Nearly-Free Electrons Model

Transcription

1 Nearly-Free Electrons Model Jacob Shapiro March 30, 2013 Abstract The following is my attempt to explain to myself what is going on in the discussion of the Nearly-Free Electron Model which can be found in Ashcroft & Mermin s SSP book. 1 Introduction We begin with some mathematical reformulation of the various functions in the system. The potential, V (r) has the periodicity of the lattice. That means that it will be equal with any arbitrary translation by an arbitrary Bravais lattice vector R. Explicitly, this means V (r + R) =V (r). Thus we may use afourierexpansionofv (r), which will contain only reciprocal lattice vectors: V (r) = P Ṽ () ei r, where is a reciprocal lattice vector, and 1 Ṽ () = volume of unity cell unity cell d3 rv (r)e i r. Observe that Ṽ (0) is the constant part of the potential (the part that does not depend on the location r) andsocanberegardedasanarbitraryclibrationoftheenergyscale.sofor convenience we can pick our energy scale so that Ṽ (0) = 0. In addition, since the potential is real, Ṽ () 1 = volume of unity cell unity cell d3 rv (r)e i r = Ṽ ( ). If, by suitable choice of the origin, the potential also has inversion symmetry (V ( r) =V (r)) thenitisclearthatṽ () = Ṽ (). The wave-function will also undergo similar treatment, though, not precisely the same it is not lattice-periodic. According to Bloch s theorem, the wavefunction in a lattice-periodic potential (such as the one defined above) can be written as k (r) =e ik r u k (r), where u k (r) are functions that also have the periodicity of the lattice, and k is any vector inside the first Brillouin zone (in fact k+ (r) = k (r), where is a recirpocal lattice vector). So we may expand u k (r) as a Fourier series, just like the potential: u k (r) = P ũk() e i r. Thus our Bloch-wave-function is: k (r) =e ik r u k (r) =e ik r P P ũk() e i r = ũk() e i(k+) r. Now our problem is not to find k (r) and E k,butrather,ũ k () and E k,for any arbitrary vector k in the first Brillouin zone and any arbitrary reciprocal lattice vector. 1

2 2 Schroedinger s Equation Schroedinger s equation is given by: ~ 2 2m r2 k(r)+v (r) k (r) =E k k (r) Putting in our expansions into the equation we get: ~ 2 2m r2 ũ k () e i(k+) r + 0 Ṽ ( 0 ) e i0 r ũ k () e i(k+) r = E k ũ k () e i(k+) r Applying the derivative and factoring by e ik r : ~ 2 2m (k+) 2 ũ k () e i r + 0 Ṽ ( 0 ) e i0 r ũ k () e i r = E k ũ k () e i r To take care of the product V (r) k (r) term we employ the following: 0 Ṽ ( 0 ) e i0 r ũ k () e i r = 0, Ṽ ( 0 ) ũ k () e i(+0 ) r Change of index: 00 := + 0,andsummingover 00 instead of 0,we get: Ṽ ( 00 ) ũ k () e i00 r 00, Change the names of the indices, to the following: 00!,! 0 and arrive at: " # Ṽ ( 0 ) ũ k ( 0 ) e i r = Ṽ ( 0 ) ũ k ( 0 ), 0 0 Going back to the Schroedinger equation we get: e i r ( apple ~ e i r 2 2m (k + )2 E k ũ k ()+ ) Ṽ ( 0 ) ũ k ( 0 ) =0 0 Since this equation is true for any arbitrary r, it follows that the expression inside the curly brackets has to be zero for any : 2

3 apple ~ 2 2m (k + )2 E k ũ k ()+ 0 Ṽ ( 0 ) ũ k ( 0 )=0 (1) Finally, we arrived at a set of equations (as many as there are s infinitely many) for each value of k that shall determine the co-efficients ũ k (). Note that we performed all the treatment where k was a fixed parameter for the Schroedinger equation. The infinitely many solutions for each value of k are labeled with the band-index n. Also note that since according to Bloch s theorem the wave-function has periodicity of reciprocal lattice vectors, we only have to describe the solutions where k lies in the first Brillouin zone. 3 Extreme Cases Zero periodic potential It can be shown (see Ashcroft & Mermin, Problem 8.1) using Bloch s theorem that the relation between the energy in the lattice and crystal momentum obeys the following relation, for any aribtrary shape of periodic 1-D potential: cos(ka) = cos( a + ) t where is related to the energy by E = ~2 2 2m, whereas if the arbitrary potential barrier s transmission co-efficient can be written as t = t e i,then t and assume those meanings then. This relationship assumes only one thing about the potential, namely, its periodicity. Any other information about the shape of the potential is embodied in the two parameters t and. At any rate, if we assume that the only effect of the potential is its periodicity, but otherwise, it s not just weak, but completely zero, we get a special situation which is similar to the free-electron, but is also periodic. In such a case we assume that t =1and =0and so our equation simply becomes: which has the solution: cos(ka) =cos( a) a = ±ka +2 n, n 2 Z Which means that we get the following relationship between the energy and momentum: E n (k) = ~2 2m (k + 2 a n)2 = ~2 2m (k + n) 2 This is quite an interesting result: we get the same free-electrion dispersion relation which we would expect, but there are additional indexed soltuions in periodicity of reciprocal lattice vectors. In other words, these are our bands (although, at the moment, there are no band gaps, you llnoticethatatthe edges of the Brillouin zones as well as in its middle we have the bands meeting ). 3

4 When turning on the potential, we would have to then apply degenerate (or nearly degenerate) perturbation theory at these places where, for a given value of k, several(in1d,onlytwo)valuesofn give the same energy. The potential will lift the degeneracy at those places. Generalizing our result to 3 dimensions we would expect to arrive at: E n (k) = ~2 2m (k + n) 2 Using this information, we go back to our momentum-space Schroedinger equation, setting the potential to exactly zero: apple ~ 2 2m (k + )2 E k ũ k () =0 (2) In this set of equations (for each k there will be as many equations as there are infinitely many), we are looking for ũ k () and E k. This set of equations will have an infinite number of solutions, which we shall conveniently index by (this is our band index, n). ~ For a given value of k, andagivenvalueof n,if 2 2m (k + )2 is special (meaning no other value of will give that same energy value), then only one of the co-efficients of ũ k () (namely, ũ k ( n ))canbefreelydetermined,while the rest of the co-efficients will be zero. Thus we arrive at a solution of the form n,k (r) =ũ k ( n ) e i(k+n) r and E n (k) = ~2 2m (k+ n) 2, where all the coefficients ũ k () =0for all 6= n. If, however, there is a set of N such values of that obey ~2 2m (k + 1) 2 = ~2 2m (k + 2) 2 =... = ~2 2m (k + N) 2,then the coefficients of the items in this set are again freely determined: k (r) = ũ k ( 1 ) e i(k+1) r +ũ k ( 2 ) e i(k+2) r ũ k ( N ) e i(k+n) r ;allother co-efficients are zero: ũ k () =086= 1, 2,..., N. As a concrete example, consider the one-dimensional lattice. In this setup, there are three special points in the Brillouin zone: k =0, ± a.anykvalue not equal to any of these three points will result in a set of equations which have only one value of equal to one particular energy. The wave-function will thus be only one plane-wave. However, for k =0, ± a,thereare,foronevalueof energy, two values of that give rise to said energy value, and as a result the wave-function will be a linear combination of two plane-waves (one travelling left and the other right): 4

5 For more complicated lattices, like 2D, and 3D setups, there will be more than two values of at the edge of the Brillouin zone resulting in the same energy. In the F.C.C., for example, at the W-point at the edge of the 1st Brillouin zone, there will be four different values of which will result in the same energy:. This is, thus, the starting point for our treatment of a weak periodic potential, which will have to employ nearly-degenerate-perturbation theory. 5

6 4 Weak Perturbing Potential Now V (r) is weak but not zero, and we can analyze equation (1) to see what happens to it. We shall distinguish between clearly non-degenerate cases and nearly-degenerate cases. 4.1 Non-Degenerate Case For any given value of k, theremightbebandsthatareveryneartoeachother, or bands that are very far apart. For example, for the 1D lattice, at k = 0, the lowest band ( =0) is far apart from all other bands, or at k = ± 1 2 a,for any value of, thebandsarequitefarapart.wewouldliketostudyhowthe this type of bands (at those positions in k-space) are affected by the perturbing potential. Let us mark that one band with the index 1. For convenience we ll define q := ~2 2m q2. In the unperturbed case, we found that ũ k ( 1 )=1and ũ k () =086= 1. We thus assume that the potential is so weak that its effect on the system is such that we could say that now, ũ k () apple Ṽk 8 6= 1, where Ṽk is a typical Fourier-series co-efficient. In addition we assume that the energy levels are spaced such that k+ E 1 (k) Ṽ k 8 6= 1. Equation (1) for the 1 band becomes: ( k+1 + E 1 (k)) ũ k ( 1 )= 0 Ṽ ( 1 0 ) ũ k ( 0 ) (3) Observe that since we defined our energy scale such that Ṽ (0) = 0, thesum on the right hand side of the equation is over such 0 values such that 0 6= 1 : ( k+1 + E 1 (k)) ũ k ( 1 )= Ṽ ( 1 0 ) ũ k ( 0 ) That same equation for any other band ( 6= 1 )wouldbe: ( ) ũ k () = 0 Ṽ ( 0 ) ũ k ( 0 )=Ṽ ( 1) ũ k ( 1 )+ Put differently: ũ k () =Ṽ ( 1) ũ k ( 1 ) + Ṽ ( 0 ) ũ k ( 0 ) If we put this result for the other bands coefficients in equation (3) back: ( k+1 + E 1 (k)) ũ k ( 1 )= Ṽ ( 1 0 ) Ṽ ( 0 ) ũ k ( 0 ) = Ṽ ( 1) ũ k ( 1 ) +O( (V k ) 2 ) "Ṽ ( 1 ) ũ k ( 1 ) # + O( (V k ) 2 ) = 6

7 = Ṽ ( 1 0 )Ṽ ( 1 0 ) ũ k ( 1 ) + O( (V k ) 3 ) ũ k ( 1 ) cancels out and we get: E 1 (k) k+1 = Ṽ ( 1 0 ) 2 + O( (V k ) 3 ) E 1 (k) k+ Since E 1 (k) k+1 is of the order of O((Ṽk) 2 ),thenobviouslyifwewantto solve for E 1 (k) up to that order, we can substitute the E 1 (k) in the denominator of the right hand side of the equation with k+1 to arrive at the result of: E 1 (k) = k+1 + Ṽ ( 1 0 ) 2 +O( (V k ) 3 ) 0 k+1 = ~2 k+ 0 2m (k+ 1) 2 + 6= 1 2m Ṽ ( 1 0 ) 2 ~ k( 1 0 ) This result is important mainly in that it shows us that the lowest order corrections to these non-degenerate states is second order! 4.2 Nearly-Degenerate Case (Handwritten: see in solution to exercises from TAU) 7