We analyze the band structure of a silicon nanotube with sp^3 bonds and variable bond lengths. This nanotube has many similarities with a carbon nanotube including a band gap at half-filling and conducting behavior which is dependent on structure. We derive a simple formula which predicts when the nanotube is metallic. We discuss our results in the context of a nanotube subject to small applied strains as this provides a means of distorting bond lengths in a predictable way and may be tested experimentally. The affects of strain on nanotube conductance has important implications for sensor technology

Bunder, J E

Hill, James M

English

Research Online

University of Wollongong Research Online Faculty of Informatics - Papers (Archive) Faculty of Engineering and Information Sciences 1-1-2009 Electronic properties of silicon nanotubes with distinct bond lengths James M. Hill University of Wollongong, jhill@uow.edu.au J E. Bunder University of Wollongong, judyb@uow.edu.au Follow this and additional works at: https://ro.uow.edu.au/infopapers  Part of the Physical Sciences and Mathematics Commons Recommended Citation Hill, James M. and Bunder, J E.: Electronic properties of silicon nanotubes with distinct bond lengths 2009. https://ro.uow.edu.au/infopapers/767 Research Online is the open access institutional repository for the University of Wollongong. For further information contact the UOW Library: research-pubs@uow.edu.au Electronic properties of silicon nanotubes with distinct bond lengths Abstract We analyze the band structure of a silicon nanotube with sp^3 bonds and variable bond lengths. This nanotube has many similarities with a carbon nanotube including a band gap at half-filling and conducting behavior which is dependent on structure. We derive a simple formula which predicts when the nanotube is metallic. We discuss our results in the context of a nanotube subject to small applied strains as this provides a means of distorting bond lengths in a predictable way and may be tested experimentally. The affects of strain on nanotube conductance has important implications for sensor technology. Keywords lengths, electronic, properties, distinct, bond, silicon, nanotubes Disciplines Physical Sciences and Mathematics Publication Details Bunder, J. E. & Hill, J. M. (2009). Electronic properties of silicon nanotubes with distinct bond lengths. Physical Review B, 79 (3), 233401-1-233401-4. This journal article is available at Research Online: https://ro.uow.edu.au/infopapers/767 Electronic properties of silicon nanotubes with distinct bond lengthsJ. E. Bunder and James M. HillNanomechanics Group, School of Mathematics and Applied Statistics, University of Wollongong, Wollongong,New South Wales 2522, AustraliaReceived 6 April 2009; revised manuscript received 14 May 2009; published 3 June 2009We analyze the band structure of a silicon nanotube with sp3 bonds and variable bond lengths. This nanotubehas many similarities with a carbon nanotube including a band gap at half-filling and conducting behaviorwhich is dependent on structure. We derive a simple formula which predicts when the nanotube is metallic. Wediscuss our results in the context of a nanotube subject to small applied strains as this provides a means ofdistorting bond lengths in a predictable way and may be tested experimentally. The effects of strain onnanotube conductance have important implications for sensor technology.DOI: 10.1103/PhysRevB.79.233401 PACS numbers: 71.10.Fd, 71.10.PmThere exists a wide variety of nanotubes which can beconstructed from inorganic materials such as silicon,1 boronnitride,2–5 and tungsten disulfide.6 While the vast majority ofnanotube research concentrates on carbon nanotubes CNTs,there is some evidence that inorganic nanotubes may be bet-ter suited to certain tasks. For example, silicon nanotubesSNTs may be preferable to CNT as hydrogen storagedevices7 and boron nitride nanotubes have several potentialapplications in electronics because, unlike CNT, the bandgap is not structure dependent and may be tuned to a desiredwidth using an electric field.8 In this Brief Report we look atthe conducting properties of a SNT derived from simple ana-lytic band-structure calculations. Numerical studies have in-dicated that SNTs are possibly metallic11 and in most casesour calculations agree with this conclusion. In some cases,however, an energy gap is formed and this may lead to semi-conducting behavior.Typically, a CNT forms sp2 bonds which result in a hex-agonal lattice structure. In SNT sp2 bonds are possible butsp3 bonds are energetically favorable and so a square or atleast quadrilateral lattice is more likely than a hexagonallattice.9–15 Therefore, we consider a SNT constructed fromrolling up a two-dimensional quadrilateral lattice. The quad-rilateral lattice allows for some interesting band-gap behav-ior which is not dissimilar to what is observed in hexagonalCNT lattices. We find that in the general case the dispersionis gapped at zero energy, except at eight points, while a CNThas six gapless points. In the special case of a parallelogramlattice the band gap vanishes and the dispersion is continu-ous.We assume that each quadrilateral in the two-dimensionallattice is identical, although possibly rotated, as shown inFig. 1. The nearest-neighbor bonds are described by fourvectors 1,2,3,4 with the constraint 4=1−2+3. We de-note the hopping strength across l by tl. Clearly this latticeis very similar to an irregular hexagonal lattice which maydescribe a CNT but with an additional bond along 4, andtherefore we will use notation which is commonly used forCNT. The quadrilateral lattice can be described by two iden-tical irregular triangular sublattices offset by 1 and withlattice vectors c1=1−2 and c2=1−3. When rolling upthe two-dimensional lattice to form a nanotube, we define thex axis as being around the circumference C and the y axis asthe longitudinal direction. We define a vector C=nc1+mc2= n+m1−n2−m3 with integer n ,m0 for which the xcomponent defines the circumference C= n+ma1−na2−ma3 and the y component is constrained by n+mb1−nb2−mb3=0, where l= al ,bl.We will first discuss the special case of a SNT constructedfrom a parallelogram lattice. We calculate the dispersion re-lation and find that this nanotube is most likely metallic.Although this case is rather simple we present it for com-pleteness as it is distinctly different from the general case ofa quadrilateral lattice. On deriving the dispersion for the gen-eral case we find that a gap opens up about zero energy. Inmost cases the Fermi energy will lie outside this gap and theSNT will be metallic. If the SNT is doped to half-filling, i.e.,one electron per lattice site, the resultant particle-hole sym-metry ensures that the Fermi energy is at zero and so must liewithin the gap. In this case the SNT behaves much like anundoped half-filled CNT with conducting behavior directlyrelated to its lattice structure. The irregular lattice structureof a SNT can either form naturally or be due to some appliedstrain. As an applied strain will distort the lattice in a pre-dictable manner and because it can be experimentally veri-fied, we discuss the effect of a small external strain on theconductance of a SNT which is doped to half-filling. Weassume that the strain is either longitudinal or torsional andthat it is applied evenly across the lattice so that the latticecontinues to comprise equal quadrilaterals. We also assumethat the strain is not large enough to cause buckling. We findthat distorting the SNT can cause it to oscillate through sev-eral metal-insulator transitions, though the energy gaps of thesemiconducting phases may be very small.If the quadrilaterals in Fig. 1 are parallelograms then eachc1c2 σ1σ2σ3 σ4FIG. 1. Color online An irregular quadrilateral lattice with thetwo sublattices represented by circles and squares.PHYSICAL REVIEW B 79, 233401 20091098-0121/2009/7923/2334014 ©2009 The American Physical Society233401-1lattice site is identical and we do not have two distinct sub-lattices. Because of this the general quadrilateral case doesnot reduce to the parallelogram case and we must considerthem separately. For parallelograms only two vectors 1,2are required as 1=−3 and 3=−4. If we define c1=1−2 and c2=1−3=21, in analogy with the general case,then the vector C must be modified C=nc1+mc2 /2= n+m1−n2 so that the circumference of the rolled-up lat-tice is C= n+ma1−na2 and n+mb1−nb2=0 for integern ,m0. Following the terminology in Ref. 16, there are twospecial cases for nanotubes constructed from parallelogramlattices: the prismatic case is n=0 for which only one latticevector 1 is required to define the circumference and theantiprismatic case is m=0 for which both lattice vectors con-tribute equally to the circumference. All other cases can bedefined as chiral.As a first approximation we consider an interactionlessHubbard HamiltonianH = − l=12xytlc†x,ycx + al,y + bl + H.c. , 1where x ,y represents every lattice site and H.c. is the Her-mitian conjugate. The operators cr and c†r are annihila-tion and creation operators, respectively, at the site r= x ,y.On rewriting the operators in terms of their momentum-space Fourier transform,H = − kxky2t1 cosk · 1 + 2t2 cosk · 2c†kck= kxkykc†kck , 2where the momentum is k= kx ,ky and k is the dispersionin momentum space.When rolling up the two-dimensional lattice the x compo-nent of the momentum must be quantized by kx=2p /C forinteger p. Let us say the Fermi energy is F with correspond-ing Fermi momentum kF= kxF ,kyF. In a completely cleanSNT there will be no free electrons and the Fermi momen-tum will lie below the dispersion k so the lattice will besemiconducting. A small amount of doping17 will introducefree carriers into the lattice, shifting the Fermi momentum,and because the dispersion is continuous in both dimensionsit should be possible to find a Fermi momentum kF for agiven Fermi energy F for which kF=F and kxF=2p /C, i.e., the system is gapless and therefore metallic.In the general quadrilateral case the two-dimensional lat-tice comprises two sublattices and the dispersion relation isdistinctly different from the simple parallelogram latticecase. The interactionless Hubbard Hamiltonian isH = − l=14xytlc2†x,yc1x + al,y + bl + H.c. , 3where the subscript of cj defines the sublattice. In Fig. 1 j=1 is represented by circles and j=2 is represented bysquares. On writing the operators in momentum space,H = − l=14kx,kyAlkc2†kc1k + Alkc1†kc2k 4with Alk= tle−ik·l. The dispersion relation is =  lAl,k =  ltl cos k · l	2 + ltl sin k · l	2. 5This dispersion, unlike the parallelogam lattice dispersion, isnot continuous over its full range as a gap has opened upabout =0.Like the two-dimensional hexagonal lattice of an unrolledCNT, the energy gap does not appear across the entire =0range and there are several points k0=  kx0 ,ky0 where thedispersion on the quadrilateral lattice can be zero k0=0.In the hexagonal lattice there are six such points, but in thequadrilateral lattice we find eight. After some algebra it canbe shown thatkx0 =b1 − b4C12 − 2q12  b1 − b2C14 − 2q14b1 − b2a1 − a4 − b1 − b4a1 − a2,ky0 = −a1 − a4C12 − 2q12  a1 − a2C14 − 2q14b1 − b2a1 − a4 − b1 − b4a1 − a2,6with q12,q14=0 ,1 andC12 = cos−1 t12 + t22 − t32 − t422t3t4 − t1t2	 ,C14 = cos−1 t12 + t42 − t22 − t322t2t3 − t1t4	 . 7The solution in Eq. 6 takes the upper sign when t3t4− t1t2t2t3− t1t40 and the lower sign in the opposite case.For either of these solutions to exist, we require that C12 andC14 are both real. In a hexagonal lattice the six zeros k0 areknown as Dirac points because the dispersion can be writtenas k0+k=vk for small k and constant v. In general thezeros of the quadrilateral lattice are not Dirac points.On rolling up the quadrilateral lattice to form a nanotubekx becomes quantized, kx=2p /C for integer p. When F isfar from the energy gap the SNT will generally be metallicif free carriers are available as it is always possible to findkF which satisfies both kF=F and the quantization re-quirement. The situation is much more interesting if the SNTis doped18–21 to half-filling so that F=0. Now the SNT willonly be metallic if kx0=2p /C for integer p. On solving thiscondition for p while making use of the constraints n+mb1−nb2−mb3=0 and 4=1−2+3 we find that whenp = n + mC12  mC14/2 8is an integer the SNT is metallic if C12 and C14 are both real.The  sign follows the same rule as the  sign in Eq. 6.With Eq. 8 we have shown that the conducting propertiesof a half-filled SNT are only dependent on the hoppingstrengths and the n ,m structure, which is also the case in aBRIEF REPORTS PHYSICAL REVIEW B 79, 233401 2009233401-2CNT. One situation where a SNT is always metallic is whenn=0 prismatic and t2= t4, as this ensures C12=C14 with thelower sign and p=0.Provided the curvature is not large tll−2. Large curva-ture ruins this simple relationship through hybridization ef-fects. So, assuming the nanotube circumference is not smallwe can write Eq. 8 solely in terms of the bond lengths. Inmost circumstances the bond lengths of a SNT will be dif-ferent, but this difference will be very slight. One way inwhich to enhance these differences is to subject the SNT to asmall applied strain.The application of strain presents an ideal way in which tocompare theory with experiment as one can distort the bondsof the nanotube in a predictable way and a wide range ofdifferent lattice structures can be obtained. However, if webegin with an unstrained parallelogram lattice and subject itto longitudinal and torsional strains we will distort the latticebut it will remain a parallelogram with a gapless dispersionand no changes in electronic properties will be observed.Therefore, we illustrate how strain may effect the electronicproperties of a SNT by using the simplest nonparallelogramlattice, a prismatic n=0 SNT, which is constructed from atrapezoidal lattice, as shown in Fig. 2 with 1=1−	a ,0,2=−	a ,1+	b, 3=−1−	a ,0, and 4=−	a ,−1−	b for some constant length . The dimensionless con-stants 	a and 	b define the distortion in the lattice prior toany external strain. This initial lattice is such that 2=4 sot2= t4 and, as discussed previously, this SNT must be metal-lic. If we apply a small strain along the longitudinal axis they components of 2,4 will increase at the same rate but inthe opposite direction and 1,3 will remain constant. In thisscenario we still have t2= t4 so the SNT remains metallic,provided C12 and C14 continue to be real.If torsional strain is applied to the SNT the changes in1,2,3,4 can be represented by the dimensionless variables 	xand 	y. After the application of torsional strain 2=−	a−	x ,1+	b+	y and 4=−	a+	x ,−1−	b−	y while 1,4remain constant. We can relate 	x and 	y to measurable pa-rameters such as the change in length of the nanotube and thetorsional angle  which defines the rotation of one end of thetube relative to the other end. In general the length is L= nqn+mqmb2−b3 / n+m for integers qm ,qn1, so thechange in length can be written as L=L0	y / 1+	b, whereL0 is the initial length of the tube. Similarly, one can find thetorsion angle, = 2L0 /C	x / 1+	b.Figure 3 is a contour plot of integer p and shows that theapplication of torsion on a SNT at half-filling may cause it tooscillate through several metal-insulator transitions. Similarbehavior has been observed in CNT.22 The values of 	a0and 	b do not have a significant effect on the shape of thecontours of integer p as can be seen by comparing Fig. 3awith Fig. 3b which are essentially the same but shiftedrelative to one another. When 	a=0 the situation is differentas the lattice reduces to a square, and so this half-filled SNTis always metallic. σ1σ2σ3σ4FIG. 2. Color online A prismatic trapezoidal lattice.(a)0.5 0.25 0 0.25 0.51.00.750.50.250∆x∆b∆y0.5 0.25 0 0.25 0.51.00.750.50.250∆x∆b∆y(b)FIG. 3. The shaded area is where p is real and the lines arewhere p is an integer, i.e., where the SNT is metallic for the trap-ezoidal SNT with n=0, m=12 and a 	a=0.0001, b 	a=0.2.(a)0 0.4 0.801x104∆xenergygap(b)0 0.4 0.800.25∆xenergygapFIG. 4. The energy gap for the trapezoidal SNT with n=0, m=12, 	b=−0.25, 	y=0, and a 	a=0.0001, b 	a=0.2. The energygap is measured in units of the proportionality constant g which isdefined by tl=g /l2.BRIEF REPORTS PHYSICAL REVIEW B 79, 233401 2009233401-3In Fig. 4 we plot the energy gap for the two SNTs con-sidered in Fig. 3. For 	a=0.0001 the contour plot in Fig. 3aclearly shows several metal-insulator transitions, but Fig.4a shows that the energy gap is very small and will possi-bly not be distinguishable from the metallic 	a=0 case. As	a increases the energy gaps of the insulator phases tend toincrease. Therefore, to experimentally distinguish the insula-tor phases from the metallic we would require that the bondlengths prior to the application of strain are significantly dif-ferent. If this is possible we may be able to tune the band gapthrough the application of torsional strain. 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Electronic properties of silicon nanotubes with distinct bond lengths

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Electronic properties of silicon nanotubes with distinct bond lengths

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