Patterson, Bailey - Solid State Physics Introduction to theory

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124 3 Electrons in Periodic Potentials

which is the precise mathematical statement of Koopmans’ theorem. A similar theorem holds for the Hartree method.

Note that the statement that εk is the negative of the energy required to remove an electron in state k is valid only in the approximation that the other states are unmodified by removal of an electron in state k. For a metal with many electrons, this is a good approximation. It is also interesting to note that

∑N ε

= E + 1

∑

i, j

(1)ψ

(2) V (1,2) ψ

(1)ψ

(2)

(3.31)

− 1

(1)ψ

(2) V (1,2) ψ

(1)ψ

∑

i,j(||)

(2) .

Derivation of Hartree–Fock Equations in Second Quantization

Notation (A)

There really aren’t many new ideas introduced in this section. Its purpose is to gain some familiarity with the second quantization notation for fermions. Of course, the idea of the variational principle will still have to be used.4

According to Appendix G, if the Hamiltonian is of the form (3.10), then we can write it as

H = ∑i, j H ij ai†a j +	1	∑i, j,k,l Vij,kla†j ai†ak al ,	(3.32)
	2

where the Hij and the Vij,kl are matrix elements of the oneand two-body operators,

V	=V	ji,lk	and a a† + a†a = δ		ij	.	(3.33)
ij,kl		ji,lk	i j	j i	ij

The rest of the anticommutators of the a are zero.

We shall assume that the occupied states for the normalized ground state Φ (which is a Slater determinant) that minimizes Φ|H|Φ are labeled from 1 to N. For Φ giving a true extremum, as we saw in the Section on the Hartree approximation, we need require only that

δΦ H Φ = 0 .

(3.34)

It is easy to see that if Φ|Φ = 1, then |Φ + |δΦ is still normalized to first order in the variation. For example, let us assume that

δΦ = (δs)ak†1 ai1 Φ for k1 > N, i1 ≤ N ,

(3.35)

4 For additional comments, see Thouless [3.54].

3.1 Reduction to One-Electron Problem	125

where δs is a small number and where all one-electron states up to the Nth are occupied in the ground state of the electron system. That is, |δΦ differs from |Φ by having the electron in state Φi1 go to state Φk1. Then

( Φ + δΦ )( Φ + δΦ )

+ Φ a

†

1δs

+ a

†

1δs Φ

= Φ

(3.36)

=1+ (δs) Φ a† a

1 Φ +δs Φ a†

1 Φ + O(δs)2

=1+ O(δs)2.

According to the variational principle, we have as a basic condition

0 = δΦ H Φ = (δs) Φ Ha† a

1 Φ .

(3.37)

Combining (3.32) and (3.37) yields

0 = ∑

i, j

Φ a† a

1 a†a

Φ +

1 ∑

Φ a† a

1 a†a†a

a Φ (3.38)

i1 k

i, j,k,l ij,kl

i1 k

j i

k l

where the summation is over all values of i, j, k, l (both occupied and unoccupied). There are two basically different matrix elements to consider. To evaluate them we can make use of the anticommutation relations. Let us do the simplest one first. Φ has been assumed to be the Slater determinant approximation to the

ground state, so:

Φ a† a

1 a†a

Φ = Φ a†

(δi

− a†a

1 )a

i1 k

= Φ a† a

Φ δ i

− Φ a† a†a

1 a

Φ .

i k

In the second term akl operating to the right gives zero (the only possible result of annihilating a state that isn’t there). Since aj|Φ is orthogonal to ail|Φ unless i1 = j, the first term is just δji1. Thus we obtain

Φ a† a	k	1 a†a	j	Φ = δ j	δi	.	(3.39)
i1	k	i	j	i1	k1

The second matrix element in (3.38) requires a little more manipulation to evaluate

Φai†1 ak1 a†jai†ak al Φ

=Φ ai†1 (δkj1 − a†jak1 )ai†ak al Φ

= δ j	Φ a† a†a a Φ					− Φ a† a†a		1 a†a a Φ
k1	i1	i	k l				i1 j k		i k l
= δ j	Φ a† a†a a Φ					− Φ a† a† (δi			− a†a	k	1 )a a Φ
k1	i1	i	k l				i1 j	k1	i	k	k l
= δ j	Φ a† a†a a Φ					− δi	Φ a† a†a a Φ
k1	i1	i	k l			k1	i1	j	k l
+ Φ a† a†a†a				1 a	a Φ .
	i1	j i	k	k		l

126 3 Electrons in Periodic Potentials

Since akl|Φ = 0, the last matrix element is zero. The first two matrix elements are both of the same form, so we need evaluate only one of them:

Φ ai†1 ai†ak al Φ = − Φ ai†ai†1 ak al Φ

= − Φ ai† (δik1 − ak ai†1 )al Φ

= − Φ ai†al Φ δik1 + Φ ai†ak ai†1 al Φ

= −δil ≤ N δik1 − Φ ai†ak (δil1 − al ai†1 ) Φ .

a†i l |Φ is zero since this tries to create a fermion in an already occupied state. So

Φ ai†1 ai†ak al Φ = −δil ≤ N δik1 + δil1δik ≤ N .

Combining with previous results, we finally find

Φ a† a

1 a†a†a

a Φ = δ j

δl

δk ≤ N − δ j

δl ≤ N δk

j i

k l

(3.40)

− δi

δl1 δkj

≤ N + δi

δlj≤ N δk1 .

Combining (3.38), (3.39), and (3.40), we have

0 = ∑i,j H i,jδ1j δi 1

∑ijkl Vij,kl

+ δ

δ1

−δ

δ1

−δ

1 δ1

0 = H k1i1

∑

i=1

+ ∑

V 1

− ∑

i=1

− ∑

,ii

j,i

j, ji

,i i

By using the symmetry in the V and making dummy changes in summation variables this can be written as

0 = H

1 1

+ ∑

−V

(3.41)

j =1

j,i

j, ji

k i

Equation (3.41) suggests a definition of a one-particle operator called the selfconsistent one-particle Hamiltonian:

H C = ∑ki [H ki + ∑Nj =1(Vkj,ij −Vkj, ji )]ak†ai .

(3.42)

At first glance we might think that this operator is identically zero by comparing it to (3.41). But in (3.41) k1 > N and i1 < N, whereas in (3.42) there is no such restriction.

3.1 Reduction to One-Electron Problem	127

An important property of HC is that it has no matrix elements between occupied (i1) and normally unoccupied (k1) levels. Letting HC = ∑ki fkiak†ai, we have

k1 H C i1 = ∑ki		fki k1 ak†ai i1
= ∑	ki	f	ki	0 a	k	1 a†		a† a†	0
	ki		ki		k		k	i i1
= ∑ki		fki		0 (a		†	a	1 −δ k1 )(a†1 a		1	−δ i1 ) 0 .
					k		k	k	i i		i

Since ai |0 = 0, we have

k1 H C i1 = + f	1 1 = 0
	k i

by the definition of fki and (3.41).

We have shown that δΦ|H|Φ = 0 (for Φ constructed by Slater determinants) if, and only if, (3.41) is satisfied, which is true if, and only if, HC has no matrix elements between occupied (i1) and unoccupied (k1) levels. Thus in a matrix representation HC is in block diagonal form since all i1|H|k1 = k1|H|i1 = 0. HC is Hermitian, so that it can be diagonalized. Since it is already in block diagonal form, each block can be separately diagonalized. This means that the new occupied levels are linear combinations of the old occupied levels only and the new occupied levels are linear combinations of the old unoccupied levels only. By new levels we mean those levels that have wave functions i|, j| such that i|HC|j vanishes unless i = j.

Using this new set of levels, we can say

H C = ∑i εi ai†ai .	(3.43)
In order that (3.43) and (3.42) are equivalent, we have
H ki + ∑Nj =1(Vkj,ij −Vkj, ji ) = εiδki .	(3.44)

These equations are the Hartree–Fock equations. Compare (3.44) and (3.24). That is, we have established that δΦ|H|Φ = 0 (for Φ a Slater determinant) implies (3.44). It is also true that the set of one-electron wave functions for which (3.44) is true minimizes Φ|H|Φ , where Φ is restricted to be a Slater determinant of the one-electron functions.

Hermitian Nature of the Exchange Operator (A)

In this section, the Hartree–Fock “Hamiltonian” will be proved to be Hermitian. If the Hartree–Fock Hamiltonian, in addition, has nondegenerate eigenfunctions, then we are guaranteed that the eigenfunctions will be orthogonal. Regardless of degeneracy, the orthogonality of the eigenfunctions was built into the Hartree–Fock equations from the very beginning. More importantly, perhaps, the Hermitian nature of the Hartree–Fock Hamiltonian guarantees that its eigenvalues are real. They have to be real. Otherwise Koopmans’ theorem would not make sense.

128 3 Electrons in Periodic Potentials

The Hartree–Fock Hamiltonian is defined as that operator H F for which
H Fuk = εk uk .	(3.45)

H F is then defined by comparing (3.24) and (3.45). Taking care of the spin summations as has already been explained, we can write

H F = H 1 + ∑ j ∫ψ j (r2 )V (1,2)ψ j (r2 )dτ2 + A1 ,

(3.46)

where

A1ψk (r1) = −∑ j(|| k) ∫ψ j (r2 )V (1,2)ψk (r2 )dτ2ψ j (r1) ,

and A1 is called the exchange operator.

For the Hartree–Fock Hamiltonian to be Hermitian we have to prove that

i H F j = j H F i .

(3.47)

This property is obvious for the first two terms on the right-hand side of (3.46) and so needs only to be proved for A1:

lA1 m = −(∑ j(||m)

=−(∑ j(||m)

=m A1 l .

∫ψl (r1)∫ψ j (r2 )V (1,2)ψm (r2 )ψ j (r1)dτ2dτ1) ∫ψl (r1)ψ j (r1)∫ψ j (r2 )V (1,2)ψm (r2 )dτ2dτ1) ∫ψm (r1)ψ j (r1)∫ψ j (r2 )V (1,2)ψl (r2 )dτ2dτ1)

5 In the proof, use has been made of changes of dummy integration variable and of the relation V(1,2) = V(2,1).

The Fermi Hole (A)

The exchange term (when the interaction is the Coulomb interaction energy and e is the magnitude of the charge on the electron) is

A1ψi (r1 ) ≡ −∑ j(|| i) ∫

ψ j (r2 )ψi (r2 )dτ2 ψi (r1 )

4πε

0 12

eψ

(r )

(r )ψ

= −∑ j(|| i) ∫

ψi (r1 )dτ2

4πε

(r )

0 12

A1ψi (r1) = ∫

(−e)

ρ(r1, r2 )ψi (r1 )dτ2 ,

(3.48)

4πε

0 12

5The matrix elements in (3.47) would vanish if i and j did not refer to spin states which were parallel.

3.1 Reduction to One-Electron Problem	129

where

ρ(r , r ) =		e∑	j(\|\| i)	ψ (r )ψ	(r )ψ	j	(r )
ρ(r , r ) =			j(\|\| i)	j 2 i	2	j	1 .
1	2			ψi (r1)
				ψi (r1)

From (3.48) and (3.49) we see that exchange can be interpreted as the potential energy of interaction of an electron at rl with a charge distribution with charge density ρ(rl, r2). This charge distribution is a mathematical rather than a physical charge distribution.

Several comments can be made about the exchange charge density ρ(rl, r2):

∫ ρ(r1, r2 )dτ2 = +e∫ ∑ j(\|\| i)ψ j (r2 )ψi (r2 )dτ2					ψ j (r1)
∫ ρ(r1, r2 )dτ2 = +e∫ ∑ j(\|\| i)ψ j (r2 )ψi (r2 )dτ2					ψ	i	(r	)	(3.49)
1.	ψ j (r1)					i	1		(3.49)
j	ψ j (r1)
= e∑ j(\|\| i) δi				= +e.
= e∑ j(\|\| i) δi	ψ	i	(r )	= +e.
		i	1

Thus we can think of the total exchange charge as being of magnitude +e.

2.ρ(rl, r1) = e∑j(||i)|ψj(r1)|2, which has the same magnitude and opposite sign of the charge density of parallel spin electrons.

3.From (1) and (2) we can conclude that |ρ| must decrease as r12 increases. This will be made quantitative in the section below on Two Free Electrons and Exchange.

4.It is convenient to think of the Fermi hole and exchange charge density in the following way: in H F, neglecting for the moment A1, the potential energy of the electron is the potential energy due to the ion cores and all the electrons. Thus the electron interacts with itself in the sense that it interacts with a charge density constructed from its own wave function. The exchange term cancels out this unwanted interaction in a sense, but it cancels it out locally. That is, the

exchange term A1 cancels the potential energy of interaction of electrons with parallel spin in the neighborhood of the electron with given spin. Pictorially we say that the electron with given spin is surrounded by an exchange charge hole (or Fermi hole of charge +e).

The idea of the Fermi hole still does not include the description of the Coulomb correlations between electrons due to their mutual repulsion. In this respect the Hartree–Fock method is no better than the Hartree method. In the Hartree method, the electrons move in a field that depends only on the average charge distribution of all other electrons. In the Hartree–Fock method, the only correlations included are those that arise because of the Fermi hole, and these are simply due to the fact that the Pauli principle does not allow two electrons with parallel spin to have the same spatial coordinates. We could call these kinematic correlations (due to constraints) rather than dynamic correlations (due to forces). For further comments on Coulomb correlations see Sect. 3.1.4.

130 3 Electrons in Periodic Potentials

The Hartree–Fock Method Applied to the Free-Electron Gas (A)

To make the above concepts clearer, the Hartree–Fock method will be applied to a free-electron gas. This discussion may actually have some physical content. This is because the Hartree–Fock equations applied to a monovalent metal can be written

−

|ψ j (r2 ) |

+ ∑I =1VI (r1) + e

∑ j =1

∫

dτ2

ψi (r1)

4πε

ψ (r )ψ

0 12

(3.50)

(r )ψ

(r )

− e∑ j(|| i) ∫

dτ2

ψi (r1) = Eiψi (r1).

4πε

(r )

0 12 i

The VI(r1) are the ion core potential energies. Let us smear out the net positive charge of the ion cores to make a uniform positive background charge. We will find that the eigenfunctions of (3.50) are plane waves. This means that the electronic charge distribution is a uniform smear as well. For this situation it is clear that the second and third terms on the left-hand side of (3.50) must cancel. This is because the second term represents the negative potential energy of interaction between smeared out positive charge and an equal amount of smeared out negative electronic charge. The third term equals the positive potential energy of interaction between equal amounts of smeared out negative electronic charge. We will, therefore, drop the second and third terms in what follows.

With such a drastic assumption about the ion core potentials, we might also be tempted to throw out the exchange term as well. If we do this we are left with just a set of one-electron, free-electron equations. That even this crude model has some physical validity is shown in several following sections. In this section, the exchange term will be retained, and the Hartree–Fock equations for a free-electron gas will later be considered as approximately valid for a monovalent metal.

The equations we are going to solve are

2		2	ψ	(r )ψ	k	(r )ψ	k′	(r )
		2	k′	2	k	2	k′	1
−		1ψk (r1) − e∑k′ ∫		4πε r ψ (r )					dτ2	ψk (r1)
−	2m	1ψk (r1) − e∑k′ ∫		4πε r ψ (r )					dτ2	ψk (r1)
				0 12 k			1
				0 12 k			1

= Ekψk (r1).

Dropping the Coulomb terms is not consistent unless we can show solutions of (3.51) are of the form of plane waves

ψ	k	(r ) = 1		eik r1 ,
	k	1	V
			V

(3.51)

that the

(3.52)

where V is the volume of the crystal.

In (3.51) all integrals are over V. Since ћk refers just to linear momentum, it is clear that there is no reference to spin in (3.51). When we sum over k′, we sum over distinct spatial states. If we assume each spatial state is doubly occupied with

3.1 Reduction to One-Electron Problem	131

one spin 1/2 electron and one spin −1/2 electron, then a sum over k′ sums over all electronic states with spin parallel to the electron in k.

To establish that (3.52) is a solution of (3.51) we have only to substitute. The kinetic energy is readily disposed of:

−	2	2ψ	(r ) =	2k 2	ψ	k	(r ) .
	2m			2m
		1 k	1				1

The exchange term requires a little more thought. Using (3.52), we obtain

)ψ

(r )ψ

)

Aψ (r

) = −

∑

k′

′

dτ ψ

(r )

4πε0V

r12ψk (r1)

1 k 1

k′

∫

ei(k −k′) (r2 −r1)

= −

∑

dτ

ψ (r

)

4πε0V

r12

k′

∫

k 1

i(k −k′) r2

= −

∑

e−i(k −k′) r1

∫

dτ

(r ).

4πε0V

r12

k′

(3.53)

(3.54)

The last integral in (3.54) can be evaluated by making an analogy to a similar problem in electrostatics. Suppose we have a collection of charges that have a charge density ρ(r2) = exp[i(k−k′) r2]. Let φ(rl) be the potential at the point rl due to these charges. Let us further suppose that we can treat ρ(r2) as if it is a collection of real charges. Then Coulomb’s law would tell us that the potential and the charge distribution are related in the following way:

	e	i(k −k′) r
φ(r1) = ∫	e			2	dτ2 .	(3.55)
φ(r1) = ∫		4πε	r		dτ2 .	(3.55)
			0 12

However, since we are regarding ρ(r2) as if it were a real distribution of charge, we know that φ(rl) must satisfy Poisson’s equation. That is,

By substitution, we see that a solution of this equation is

φ(r ) =	ei(k −k′) r1	.	(3.57)
	ε0 \| k − k′\|2
1

Comparing (3.55) with (3.57), we find

∫	ei(k −k′) r2 dτ2 =	ei(k −k′) r1	.	(3.58)
		ε0 \| k − k′\|2
	4πε0r12

132 3 Electrons in Periodic Potentials

We can therefore write the exchange operator defined in (3.54) as

A1ψk (r1) = −	e2	∑k′	1	ψk (r1) .	(3.59)
	ε0V		\| k − k′\|2

If we define A1(k) as the eigenvalue of the operator defined by (3.59), then we find that we have plane-wave solutions of (3.51), provided that the energy eigenvalues are given by

E	k	=	2k 2	+ A (k) .	(3.60)

			2m	1

If we propose that the above be valid for monovalent metals, then we can make a comparison with experiment. If we imagine that we have a very large crystal, then we can evaluate the sum in (3.59) by replacing it by an integral. We have

A1(k) = −	e2 V	∫	1	d3k′ .	(3.61)

	ε0V 8π3		\| k − k′\|2

We assume that the energy of the electrons depends only on |k| and that the maximum energy electrons have |k| = kM. If we use spherical polar coordinates (in k′-space) with the k′z-axis chosen to be parallel to the k-axis, we can write

2π

k′2 sinθ

A1(k) = −

∫

M ∫

∫

dφ

+ k′

− 2kk′cosθ

8π ε

k′2

= −

∫

M ∫

d(cosθ

′

4π

−1

+ k

− 2kk

cosθ

2kk′f ) f =+1

M k′2

ln(k 2

+ k′2 −

= −

∫

− 2kk′

4πε0

−1

=−1

k 2 + k

′2 − 2kk

′

∫0

k′ln

+ k

′

+ 2kk

dk′

8π ε0k

′

k + k′

dk′.

= −

∫0 M k′ln

4π 2ε0k

k − k′


	dθ dk′

) dk′

dk′ (3.62)

But ∫x(ln x)dx = (x2/2)ln x − x2/4, so we can evaluate this last integral and finally find

e2k

k 2

− k 2

k + k′

(k) = −

(3.63)

4π

ε0

kkM

k − k′

3.1 Reduction to One-Electron Problem	133

The results of Problem 3.5 combined with (3.60) and (3.63) tell us on the Hartree– Fock free-electron model for the monovalent metals that the lowest energy in the conduction band should be given by

E(0) = −	e2	kM	,	(3.64)
	2π 2
		ε0

while the energy of the highest filled electronic state in the conduction band should be given by

E(kM ) =	2kM2	−	e2kM	.	(3.65)

	2m		4π 2ε0

Therefore, the width of the filled part of the conduction band is readily obtained as a simple function of kM:

[E(kM ) − E(0)]=	2kM2	+	e2kM	.	(3.66)
	2m
			4π 2ε0

To complete the calculation we need only express kM in terms of the number of electrons N in the conduction band:

N = ∑k (1) = 2	V	∫kM d3k =	2V	4π	kM3 .	(3.67)
	8π 3		8π 3	3
		0

The factor of 2 in (3.67) comes from having two spin states per k-state. Equation (3.67) determines kM only for absolute zero temperature. However, we only have an upper limit on the electron energy at absolute zero anyway. We do not introduce much error by using these expressions at finite temperature, however, because the preponderance of electrons always has |k| < kM for any reasonable temperature.

The first term on the right-hand side of (3.66) is the Hartree result for the bandwidth (for occupied states). If we run out the numbers, we find that the Hartree–Fock bandwidth is typically more than twice as large as the Hartree bandwidth. If we compare this to experiment for sodium, we find that the Hartree result is much closer to the experimental value. The reason for this is that the Hartree theory makes two errors (neglect of the Pauli principle and neglect of Coulomb correlations), but these errors tend to cancel. In the Hartree–Fock theory, Coulomb correlations are left out and there is no other error to cancel this omission. In atoms, however, the Hartree–Fock method usually gives better energies than the Hartree method. For further discussion of the topics in these last two sections as well as in the next section, see the book by Raimes [78].

Two Free Electrons and Exchange (A)

To give further insight into the nature of exchange and to the meaning of the Fermi hole, it is useful to consider the two free-electron model. A direct derivation

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