The nearly-free electron model PDF

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The nearly-free electron model - Week 11...

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QUEEN MARY, UNI VERSITY OF LONDON SCHOOL OF PHYSICS AND ASTRONOMY

Condensed Matter A Week 11: The nearly-free electron model Key reference for weeks 10–12: Sidebottom, Fundamentals of Condensed Matter and Crystalline Physics, chapters 12–13 So far we have seen that the free electron gas model can explain a surprising number of the properties of metals, including their heat capacity and bulk modulus. However, in order to deal with real systems, we do need to consider the effect of the metal ions. Thus we reach the nearly-free electron gas model, in which the electrons move in a weak, periodic potential (still without interacting with each other except via the Pauli exclusion principle). This has two principal effects: 1. The dispersion curves become periodic, with the same periodicity as the reciprocal lattice – just as we saw with phonons. Specifically, there are now dispersion relations E=

h¯ 2 (k − G)2 2m

(1)

for any reciprocal lattice vector G. 2. A small band gap opens up at the Brillouin zone boundary. This can be understood by considering the consequences of a wavefunction with wavevector k = 21 a∗ in a periodic potential, shown in the top figure below. If we choose one value of the phase, we find the resulting|ψ|2 concentrates electron density around the metal ions (middle figure;ψ shown as dashed line, |ψ|2 as solid line). If we choose another, then electron density is concentrated away from the metal ions. Thus at this particular value of k we have a low- and a high-energy solution, without any possible electronic states in between.

V

1 0.5

ψ1

0 1 0

ψ2

−1 1 0 −1 −3

−2

−1

0 x

1

2

3

We will spend the next few lectures exploring the consequences of these features of the model, as well as the mathematics that leads up to them. Let’s begin with the maths, which is probably the most difficult derivation of this course. The derivation in the box below is for interest and not examinable, although the key results (i.e., the bits out of the box) are! 1

We start with the Schrödinger equation itself. For simplicity I’ll work in 1D but the results all extend easily to 3D: −

h¯ 2 ∂ 2 ψ +V ψ = Eψ 2m ∂ x2

(2)

We apply a periodic potentialV (x): in other words, we require thatV (x) = V (x + T ) for all real lattice vectorsT = na, n ∈ Z . This means that we can express V (x) as a Fourier series over the reciprocal lattice vectors: V (x) = ∑ VG exp(iGx)

G = ha∗ , h ∈ Z.

(3)

G

Furthermore, we express our solutions ψ(x) in terms of the ansatz ψ(x) = ∑Ck exp(ikx).

(4)

k

Although this looks very similar to the expression for the potential, there is an important difference: since the potential has the periodicity of the lattice, we know that the reciprocal space vectorsG are all reciprocal lattice vectors, whereas we do not yet know anything about the wavevectorsk in the ansatz for ψ(x). We now have all the key ingredients needed to solve for the wavefunctions of electrons in this model (the eigenfunctions of the Schrödinger equation) and their energies (the eigenvalues). Substituting everything into the Schrödinger equation gives

∑ k

 ′  ¯h2 k2 Ck exp(ikx) + ∑ Ck′ VG exp i(k + G)x = E ∑ Ck exp(ikx). 2m ′ k ,G k

(5)

(Note that I have played a small notational trick in that I have written the index of the Fourier series forψ(x) as k in the first and third terms butk′ in the second. Of course it doesn’t matter what we label the index of a sum (just like it doesn’t matter what we label the dummy variable of an integral) but this will make the following manipulation easier.) We now set k′ = k − G. Again, this is valid since the sums in our equation remain the same regardless of any relationship between their index variables. But this allows us to rewrite the equation as " # ! h¯ 2 k2 (6) ∑ exp(ikx) 2m − E Ck + ∑ VGCk−G = 0. k G The important thing to note here is that this has to be true for allx. But since the term in square brackets is independent ofx, the only way this can happen is if this term vanishes for all k, giving us ! h¯ 2 k2 (7) − E Ck + ∑ VGCk−G = 0. 2m G There are two points to make at this stage: the first is that we haven’t actually made any assumptions about the physics except to specify that the potential is periodic. In other words, equation (7) is just the Schrödinger equation in a periodic potential. The second point is that we can already see that the coefficientsCk depend only on the coefficientsCk−G for reciprocal lattice vectors G, anticipating the periodicity of the dispersion curves.

2

Given this second result, we can rewrite our expression (4) for the solutions ψ(x), labelling each solution by its characteristic value of k:   ψk (x) = ∑ Ck−G exp i(k − G)x

(8)

G

=

"

∑Ck−G exp(−iGx) G

#

exp(ikx)

= uk (x) exp(ikx).

(9) (10)

Here we have defined the Fourier series in square brackets asuk (x), which by definition has the periodicity of the lattice.

Equation (10) is known as Bloch’s theorem, which can be restated as the fact that every solution to the Schrödinger equation in a periodic potential is a plane wave modulated by some function with the periodicity of the lattice. This also justifies assigning a characteristick value to each state, which in turn means that dispersion curves still make sense! Anyway, so much for the equation itself: how do we solve it given the further assumption, in the nearly free electron gas model, that the potential V (x) is weak? The easiest way turns out to be to consider what happens when we translate our equation (7) in reciprocal space by a reciprocal lattice vectorG; that is, if we replace every reference to k by k − G: ! ¯h2 2 (11) E− |k − G| Ck−G = ∑ VG′ Ck−G−G′ 2m G′ = ∑VG′′ −GCk−G′′

(12)

G′′

where in the first step we have relabelled the index of the sum G′ , since G is now our fixed translation, and in the second we have changed the index yet again by defining G′′ = G + G′. We want to solve (12) for the coefficientCk−G. Ck−G (It might look like there is a complication here, since the coefficient turns up on both the left and (in the sum) right sides of the equation. Can you explain why this is not a problem?) Now we introduce our key assumption, that the potential is weak. This being so, all of theVG′′ −G are small. From (12), we therefore have two options: either Ck−G is small orE − (¯ h2 /2m)|k − G|2 is small. Now certainly there is one case in which the second option holds: because we expect that in a weak potential the eigenvaluesE will be approximately the same as in no potential, E ≈ ¯h2 k2 /2m and hence the second factor is small whenG = 0. If this were the only case where the second factor is small, then we could conclude that Ck−G is small for allG 6= 0 , and hence thatψk (x) ≈ Ck exp(ikx), which is just the same solution as for the free electron gas. In fact, though, there is one other case when this second option will be true. If we are at a particular k value such that E≈

h¯ 2 |k − G|2 2m

(13)

then equation (12) shows us that, even though theVG′′ −G are small,Ck−G can be large. This occurs where two branches of the dispersion relation, arising from different reciprocal lattice points, intersect – precisely, in other words, at the Brillouin zone boundary. This is illustrated below for the particular case G = a∗ .

3

h¯ 2 ∗ 2 2m (k − a )

E=

h¯ 2 2 2m k

E

E=

π a

0

a∗ =

2π a

k In this case Ck−G may be big, for the particular G that is relevant. But the coefficients for all otherG will certainly be small. So the sums in the Schrödinger equation (7) turn into just two simultaneous equations to solve in the only two coefficients that are too big to ignore, Ck and Ck−G : ! h¯ 2 k2 − E Ck +VGCk−G = 0 (14) 2m ! ¯h2 (k − G)2 − E Ck−G + V−GCk = 0 (15) 2m We use the same trick for solving these homogenous linear equations as we did when we met similar equations in the case of phonons: we rewrite the equations together as a matrix equation: ! ! ! ¯h2 k2 − E VG 0 Ck 2m (16) = h¯2 (k − G)2 − E 0 Ck−G V−G 2m Certainly, this equation has the trivial solutionCk = Ck−G = 0 . The only way it can have any other (more interesting) solution is if the square matrix is singular – that is, if it is not invertible. So we set the determinant to zero. At this stage it is convenient to simplify the notation somewhat, by writing E0 to represent the energies of the free electron gas model. Thus we write E 0k =

h¯ 2 k2 2m

E k−G = 0

h¯ 2 (k − G)2 . 2m

(17)

In this notation, the square matrix becomes E 0k − E V−G

VG −E E k−G 0

!

(18)

and setting its determinant to zero gives (E0k − E )(E k−G − E) −VGV−G = 0 0 E

2

− (E 0k

+ E0k+G)E + (E0kE k−G 0

−VGV−G ) = 0.

(19) (20)

This is just a quadratic in E . Using the quadratic formula shows that the energies of the two intersecting dispersion curves near the Brillouin zone boundary are given by h 1 i  E = 12 E0k + E0k−G ± (E0k + E 0k−G )2 + 4(VGV−G − E 0k E0k−G) 2 (21) q (22) = 12 (E 0k + E0k−G) ± 41 (E 0k − E 0k−G )2 + |VG |2 .

Comparing this to the free-electron result shows, just as our qualitative

4

argument did, that a band gap has opened at the Brillouin zone boundary, but that away from this boundary the curves are much the same as the free-electron case:

E

E+

E− π a

0

a∗ =

2π a

k It is particularly useful to consider the difference in energy between the two curves, which is q E+ − E− = (E0k − E 0k−G )2 + 4|VG |2 . (23)

Away from the Brillouin zone boundary, the second term is negligible compared to the first (remember that ourVG are by assumption small) and we find that the difference between the curves is the same as in the free electron case. But at the Brillouin zone boundary, the first term becomes zero, and so the band gap is just given by ∆E = 2|VG |.

There are two useful ways of visualising the results of these calculations. First, just as for phonons, we can draw a dispersion curve, hereE against k. As for phonons, since this is now periodic, we will often plot just the first Brillouin zone. And again as for phonons, when we deal with multiple dimensions we need to choose a particular path through the Brillouin zone in order to sketch the dispersion curve. As an example, consider the dispersion curve for a two-dimensional square lattice with zero potential as we travel from Γ = (0, 0) to X = ( 12 , 0) to M = ( 12 , 12 ) and back to Γ: M X Γ

Γ 10

E/(h2 /8ma2 )

8 ky M 1 ∗ 2a

6

4

X Γ

1 a∗ 2

kx

2

0

0

0.2 k=

0.4

(δ , 0)a∗

0

0.2

0.4 0.5

(0.5, δ )a∗

0.4

0.3

0.2

(δ , δ )a∗

5

0.1

0

This looks rather complicated but is simply a result of applying equation (1) not just for G = Γ but also for neighbouring reciprocal lattice vectors. To begin interpreting it, consider the left-hand section, wherek = (δ a∗ , 0) for δ ranging from 0 to 0.5. The bottom curve is obviously the usual curve E = h¯ 2 |k|2 /2m, originating from G = Γ. The next curve up originates from G = (1, 0) , as we can see by substituting this value into (1) to give E = ¯h2 /2m|k − (1, 0)a∗ |2 = h2 /2ma2 (δ − 1)2. The curve above that originates from the points G = (0, ±1), and so forth. As an exercise, see if you can identify the reciprocal lattice points that give rise to a few more of the curves in this diagram. The second useful visualisation method is to sketch the occupied states in reciprocal space. The shape of the Fermi surface will change from its form of a sphere in the free-electron gas model, depending on how wide the band gap is. In order to understand these sketches, we will need a few results about the Brillouin zone itself. Recall that the (first) Brillouin zone is the region closer to Γ, the origin of reciprocal space, than to any other reciprocal lattice point. Similarly, we can define the second Brillouin zone, for whichΓ is the second-nearest reciprocal lattice point, and so on for the third, fourth, and following Brillouin zones. There is a quick way of constructing these zones, which is to draw planes that bisect the reciprocal lattice vectors. For instance, consider again a two-dimensional square lattice:

first Γ

second third

An online introduction to Brillouin zones is available athttp://www.doitpoms. ac.uk/tlplib/brillouin_zones/index.php(link also on QM+) which may

help in understanding these concepts. Another key result comes from our calculation of the density of the grid of allowed k -points. We have seen that the number of k-points per unit reciprocal-space volume isV /8π 2 where V is the real-space volume of the sample. On the other hand, we know that the Brillouin zone has the volume of a unit cell of the reciprocal lattice, so that its volume is 8π 2 /Vc where Vc is the real-space volume of the unit cell. Comparing these results, we find that the number of states in the first Brillouin zone is equal to the number of unit cells in the sample. Since each state can hold two electrons, this means that monovalent metals (where each atom contributes a single electron to the gas) fill up half of the first Brillouin zone while divalent metals fill up a reciprocal space volume equal to the entire first Brillouin zone. Exactly where the occupied states sit in reciprocal space is determined by the band structure of the particular material. Once we introduce the band gap, then, we find that because of the number of states in the first Brillouin zone, monovalent systems in 1D are metals, but divalent systems are insulators or semiconductors. Of course in 3D there are many examples of divalent metals, including Mg and Ca. To understand this we consider a 2D electron gas based on a square lattice. The diagram below shows the occupied levels in the Brillouin zone and the 6

E

dispersion curves from Γ to X = ( 12 , 0) and M = ( 12 , 12 ) in three separate cases: with zero potential (top), with a weak potential (middle), and with a stronger potential (bottom): M Γ X

EF

0.5

0.4

0.3 k=

0.1

(δ , δ , 0)a∗

0 Γ

0.2

0.4

k = (δ , 0, 0)a∗

X

E

M

0.2

EF

0.5

0.4

0.3 k=

M

0.2

0.1

(δ , δ , 0)a∗

0 Γ

0.2

0.4

k = (δ , 0, 0)a∗

E

EF

0.5

0.4

0.3 k=

0.2

(δ , δ , 0)a∗

0.1

0

0.2

0.4

k = (δ , 0, 0)a∗

In the free-electron case, the occupied states extend into the second Brillouin zone (or equivalently, the second band) in the X direction, while they don’t quite reach the Brillouin zone boundary in the M direction. If we introduce a weak potential, band gaps open, which means both that the Fermi level has to rise slightly to fit the right number of electrons into the system 7

X

and that more electrons are now near the corners of the first Brillouin zone while fewer cross its boundary. Finally, if the potential is strong enough, so the band gap is large enough, the system becomes an insulator, as in 1D. The occupied states are squeezed right into the corner of the Brillouin zone at M, and no longer cross the Brillouin zone boundary at X. It is worthwhile considering what would happen if we took the opposite approach to the one we’ve looked at in the past few lectures. That is, rather than starting with a gas of electrons and gradually introducing a potential for them to move in, suppose we instead start with atoms and gradually allow them to interact. This is known as the tight-binding model, since we begin by supposing that electrons are tightly bound to the atoms from which they originate. The good news is that this gives qualitatively the same sort of dispersion curves as we’ve already seen. Just as in the case of the nearly-free electron gas, we have bands that electrons can occupy (often arising from one or a few atomic orbitals that overlap from each atom in the solid) with band gaps between them in which the density of states is zero. For this reason, the nearly-free electron gas is a good model of many systems even when the assumptions are not entirely justified – for instance when there is a substantial potential. Next week: we finish this course by applying these ideas to real semiconductors.

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