x + a)Tˆ (a) x .
Thus, Tˆ (a) x is the eigenket of x̂ with the eigenvalue (x+a).
(B.5)
or
Tˆ (a) x = x + a ,
(B.6)
or
Tˆ + (a)Tˆ (a) x = Tˆ + (a) x + a = x .
When x is replaced by x-a in Eq.(B7), we get
x − a = Tˆ + (a) x ,
or
x − a = x Tˆ (a ) .
Note that
x ψ ' = x Tˆ (a) ψ = x − a ψ = ψ ( x − a) .
(B.7)
(B.8)
(B.9)
(B.10)
The average value of p̂ in the new state ψ ' is equal to the average value of p̂ in
the old state ψ under the translation of the system
ψ ' pˆ ψ ' = ψ pˆ ψ ,
(B.11)
or
ψ Tˆ + (a) pˆ Tˆ (a ) ψ = ψ pˆ ψ ,
or
Tˆ + (a) pˆ Tˆ (a) = pˆ .
So we have the commutation relation
[Tˆ (a ), pˆ ] = 0̂ .
From this commutation relation, we have
pˆ Tˆ (a) p = Tˆ (a) pˆ p = pTˆ (a) p .
Thus, Tˆ (a) p is the eigenket of p̂ associated with the eigenvalue p.
43
(B.12)