Electric and Magnetic Response of Multi-Wall Carbon Nanotubes

Carbon nanotubes are rolled graphene sheets and exhibit unique electronic properties because of their topological structure. In particular, theoretical studies have revealed interesting features of the response of nanotubes in electric and magnetic fields. The purpose of this paper is to study effects of such external fields on multi-wall nanotubes. The electronic states of a nanotube change from metallic to semiconducting depending on the tubular circumferential vector. The characteristic properties, first predicted in a tight-binding model, can also be described well in a k·p scheme or an effective-mass approximation. It has been shown, for example, that the depolarization effect plays a crucial role in an electric field perpendicular to the axis and drastically affects optical absorption spectrum.2−4) In a magnetic field perpendicular to the axis, nanotubes exhibit a large diamagnetism. This leads to a strong tendency of nanotubes to align in the direction of a magnetic field, which has been used for experimental observation7−11) of the Aharonov-Bohm effect on the band gap predicted theoretically.12−14) Response of a single-wall nanotube to a magnetic field was studied also in a tight-binding model and ab-initio calculation, and that to an electric field in a tight-binding model and in a k·p model. In this paper, we shall calculate the distribution of induced electric and magnetic fields of single-wall carbon nanotubes and use the results for the study of dielectric screening effect and NMR lineshape in multiwall nanotubes. It is organized as follows: In §2, a very short review is given on the k·p scheme for the electronic states and the dielectric function. Then, a set of linear equations for the effective electric field for multi-wall tubes is derived and some examples of explicit results are presented. In §3, the induced current due to an external magnetic field is calculated in single-wall nanotubes and used for determining the field distribution giving NMR lineshape. A short summary is given in §4.


§1. Introduction
Carbon nanotubes are rolled graphene sheets and exhibit unique electronic properties because of their topological structure.In particular, theoretical studies have revealed interesting features of the response of nanotubes in electric and magnetic fields.The purpose of this paper is to study effects of such external fields on multi-wall nanotubes.
The electronic states of a nanotube change from metallic to semiconducting depending on the tubular circumferential vector.−4) In a magnetic field perpendicular to the axis, nanotubes exhibit a large diamagnetism. 5,6)−14) Response of a single-wall nanotube to a magnetic field was studied also in a tight-binding model 15) and ab-initio calculation, 16) and that to an electric field in a tight-binding model 17) and in a k•p model. 18)n this paper, we shall calculate the distribution of induced electric and magnetic fields of single-wall carbon nanotubes and use the results for the study of dielectric screening effect and NMR lineshape in multiwall nanotubes.It is organized as follows: In §2, a very short review is given on the k•p scheme for the electronic states and the dielectric function.Then, a set of linear equations for the effective electric field for multi-wall tubes is derived and some examples of explicit results are presented.In §3, the induced current due to an external magnetic field is calculated in single-wall nanotubes and used for determining the field distribution giving NMR lineshape.A short summary is given in §4.§2.Electric Response

Effective-mass description
In a graphene sheet the conduction and valence bands consisting of π orbitals cross at K and K' points at the corners of the Brillouin zone, where the Fermi level is located. 19,20)Electronic states of the π bands near a K point are described by the k•p equation: 1,21,22) where γ is a band parameter, σ = (σ x , σ y ) is the Pauli spin matrix, and k = ( kx , ky ) = −i∇ is a wave-vector operator.
The structure of a nanotube is specified by a chiral vector L corresponding to the circumference.Electronic states of a nanotube with a sufficiently large diameter are obtained by imposing the boundary conditions around the circumference direction: 1) where ϕ = φ/φ 0 with φ being a magnetic flux passing through the cross section and φ 0 being the flux quantum given by φ 0 = ch/e, and ν is an integer (ν = 0 or ±1) determined by the structure.Metallic and semiconducting nanotubes correspond to ν = 0 and ±1, respectively.As shown in Fig. 1 (a), we shall choose the x axis in the circumference direction and the y axis in the axis direction.
The energy bands are specified by s (s = −1 and +1 for the valence and conduction band, respectively), integer n corresponding to the discrete wave vector along the circumference direction, and the wave vector k in the axis direction.The wave function for a band associated with the K point is written as where A is the length of the nanotube, L = |L|, and )

Submitted to Journal of Physical Society of Japan
The corresponding energy is given by (2.7) For the K' point, the k•p Hamiltonian is obtained by replacing ky with − ky in eq.(2.1) and the boundary conditions (2.2) by replacing ν with −ν.Therefore, the energy band is given by eq.(2.7) in which κ νϕ (n, k) is replaced with κ −νϕ (n, k), and the wave function is given by eq.(2.3) in which b νϕ (n, k) is replaced with b −νϕ (n, k) * . 1)The wave functions F ϕ,K and F ϕ,K with flux ϕ for the K and K' points, respectively, are related to each other through F ϕ,K = σ z F * −ϕ,K e iψ and F ϕ,K = σ z F * −ϕ,K e iψ by the time reversal, where σ z is the Pauli matrix and ψ is an arbitrary phase. 23,24)

Dielectric Function
In the presence of a uniform electric field E 0 in the direction perpendicular to the nanotube, the effective field within the cylindrical surface is written as and the corresponding potential is given by (2.9) It is possible to consider a perturbation with more general form, with q x = 2πl/L and q y = q with integer l, where c.c. represents a complex conjugate.Within the linear response, this potential induces polarization and the resulting electron density is given by δn(r) = δn(q)e iq•r +c.c., (2.11)   with δn(q) = −Π(q)Φ 0 (q), (2.12) where the polarization function Π(q) is the sum of the contributions of the states in the vicinity of the K and K' points.The contribution near the K point is given by 13) (2.13) with α ≡ (s α , n), β ≡ (s β , m), spin degeneracy g s = 2, and g 0 (ε) being the cutoff function given by .14)This function contains two parameters α c and ε c and should be chosen in such a way that only the contributions from states in the vicinity of the Fermi level, where the k•p scheme is valid, should be included.The results are independent of their values as long as the cutoff function decays smoothly.The similar expression can be obtained for the K' point.In the following, we shall exclusively consider the case of absolute zero of temperature where the Fermi level lies at ε = 0.−29) Therefore, the total potential for an electron becomes Φ(q) = Φ 0 (q) − V (q)Π(q)Φ 0 (q). (2.18) In a self-consistent approximation corresponding to a random-phase approximation, the effective potential Φ 0 in the second term in the right hand side of eq.(2.10) should be replaced by the total potential Φ(q).Further, we have to add the contribution of polarization of core states, σ bands, π bands away from the K and K' points, and the surrounding material if any.Denoting this contribution by a static dielectric constant κ, we have giving (2.21)

Flux Average and Graphene
In order to give a rough estimate of the dielectric function, we shall first consider an flux average, i.e., Π av (q) = 1 0 dϕ Π(q). (2.22) Obviously, this average converts the summation over n into a continuous integral over k x = κ νϕ (n).Therefore, the result is independent of whether the nanotube is metallic or semiconducting and becomes the same as that in graphene.Therefore, we have (2.23) with valley degeneracy g v = 2. 30−33) In graphene, we have Then, we have This dielectric function is independent of |q|.In usual semiconductors in three dimension with nonzero bandgap, the dielectric function due to inter-band transitions is usually given by a constant independent of the wave vector.The above result shows that the same is true even in two-dimensional graphene with vanishing gap.

Metallic and Semiconducting Nanotubes
In the following, we shall confine ourselves to the case of the uniform electric field perpendicular to the tube axis given by eq.(2.9), i.e., we set q x = 2π/L and q y = 0.For n> 0 and t 1, we have Therefore, V (q) in eq. ( 2.17) exactly agrees with eq.
(2.24) when q y = 0.This is not so trivial, but is the consequence of the fact that the Coulomb potential for two line charges is given by the logarithmic function, as is discussed in Appendix A.
It is straightforward to numerically calculate the polarization function in metallic and semiconducting nanotubes.The results are with q = 2π/L and dimensionless quantity This shows that the polarization function is slightly (less than 10 %) larger for semiconducting tubes than that for metallic tubes and consequently that the dielectric function ε(q) in nanotubes is essentially the same as that in graphene, irrespectively of metals or semiconductors.
As has been discussed previously, 1,2,3,34) there is a severe restriction on interband transitions due to electric field perpendicular to the axis.The momentum conservation gives the selection rule Δn = ±1.First, we shall consider a semiconducting nanotube with ν = +1.At a spectral edge k = 0, the wave functions are all an eigen function of σ x .The potential due to the applied electric field is given by a diagonal matrix in the pseudo-spin space and therefore has a nonzero matrix element between states with the same eigen value of σ x .Therefore, the matrix element becomes nonzero only between the valence band with n = 0 and the conduction band with n = +1 and between the valence band with n = +1 and the conduction band with n = 0.The corresponding band gap is 2πγ/L.These allowed transitions are illustrated in Fig. 2. The same is applicable to the case with ν = −1 except that n = +1 is replaced by n = −1.
In a metallic nanotube with ν = 0, on the other hand, the wave function at k = 0 is an eigen function of σ y for n = 0 and that of σ x for n = 0. Therefore, the matrix element becomes nonzero between n = 0 and n = ±1 with intensity half of that for n = 0 and n = ±1 in semiconducting nanotubes.The corresponding band gap is 2πγ/L, the same as in semiconducting nanotubes.As a result the polarization function is essentially independent of whether the nanotube is metallic or semiconducting.This near independence is qualitatively the same as that obtained in previous calculations. 17,18)

Field Distribution
In the following, we shall completely neglect the small difference between metallic and semiconducting nanotubes and use C ν ≈ 1 in order to consider the electric field distribution inside and outside of singleand multi-wall nanotubes.Further, we shall use κ ≈ 2.5 corresponding to the bulk graphite. 35)Then, we have (2.29) The induced charge becomes (2.30) In order to calculate the corresponding electric field, we shall choose the coordinates (X, Y ) and the corresponding polar coordinates (R, Θ) as shown in Fig. 1  (b).We set +Y to the electric field direction.The total electric field including the effect of the induced charge is calculated self-consistently as ) where E in and E out represent the field inside and outside the tube, respectively, R 0 = L/2π is the tube diameter, e Y is the unit vector in the Y direction, e R is that in the radial R direction, and e Θ is that in the Θ direction.This shows that the field is reduced by about factor six due to the screening by the carbon nanotube.The field distribution is exactly the same as that of a dielectric material with a cylindrical form.

Multi-Wall Nanotubes
The lattice structure of adjacent walls in usual multi-wall nanotubes is incommensurate.−38) It is possible, therefore, to discuss the dielectric screening of multi-wall nanotubes without consideration of virtual inter-wall charge transfer.
The field distribution inside and outside a nanotube given by eq.(2.31) shows that the effective field on the surface of coaxial cylinder is given by eq.(2.8) with E 0 replaced by an appropriate value.Consider a multi-wall tube with M walls.Let R j be the radius of the jth wall and E j be the effective electric field.Then, we have where the second term in the right hand side represents the field associated with tubes lying inside tube j and the third term that associated with tube j and those outside.This constitutes a set of linear equations uniquely determining E j for a given E 0 .
It is instructive to convert the above equation into a continuum model.Let E(R) be the effective in-plane field of a tube at distance R from the center and Δ be the inter-tube distance.Then, the continuum version of the equation becomes where R min = R 1 −(Δ/2) and R max = R M +(Δ/2).First, multiply the above equation by R 2 and take the derivative with respect to R. Next, divide it by R and take another derivative with respect to R.Then, we have the following differential equation with Then, the above is rewritten as which is the modified Bessel equation.The solution can be written as in terms of modified Bessel functions I 2 (t) and K 2 (t).The coefficients A 1 and A 2 are determined by the condition that this actually satisfy the integral equation.We have with t min = R min / Δ and t max = R max / Δ.The corresponding electric field becomes (2.41) where E mid represents the effective in-plane field in the multi-layer, i.e., R min < R < R max .Consider a sufficiently thick nanotube.Near the outer edge the term proportional to I 2 dominates the contribution.The asymptotic expansion I 2 (t) ∝ e t /t 5/2 for t 1 gives for R < ∼ R max .This shows that the effective penetration depth is given by . (2.43) The penetration depth is smaller than the wall separation Δ for multi-wall tubes with R M < ∼ 10×Δ and therefore usually less than a few walls except in extremely thick nanotubes.
Let us consider a dielectric material with a cylindrical form.Let its inner and outer radius be a and b, respectively, and its dielectric constant κ.When it is in a uniform external electric field E 0 in the Y direction, the field distribution is calculated as (2.44) with where E in denotes the field for R < a, E mid for a < R < b, and E out for R > b.
Let d = b − a and take the limit d → 0 and κ → ∞ with their product κd being a constant.Let us define Then, we have (2.47) Therefore, the response of a single-wall nanotube in a uniform electric field is exactly the same as that of the thin cylindrical dielectric material.However, the polarization of a carbon nanotube is determined by the field component parallel to the surface, while that in the dielectric material is induced by the field in every direction.As a result, the polarization in multi-wall nanotubes becomes completely different from that of a cylinder with a uniform dielectric constant.

Numerical Results
For numerical calculations we shall choose Δ = 0.34 nm and α E = 6 independent of whether tubes are metallic or semiconducting.Figure 3 shows the calculated effective electric field of nanotubes with inner diameter R min = 1 nm for layer number M =1, 2, 3, 4, 5, 10, an 20.The dots show E j and the dotted lines show E(R) obtained in the continuum approximation.Figure shows clearly that the electric field is screened out by a few outer layers.Figure 4 shows the results for thicker nanotubes with R min = 10 nm.The field penetrates into more layers than in the case of R min = 1 nm in agreement with eq.(2.43).However, the field inside multi-wall nanotubes is reduced considerably.We can conclude therefore that it is hardly possible to apply an electric field inside the tubes in thick multi-wall nanotubes.§3.Magnetic Response

Induced Current
In a uniform magnetic field B 0 perpendicular to the axis, the effective field for an electron in the nanotube is given by B 0 cos(2πx/L) and the corresponding vector potential is Within a linear response, the induced current is in the axis direction and has the same spacial dependence, i.e., It is straightforward to calculate j 0 using the lowest order perturbation with respect to B 0 .In the following, however, we shall use the known result for the magnetic susceptibility. 1,5)he change in the total energy due to the magnetic field is written as On the other hand, it is also written in terms of the susceptibility χ as The susceptibility is given by 1,5) with where k = kL/2π and the cutoff function g 1 (ε) is given by It has been shown that the dependence of the susceptibility on ϕ and ν is weak (at most less than 10 %) and therefore we can replace W 2 by its flux average (3.8) Then, we have The magnetic field due to the induced current is easily calculated as where B in and B out represent the field inside and outside the tube, respectively, and This field distribution is the same as that of a diamagnetic cylinder.Because α B 1, we can safely neglect this induced field in considering the magnetic response of the nanotube.

Magnetic-Field Distribution
With the use of the above result, we can determine the magnetic-field distribution on the cylinder surface and the NMR lineshape.−50) First, we should note that the field varies discontinuously on the cylinder surface due to the induced surface current.The field at a carbon nucleus is given by the average of that of the outside and inside of the cylinder surface.Therefore, the magnetic field becomes (3.13) To the lowest order in α B , the field strength is determined by B Y and given by The NMR intensity I(B)dB is proportional to the number of nuclei having the field strength lying B and B+dB, i.e., Consider a multi-wall nanotube consisting of M walls.The field of the jth layer in the Y direction is given by with (3.17) The corresponding NMR intensity becomes The corresponding result for the multi-wall tube becomes The innermost nanotube is affected most strongly by the field due to diamagnetic current in the direction opposite to the applied field and therefore the minimum field is given by It shows that the minimum field decreases in proportion to wall number M .The width of the field distribution is largest for the outermost nanotube.The maximum field is given by In sufficiently thick and nearly-filled multi-wall nanotubes, i.e., those with small R min , the maximum field is approximately given by B max ≈ B 0 [1+α B (M/3)] and therefore increases in proportion to M .In a rough approximation, the field distribution in each wall, I j (B), can be approximated by a constant continuing from its minimum to maximum, as illustrated in Fig. 6.Then, the field corresponding to the maximum of the distribution function is given by the minimum magnetic field of the outermost wall,

Numerical Results
(3.22) In sufficiently thick and nearly-filled multi-wall nanotubes, the peak field is approximately given by B max ≈ In thick but near-empty multi-wall nanotubes, on the other hand, this peak field is close to B min .This difference manifests itself in Figs. 5 (a) and (b) for M > ∼ 10. §4.

Summary and Conclusion
The electric field induced by a uniform external field perpendicular to the axis has been calculated within an effective-mass scheme and used for the evaluation of the field distribution in multi-wall nanotubes.The induced field is essentially independent of whether the nanotube is metallic or semiconducting and reduces the effective field down to ∼ 1/6 of the applied field in a single-wall nanotube.A single-wall nanotube in a uniform fields is nearly equivalent to a dielectric medium with a cylindrical form.In multi-wall nanotubes, this near equivalence becomes invalid and the field is almost completely screened out inside sufficiently thick multiwall nanotubes.The penetration depth determining a near exponential decay of the field is usually only 1∼2 walls but increases slowly with the increase of the thickness.
In a uniform magnetic field perpendicular to the axis, the induced field is also nearly independent of whether the nanotube is metallic or semiconducting but is much smaller than the applied field.A single-wall nanotube is therefore nearly equivalent to a diamagnetic medium with a cylindrical form.The resulting field distribution has been used for the evaluation of the NMR lineshape in multi-wall nanotubes.The distribution becomes wider with width nearly proportional to the wall number.NMR experiments in nanotubes aligned in directions perpendicular to magnetic field are highly expected.

Figure Captions
Fig. 1 (a) The coordinate system in the nanotube with an Aharonov-Bohm magnetic flux φ.(b) The coordinate system used for denoting the electric and magnetic field distribution.
Fig. 2 A schematic illustration of allowed transitions in the vicinity of the band edges for an electric field perpendicular to the axis.Others are all forbidden. )

Figure 5 (
Figure 5 (a) shows some examples of I(B) of nanotubes with R min = 1 nm with walls M = 1, 2, 5, 10, 20, and 50.We have introduced a Gaussian broadening ∝ exp(−B 2 /Γ 2 B ) with Γ B /α B B 0 = 0.1.Figure 5 (b) shows the corresponding results for thicker tubes with R min = 10 nm.The field distribution becomes broader with the increase of the wall number M as has been discussed in the previous section.In a rough approximation, the field distribution in each wall, I j (B), can be approximated by a constant continuing from its minimum to maximum, as illustrated in Fig.6.Then, the field corresponding to the maximum of the distribution function is given by the minimum magnetic field of the outermost wall,

Fig. 3
Fig. 3 Calculated electric field around multi-wall nanotubes with R min = 1 nm and wall numbers M = 1, 2, 3, 4, 5, 10, and 20.The dots represent the field obtained by solving the set of self-consistent equations for the field of each layer and the dotted lines the field obtained in the continuum model.(a) The field in the wide range.(b) The field in the vicinity of the nanotubes.

Fig. 4
Fig. 4 Calculated electric field around multi-wall nanotubes with R min = 10 nm.

Fig. 6 A
Fig.6 A schematic illustration of the field distribution I j (B) in each layer and the total distribution giving the NMR lineshape.When I j (B) is approximated by a rectangular form, the total distribution becomes close to a triangular form.

Fig. 7 Fig. 6 Fig. 7
Fig.7 The potential at point P due to charges on the cylinder surface ρe ilθ homogeneous in the axis direction is calculated.