Lesson 1: Molecular electric conduction. I. The density of states in a 1D quantum wire

Transcription

1 Lesson 1: Molecular electric conduction I. The density of states in a 1D quantum wire Consider a quantum system with discrete energy levels E n, n = 1,2, Take an energy (not necessarily an energy level) E and ask: how many states N(E) have an energy less than E. One can use the Heaviside function to answer this (see box on right), and then: N(E) = θ(e E n ) (1.1) n The derivative of the accumulative number of states is the density of states (DOS) at energy E: ρ(e) = N (E). Clearly: ρ(e) = δ(e E n ) (1.2) Since 0 n δ(x)dx = 1 the DOS has the property that for any smooth function f(e) of the energy: The Heaviside θ(x) and Dirac δ(x) functions. These have the property that for any f(x): f(x)θ(y x)dx y = f(x)dx f(x)δ(y x)dx = f(y) The first integral shows that the definition of the Heaviside function is: θ(x) = 1 if x > 0 θ(x) = 0 if x 0 The second integral shows that δ(x) is zero for all x 0 but, putting f(x) = 1: δ(y x)dx = 1 Taking the derivative of the 1 st integral wrt y of shows the relation between the two functions: θ (x) = δ(x) f(e)ρ(e)de = f(e n ) n (1.3) Thus we can replace a summation over the states by integration: n ρ(e)de. We never really use delta-functions when we speak of DOS. Almost always there are decay mechanisms that widen these delta spikes. Thus the delta functions are usually replaced by lorentzian functions or gaussians.

2 Let us compute the DOS of an important basic system: a particle on a ring of circumference L. As with any free particle the energy is E m = p m 2 where m = 0, ±1, ±2, and p 2μ m is the linear momentum of the particle corresponding to an angular momentum ħm. For a particle on a ring of radius a the angular momentum r p is simply ap. Thus p m = ħm L/2π = m h L. We see that the linear momentum p m = m Δp is equally spaced where the spacing is Δp = h. Compute N(E). We find the momentum corresponding to E: m p(e) 2 = E p(e) = ± 2μE 2μ And thus the number of states with energy less than E is the number of states m with momentum p m in the interval p(e) < p m < p(e). So: N(E) = 2 m(e) 1 where m(e) = [ p(e) ] and we use the notation that [x] is the largest integer smaller than x. If we Δp think of a long wire, we can imagine that the integer aspect is smoothed out, so we can write: N(E) = 2L p(e) h The density of states is just the derivative of this. From p 2 = 2μE, and taking derivative w.r.t. E we have pp = μ so that: ρ(e) = N (E). Thus, in terms of momentum: ρ(e) = 2 μl hp(e) (1.4) Note that the density of states is proportional to the length of the wire and inverse proptional to the square root of the energy. II. Theory of electric conduction through molecules Consider an electron on the ring discussed above and ask: for a given state m what is the electric current? Since the velocity is v m = p m, the time the electron completes a rotation μ

3 is τ m = L and so I v m = e = ev m m τ m L functions of E = ep m. For a long wire we can write these quantities as μl I(E) = e μl p(e) (2.1) Now, let us try to connect the current and the density of states ρ. Using (1.4), we obtain a relation between the current going left (say) and the density of states of the wire : I(E)ρ(E) = e h (2.2) Note we lost a factor 2 because for each energy we consider only the left going states. This relation is remarkable: when an electron is transported through an energy level E of a device, the product of the current by the density of states is independent of E, of the electron mass or the wire length! In fact the result is a constant of nature e and is a h purely quantum effect (due to the presence of h)! What does this inverse proportionality of the current to the density of states mean? It means that wen you calculate the current resulting from all energy levels in the interval ΔE ΔE = E 2 E 1 then the current is the simple result: I(E 2, E 1 ) = e E 2 h I(E)ρ(E)dE = e h (E 2 E 1 ) = e h ΔE E 1 The total current of electrons in an energy range ΔE is just e ΔE: indeed a remarkably h simple rule! This rule is now going to be used to compute the current in a molecular junction. Now, consider a wire connected to two electrodes. We focus on non-interacting electrons. This means that each electron is independent (except for Pauli principle) from each other electron. The electrodes inject electrons into the wire and after the electron passes through the wire it is absorbed by the electrodes. The total current in the wire is the difference between currents going from the left to the right and from the right to the left: I = I L R I R L (2.3)

4 To get the current from left to right we have to sum over all states with positive momentum: I L R = 2 R n,p>0 w n i L R (E n ) where n enumerates the states in the left lead and R i L R (E n ) is the electric current in the right lead as a result of an impingement of an electron of energy E n coming from the left lead. Note that such an electron coming from the R left can either pass to the right side and contribute to i L R or to be reflected backwards. Clearly there is a probability T(E n ) that the impinging electron passes to the right. Because the number of electrons impinging on the wire coming from left is: e R (E) = w(e)t L R (E) ( hρ L (E) ) i L R T L R (E) is a transmission coefficient. w(e) is the joint probability that the left lead actually has an electron in state n and that on the right the corresponding level of the same energy is vacant (because of the Pauli principle, if the level is occupied the electron cannot flow there since a level cannot be occupied by 2 electrons or more). The factor 2 comes from the two possible spin states. Clearly, because electrons are fermions the population of levels is determined from the Fermi Dirac distributions in the two electrodes, thus: 1 w(e) = f L (E)(1 f R (E)) (2.4) Where f j (E) = 1+e β(e μ ), j = L, R is the Fermi-Dirac distribution, μ j is the chemical j potential of the left or right electrode and (k B β) 1 is the temperature (k B is Boltzmann s constant). All these give: R I L R = 2 f L (E n )(1 f R (E n ))i L R (E n ) n,p>0 (2.5) R = 2 f L (E)(1 f R (E))i L R (E)ρ L (E)dE = 2e h f L(E)(1 f R (E))T L R (E)dE T L R (E) is a transmission coefficient. It answers the question: what is the probability that an electron of energy E moving to the right in the left lead ends up in the right lead. Posed this way, such a question is ill-defined, but it is intuitively clearer.

5 A similar expression will be used for I R L. Thus I R L = 2e h f R(E)(1 f L (E))T R L (E)dE (2.6) We will show in the next section that T R L (E) = T L R (E) and we thus call both the transmission coefficient T(E). From Eq. (2.3) we then find: I = 2e h [f(e μ L) f(e μ R )]T(E)dE (2.7) This is Landauer s equation for the current in a junction. It is very simple. The electronic structure of the leads and the constriction go in only through the transmission coefficient. At zero temperature, with no barriers (T(E) = 1) we get the almost similar rule as we had above: I = 2e h (μ L μ R ) = 2e2 h ΔV (2.8) Where ΔV = e 1 (μ L μ R ) is the voltage bias between the two electrodes. The conductance is the derivative of the current by the bias. The conductance in the above case is thus G = di dδv = 2e2 h. III. 1D model of a molecular junction We now build a model for a molecular junction. In the junction there are 3 entities: the molecule, and the left/right metallic leads. Our model will assume that electrons are noninteracting. This may sound strange, since electrons interact quite strongly, however, it is well known that a good approximation for the behavior of electronic systems is that by modifying the overall potential suitably they can be approximately assumed to be noninteracting. In some phenomena interactions are extremely important, however, such phenomena are beyond the scope of our treatment. The model for the molecule is a well, for the present we will assume a square well, but the methods we develop below are suitable for any shape. So our molecule is schematically shown in this diagram:

6 E Vacuum level E 3 E 2 E 1 The molecule is a potential well with several discrete energy levels up to a vacuum level. The left lead has the following schematics: E Vacuum level μ L When a molecular junction is formed one has: E μ L E 3 E 2 μ R E 1

7 Due to the double barrier tunneling phenomenon, we will discuss in the next section, one can think of conduction as "going through" the energy levels of the molecule (this is a good assumption when the molecule is weakly coupled to the metal). At zero temperature one can see in the figure that conductance through molecular level 1 is impossible since the corresponding energy levels on both sides are occupied (the Fermi-Dirac difference terms at this energy give zero). In fact due to this, this level of the molecule will itself be occupied by 2 electrons. I say it is nearly impossible because one can still imagine a process in which one electron in level 1 of the molecule will jump up in energy to the lowest unoccupied state of the right lead (energy μ R ) and an electron from the left lead at the same energy will replace it. But as we shall see in calculations such events have very small probability since they are essentially tunneling events (violate the energy conservation for a short period of time). Conduction through level 3 is also nearly impossible for the same but opposite reason: there are no electrons of energy E 3 in the metals and thus conductance will occur only through tunneling which has a small probability. However conduction may occur more readily through level 2 since the left lead can supply electrons that go through the molecule and come out in the vacant levels of the right lead. IV. Calculating T(E): Transfer Matrix Method In order to estimate the conductance we need a way to compute the transmission coefficient in the Landauer formula T(E). We develop such a method, for 1D systems now. Consider a potential step, going from potential V L, V R at x = a. The wave functions with energy E on the left and right side of this point are: ψ L (r) = L + e ikr + L e ikr ψ L (r) = ik(l + e ikr L e ikr ) (3.1) ψ R (r) = R + e iqr + R e iqr ψ R (r) = iq(r + e iqr R e iqr ) (3.2) Where ħ2 k 2 = E V L, 2μ 2 q 2 2 E V. ψ L (r) is composed of two states with exact mo- R mentum k: one going to the right with amplitude L + and the other goinf to the left, with amplitude L. Similarly, the other wave functions.

8 Consider the wave function φ(r) = e ikr. It is certainly non-square integrable, so it is not a usual wave function. We can always imagine it as the limit of a very wide square integrable wave function, for example ψ k,σ (r) = e ikr e r2 2σ 2 / πσ which is normalized when is extremely large this function looks like a plane wave in the vicinity of the origin. All calculations can be done with finite and the limit can be taken after the proper observables are calculated. For example, suppose we want to compute the density of the particle at a point r 0. Then: n σ,k (r 0 ) = ψ(r 0 ) 2 = e r 2 0 σ 2 πσ (3.3) Keeping r 0 fixed and increasing σ indefinitely, we obtain: n σ,k (r 0 ) = 1 πσ (3.4) The fact that n becomes zero in the limit of σ is expected since we have one particle spread out all over space. Now we want to compute the current density j σ,k (r 0 ) ħ m Imψ (r 0 )ψ (r 0 ) as follows: j σ,k (r 0 ) = ħ m Im {e ikr o e r 2 0 σ 2 πσ eikr o e r 0 2 σ 2 (ik r 0 πσ σ 2)} = ħk m 2 e r 0 σ 2 πσ (3.5) Keeping r 0 fixed and increasing σ we obtain: j σ,k (r 0 ) = ħk m n σ,k(r 0 ) (3.6) While each quantity goes to zero when σ 0, the ratio of the current density to the density is ħk m which is independent of σ: it is a physical result. Let us go back to the wave function which on the left is ψ L and on the right is ψ R. What are the conditions we must demand the the fragments meet? The function must be continuous and so must the 1 st derivative. Thus: ( eika e ika ke ika ke ika) (L ) = ( eiqa L + qe iqa e iqa qe iqa) (R ) (3.7) R + Define:

9 M(a; k q) = ( eiqa qe iqa e iqa 1 qe iqa) ( eika e ika ke ika ke ika) (3.8) And so: M(a; k q) ( L L + ) = ( R R + ) (3.9) M(a; k q) is the transfer matrix. It determines the way the wave with wave-number k interacts with the interface at a, after which the wave must have wave-number q. Note that: Thus, M(a; k q) = 1 ( eiqa e iqa 1 qe iqa qe iqa) = 1 2q (qe iqa M(a; k q) = 1 2q e iqa 2q (qe iqa qe iqa e iqa) ( eika e ika ke ika ke e iqa qe iqa eiqa) (3.10) ika) and: + k)ei(k q)a (q k)e i(k+q)a ((q (q k)e i(k+q)a (q + k)e i(k q)a) (3.11) The transfer matrix can be used as in "Lego-land": You simply calculate the matrix at each junction and then multiply them. For example, a wave hitting two consecutive interfaces. One at a 1 and the other at a 2. The potential at left is V 0, after a 1 it is V 1 and after a 2 it is V 2. Then the TM of the entire "device" is: M = M(a 2 ; q 1 q 2 )M(a 1 ; q 0 q 1 ) (3.12) To calculate the probability to transfer from left to right, we impose the boundary conditions that we have an incoming wave from left (amplitude 1) and an outgoing wave from the right. There is no incoming wave from the right thus: R = 0. There may be a reflected wave, of amplitude L. Thus the scattering is described by: M ( 1 L ) = ( R + 0 ) (3.13) There are 2 equations with 2 unknowns. After rearrangement, we get: From the second equation the L_ amplitude is: ( M 11 + M 12 L M 21 + M 22 L ) = ( R + 0 ) (3.14)

10 L = M 21 M 22 (3.15) And the right amplitude is: R + = M 11 M 21M 12 M 22 = det M M 22 (3.16) We now consider the meaning of these amplitudes. Our physical problem consists of an electron coming in from the left with a wave e ik Lr and this causes two scattering waves: L e ik Lr on the left and R + e ik Rr on the right (k L and k R may not be equal if the left and right potential difference is non-zero). The current density on the left is the sum of currents of the incoming and reflected currents: J L = ħk L m ħk L m L 2 = ħk L m (1 L 2 ) (3.17) While on the right is just one wave: J R = ħk R m R + 2 (3.18) Now, the two current densities must be equal, J L = J R, since there is no accumulation of particles anywhere, thus k L = k R R k L L 2 or: 1 = k R k L R L 2 (3.19) Since L 2 is the probability to be reflected, this equation shows that the probability to be transmitted is given by: T(E) = 1 L 2 = k R k L R + 2 (3.20) In the appendix at the end of this document we show that the probability to be reflected L 2 for the R L direction is the same as for the L R direction and similarly for the transmittance: T(E) = k R k L R + L R 2 = k L k R R + R L 2 (3.21)

11 V. Numerical Integration We want to discuss the way to compute a controlled approximation to the definite integral of a function f(x): b I(f) = f(x)dx a (4.1) Formulas which give approximations that can be used for any function are called a quadrature rule. To be controlled, they must give an estimate for the error. A. The Rectangle Rule The simplest way to perform numerical integration I(f) is to divide the interval [a, b] to many (large N) equally spaced intervals, with spacing h = b a N : And then: a = x 0, x n+1 = x n + h, x N = b (4.2) N 1 I(f) f n h = (f 0 + f 1 + f N 1 )h n=0 (4.3) (we use: f n f(x n )). The error associated with this procedure can be estimated using Taylor expansion to first order: b I(f) = f(x)dx a N 1 x n+1 = f(x)dx = (f(x n ) + O(h))dx N 1 n=0 x n N 1 n=0 x n+1 = f n h + O(h 2 )N = f n h + O(h) n=0 N 1 n=0 x n (4.4) Thus, when N is increased 2 fold, making the use of (4.3) twice as expensive, the error is decreased by a factor of 2. This is a first order formula. B. The Trapeze Rule There is a simple way to improve. For x [x n, x n+1 ]. Instead of using the Taylor expansion to first f(x) = f n + O(h). Let us use it to second order:

12 where f n f (x n ), thus: f(x) = f n + f n (x x n ) + O(h 2 ) (4.5) x n+1 f(x)dx = f n h + 1 x n 2 f n h 2 + O(h 3 ) (4.6) We need a finite-difference expression for f n which is at least O(h). This is taken again from the same Taylor expansion: f n = f n+1 f n h + O(h) (4.7) Plugging this into Eq. (4.6) we obtain after rearrangement: x n+1 f(x)dx x n = f n + f n+1 h + O(h 3 ) 2 (4.8) The error in this formula can be shown to be bounded by h3 12 f where f is the largest value of f in the interval. Using this this in I(f), we arrive at the "Trapezoidal Rule": I(f) = [ f (f 1 + f f N 1 ) + f N 2 ] h + O(h2 ) (4.9) The error is opbviously bounded by Nh3 12 f = h2 (b a) f. Thus, at negligible cost (just 12 2 divisions by 2), we obtain a quadrature rule that gives a higher order integration: twice as much work reduces the error by a factor 4. This formula approximates the integral as a sum of trapeze pieces:

13 Figure 1: The Trapez Rule (left) where a line is stretched between consecutive points. Simpson's rule (right) where a parabola is cast between three consecutive points. C. Simpson's rule Next, we can use the Taylor theorem to the 4 th order around an odd indexed point: f(x) = f 2n+1 + f 2n+1 (x x 2n+1 ) f 2n+1(x x 2n+1 ) 2 (4.10) f 2n+1(x x 2n+1 ) 3 + O(h 4 ) Upon integration in the interval [x 2n, x 2n+2 ] the odd derivatives cancel so we have up to 5 th order (two orders better than the trapez rule!): x 2n+2 f(x)dx x 2n = f (2n+1) 2h f 2h 3 2n O(h5 ) (4.11) In order to continue, we must find an approximation to f 2n+1 which is good to order Using the same Taylor expression (4.10) twice, once sticking x 2n and once x 2n+2 : 2 h. f 2n = f 2n+1 f 2n+1 h f 2n+1h f 2n+1h 3 + O(h 4 ) (4.12) f 2n+2 = f 2n+1 + f 2n+1 h f 2n+1h f 2n+1h 3 + O(h 4 ) Adding these two expressions and rearranging, we have: f 2n+1 = f 2n 2f 2n+1 + f 2n+2 h 2 + O(h 2 ) (4.13)

14 Plugging into Eq. (4.11) and rearranging gives: x 2n+2 f(x)dx = (f 2n + 4f (2n+1) + f 2n+2 ) h x 2n 3 + O(h5 ) (4.14) In this case, the error term is bounded by (2h) f = h5 90 f. Summing over n, and assuming N is even (N = 2M) we have Simpson's rule: M 1 x 2n+2 h I(f) = f(x)dx = (f 2n + 4f (2n+1) + f 2n+2 ) 3 + O(h4 ) n=0 x 2n M 1 n=0 (4.15) = [(f 0 + 4f 1 + f 2 ) + (f 2 + 4f 3 + f 4 ) + ] h 3 + O(h4 ) Thus, Simpson's rule is: = [f 0 + 4f 1 + 2f 2 + 4f 3 + 2f 4 + ] h 3 + O(h4 ) M M 1 I(f) = [f f 2n f 2n + f 2M ] h 3 + O(h4 ) n=1 n=1 (4.16) With minute amount of additional work we obtain a much higher degree of approximation. Now, increasing work by a factor of two improves the accuracy by a factor 16. One can continue this procedure to higher orders, but one has to always make sure that the N th derivatives of the integrand do not grow faster than the order of the method. Since on computer we have finite arithmetic, numerical noise grows as you take higher and higher order derivatives. Going one step further in this analysis gives Boole s rule: M M M M 1 I(f) = [7f f 4n f 4n f 4n f 4n n=1 n=1 n=1 n=1 (4.17) + 7f 4M ] h 90 + O(h6 )

15 VI. Appendix: Notice also that for a particle coming in from the opposite side the same method will involve instead of M(a 1, k 1 k 2 )M(a 0, k 0 k 1 ) the following matrix product: M( a 0, k 1 k 0 )M( a 2, k 2 k 1 ) (just redraw the junction using a mirror image). Notice the following: M(a; k q) = 1 2q + k)ei(k q)a (q ((q (q k)e i(k+q)a k)e i(k+q)a (q + k)e i(k q)a) (4.18) M( a; q k) = 1 2k Note the special structure of the M matrices: When one multiplies two matrices one obtains: (q + k)ei(k q)a (q k)e i(k+q)a ( (q k)e i(k+q)a (q + k)e i(k q)a ) M(a, k q) = 1 2q ( A B B A ) (4.19) A B C D ( B A ) ( D C ) = ( AC + BD AD + BC B C + A D B D + A C ) (4.20) The product is also of the same structure. These then form a group. The mirror change moves from: We define a mirror of a matrix : Clearly (Q M ) M = Q. It is easy to check that: One further easily sees that A B A B ( B A ) ( B A ) (4.21) A B A B ( B A )M = ( B A ) (4.22) (PQ) M = Q M P M (4.23) det Q = det Q M (4.24) Q 11 = (Q M ) 11 Q 22 = (Q M ) 22 From this we have:

16 M(a, k q) = 1 2q ( A B B A ) (4.25) M( a, q k) = 1 2k ( A B B A )M = q M(a, k q)m k We thus have that the total TM of the mirror obeys: M R L = k R k L (M L R ) M (4.26) Thus one sees that the mirror barrier has: R + M = k R k L R + L M = M 12 M 22 = L (4.27) Thus: ( k L k R ) R + M 2 = ( k R k L ) R + 2 L M = M 12 M 22 = L (4.28)