Homework 2 - Solutions

Transcription

1 Homework 2 - Solutions Carbon Nanotubes Part 4) We can write the wavefunction of the grahene sheet within the tight-binding model in the usual way ( ~ R)= e i~ k ~R Then, to imose eriodic boundary condition we want to ensure that our wavefunction satisfies the constraint ( ~ R + ~ W )= ( ~ R) (2) That is, the winding vector ~ W is a vector which wras comletely around the eriodic axis of the carbon nanotube so that when we move from a site ~ R by the vector ~ W, we end u at the same site ~ R.Sothen,we write (1) ( ~ R + ~ W ) ( ~ R) () ei ~ k ( ~ R+ ~ W ) = e i~ k ~R ) ~ k ( ~ R + ~ W )= ~ k ~R +2 ` ` 2 Z (5) ) ~ k ~W =2 ` ` 2 Z (6) Then, enforcing the eriodic boundary conditions of the carbon nanotube means the our recirocal lattice vectors ~ k must satisfy the condition ~ k ~W =2 `. (4) Part 5) a) The 2d honeycomb lattice of grahene has the same lattice structure as the hexagonal lattice, but with a two atom basis. The lattice is biartite, so it can be slit into A and B sublattices, with each sublattice forming a hexagonal lattice. The three unit vectors which oint between the two sublattices are given by: ~e 1 = a( 2, 1 2 ), ~e 2 = a(0, 1) and ~e = a( 2, 1 2 ) (7) Then, the rimitive basis vectors are given by the two vectors ~a 1 and ~a 2 ~a 1 = ~e 1 ~e = a(1, 0) (8) ~a 2 = ~2 ~e 2 = 1 a( 2, 2 ) () Then, we can find our recirocal sace basis vectors by finding the vectors which satisfy the relationshi These vectors are ~a i ~b j =2 i,j (10) ~ b1 = 2 a ( 1, 1 ) (11) ~ b2 = 2 a (0, 2 ) (12) 1

2 Using these vectors we can draw the 1st Brillouin zone by bisecting the vectors {± ~ b 1, ± ~ b 2, ±( ~ b 1 ~ b2 } Now, in class it was seen that the two sublattice structure of grahene leads to an energy disersion of the tye X " ± (k) =" 0 ± f(k) with f(k) = e i~ k e i (1) Using our values of e 1,e 2 and e above, setting the lattice constant a = 1, gives f(k) = [e iky + e i( 2 kx+ 1 2 ky) + e i( 2 kx+ 1 2 ky) (14) = [e ıky + 2 cos( 2 k x)e i 1 2 ky ] (15) = e i ky 2 ale e 2 ıky + 2 cos( i=1 2 k x) (16) Clearly, the energy given above is slit into two bands, one each for the lus and minus values of f(k). Now, the grahene lattice has two atoms er rimitive unit cell, since the honeycomb lattice contains a two lattice basis. This tells us that the first energy band is comletely filled while the second band is emty. Then, the Fermi energy is the energy which slits the two bands, which occurs when f(k) = 0. To find when f(k) = 0, we must set the real and imaginary arts of the r.h.s. of Eq. (16) to zero and solve. This is done in the lecture notes and the result is the two equations sin( 2 k y)=0 ) k y =0, ± 2 (17) subbing in our solutions for k y gives cos( 2 k y) + 2 cos( 2 k x)=0 (18) ±1 + 2 cos( 2 k x)=0 ) k x = ± 4, ± 2 (1) Therefore, this gives six oints where f(k) = 0 so that "(k) =" F (k x,k y )=(± 4 2, 0), (±, ±2 ) (20) To verify that these oints lie on the corners of the first Brillouin zone, note that the corners of the 1st B.Z., are at the oints {± 2 (2~ b 1 ~ b2 ), ± 2 (~ b 1 + ~ b 2 ), ± 2 (2~ b 2 ~ b1 )}. That is, the corners of the B.Z. lie two thirds the way to the second nearest neighbor lattice oints. Plugging in the values of ~ b 1 and ~ b 2 shows that the six Fermi oints given above coincide with the corners of the 1st B.Z. Note: this solution is written for the *first* version of the homework. The solution for the corrected homework (with definition of m,n changed) is easier. It is left b) as an exercise. The lines given in art 4 are the values of ~ k which satisfy the equation ~ k ~W =2 ` for any integer `. For the winding vector (m, n) =(, ), we have ~ W =6~a 1 +~a 2 = ( 2, 1 2 )Then ~ k W =2 ` ) kx + k y = 4 ` ) k y = k x + 4 ` We can lot these lines within the first B.Z. (for all values of ` such that the lines fall within the B.Z.) and find that the allowed k-oints fall along the lines in the figure below: (21) (22) 2

3 For the winding vector (m, n) =( 2, 2), we have ~ W =2~a 2 =(, ) Then ~ k W =2 ` ) kx +k y =2 ` (2) ) k x = 2 ` k y (24) ) k y = k x + 2 ` (25) Again, lotting these lines for di erent ` within the first B.Z. looks like: Part 6) For the (m, n) =(, ) tube, the allowed k y wavevectors for a given k x must obey the restriction k y = kx +4 `/. These lines fall within the first B.Z. for ` 2{0, ±1, ±2, ±}. The energy is found by taking

4 the absolute value of Eq. (16) to get s "(k) =± f(k) = cos( k x /2) cos( k x /2) + cos( k y /2) (26) Subbing in our equation for k y and lotting in Mathematica gives The two disersion bands at individual oints. Therefore, the (,) nanotube is metallic. For the (m, n) =(2, 2) tube, the allowed k x wavevectors are given by k x = k y 2 `/. Subbing this into the equation for energy and lotting "(k) as a function of k y gives: Here, the uer and lower band never touch, and so the (2, to band theory. 2) carbon nanotube is insulating according 4

5 art 7) For a general winding vector (m, n), the vector ~ W has the form ~W =(m + n)~a 1 + n~a 2 = (m + n n 2, n 2 )= (m + n 2, n 2 ) (27) The only allowed wavevectors in recirical sace, ~ k, are those which satisfy the condition condition ~ k W ~ =2 `. Note, that there are only two indeendent corners of the Brillouin zone, those at wavevectors ( 2, 2 2 ) and (, 2 ). The other four corners of the B.Z. are translations of these corners by some combination of basis vectors, so that if ~ k ~W =2 ` for one of these two corners, then veck 0 ~W will also equal 2 ` for the other four corners of the B.Z. (since ~ b i ~a j =2 i,j and W ~ =(m + n)~a 1 + n~a 2 ). Then, for the corner oint ~ k =( 2, 2 )wehave ~ k W ~ =2 ` (28) 2 ) (, 2 ) (m + n n 2, )=2 ` (2) 2 m n ) 6 + n 2 = ` (0) ) 2m n +n =6` (1) ) (n m) =` (2) And, for the corner oint ~ k =( 2, 2 )wehave ~ k W ~ =2 ` () 2 ) (, 2 ) (m + n n 2, )=2 ` (4) 2 m ) + n 6 + n 2 = ` (5) ) 2mn +n =6` (6) ) (2n + m) =` (7) But, now notice that these are actually the same condition. Start with 2n + m = ` (8) ) n n + m = ` () ) n + m = (` n) (40) ) n m = `0 where `0 =(n `) is an arbitrary integer (41) Therefore, in order for an allowed wavevector for a nanotube cylinder with ~ W given by (m, n), to lie on a Brillouin zone corners, (m, n) must satisfy the conditions 2n + m =`0 (42) It is obvious that the tubes satisfying this condition are metallic as we have already shown in art 5) that the corners of the Brillouin zone corresond to the oints where f(k) = 0. Since grahene has a two article basis, this imlies that the rimitive unit cell contains two atoms and so the first Brillouin zone is exactly filled with two articles er state. This means that the first band is exactly filled and the second band is emty. Since the energy disersion is given by the relation "(k) =" 0 ± f(k) (4) 5

6 Then, the system is metallic if and only if an allowed k vector lies on the oint where f(k) = 0. So then, any nanotube where the Brillouin zone corners are allowed wavevectors contains a oint where f(k) = 0 and so is metallic. art 8) Now, we have f(k) = t 0 X vert.e i e i~k ~e i t X diag.e i e i~k ~e i Both the armchair and the zig-zag tubes di erentiate between the ~e 1 and ~e 2 /~e bonds (since the vector W is either arallel or erendicular to the ~e 1 bond in these cases. The band structure of the nanotubes is then given by (44) " = " 0 ± f(k) (45) To solve for the oints when f(k) = 0, let s let s assume that the vertical ~e i bond with hoing t 0 is the bond e 2 =(0,a). This will make solving for f(k) = 0 convenient and ocne we find these oints we can rotate our axes so that ~e 1! ~e 2. Using this choice we can write f(k) = e( t 0 e 1 2 ıky te ı( 2 kx+ 1 2 ky) te ı( 2 ıkx+ 1 2 ky) =0 (46) 0 = e 1 2 ıky t 0 e 2 ıky te 2 ıkx te 2 ıkx (47) Im[f(k)] = 0 ) sin( 2 k y)=0 (48) Re[f(k)] = 0 ) t 0 cos( 2 k y)+cos( 2 k x)=0 (4) The solutions to Eq. (48) are We can sub this into Eq. (4) to get: k y =0, ± 2 (50) ±t 0 +cos( 2 k x)=0 (51) ±1+ t 0 cos( 2 k x)=0 (52) 2 k x = cos 1 t0 (5) ) k x = 2 cos 1 t0 (54) Eq. s (50) and (54) give us the Dirac oints for our modified Hamiltonian, when e 2 has hoing t 0.Tofind the solution when bond ~e 1 has hoing t 0 we want to rotate our axes, so that what we defined as k y above becomes ~e 1 = 1 2 k x + 2 k y. That is we want to rotate our axes clockwise by angle /6, to let k x! 1 2 k x 2 k y and k y! 2 k x k y (55) 6

7 Then in our new basis, our Dirac oints given by (50) and (54) must satisfy: 2 k x k y =0 or 2 (56) 1 2 k x 2 k y = 2 t cos 1 0. (57) Solving this system of equations for k x and k y gives k x = 1 t cos 1 0 t k y = cos 1 0 (58) (5) Now, recall that for the (, ) armchair nanotube, our wavevectors k x, k y had to satisfy the requirement k y = k x + 4 ` (60) Subbing Eq. (58) into this condition gives ale 1 t k y = cos 1 0 t = cos ` + 4 ` (61) (62) For ` = 0 this is exactly the k y coordinate of the Dirac oint we just found above. Therefore, for the armchair nanotube, the Dirac oints alway fall along one of the quatized wavevectors. Therefore, the modifications of the hoing term doesn t a ect the metalicity of the nanotube in this case. Now, for the( 2, 2) zig-zag nanotube, our wavevectors had to satisfy the condition k x = k y 2 ` (6) Subbing in Eq. (58) and (5) gives 1 t cos 1 0 = t cos 1 0 t ) 4 cos 1 0 ` ) t 0 = cos 2 2 ` (64) = 2 ` (65) Therefore, the allowed wavevectors of the zig-zag nanotube will not in general cross the Dirac oints unless the hoing arameters satisfy the above condition. If the above condition is satisfied by the secific values of t and t 0 then the system will be metalic, otherwise it will be insulating. For ` = 1, the condition above is t 0 1. must be satisfied for the nanotube to be metallic. (66) 7