11. Excitons in Nanowires and Nanotubes

Transcription

1 Excitons in Nanowires and Nanotubes We have seen that confinement in quantum wells leads to enhanced excitonic effects in the optical response of semiconductors The bindin enery of the stronest bound excitons increase by a factor of 4 in the ideal D case Consequently, one expects this trend to continue to D-structures with even stroner bindin of excitons As we will see, this is precisely what happens even to the point, where excitons completely dominate the response Followin the previous chapter, we will limit ourselves to variational calculations of excitons in D-structures These structures are assumed to be infinite alon the lon-axis direction and stronly confinin in the two transverse dimensions Also, the system is excited alon the lon-axis so that centre-of-mass momentum for this direction is to remain zero throuhout We take the z-axis as the lon-axis and so the confined electronhole pair is described by the Hamiltonian ˆ ˆ ˆ d Heh = he( xe, ye ) + hh( xh, yh ) + E x x + y y + z () ( e h) ( e h) Here, ˆ (, ) he xe ye = me ( d / dxe + d / dye ) + Ve ( xe, ye ) is the Hamiltonian for the transverse motion of the electron confined by the potential V e and h ˆh is the analoous term for the hole In this expression, the effective electron and hole masses should be taken in units of the reduced electron-hole pair mass, which is taken as the unit of mass In a purely variational treatment, we attempt to describe the exciton state by the variational ansatz Ψ exc( re, rh ) = ϕe( xe, ye ) ϕh( xh, yh ) Ψ( z) The correspondin expectation value for the enery is then d E= E + Ψ( z) + V( z) Ψ( z), () e h with E = E + E+ E and an effective potential ϕe( xe, ye) ϕh( xh, yh) V( z) = dxedyedxhdyh (3) ( x x ) + ( y y ) + z e h e h At this point, it is instructive to consider some specific examples of transverse confinement:

2 A Rectanular nanowire with transverse dimensions d d (Ref []): x y ϕ( xy, ) = sin sin d d d B Hexaonal nanowire with diameter d (Ref []): 8 x ( x 3 y) ( x + 3 y) ϕ( xy, ) cos cos cos /4 / 3 (6+ 5 ) d d d d C Circular nanotube with radius r (Ref [3]): r ϕ( x, y) = δ( x + y r) We note that in the nanotube model, the wave function is completely localized to the cylinder wall and rather than ivin the wave function itself, we ive the normalized square In cases A and B, the effective potential can only be computed numerically However, for the nanotube model an analytic result can be found To this end, we introduce polar transverse coordinates x = ρcosθ and y = ρsin θ Due to the complete localization on the cylinder wall, we always have ρ = r In this manner, ( x x ) + ( y y ) = 4r sin (( θ θ )/) Hence, in polar coordinates, e h e h e h δρ ( e r) δρ ( h r) e h + Vz ( ) = ρρdρdθdρdθ ( r) 4r sin (( θ θ )/) z dθ 4r sin ( θ /) + z (4) = 4 4r = K z z e h e e h h In the last line, K is a so-called complete elliptic interal of the kind defined by Kx ( ) / dθ x sin θ From this definition, it is seen that K() = / and the potential approaches the bare D Coulomb potential Vz ( ) = / z in the limit r From Eq(3) it is clear that this must always be the limit of a D effective potential whenever the confinin potential becomes sufficiently narrow that xe xh and ye yh because of the confinement

3 Fiure Effective Coulomb potential for three different D confinements compared to the bare potential The actual effective potentials for the three models listed above are illustrated in Fi When compared to the bare Coulomb potential, it is clear that the behaviour as z is must less sinular Hence, for models A and B, the sinularity is completely removed and for C, the sinularity is now loarithmic instead of / z However, in all cases the bare potential / z is found as a limit when the diameter of the nanowire or tube becomes very small It miht then be thouht that a viable and simple model for D excitons would result from usin the pure D potential Vz ( ) = / z in Eq() similarly to the D quantum well case To look at the properties of this simple model we need to start with a reularized potential, however For this purpose, we will take Vz ( ) = /( z + c) as our potential, where c is a positive constant, which should eventually be taken to zero This Loudon model was oriinally analyzed by R Loudon [4] The form is mathematically simpler than any of the alternatives above As our variational ansatz we will, as usual, try the exponential Ψ ( z) = αe α z Differentiatin twice leads to d Ψ ( z)/ = α Ψ( z) αψ () δ( z) Thus, the expectation value Eq() is z e α E= E + α 4α z+ c Evaluatin this interal leads to yet another complicated function: the exponential interal Ei : α 4 α ( α ) c E= E + + e α Ei c 3

4 However, as c should o to zero we can use an expansion based on partial interation valid for small c z e α E= E + α 4α αz z E α α 4α e ln( z c) 8α e ln( z c) z E α α 4αln( c) 8α e ln( z c) z E α α 4αln( c) 8α e ln( z) = E + α + 4 α( γ+ ln( αc)), z+ c = = (5) βz where the interal e ln( z) = ( γ + ln( β))/ β with γ = 577 as Euler s constant has been applied Differentiatin, one finds = α+ 4( γ+ ln( αc)) + 4 With a few manipulations this condition can be reformulated as α α exp = 4ce + γ The formal solution to the equation wexp( w) = x is sometimes called the product loarithm (pl), ie w= pl( x) Thus, the solution for α is = pl ln ln ln 4ce 4ce 4ce, α + γ + γ + γ where the second expression is the expansion for low c Accordinly, the enery is E= E 8pl 4pl + γ + γ 4ce 4ce As the plot of the product loarithm to the riht shows, it is a monotonically increasin function that diveres loarithmically as the arument increases The extremely important conclusion is this: as c oes to zero, we find α and E Hence, the wave function becomes completely localized to the point z = This is not an artefact of the variational approach because the variational estimate for the enery is always hiher than the true value We therefore conclude that the D Coulomb model is patholoical in that 4

5 the round state collapses and the round state enery diveres In Fi, we illustrate the behaviour of the enery as c becomes smaller Fiure Bindin enery and variational parameter (inset) in the Loudon model as a function of the potential cut-off Even thouh the pure D Coulomb model is clearly unphysical, it is still correct that the true potential for all realistic models approaches this strane situation as the confinement becomes stroner Hence, even if actual exciton bindin eneries obviously do not divere, they can still row extremely lare compared to bulk values As an example of this, we now consider the true nanotube potential iven by Eq(4) With the same exponential ansatz as above, we find the enery expectation value 3α r E= E + α + αj( αr) Y( αr) F 3,;,, ; 4α r, where J and Y are Bessel functions of first and second kind, respectively, and F 3 is a eneralized hypereometric function [5] The similarity with the Loudon model above becomes apparent if we aain use partial interation to approximate to lowest order in r Usin the definition of the elliptic interal, we have / αz 8α e E= E + α d z + 4r sin θ θ The indefinite interal of / z + x is ln( z z x ) + + and it follows that 5

6 / 8α αz αz E= E + α e ln( z z 4r sin θ) α e ln( z z 4r sin θ) dθ / 8α αz = E + α ln(rsin θ) α e ln( z z 4r sin θ) dθ / 6α αz E + α + 4αln( r) e ln( z) dθ = E + α + 4 α( γ+ ln( αr)) Comparin to Eq(5), we see that r takes the place of c We therefore expect to find precisely the same behavior as above when r oes to zero It should be noted that this similarity is obtained even thouh the nanotube potential is actually (loarithmically) diverent at the oriin whereas the Loudon model potential is finite The exciton bindin enery for the nanotube model is illustrated in Fi 3 Fiure 3 Bindin enery and variational parameter (inset) of excitons in carbon nanotubes as a function of nanotube radius, all in natural exciton units From Fi 3 it is apparent that the exciton bindin enery may become very lare if the nanotube radius is sufficiently small The question is then: what are the actual values of ab and Ry? We recall that a B = 59Å εm/ meh and Ry = 36eV meh / mε Hence, to answer this question we need the reduced effective electron-hole pair mass and a value of the dielectric constant The latter is relatively straiht-forward since most experiments are performed in liquid suspensions and a reasonable value describin the screenin in this case is ε = 35 [3] To compute m eh we need to consider the band structure For nanotubes excited alon the lon-axis the allowed transitions are between bands symmetrically 6

7 positioned above and below the Fermi level In a simple nearest-neihbor tiht-bindin model the transition enery is iven by ka y ka y 3ka x Ecv( k) = γ + 4cos + 4cos cos, (6) where γ 3 ev is the hoppin interal and a = 46 Å is the lattice constant The nanotube is characterized by the chiral indices ( nm, ) and in terms of these, the components of the k- vector are 3( n+ m) n m kx = k q L L n m 3( n+ m) ky = k + q L L Here, k is the continuous lon-axis component of the k-vector and q is the quantized shortaxis component iven by q= p/ r, where p is an inteer and r is the radius Also, L= n + m + nm is the radius in units of a/, ie r= al An important point about the enery dispersion Eq(6) is that Ecv( K ) = with K= ( 3,) /3a Thus, the band ap is found at the allowed k-point closest to K To simplify the analysis, we expand the dispersion in the vicinity of K and find Ecv( k) 3 aγ k K We introduce ˆl and ŝ as unit vectors for the lon-axis and short-axis, respectively If we express K and k in terms of the projections alon these directions we find k = klˆ + qsˆ and similarly K= Klˆ + Qsˆ, where it can be shown that Q= ( n+ m)/3r Hence, settin q = Q, leads to the condition p= ( n+ m)/3 This condition can only be fulfilled if ( n+ m)/3 is an inteer in which case the nanotube is a metal If not, the minimum difference becomes q Q min= /3r In this case, the minimum transition enery, ie the band ap E, then becomes a E γ = 3 aγ q Q min = 3r To find the effective mass, we consider an approximately parabolic dispersion Ecv( k) E + ( k K) m eh 7

8 so that, to lowest order, the square becomes Ecv( k) E + ( k K) E / meh On the other hand, the square of the enery dispersion for the nanotube near the minimum is E { cv( k) 3 a γ ( k K) + q Q min} = E + 3 a γ ( k K) Hence, a comparison demonstrates that meh = E /3a γ Consequently the effective mass is proportional to the band ap Pluin in the numbers it turns out that E = 4 evå / r and m / m= 9Å / r An extremely important point is then that the eh effective Bohr radius a B = 59Å εm/ meh becomes a linear function of r iven by a B = 97r As a consequence, the radius r measured in units of a B is always rouhly! At this value, the exciton bindin enery as computed above and illustrated in Fi 3 is around -744 Ry A more accurate calculation [3] finds a bindin enery of approximately -8 Ry It is noted that these values are substantially hiher than the maximum value - 4 Ry found for D structures In analoy with the effective Bohr radius, the effective Rydber also depends on r and insertin values we find Ry = evå / r It therefore follows that the ratio between exciton bindin enery and band ap is a near constant of around -4% This is obviously a hue value, which will completely rearrane the optical response As an example, we consider the (7,6) nanotube with a radius of r = 44 Å For this structure, the exciton bindin enery is then -39 ev Unfortunately, there is no simple way to sum all the contributions to the optical response analytically Instead, a numerical calculation of bound and unbound excitons can be made usin a finite basis set [6] Summin the different terms leads to the spectrum shown in Fi 4, where the independent-particle result is included for comparison The very lare red-shift of the resonance corresponds to the value of the exciton bindin enery Also, it is noticed that the peak is now much more symmetric than the inverse square-root of the independentparticle spectrum 8

9 Fiure 4 Normalized absorption spectra of a (7,6) carbon nanotube The curves show the spectra includin and nelectin excitonic effects, respectively Exercise: p nanotube excitons In this exercise, we will attempt to compute the enery of the p exciton in a nanotube and 3 z for this purpose the ansatz Ψ ( z) = β ze β will be used Note that it is always orthoonal to the round state a) show that Ψ ( z) is normalized and that the kinetic enery is The difficult part lies in determinin an approximate expression for the Coulomb enery valid for small but finite r It is iven by β 8 U= / Ψ ( z) dθ z + 4r sin θ As a start, we will consider the definite interal W = f( z) z + x The first few indefinite interals of the square-root are denoted S ( z, x ) so that n 9

10 S( z, x) z + x S ( z, x) S ( z, x) = ln z+ z + x ( ) ( ) ( ) S ( z, x) S ( z, x) = zln z+ z + x z + x 3 S z x S z x z x z z x z z x 4 4 3(, ) (, ) = ( )ln b) Under the assumption that f ( ) =, show by repeated use of partial interation that W = f()ln x f ( z) S ( z, x) = f()ln x f () x+ f ( z) S ( z, x) ()ln () () ln ( ) 3(, ) = f x f x+ f x x f z S z x 4 At this stae, no approximations have been made However, to actually calculate the interal, we will now use the small-x expansion S3( z, x) ( z x )ln( z) (3 z + x ) 4 4 In the present example, f is the function f( z) ( z) β z e 3 βz =Ψ = and so f z = β z βz e f z = β βz+ β z e f z = β βz+ β z e 3 βz 3 βz 4 βz ( ) 4 ( ), ( ) 4 ( 4 ), ( ) 8 (3 6 ) With this form and the approximate S (, ) 3 z x we find 3 β β x f ( z) S3( z, x) { + γ+ lnβ} 3 β β x In turn, W becomes W + { + γ+ lnβx} and we then have / 8 β 3 U= + r sin + + ln( rsin ) d { r } = β β + γ+ β 3 8 r ln( ) { } β θ γ β θ θ

11 p 8 ln( ) 3 The total enery is therefore E = E + β β β r { + γ+ βr } c) Show that the minimum enery is approximately E p E 8r { + γ + lnr} This result is plotted below References Fiure 5 Variational p exciton bindin enery [] TG Pedersen, PM Johansen and H C Pedersen, Phys Rev B6, 54 () [] TG Pedersen, Phys Stat Sol (c), 46 (5) [3] TG Pedersen, Phys Rev B67, 734 (3) [4] R Loudon, Am J Phys 7, 649 (959) [5] IS Gradsteyn and IM Ryzhik Table of Interals, Series and Products (Academic Press, San Dieo, 994) [6] TG Pedersen, Carbon 4, 7 (4)