Patterson, Bailey - Solid State Physics Introduction to theory

74 2 Lattice Vibrations and Thermal Properties

Using (2.106), we have

ω2 2γ

1

M

	2πs		4e		2	πs
− cos		+		sin		.	(2.117)
	N		N
						N

Using the half-angle formula sin2 θ/2 = (1 − cos θ)/2, we can recast (2.117) into the form

ω ωc		πs				e
	sin	N		1	+		.	(2.118)
						N

We can make several physical comments about (2.118). As noted earlier, if the description of the lattice modes is given by symmetric (about the impurity) and antisymmetric modes, then our development is valid for symmetric modes. Antisymmetric modes cannot be affected because u0 = 0 for them anyway and it cannot matter then what the mass of the atom described by u0 is. When M > M 1, then e > 0 and all frequencies (of the symmetric modes) are shifted upward. When M < M 1, then e < 0 and all frequencies (of the symmetric modes) are shifted downward. There are no local modes here, but one does speak of resonant modes.13 When N → ∞, then the frequency shift of all modes given by (2.118) is negligible. Actually when N → ∞, there is one mode for the e > 0 case that is shifted in frequency by a non-negligible amount. This mode is the impurity mode. The reason we have not yet found the impurity mode is that we have not allowed the φ defined by (2.106) to be complex. Remember, real φ corresponds only to modes whose amplitude does not diminish. With impurities, it is reasonable to seek modes whose amplitude does change. Therefore, assume φ = π + iz (φ = π corresponds to the highest frequency unperturbed mode). Then from (2.111),

N				1
tan		(π + iz)	= e tan		(π + iz) .	(2.119)
	2			2

Since tan (A + B) = (tan A + tan B)/(1 − tan A tan B), then as N → ∞ (and remains an even number), we have

Nπ

iNz

iNz

tan

+

= tan

= i .

(2.120)

2

2

2

Also

π + iz

=

sin(π / 2 + iz / 2)

= −

sin(π / 2) cos(iz / 2)

tan

2

cos(π / 2

+ iz / 2)

sin(π / 2) sin(iz / 2)

(2.121)

iz

z

= −cot

= +i coth

.

2

2

13 Elliott and Dawber [2.15].

2.2 One-Dimensional Lattices (B) 75

	Combining (2.119), (2.120), and (2.121), we have
		e coth			z	= 1.		(2.122)
			2
	Equation (2.122) can be solved for z to yield
		z = ln			1 + e		.	(2.123)
					1 − e
	But
	cosφ = cos(π + iz) = cosπ cos iz
	= −	1	(exp z + exp− z)					(2.124)
		2
			+ e2
	= −	1
		1	− e2
	by (2.122). Combining (2.124) and (2.106), we find
	ω2 = ωc2 /(1 − e2 ) .							(2.125)

The mode with frequency given by (2.125) can be considerably shifted even if N → ∞. The amplitude of the motion can also be estimated. Combining previous results and letting N → ∞, we find

un = (−)	|n| M − M 1	ωc2		1− e |n|	u0 = (−1)	n 1		− e |n|		(2.126)
	2γ	2e					1		u0 .
				1+ e				+ e

This is truly an impurity mode. The amplitude dies away as we go away from the impurity. No new modes have been gained, of course. In order to gain a mode with frequency described by (2.125), we had to give up a mode with frequency described by (2.118). For further details see Maradudin et al [2.26 Sect. 5.5]

2.2.6 Quantum-Mechanical Linear Lattice (B)

In a previous Section we found the quantum-mechanical energies of a linear lattice by first reducing the classical problem to a set of classical harmonic oscillators. We then quantized the harmonic oscillators. Another approach would be initially to consider the lattice from a quantum viewpoint. Then we transform to a set of independent quantum-mechanical harmonic oscillators. As we demonstrate below, the two procedures amount to the same thing. However, it is not always true that we can get correct results by quantizing the Hamiltonian in any set of generalized coordinates [2.27].

76 2 Lattice Vibrations and Thermal Properties

With our usual assumptions of nearest-neighbor interactions and harmonic forces, the classical Hamiltonian of the linear chain can be written

H ( pl , xl ) =

1

∑

l

p 2

+

γ

∑

l

(2x 2

− xl xl+1 − xl xl−1 ) .

(2.127)

2M

2

l

l

In (2.127), p1 = Mx·1, and in the potential energy term use can always be made of periodic boundary conditions in rearranging the terms without rearranging the limits of summation (for N atoms, xl = xl+N). The sum in (2.127) runs over the crystal, the equilibrium position of the lth atom being at la. The displacement from equilibrium of the lth atom is xl and γ is the spring constant.

To quantize (2.127) we associate operators with dynamical quantities. For (2.127), the method is clear because pl and xl are canonically conjugate. The momentum pl was defined as the derivative of the Lagrangian with respect to x·l. This implies that Poisson bracket relations are guaranteed to be satisfied. Therefore, when operators are associated with pl and xl, they must be associated in such a way that the commutation relations (analog of Poisson bracket relations)

[xl , pl′ ] = i

δll′

(2.128)

are satisfied. One way to do this is to let

pl →

∂

,

and

xl → xl .

(2.129)

i

∂xi

This is the choice that will usually be made in this book.

The quantum-mechanical problem that must be solved is

∂

xn ) = E(x1

xn )

.

(2.130)

H

∂xl

, xl ψ(x1

i

In (2.130), ψ(x1 xn) is the wave function describing the lattice vibrational state

with energy E.

How can (2.130) be solved? A good way to start would be to use normal coordinates just as in the Section on vibrations of a classical lattice. Define

X q = 1	∑l eiqla xl ,	(2.131)
N

where q = 2πm/Na and m is an integer, so that

xl	= 1	∑q e−iqla X q .	(2.132)
	N

The next quantities that are needed are a set of new momentum operators that are canonically conjugate to the new coordinate operators. The simplest way to get these operators is to write down the correct ones and show they are correct by the fact that they satisfy the correct commutation relations:

P	′ =	1 ∑	l	p	e−iq′la ,	(2.133)
q		N		l

2.2 One-Dimensional Lattices (B) 77

or
pl = 1	∑q′′ Pq′′eiq′′la .	(2.134)
N

The fact that the commutation relations are still satisfied is easily shown:

[X q , Pq′] =	1	∑l,l′[xl′, pl ]exp[ia(ql′ − q′l)]
	N
=	1	∑l,l′i δll′ exp[ia(ql′ − q′l)]	(2.135)
	N

= i δqq′ .

Substituting (2.134) and (2.132) into (2.127), we find in the usual way that the Hamiltonian reduces to a fairly simple form:

	H =	1	∑q Pq P−q +γ ∑q X q X −q (1−cos qa) .	(2.136)
		2M

Thus, the normal coordinate transformation does the same thing quantummechanically as it does classically.

The quantities Xq and X−q are related. Let † (dagger) represent the Hermitian conjugate operation. Then for all operators A that represent physical observables (e.g. pl), A† = A. The † of a scalar is equivalent to complex conjugation (*).

Note that

Pq† = 1	∑l pl eiqla = P−q ,
N

and similarly that

X q† = X −q .

From the above, we can write the Hamiltonian in a Hermitian form:

	H = ∑q	1	Pq Pq† +γ (1−cos qa)X q X q† .	(2.137)
		2M

From the previous work on the classical lattice, it is already known that (2.137) represents a set of independent simple harmonic oscillators whose classical frequencies are given by

ωq = 2γ (1 − cos qa) / M = 2γ / M sin(qa 2) .

(2.138)

However, if we like, we can regard (2.138) as a useful definition. Its physical interpretation will become clear later on. With ωq defined by (2.138), (2.137) becomes

	1		1
H = ∑q		Pq Pq† +		Mω 2 X q X q† .	(2.139)
	2M
		2

78 2 Lattice Vibrations and Thermal Properties

The Hamiltonian can be further simplified by introducing the two variables [99]

a		=	1	P − i	Mωq	X † ,
	q		2M ωq	q	2		q
a† =			1	P† + i	Mωq	X		.
	q		2M ωq	q	2		q

Let us compute [aq, a†q1]. By (2.140) and (2.141),

[a

q

, a†

] =

i

Mωq {[P

, X

q

1

] −[X †, P†

]}

q

1

2M ωq

2

q

q

q

1

=

i

(−i δqq1

− −i δqq1 )

2

= δqq1 ,

or in summary,

[aq , aq†1 ] =δqq1 .

(2.140)

(2.141)

(2.142)

It is also interesting to compute ½Σq ωq{aq, a†q}, where {aq, a†q} stands for the anticommutator; i.e. it represents aq a†q + a†qaq:

12 ∑q ωq{aq , aq†} =

∑

1

−i Mωq

1

P† + i Mωq

1

q

ω

P

X

†

X

2

q

2M ωq

q

2

q

2M ωq

q

2

q

+ 1

∑

1

P† + i Mωq

1

P −i Mωq

q

ω

X

X

†

2

q

2M ωq

q

2

q

2M ωq

q

2

q

= 12 ∑q

Observing that

1

Mωq

i

i

ω

P P† +

X † X

q

−

X †P† +

P X

q

2M

ω

2

2

2

q

q q

q

q q

q

q

Mωq

1

P†P

X †

i

i

P† X †

+

+

X

q

+

X

P

−

.

2M ωq

q q

2

q

2

q q

2

q q

X q Pq + Pq X q − X q†Pq† − Pq† X q† = 2(Pq X q − Pq† X q† ), Pq† = P−q , X q† = X −q ,

2.2 One-Dimensional Lattices (B) 79

and ωq = ω–q, we see that

∑q ωq (Pq X q − Pq† X q† ) = 0 .

Also [X †q, Xq] = 0 and [P †q, Pq] = 0, so that we obtain

1			1		1
	∑q	ωq {aq , aq†} = ∑q		Pq Pq† +		Mωq2 X q X q†	= H . (2.143)
2			2M
				2

Since the aq operators obey the commutation relations of (2.142) and by Problem 2.6, they are isomorphic (can be set in one-to-one correspondence) to the step-up and step-down operators of the harmonic oscillator [18, p349ff]. Since the harmonic oscillator is a solved problem so is (2.143).

By (2.142) and (2.143) we can write

H = ∑q ωq (aq†aq +	1	) .	(2.144)
	2

But from the quantum mechanics of the harmonic oscillator, we know that

aq† nq =	(nq +1) nq +1 ,	(2.145)
aq nq	= nq nq −1 .	(2.146)

Where |nq is the eigenket of a single harmonic oscillator in a state with energy (nq + ½) ωq, ωq is the classical frequency and nq is an integer. Equations (2.145) and (2.146) imply that

a†qa q nq = nq nq .

(2.147)

Equation (2.144) is just an operator representing a sum of decoupled harmonic oscillators with classical frequency ωq. Using (2.147), we find that the energy eigenvalues of (2.143) are

E = ∑q ωq (nq +	1	) .	(2.148)
	2

This is the same result as was previously obtained.

From relations (2.145) and (2.146) it is easy to see why a†q is often called a creation operator and aq is often called an annihilation operator. We say that a†q creates a phonon in the mode q. The quantities nq are said to be the number of phonons in the mode q. Since nq can be any integer from 0 to ∞, the phonons are said to be bosons. In fact, the commutation relations of the aq operators are typical commutation relations for boson annihilation and creation operators. The Hamiltonian in the form (2.144) is said to be written in second quantization notation. (See Appendix G for a discussion of this notation.) The eigenkets |nq are said to be kets in occupation number space.

80 2 Lattice Vibrations and Thermal Properties

With the Hamiltonian written in the form (2.144), we never really need to say much about eigenkets. All eigenkets are of the form

mq	= 1 (aq† )mq 0 ,
	mq !

where |0 is the vacuum eigenket. More complex eigenkets are built up by taking a product. For example, |m1,m2 = |m1 |m2 . States of the |mq , which are eigenkets of both the annihilation and creation operators, are often called coherent states.

Let us briefly review what we have done in this section. We have found the eigenvalues and eigenkets of the Hamiltonian representing one-dimensional lattice vibrations in the harmonic and nearest-neighbor approximations. We have introduced the concept of the phonon, but some more discussion of the term may well be in order. We also need to give some more meaning to the subscript q that has been used. For both of these purposes it is useful to consider the symmetry properties of the crystal as they are reflected in the Hamiltonian.

The energy eigenvalue equation has been written

Hψ(x1 xN ) = Eψ(x1 xN ) .

Now suppose we define a translation operator Tm that translates the coordinates by ma. Since the Hamiltonian is invariant to such translations, we have

[H ,Tm ] = 0 .

(2.149)

By quantum mechanics [18] we know that it is possible to find a set of functions that are simultaneous eigenfunctions of both Tm and H. In particular, consider the case m = 1. Then there exists an eigenket |E such that

H E = E E ,		(2.150)
and
T E = t1	E .	(2.151)
1

Clearly |t1| = 1 for (T1)N |E = |E by periodic boundary conditions, and this implies (t1)N = 1 or |t1| = 1. Therefore let

t1 = exp(ikq a) ,

(2.152)

where kq is real. Since |t1| = 1 we know that kqaN = pπ, where p is an integer. Thus

	kq =	2π	p ,	(2.153)
		Na

and hence kq is of the same form as our old friend q. Statements (2.150) to (2.153) are equivalent to the already-mentioned Bloch’s theorem, which is a general theorem for waves propagating in periodic media. For further proofs of Bloch’s theorem and a discussion of its significance see Appendix C.

2.2 One-Dimensional Lattices (B) 81

What is the q then? It is a quantum number labeling a state of vibration of the system. Because of translational symmetry (in discrete translations by a) the system naturally vibrates in certain states. These states are labeled by the q quantum number. There is nothing unfamiliar here. The hydrogen atom has rotational symmetry and hence its states are labeled by the quantum numbers characterizing the eigenfunctions of the rotational operators (which are related to the angular momentum operators). Thus it might be better to write (2.150) and (2.151) as

H E, q = Eq	E, q	(2.154)
T E, q = eikqa	E, q .	(2.155)
1

Incidentally, since |E,q is an eigenket of T1 it is also an eigenket of Tm. This is easily seen from the fact that (T1)m = Tm.

We now have a little better insight into the meaning of q. Several questions remain. What is the relation of the eigenkets |E,q to the eigenkets |nq ? They, in fact, can be chosen to be the same.14 This is seen if we utilize the fact that T1 can be represented by

T1 = exp(ia∑q′q′aq†′aq′) .

(2.156)

Then it is seen that

T1 nq	= exp(ia∑q′q′aq†′aq′) nq	(2.157)
	= exp(ia∑q′q′nq′δqq′) nq = exp(iaqnq ) nq .

Let us now choose the set of eigenkets that simultaneously diagonalize both the Hamiltonian and the translation operator (the |E,q ) to be the |nq . Then we see that

kq = q nq .

(2.158)

This makes physical sense. If we say we have one phonon in mode q (which state we characterize by |1q ) then

T1 1q = eiqa 1q ,

and we get the typical factor eiqa for Bloch’s theorem. However, if we have two phonons in mode q, then

T1 2q = eiqa(2) 2q ,

and the typical factor of Bloch’s theorem appears twice.

14 See, for example, Jensen [2.19]

82 2 Lattice Vibrations and Thermal Properties

The above should make clear what we mean when we say that a phonon is a quantum of a lattice vibrational state.

Further insight into the nature of the q can be gained by taking the expectation value of x1 in a time-dependent state of fixed q. Define

| q ≡ ∑n	Cn exp[−(i /	)(En )t] | nq .	(2.159)
q	q	q

We choose this state in order that the classical limit will represent a wave of fixed wavelength. Then we wish to compute

q | x p | q = ∑

1

Cn*

C

1 exp[+(i /

)(En

− E

1 )t] nq | x p | n1q .

(2.160)

n ,n

q

n

q

n

q q

q

q

By previous work we know that

x

p

= (1/

N )∑ 1 exp(−ipaq1)X

1 ,

(2.161)

q

q

where the Xq can be written in terms of creation and annihilation operators as

X q

=

1

2

(aq† − a−q ) .

(2.162)

2i

Mωq

Therefore,

x p = 1

2

∑q1 exp(−ipaq1)(a†1 − a

1 )

1 .

(2.163)

2i

NM

q

−q

ωq1

Thus

nq xp n1q

=

1

2

∑q1 (ωq1 )−1 2 exp(−ipaq1)

nq

a†1 n1q

(2.164)

2i

NM

q

1

1

− ∑q1 exp(−ipaq ) nq

a−q1 nq

.

By (2.145) and (2.146), we can write (2.164) as

1

1

2

−ipaq

1

nq

+ipaq

1

n1q

−1

nq x p nq

=

2i

NMωq

e

nq +1δn1q +1

− e

nq

δnq

.

(2.165)

Then by (2.160) we can write

q xp q =

1

2

∑n

*

nq e

−ipaq

e

+iωqt

2i

NMωq

q

Cnq Cnq −1

(2.166)

−iωqt

− ∑n

Cnq Cnq +1 nq +1 e

+ipaq

e

.

q

2.2 One-Dimensional Lattices (B) 83

In (2.166) we have used that

Enq = (nq + 12 ) ωq .

Now let us go to the classical limit. In the classical limit only those Cn for which nq is large are important. Further, let us suppose that Cn are very slowly varying functions of nq. Since for large nq we can write

		nq		nq +1 ,
q x p q =	2	(∑∞n	=0	nq | Cn	|2 ) sin[ωqt − q( pa)] . (2.167)
	NMωq	q		q

Equation (2.167) is similar to the equation of a running wave on a classical lattice where pa serves as the coordinate (it locates the equilibrium position of the vibrating atom), and the displacement from equilibrium is given by xp. In this classical limit then it is clear that q can be interpreted as 2π over the wavelength.

In view of the similarity of (2.167) to a plane wave, it might be tempting to call q the momentum of the phonons. Actually, this should not be done because phonons do not carry momentum (except for the q = 0 phonon, which corresponds to a translation of the crystal as a whole). The q do obey a conservation law (as will be seen in the chapter on interactions), but this conservation law is somewhat different from the conservation of momentum.

To see that phonons do not carry momentum, it suffices to show that

		nq Ptot nq			= 0 ,		(2.168)
where
		P	= ∑ p		l	.	(2.169)
		tot		l
By previous work
p	l	= (1/ N )∑ 1		P 1 exp(iq1la) ,
			q	q

and

Pq1 = 2M ωq1 (aq1 + a−†q1 ) .

Then

nq Ptot nq =	M	∑ ∑ 1	ω 1	exp(iq1la) nq (a 1	+ a†	) nq = 0 (2.170)
	2N	l q	q	q	−q1

by (2.145) and (2.146). The q1 → 0 mode can be treated by a limiting process. However, it is simpler to realize it corresponds to all the atoms moving together so it obviously can carry momentum. Anybody who has been hit by a thrown rock knows that.