l08-ece453.pdf l09-ece453.pdf l10-ece453.pdf l11-ece453.pdf l12-ece453.pdf

Transcription

1 4.9.3-l8-ece453.pdf l9-ece453.pdf l-ece453.pdf 4.9.-l-ece453.pdf 4.9.-l-ece453.pdf l3-ece453.pdf 4..-l6-ece453.pdf 4..4-l7-ece453.pdf 4..6-l8-ece453.pdf 4..8-l9-ece453.pdf 4..8-l-ece453.pdf 4..-L.pdf 4..-l-ece453.pdf 4..-L.pdf 4..-l-ece453.pdf 4..5-l3-ece453.pdf 4..7-l4-ece453.pdf 4..9-l5-ece453.pdf 4..-l6-ece453.pdf 4..3-l7-ece453.pdf l-ece453.pdf l-ece453.pdf l3-ece453.pdf l4-ece453.pdf 4.9.-l5-ece453.pdf l6-ece453.pdf 4.9.-l7-ece453.pdf

2 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University Lecture 8: Schrödinger quation Ref. Chapter. Networ for Computational Nanotechnology

3 Schrödinger quation and Hydrogen Atom :5 U + Schrödinger equation was introduced in 93 s and helped to describe the energy levels of H atom. ih Ψ t h m Ψ + U This equation provided a conceptual frame wor for which wave nature of electrons comes naturally. The potential U(r) that appears in the Schrödinger equation, loos lie the figure for H atom. The is the Coulomb potential: r v ( r )Ψ ( n 3) ( n ) ( n ).5eV 3.4eV 3.6eV U ( r) q / 4πε r Schrödinger showed that for this potential the equation can be solved analytically and he found solutions, which agreed with the eperiments. In the net few lectures we want to understand how to solve for the energy levels of a material using Schrödinger equation. For H atom, the problem is analytically complicated for the eact shape of potential but if we approimate the potential as a confining bo for electron we can get a relatively easy answer that helps us to understand the physics. +

4 D Solutions 5:54 ih Ψ t h Ψ + U m + + y z v ( r )Ψ This equation in general is a function of time and space (3 dimensions). For the purpose of simplicity we ll solve it in dimension. This is the same as assuming that nothing varies in y and z dimensions. In D: i Ψ h h Ψ U t m + ( ) Ψ If U is a constant (for instance ), solutions can be written by inspection. For constant coefficients, eponentials (plane waves) satisfy the equation: it / h Ψ Ψ e i e To show this, substitute the answer in the equation. α t α t Note: e α e ih i h Ψ t Ψ h m h m h m ( i ) Ψ Ψ As long the above equation is satisfied, our guessed solution will satisfy the Schrödinger equation. Note that for a given value of, the value of is fied but the equation given above.

5 - Relationship :5 What if U was a non constant (U)? The answer is that the same solution will satisfy the equation: i Ψ h h Ψ U t m + h + U m We have the same solution but a new - relationship. If U was a function of, this solution wouldn't wor: h + U m For a given, both and are constants but can change so will change and this is a contradiction. ( ) Ψ The general rule is that as long as the coefficients are not function of time, the time portion of the solution is (phi is some function of ): it / h Ψ φ ( ) e The - relationship loos lie: U Notice that there is no quantization involved. All values of energy are allowed. The quantization comes in play when we impose a bo.

6 Particle in a Bo 8: Consider a U() lie: U() 3 U L The question is: what are the allowed energies for this problem? Notice that since the potential is infinite at the two sides the wavefunction cannot get out of the bo and has to go to at the two sides of the bo. We have: Ψ( ) B.C. Ψ( L) The solution it / h Ψ Ψ e i e will not wor here because it is never. Since i Ψ h h Ψ U t m + is a linear equation, we can use a superposition solution. e e i i i + e cos i e i sin To satisfy the first Boundary Condition (BC), we choose sin function as an answer: it / Ψ Ψ sin To satisfy the second BC: Ψ h ( ) e i e nπ n L

7 From Confinement to Discreteness 36:6 Only certain value of are allowed: n nπ / Which if it is put in L h U m + results in discrete allowed values of : n U + h n π m L The discreteness of energy levels is a consequence of trying to confine the wavefinction in a potential bo. If confined in a smaller bo, the energy values will be further apart. An eample for the level spacing: 4 h 4 m 34 ( 6.6 J sec ) 8 mev π L 9. 3 g /.6 8 ( m ) 9 π L π L The physical sense is that when an electron is confined in a bo of size Å, we ll get discrete levels with the energy spacing of the order few mev.

8 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University Lecture 9: Schrödinger quation Ref. Chapter. Networ for Computational Nanotechnology

9 D Schrödinger quation, Dispersion Relation :3 i Ψ h h Ψ U t m + ( ) Ψ We ve been taling about solutions to D Schrödinger equation. A simple eample is the case of constant potential. For harder cases, we we ll learn a numerical method that would help us to solve for the energy levels of a material with arbitrary potential. First we ll consider the D case and then we ll get into 3D. The simplest case is: i Ψ h h Ψ U t m + () The solutions to a differential equation with constant coefficients (lie above) can be solved by plane waves: Ψ Ψ / h (, ) Ae i t e it () Substitution of this solution into the equation leads to an - relationship. () in () > i h ih Ψ ( i) Ψ + U Ψ h m h n n U + Dispersion Relation m U

10 Vibrating String 6:9 Schrödinger equation is a wave equation. A common eample of a wave equation is the acoustic waves on a string. There the quantity that is used to describe the wave is the displacement of each point on a string from an equilibrium point at a particular time. Harmonically oscillating string Again by substituting the solution in the wave equation we can get the dispersion relation. () in () > ω v ( i ω) u v ( i) Dispersion Relation ω U (, t) L The equation describing such waves is: u t u u v () Solutions can be written as: iωt U (, t) e i e () You can see the analogy between the two wave equations. One point is in order and that is: there are times where people start from a dispersion relation and deduce form it a differential equation; reveres of what we ve done here.

11 Waves in a Bo - Allowed Values of :5 The net thing to understand is solving for the energy levels having more complicated potential functions. The first eample that was presented was that of a particle in a bo. This way we can have solutions lie: Infinite Square Well U() 3 L U L Assuming that the potential is very high at the two ends, one can show that the wave function has to go to at the two ends: Ψ( ) & Ψ( L) This allows only certain values of which leads to discreteness in the energy levels shown in the left figure. The reason is that the wavelength must be such that the wave could fit inside the bo. ( is related to the wavelength by ) π λ

12 Discrete nergy Levels 5:58 What we saw last day was that for the wave that goes to at the two ends we cannot use: Ψ it / h (, ) Ae i t e because this is never. Notice that there are two values of for a given energy. We can use a superposition of solutions with and to write the proper solution. The corresponding energy also becomes quantized: n U + h m π L This shows that whenever an electron is confined in a certain region, the allowed energy values become quantized. n We have: e e i i i + e cos i e i sin Since in this problem the wave has to go to at, we choose the sin. Further, the boundary condition at L requires: n L nπ

13 Particle in a Finite Bo 8:5 Suppose we have potential bo lie: U ( ) U (a) U ( ) U We are trying to find a solution with a given. A solution lie Ψ (b) will not satisfy the equation because we have two different values for the potential; hence two different values for. To see this consider: L it / h (, ) Ae i t e h U + h m U ( ) U (c) m( U ) () In region b we can write the solution as: Ψ i i Ae + Be This is because we get two possible values of for a given U from equation () so we can use a linear combination as a general solution. In regions b & c we can write the solution as: i i Ψ Ae + Be h U m( ) If <U, then Gives an imaginary value for and we get a decaying eponential in region a & c: Ψ a De + γ Ψ c Ce γ For region (c), ep(γ) is not allowed because the wavefunction cannot go to infinity for large. For region (a), ep(-γ) is not allowed because the wavefunction cannot go to infinity for large negative.

14 Boundary Conditions 5:34 The solutions in the regions a, b and c need to match at the boundaries o and L. Taing this requirement into account puts a restriction on the allowed values of constants A,B,C and D. We can use this restrictions to find the constants. These restrictions are called Boundary Conditions (BC s). De + γ i (a) Ae Be (b) L (c) The first BC is that the wavefunction has to be continuous across the boundaries. The second BC is that the derivative of wavefunction has to be continuous across the boundaries. i + Ce γ What we mean is that discontinuities Discontinuous ψ ( ) or dψ / d () are not allowed because then the Schrödinger equation will not be satisfied at the point of discontinuity. Same is true for the derivative of the wavefunction. The reason is that in each case there will be an unmatched delta function (infinite height) in the Schrödinger equation where all other entities are finite: i Ψ h h Ψ U t m + ( ) Ψ Note: if Ψis discontinuous, then dψ/ d is a delta function. If dψ/ d is discontinuous, then d^ψ/d^ is discontinuous.

15 Utilizing BC s to Determine the Constants 7:58 (a) Ae De + γ i Be i (b) L + Ce γ (c) We will not get into the details of this problem. Instead we ll use a more convenient method of solving and that is the numerical method. We can utilize BC s as follows: Continuity of Ψacross gives us one equation that relates A, B and D: D A + Continuity of Ψacross L gives us one equation that relates A, B and C. Similarly, continuity if dψ/d gives us two equations at the two boundaries. One gets a set of equations that could be used to eliminate the unnown constants. B

16 Ψand lectron Density n 3:4 In electronic devices we are interested in two things the most: electron density inside a device and current flow. If an electron has a wavefunction, the associated electron density is: n Ψ * Ψ For eample tae the potential well: Infinite Square Well Notice that for electron inn the bo, the shape of electron density n may indicate that there is a fraction of electron at each point. This is not right. The correct view is that to loo at these fractional values as probabilistic values of finding the electron at each point. Notice for adding electrons in different states we have two ways to come up with n: * * * ΨΨ+ΨΨ or Ψ +Ψ Ψ+ Ψ ( ) ( ) L The correct answer is the first one. With electrons what we have to add are the electron densities not the wavefunctions. The associated electron density would come from squaring the sin wave inside the bo. (solid:ψ & dashed :n)

17 n, I and quation of Continuity 39:4 For an electron that has the wavefunction: lectron density is: Current can be written as (the reason comes later): () in () > () in (3) > n I n Ψ * Ψ () A ih m Ψ it / h (, ) Ae i t e () ih m h A m ( ) ( ) q ia ia q dψ d * ( q) ψ ψ (3) Where did equation (3) come from? If one accepts (), then (3) is the only consistent epression for current I considering the fact that Ψhas to satisfy the Schrödinger equation. Accepting () and (3) will satisfy the continuity equation and since the equation of continuity is general true argument (3) must be the right choice for I. This is a general argument and states that the electron density in a region cannot change over time unless the gradient of current is non- in the region. Moreover, the amount of current leaving the region has to be equal to the rate of change of electron density. I Continuity quation: * dψ d ( qn ) t I

18 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University Lecture : Finite Difference Method Ref. Chapter. Networ for Computational Nanotechnology

19 Time Independent Schrödinger quation :3 Today we want to tal about numerical solutions to Schrödinger equation. Consider the time dependent Schrödinger quation i Ψ h h Ψ U t m + ( ) Ψ If the potential is independent of, then the it / h i solution to this equation can be written as: Ψ(, t) Ae e The only thing that remains to be done is the relationship between and called the dispersion relation which can be found by substituting the solution in the equation. Generally U() is a complicated function and analytical solutions are not achievable. There we have to rely in numerical solutions. Notice that as long as U is independent of time, the time portion of the solution above is it / h acceptable. So we can write the solution as: Ψ(, t) Ae Φ( ) () Φ() is a complicated function that remains to be found. By substituting () in the Schrödinger equation we get the time independent Schrödinger equation: i ih e h it / h Time Independent Schrödinger quation: it / h h Φ Φ( ) e m h d Φ + U ( ) Φ m d + U ( ) Φ

20 Overview - igenvalue Problem 7:4 What we want to do net is to solve: The basic idea for any method of numerical solution to the differential equation is to turn it into a matri equation. We ll consider the finite difference method here. We ll end up with: What we ll learn for the rest of the class is how the finite difference method turns the differential equation into a matri equation. Once one has a matri equation lie above, The eigenvalues of the N by N matri can be evaluated. There will be N eigenvalues and N eigenvectors. An eample of eigenvalue problem: ()> To see this write the left hand side of () as: And subtract from it the matri: Φ + Φ ) ( U d d m h N N φ φ φ φ φ φ M M N by N () φ φ φ φ d c b a φ φ b c b a φ φ d c b a

21 igenvalue Problem Continued 7:9 From a c The eigenvalues can be found using the argument that the matri on the left must be singular. If it wasn t then multiplying both sides by its inverse would result in a eigenvector which is not the correct answer. The matri won t have an inverse if its determinant is not. We have: To see hoe the procedure wors, consider a special case: From ( )> b φ b φ ( a)( d ) bc (), igenvalues φ φ φ φ ( )( ) +, Corresponding to each eigenvalue, there is an eigenvector: +, then φ φ ( + ) φ φ φ φ +, then φ φ ( ) φ φ φ φ - Notice that if the eigenvectors are multiplied by an arbitrary scalar, the result is still an eigenvector. In general for an N by N matri we ll have N eigenvalues and N corresponding eigenvectors. For large value of N, Matlab can be used to find eigenvalues and eigenvectors: [ V, D] eig( H ) D has the eigenvlaues of matri H as its diagonal elements. V has normalized eigenvectors of H as its columns.

22 Function Becomes a Column Vector 9:3 How can we describe a function as a vector? We set up a lattice of discrete points and record the value of the function at each lattice point. Figure shows the discrete points. As it can be seen corresponding to each lattice point there is a value for the wave function. This can also be viewed as sampling of a continuous function into discrete values. Remember that in order to be able to perform a numerical method we have to a have a finite number of equations so that we can solve them. First create a lattice for the -D problem. a Φ (,t) therefore becomes a column vector telling the value of Φ at different points. i.e. φ n 3 n- n n+ N φ (, t ) φ ( φ ( φ ( M φ ( M φ ( 3 n N ) ) ) ) )

23 Writing Potential Function U() as a Matri 3:5 Φ h d + U ( ) Φ 4 m d Hamiltonian Net question is: How do we convert the Hamiltonian operator into a matri? First try writing the matri for U() and then the matri for h d m d The total Hamiltonian should be a sum of these two. For now concentrate on the easy part which is U() (notice that a term is being neglected in the net equation at this point) [ ] Consider Φ U () Φ Since U() is a potential function, on a discrete lattice U would tell us the potential at ach lattice point, hence it will be diagonal: Writing above as a matri equation: φ φ M φ N U ( ) U M U( ) M M M M L ( ) U() L O L L L O L φ U ( ) φ n n φ φ M φ N L L M M M O U( ) N n

24 Differential Operator s Discrete Representation 38:4 a φ n 3 n- n n+ N At each particular point the Schrödinger equation (after dropping U) can be written as: h d Φ [ ] n m d φ Now start from left and right everything at point n. The constants remain the same. For the wavefunction we right its value at that particular point. We will have: n h φ n m What goes in the parenthesis is the discrete representation of the differential operator. So to turn the differential equation to a difference equation, the most important step is to write the second derivative as a difference epression. Let s state the answer first and then derive it: d φ d n φ Based on what we ve done so far, we can write: φ t n n+ φ n + a φ n ( φ φ φ ) U ( n) φn t n n + n+ h ma

25 Differential quation Becomes a Matri quation 4:57 Accepting,how can we write the matri equation? This will be a tridiagonal matri: To see this consider for eample one of the equations from the series of equations in (): If we d want to also include the potential to the matri we can add its corresponding values to the diagonal elements: ( ) () + + n n n n t φ φ φ φ N N t t t t t t t t t φ φ φ φ φ φ...??... L L L O M M M O O M M O O M L L L ( ) 3 φ φ φ φ + t ( ) ( ) ( ) N N N U t t t U t t t U t φ φ φ φ φ φ...??... L L L O M M M O M M M O M M M L L L L L L ( ) ) ( + + n n n n n t U φ φ φ φ

26 Boundary Conditions 46: How do we determine the far elements on the anti-diagonal? In other words how do we handle the boundaries? Dropping the two terms is equivalent to setting the wavefunction to at the two ends: This would be appropriate for the particle in a bo problem where the wavefunction is not allowed to penetrate outside the bo. Net day we ll start with the finite difference method. ( ) ) ( + + n n n n n n t U φ φ φ φ φ ( ) ( ) ( ) N N N U t t t U t t t U t φ φ φ φ φ φ...??... L L L O M M M O M M M O M M M L L L L L L 3 n- n n+ N a φ n + φ N φ

27 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University 9..4 Lecture : Finite Difference Method Ref. Chapter. Networ for Computational Nanotechnology

28 Time Independent Schrödinger quation and Discretization :8 h d Φ + U ( ) Φ m d φ φ M φ N a 3 n- n n+ N φ n φ φ M φ N As it was mentioned before all numerical methods for solving differential equations involve some scheme to covert the original equation into a matri equation. What we do here is to discretize the lattice and then we write the second derivative as a difference equation. Figure on the left shows the discrete points. As it can be seen corresponding to each lattice point there is a value for the wave function. This can also be viewed as sampling of a continuous function into discrete values. Remember that in order to be able to perform a numerical method we have to a have a finite number of equations so that we can solve them. To mae things easier, for now we set U(). Note that the matri for U can be written down easily and it can later be added to the matri that represents the differential operator which we now we ll try to find

29 From Derivatives to Differences 6:54 φ n a 3 n- n n+ N h φ n m At each particular point the Schrödinger equation (after dropping U) can be written as: h d Φ [ ] n m d φ Now start from left and right everything at point n. The constants remain the same. For the wavefunction we right its value at that particular point. We will have: n What goes in the parenthesis is the discrete representation of the differential operator. So to turn this differential equation to a difference equation, the most important step is to write the second derivative as a difference epression. We start with the difference equation for the first derivative

30 From Derivatives to Differences 9:39 Consider the lattice in the a vicinity of the nth point. a n- n n+ In order to right the derivative of the function at n we need to now its values close to n. This can be done by calculating the amount by which the value of wave function changes as moving from one lattice point to another. This gives us the first derivative. The second derivative is of course the difference of the first derivatives. The amount by which the function changes going from n- to n divided by the distance between the two points is the first derivative. dφ d n+ dφ d n+ n φ φn a φn φn a Notice that the closer the lattice points the better the approimation. What do we do now to get the second derivative at n? Subtract the first derivative at n-/ from the first derivative at n+/ and again divide by the distance (a) between them

31 From Derivatives to Differences 3 4:9 The second derivative is: dφ d φ d n + d a Where So n dφ φ n+ d n+ a d φn d dφ d n φ dφ d φ φn φn a n+ φ n + a φ n n n and And finally our Schrödinger equation becomes: h φn φ n+ φ n + φn ma ( ) This was the essence of the finite difference method. The equation above is the Schrödinger equation at a particular point n. There are N of these equations. All together they can be written as a matri equation. To mae things more compact and elegant define: t h ma and tae the negative sign inside to get: φ n ( φ φ φ ) t n n+ n

32 We now want to see how these equations Become a matri. To see this observe that the nn element in the matri is what multiplies There are two more non-zero entries to the left and the right of nn element. They are the factors that multiply and are both equal to t. Same analysis goes for every row to get the matri equation on the right: At this point we can add the matri of U to the matri above: this is basically adding the values of U() at each point of the lattice to the corresponding diagonal element of the matri above. (Remember that U is diagonal) There is one point that needs to be taen care of and that is the boundary conditions mared by a? in the above matri N N φ φ φ φ φ φ M M ( ) + n n n n t φ φ φ φ Row n Column n : t φ n and n φ φ n+ N N t t t t t t t φ φ φ φ φ φ...??... L L L O M M M O M M M O M M M L L L L L L n n n + The Hamiltonian Matri 8:3

33 Boundary Conditions (BC) :3 φ t φ t. M. M. M φ N? t t M M M L L L L L O O L L L L O t? φ M φ M. M. t. t φ N The easiest boundary condition to implement is to assume that the wavefunction is at the two ends of the - D lattice. Notice that the issue of boundary conditions comes in play because we want to have a finite dimensional matri. So we have to end it somewhere and that brings about boundaries. 3 n- n n+ N This assumption is an eact one if we were to describe the energy levels of a particle with infinite potential walls. In that case there would be no way for the wavefunction to lea outside of the bo. However if the potential walls were not infinite, there would be some leaage. In that case we increase the lattice size to incorporate non-zero values of the wavefunction. One point to notice is that the potential would have different values inside and outside of the bo which can easily be entered in the Hamiltonian matri. (We just add what ever potential we have the diagonal element corresponding to the lattice points.) Ψ V()

34 ffective Mass 7:8 In the following wees we ll be taling about the energy levels of solids. One very interesting point about metals and semiconductors is that the electronic levels can be obtained from a slightly modified Schrödinger equation. The modification is in the value of m. h d Φ + U ( ) Φ m d Whereas the value of m in the equation above is the actual mass of electron, the * modified m denoted by m is not. It is called the effective mass. It incorporates the potential of the solids who have periodic structure. Using this concept, one can obtain the energy levels of such solids using what s called an effective mass equation without having to enter the value of U() which corresponds to the periodic nuclear potentials that eist in a solid. h d Φ * m d Φ What this means is that given a piece of silicon, you could estimate the energy levels by solving a particle in a bo problem which leads us to another point and that is the usage of periodic boundary conditions. Although the right boundary conditions to be used are infinite wall BCs, the periodic BCs are often used because of reasons:. It maes the math easy.. When trying to understand the energy levels deep in the middle of a solid, then what happens at the boundaries are not important. But what is periodic BC?

35 Periodic Boundary Conditions 3: 3 n- n n+ N The idea is as the name suggests. Considering the above lattice, on thins of the Nth point as having the eact potential as the th point. The N+ point would have the eact potential as the first point and so on. Notice that we ve assumed that the period is N. Knowing this, how can we modify the Hamiltonian matri to obey the boundary conditions? Remember that what ever the boundary conditions are, they enter at the points indicted here. H To implement the periodic boundary conditions thin lie this: We ve assumed the solid to be lie a ring meaning the Nth point is the same as the th point. Ordinarily point one is connected to the th point described by t. But now point is point N. Indeed what happens is that point N is connected to point via t. As the result this value seats in the place of the matri element that connects N to. H quation now becomes: How about the other diagonal element? t φ tφ tφ tφn

36 Periodic Boundary Conditions 36: Using the same idea, the other element will also be turn out to be t. This is consistent with postulates of quantum mechanics: We want the eigen values of the Hamiltonian matri to be real. For this to happen, the H matri must be Hermitian. If all the matri elements are real then being Hermitian comes down to being symmetric. And symmetric means that we have to have t for both of our diagonal elements that describe the boundary conditions: H t t In general H must be Hermitian. This means that the matri is the same as its conjugate transpose. But why does this periodic boundary condition mae things simpler? The answer lies in the properties of the eponential function as a solution of Schrödinger equation: h d Φ * m d Φ For the bo boundary conditions we had to consider the function sin() because the eponential doesn t go to at the ends. However the situation is different for periodic boundary conditions

37 ponentials as Solutions of Schrödinger quation 4:4 i Indeed, the eponential can be considered as a solution to Schrödinger equation with periodic boundary. The advantage of eponentials is that taing their derivatives is very easy, their magnitude squared is, ect. Let s see what this Per. BCs means for eponentials: i i L i e e ( + ) e For the equation to hold we have to have: L n π e π n L One last point: It seems as if with eponentials we have half as much allowed values of because they are π apart whereas for sin() values of are only π apart. This is not true though. Because in the case of sin(), the and values produce two wavefunctions that are different just by a constant factor. This means that we don t have two independent functions. For eponentials on the other hand, and values produce two independent solutions. Altogether we conclude that the number of values in both cases are the same. For sin() all values are positive and π i apart. For e are both positive and negative and apart. π

38 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University 9..4 Lecture : Separation of Variables Ref. Chapter. &.3 Networ for Computational Nanotechnology

39 Schrödinger equation (Matri Form) :7 Remember time independent Schrödinger equation: In one dimension (say in ) if we neglect the potential U, we ll have: But is there any difference between the solutions of the matri equation and the analytical solution? In particular how do the eigenvalues of the two methods compare to each other? The answer depends on the range of energy which is considered. For low energies the two answers match well whereas for high energies the two deviate from each other and of course it is the numerical answer obtained from the matri on the left that is not quit right. Let s discuss this more in depth: (See net page) Φ + Φ U m Φ Φ d d m Using finite difference method, we have this matri equation: N N t t t t t t t t t φ φ φ φ φ φ / ma t 3 n- n n+ N a

40 Numerical vs. Analytical eigenvalues 5:49 You can set up a matri and as Matlab to find the eigenvalues. The number of eigenvalues will be the same as the number of rows (or columns) of the matri. For a by matri, we get eigenvalues. Plotting them versus energy gives us the figure below. How does the answer loo lie analytically? For a constant potential, we now the answer analytically and that is a good bench mar for checing the numerical solution. So what are the analytical eigenvalues for potential? Schrödinger equation reads: d The solutions to this equation are the (ev) 3 4 Numerical: Blue Analytical: Red m d Φ α Φ plane waves: ep (i) The eigenvalues are given by: / m/ m Periodic boundary conditions require the solutions at the beginning and the end of the lattice to match: i ( + L) i e e e il π υ ( υ is an integer) L in the eigenvalue equation L πυ Putting this m The analytical eigenvalues loo lie a parabola whereas the numerical ones level off. Why do the answers match at low energies but not high energies? π ν L

41 Wavefunction varies rapidly at high energies : As it can be seen form the figure, the numerical solution deviates from the analytical one. The reason is that at low energies the wavefunction varies slowly whereas at high energies it changes very fast. If the lattice points are not close enough then the numerical solution fails to capture the whole physics and fails to give the right answer. Therefore we can trust the numerical eigenvalues at low energies and ignore them for higher values of energy. It turns out that for current flow the energy range of interest is a small region around the chemical potential. This is why we re o to use the numerical solution. High nergy Wave: Infinite Square Well L Why do the two answers deviate? The reason relies in turning the second order derivative to a difference equation. d Φ Φ n Φn + Φ n+ d a We will discuss this issue more when we tal about bandstructure.

42 ffect of boundary conditions on solutions 5: How do the boundary conditions affect the results? More specifically what is the difference between the eigenvalues for periodic boundary conditions versus bo boundary conditions? For period boundary conditions we have the solutions: e i e i Both of these solutions give the same eigenvalue because and give the same energy: / m The eigenvalues in this case come in pairs: at each energy there are two eigenvalues., Using the bo boundary conditions, the situation will be different. In that case there will be no pairs but the eigenvalues will be closer together. In this case they will be π / L apart whereas in the previous case they were apart. Why? π / L For the solutions we can have any linear combination of eponentials that satisfy bo boundary conditions: i i ( e e ) / i sin( ) The bo boundary conditions tells us that the wavefunction has to go to at the ends. sin( L) L υπ Notice that and values give two wavefunctions that are not linearly independent and hence they are not separate solutions. In both cases we have the same number of eigenvalues.

43 Separation of variables (reasons/conditions) 3:58 How do we solve Schrödinger equation in two dimensions? The equation reads: Φ + + U Φ m y Just as we set up a lattice in dimension, we can set up a lattice in two dimensions. N Then we have to turn the equation into a matri equation. N The wavefucntion vector in the matri form has the value of the wavefincion at each discrete lattice point. For this lattice we will have N^ values. This means that the matri size will be N^ by N^. As N gets large, it will be hard for the computer to calculate its eigenvalues. We can see that this maes it hard to solve problems in higher dimensions because of the limited power of computers. For higher dimensions we use another method called separation of variables. This method can be used only if the potential satisfies certain conditions. If it is so, then the big problem in higher dimensions can be broen into small problems in lower dimensions. How does this method wor and what condition should the potential satisfy?

44 Method of Separation of variables 3: If the potential has the following property: U (, y) U ( ) + Uy( y) Then separation of variables can be used to give the wavefunction as: Φ(, y) X ( ) Y ( y) We can find X and Y by solving the following equations: X ( ) m yy ( y) m y + U () ( ) X( ) () One thing that we can show easily is that if we combine these two equations we can get bac the original -D Schrödinger equation. + Uy( y) Y ( y) () To show this multiply () by Y(y) and () by X(). In both cases we can put X or Y in front of the derivative operator because the derivative is with respect to another variable. + y XY XY + U( ) XY m Φ m y m + Uy( y) XY From above we can see that: y What we have gained is that if the potential can be separated into to parts the big -D problem can be broen into two -D problems which are a lot easier to solve. + y + U Φ by by & by

45 An eample: particle in a bo problem 36:3 For particle in a -D bo we had: υ U m π L How do we solve the problem of a particle in a -D bo? υ L υπ ν y Y(y U ) X( ) What are the solutions to the problem of a particle in a -D bo? The eigenvalues become υ π ν m L π yn n m L n L nπ We ll have product solutions of X() and Y(y) for eigenfunctions: Φ(, y) sin υ sin υ The eigenenergies are υ L υπ π υ, n ( υ + n ml y )

46 Separation of variables nergy Levels 43: If we had solved the problem directly by setting up lattice with points in each direction, we d have a by matri, which would give us eigenvalues. Here we have two lattices each with points. That gives us a total of eigenvalues. So what happened to the? To answer this question have a loo at: π υ, n ( υ + n ml The point is you can pair a single eigenvalue in the direction to any of the values in the y direction and vice versa. This gives you a total eigenvalues. It is important to remember that the method of separation of variables could be used only if the potential has the form: U (, y) U ( ) + Uy( y) ) nergy levels in -D (,) (,) or (,) (,) 8 5 One point to notice is that with the method of separation of variables, one needs two indices to label the energy levels in -D and 3 indices to label the energy levels in 3-D: nergy levels in 3-D (,,) (,,) (,,) 9 6 3

47 Introduction to Nanoelectronics Prof. Supriyo Datta C 453 Purdue University Lecture 3: Atomic nergy Levels Ref. Chapter.3 & 3. Networ for Computational Nanotechnology

48 3D Schrödinger quation :3 3D Schrödinger equation is: Φ + U ( r ) Φ m Considering only one dimension: d Φ + U ( ) Φ m d We ve learned how to turn this into a matri equation using method of finite differences through which one can find the eigenvalues and eigenvectors. In 3D, we ve discussed that the size of matrices can get very large if one tries to turn the whole 3D equation into a matri equation. But under certain conditions we can use the idea of separation of variables which is very powerful in maing the problem more tractable. (discussed last day) We can use Separation of variables if the potential can be written as: U (, y, z) U ( ) + Uy( y) + Uz( z) If that is satisfied the one 3 dimensional problem can be written as 3 one dimensional problems which is easier to handle. What we want to tal about today is how the energy levels loo lie for atoms. We start with the simplest of all atoms: hydrogen atom. The potential that the electron sees in the hydrogen atom is: q U ( r ) 4πε r r + y + + The epression for r shows that this potential is not separable in Cartesian coordinates but it will be in spherical coordinates z

49 Spherical Coordinates 8:46 In spherical coordinates the variables are: z φ θ r theta the y The potential in spherical coordinates depends only on r and it is separable. It can be written as 3 function of r, theta and phi where the function of and phi are basically zero. The hard thing about this coordinate system is that operators are much more complicate; hence more complicated to wor with. y z In Cartesian coordinates r r r + r sinθ + sinθ θ θ sin θ φ + In Spherical coordinates

50 Spherical Harmonics :3 For potentials that are symmetric in r and do not depend on theta or phi one can write the solutions to the Schrödinger equation as: q U ( r ) 4πε r Φ(r, θ, φ) f ( r) l m Y (θ, r φ) Radial Angular The function f can be found by solving a D equation: d r) U( r) +... f(r) m dr ( The angular part is already nown for any atom whose potential does not depend on theta or phi. One only has to find the redial part. The Y(l,m) are called spherical harmonics. Let s discuss them more: L r + + r r r If we act operator L on Y(l,m) s we get: l l We see that the operator acts on the function and generates the same function with a multiplicative constant. In such cases the functions are called to be the eigenfunctions of the operator. The property of Y(l,m) s is that they are eigenfunctions of L. f sinθ + sinθ θ θ sin θ φ m L( θ, φ)y l( l + ) Y m

51 Spherical Harmonics 7:3 The simplest Y(l,m) is Y(,) and is just a constant: Y(,). The general rule is: If m n l n then n, n,,,... +,...,,..., So: if l then m -,, and the Y(l,m) s are: Y Y Y ± cos( θ ) sin( θ ) e n, n ± iφ Let s see if Y(,) is an eigenfunction of the operator L () & () > LY (some constant) Y? L r sinθ + sinθ θ θ sin θ φ θ sin θ φ ( cos ) () sinθ ( cosθ ) sinθ θ θ ( sinθ sinθ ) sinθ θ sinθ cosθ cosθ sinθ where ( ) () LY Y l( l + ) ( + )

52 Solution of Schrödinger quation 6:3 Bac to the solution of Schrödinger equation. Consider the function: Φ(r, θ, φ) g( r) Yl (θ, φ) Putting this answer into the Schrödinger equation, we get: gy m l Note : m g ( r) f ( r) / Φ + U ( r ) Φ m d d m g l l + g + UgYl m + dr r dr ( ) mr L( θ, φ)y l m l ( l + )Y l m has been used. r Using g(r)f(r)/r, we get the final result: d l( l + ) f ( r) + U ( r) + m dr mr f ( r)

53 Hydrogen nergy Levels 36: f d l( l + ) ( r) + U ( r) + m dr mr f ( r) We can now tal about energy levels in H atom: Since the potential in Hydrogen atom has the form of /r, the levels s and p are at the same energy. This does not have to be the case for other atoms. Also note that for each value of l, there are l+ values of m that have the same energy. So when l, m-,, and the levels comes in triplets. When l, there will be 5 degenerate levels. One thing to remember is that anytime the Potential is spherically symmetric, the solutions to Schrödinger equation can be written as: m Yl Φ(r, θ, φ) g( r) (θ, nergy φ) l l l 4s 3s s s 4p 3p p 4d 3d

54 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University..4 Lecture 6: Bandstructure Ref. Chapter 5. Networ for Computational Nanotechnology

55 Basis Functions :3 What we are trying to learn are the energy levels of a given solid. The concept of bandstructure is very useful in understanding the energy levels of periodic solids. Note that the atoms in solids we are interested in are arranged periodically. The starting point is the Schrödinger equation: Φ + U ( r ) Φ m The direct war of solving this equation is to set up a lattice of points and turn the equation into a matri equation. But this direct approach gets intractable because the matrices get huge for problems dealing with more than dimension. Instead what is widely used is the method of basis functions: one uses his/her nowledge of the material to write the function Φ in terms of a set of basis functions...4 ample: suppose you are interested in the lowest energy levels of a hydrogen molecule. There, one nows the wavefunctions of s levels of each H atom. When the two atoms are brought close to each other the wavefunctions mi up somewhat but the overall wavefunction is composed of the two individual wave functions: Φ r φ u r ( ) ( ) Instead of looing for a function we only need to find the proper coefficients which in the case of H molecule there are only of them. u L ( r ) R u R m ( r ) m m

56 ..4 Basis Functions and the Overall Wavefunction 4:9 After writing the total wavefunction as some of a set of basis functions one can turn the Schrödinger equation into a matri equation: The elements of H can be found by first principles (ab initio) or by the semi empirical method which is easier to deal with. In the case of H molecule we want to find the elements of a by matri. From the symmetry of the problem we can name the elements as: The parameters t and ε can be found by choosing them in a way that the wavefunction calculated this way would fit the eperimental data. The advantage of this method is that it is a lot faster. The net step is to find the eigenvalues of our by matri. Let λ represent the eigenvalues of the matri. Then And eigenvectors are: N N φ φ φ φ φ φ H ε ε t t H ± t ε λ ( ) ( ) r u r m m φ Φ + t ε t ε + ) ( t t t ε ε ε ) ( t t t ε ε ε s H -3.6eV s H -3.6eV A B R

57 A Note on Matrices 4:5 A note about matrices, their eigenvalues and eigenvectors: If a constant number is added to the diagonal elements, then all the eigenvalues are modified by adding that constant value to them and the eigenvectors remain unchanged. [ A]{} φ λ{ φ} [ A + c I ]{} φ [ A]{} φ + c{} φ ( λ + c) {} φ..4

58 Periodicity and the Principle of Bandstructure 8:58 Let s now move on to bandstructure and start with a simple eample. Consider a hypothetical solid consisted of an array of hydrogen atoms: H H H H H H H N ε t Since the solid is periodic, all of the t s are actually the same. Notice that in general the rest of the matri elements are not zero but because the coupling between atoms gets weaer the farther they are apart, in some cases people only consider the nearest neighbors...4 t ε t t ε t One can include more terms to get higher order answers. The important thing to note is that the principles of bandstructure will wor regardless. For that to wor each row in the matri should loo the same as others (although the elements might be shifted) What we ll see in the net couple of lectures is that if H would from a solid lie the left figure it would be a metal and conduct very well. But in practice if Hydrogen atoms are put in array together the configuration that minimizes energy will be more lie: H H H H H H This is what is called a dimerized solid. In this configuration the solid will be a semiconductor.

59 Solution of Schrödinger quation and The principle of Bandstructure 33:5 We want to find the eigenvalues of the Hamiltonian matri: φ ε t φ t ε t t ε φ N φn Because the solid is periodic and the matri loos lie above, one can use the principle of bandstructure to find the answers analytically. The basic idea is that the matri equation above can be written as N algebraic equations. For eample the nth row gives us the nth equation: φ n tφ..4 n + εφn + tφn+ () Now if the solid is periodic, then any row s equation will loo lie and the following ina solution will satisfy : a φ φ To chec the answer plug it in the equation: ina i( n ) a ina i( n+ )a φ e This is called the - relationship and our solution satisfies if this relation is satisfied. n φ n 3 n- n n+ N tφ e + εφ e + tφ e + ia ia te + ε te ε + t cos a e

60 Dispersion Relation 43:5 We can plot the - relationship: But the above seems to suggest that - relationship is a continuous function; hence infinite amount of eigenvalues not N of them. So how do we get N eigenvalues? The point is that there are only specific values of that are allowed. This is coming form the Periodic Boundary Conditions (PBC):..4 ε + t cos a 3 n- n n+ N ε t ε + t -π π φ a π N e e ina ina e i( n+ N ) a e iπ (integer) Na πν ν : integer π ν Na () quation () clearly shows that cannot assume any values and since is related to via - relation, energy eigenvalues can only assume particular values.

61 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University.4.4 Lecture 7: Bandstructure Ref. Chapter 5. Networ for Computational Nanotechnology

62 Review :3 The principle of bandstructure helps us to find the eigenvalues of a periodic matri. This is important for us because solids that we are interested in have periodic structures. Consider the eample of last day: ε t φ φ t ε t φ φ N s s t ε 3 n- N φ φ N Regardless of the details of the problem, the matri that we write will be periodic. t Notice that as long as every row loos the same, the principle of bandstructure applies regardless of how each row loos. The nth row is: This solution satisfies the Schrödinger equation if the - relationship below is met. in φ n tφn + εφn + tφn+ φ n e ina φ () ia te + ε + te ε + t cos a () ia As long as the above - relationship is met, the solution we chose will satisfy the Schrödinger equation.

63 Review 5:3 What we ll learn today is a generalization of bandstructure method to a solid where the unit cell consists of two atoms. a 3 3 N N The above dimerized solid is semiconductor whereas for the old case shown below the solid is metallic. To now why consider: ε t ε + t -π π φ a π L 3 n- N For the above solid the eigenvalues are given by: φ + φ e ina Na πυ n Periodic Boundary Condition N n ε + t cos a π υ ν : integer Na

64 Array of H Atoms - Metallic Conduction 3:58 How can we tell if a solid is metallic or not? Well, we d have to investigate the energy levels around the Fermi level. If there are lots of levels around the Fermi energy, then the solid conducts well, otherwise it will conduct poorly. Where would the chemical potential be for this simple structure? s s 3 n- N Fermi Level N N ε t ε + t -π π φ a π N By definition, the Fermi energy is where, above it all levels are empty and below it all levels are full (at absolute ). Since there are lots of levels around Fermi level, in this case the solid will conduct well.

65 Dimerized Solid - Semi-Conduction 6:3 In the case of dimerized solid, there are two atoms per unit cell. So N unit cells will consist of N Hydrogen atoms. a 3 3 N N What we will see shortly is that the energy levels of a dimerized solid loo lie: Fermi Level N electrons In this case, the Fermi energy lies between the two bands of allowed energy levels. Since there are not levels around the Fermi energy, the solid will not conduct.

66 Dimerized Solid - Schrödinger quation 8:5 Net we want to find the dispersion relation for the dimerized solid: We see that now every other row in the matri is the same not every row. To use the principle of bandstructure every row must loo the same. To get this configuration we can do this: What has happened is that by combining two elements of the wavevector into a new one we get: Similarly: a N N N N t t t t φ φ φ φ ε ε ε φ φ φ φ t t 3 3 N N + + N N ψ ψ ψ α β β α β β α ψ ψ ψ φ φ ψ ε ε α t t t β

67 Dimerized Solid: Dispersion Relation 7:8 We can now use the principle of bandstructre to write the solution to the set of equations: ψ ψ ψ N α + β β α β The claim is that the following will satisfy the equations: quation n reads: + β α ψ ψ ψ N ina { ψ } { ψ } e () n { ψ n} () [ ]{ } [ + α ψ + β ]{ ψ } + [ β ]{ ψ } n n n+ Putting () in () ina ina { ψ } e [ α ]{ φ } e + [ + i( n ) a i( n+ ) a β ]{ ψ } e + [ β ]{ φ } e Canceling the same factors from the two sides: Fermi Level { ψ } [ h]{ } ψ [] [ ] [ ] + ia + ia h α + β e + [ β] e π π a

68 Dimerized Solid: - curve / Conduction 36:8 Fermi Level π N lectrons a There are N states available. There are total of N electrons. ach state can tae two electrons. At temperature, the lower band is filled and the upper band is empty. Fermi levels lies between the two bands. Since there are no states around the Fermi level, the dimerized solid will not conduct well. π N states N states The fact is that if we had a linear array of H atoms, it would be arranged lie a dimerized solid and it won t conduct. a

69 Dimerized Solid: igenvalues 39:56 Net we want to find the eigenvalues. First, let s write matri h(): [ ] [ ] [ ] + ia [ ] + ia h( ) α + β e + β e ε t + α β β t ε t ia ε t + [ ] te h( ) + ia + te t ε t t t t t Δ Δ t + t t + t ε igenvalues are: ε ± Δ Δ t + t e ia π π a

70 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University.6.4 Lecture 8: Bandstructure 3 Ref. Chapter 5. & 5. Networ for Computational Nanotechnology

71 Review :3 a 3 3 N N The dispersion relation loos lie: Last time we taled about the principle of bandstructure which allows one to calculate the energy levels of a periodic structure. In particular we taled about a dimerized solid which has two atoms per unit cell. ina { ϕ } { } e { ϕ } [ h( ) ]{ } n ϕ ϕ N φ φ φ N very row loos the same φ φ φ N Today we want to epand this idea to a higher dimension: D. In particular we ll use the principle of bandstructure to describe the energy levels of graphene

72 Graphene: Atomic Orbitals 5:55 GRAPHN Heagonal Lattice The importance of this material comes in play when they role this up to become a carbon nanotube which is of great interest nowadays. Today we ll tal about the energy levels of a Graphene sheet. Remember that the starting point is the Schrödinger equation which can be turned into a matri equation once a proper set of basis functions are has been chosen. How many basis functions do we need per atom? The basis functions are the atomic orbitals. There are 4 orbitals for the valence electrons of a carbon atom. 4 p C6: s s As it turns out we can tal only one of these four orbitals for our basis set. This maes the problem easier and tractable analytically. The reason that we can ignore the other 3 is that the Pz orbital does not mi with the other ones (the structure is planar). So the problem can be separated into two parts: the one for Pz orbital; and the one for S, P, Py. At the end of the day it turns out that only the Pz orbitals play a major role in determining the electronic and optical properties. mpty f Filled Two Carbon Atoms in a Bond S, P, Py Pz Pz S, P, Py Levels around Fermi Level

73 Graphene: Unit Cell / Basis Vectors 4:5 To construct a basis set, the net question is: how many atoms are there per unit cell? In other words what is the minimal number of atoms that can be put together to construct a cell from which, the whole lattice can be constructed? The answer is: we need two atoms Basic Unit Cell R FG G a a a a b The lattice structure only repeats in pairs of! F 3a 3a a ˆ ˆ + y a a 3a 3a ˆ yˆ a a

74 Graphene: Schrödinger quation 3: Schrödinger equation reads: 3 4 n m { φ } [ ]{ φ } () n H nm m quation can be solved by: Where vector dn is: Putting () in (): m i d { φ } { φ } e n () n Unit cell n d n origin { } i d n i d m e [ H ]{ } e φ nm φ m { } i [ ] ( d d ) m n φ H nm e { φ } m [ h( )] d m Unit cell m

75 Graphene: h() 9: Let s rewrite h(): [ ( )] i [ ] ( dm dn h H ) nm e () m To evaluate h(), we choose any unit cell n and then perform the summation over its nearest neighbors including n itself: 3 a n a 4 m n, n, To write the phase factor notice that dm-dn for m is Actually a. Continuing lie this, we can write all of the 4 terms that run through neighbors to 4: m, m, t n d n origin d m d n d m m Phase factor is mn n, n, n, n, ε t t ε t e t e i a i + + t t e i e ( a ) ( a ) i a +

76 Graphene: h() / - relationship 4:5 All together we have the following by matri for Graphene: ε Δ h ( ) i a i a * Δ t e + e + Δ ε [ ( )] - Relationship eigenvalues : ε ± Δ ε

77 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University.8.4 Lecture 9: Bandstructure Ref. Chapter 5. & 5. Networ for Computational Nanotechnology

78 Review :5 The bandstructure principle helps us to understand the energy levels of large solids and serves as a bacground to understand the energy levels of nanostructures. For electrical conduction we need to understand the energy levels and we need to now where the Fermi level lies because the energy levels around the Fermi level determine the I-V characteristics. We always start from Schrödinger equation: Ψ ( r ) m As long as there is spherical symmetry for the potential we can reduce this equation to three D equations. Solving each equation numerically then involves turning the equation to a matri equation. To so we first set up a lattice of points. + U ( r ) Ψ ( r ) By doing this we have discretized the quantities we are dealing with. For eample the wavefunction will have a value at each lattice point. a Ψ n 3 n- n n+ N Notice that without the spherical symmetry the problem is not separable; hence we have to deal with the 3D equation which easily gets out of hand. To solve the problem we turn into another method, namely the idea of basis functions. Ψ( r ) ψ mum( r ) m

79 Basis Functions 6:7 A good eample for use of basis functions is to use it for the Hydrogen molecule. The separate Hydrogen atoms are have these s orbital. We can use these as our basis functions eeping in mind that the wavefunction of the Hydrogen molecule will be some linear combination of these two. u s,p ( r ) u ( r ) U s s,p s ( r ) U ( r ) + H + H Again Schrödinger equation becomes a matri equation: bn*bn With this bacground we get into the concept of bandstructure. Consider a D solid: y a t ε a n m The number of basis functions is b*n where b is the number of atoms and N is the number of orbitals per atom.

80 Periodic Matri :4 The principle of bandstructure helps us to find the eigenvalues of a periodic matri in a simple way. This is important for us because solids that we are interested in have periodic structures so the Hamiltonian matri loos periodic. For eample: φ φ φ N φ φ φ N This matri equation is a set of algebraic equations each of which can be written as: b*b b*b { φ } [ ]{ } n H nm φ m b m b b b N*N b orbitals per site a t a We can now thin of the matri as an N by N matri where each element is b by b. In doing so what we ve gained is the periodicity of the matri. Once the matri is periodic we can use the principle of bandstructure and find the energy levels in a relatively easy manner. To see this start with the wavefunction: we can write the solution to the Schrödinger equation as: n m { } { } i d m φ φ e m

81 Principle of Bandstructure 6:36 { φ } [ ]{ } n H nm φ m Substituting in, we have: Where: () b m b b b [ ]{ } { φ } h( ) { } { } i d φ φ e m () m φ [ ( )] i [ ] ( d m d n h H ) nm e m What has happened is that finding the eigenvalues of h() has now become more tractable. Since for a particular, h() is b*b we can easily find its eigenvalues. runs from to N based on the number of lattice points. So for each we get b*b eigenvalues. Altogether we have Nb*Nb eigenvalues which is eactly what we were after.

82 ample: D Solid 9: Let s apply what we ve learned to a special case. Consider a D lattice with orbital per unit cell. s s 3 n N The general epression for h() is: [ ( )] i [ ] ( dm dn h H e ) m nm Fi a point n and write the summation: h() becomes: [ ] () ia ia h ε+ te + te ε+ cos t a We can now draw the dispersion relation: a mn- mn mn+ N t ε a

83 ample: D Dimerized Solid 5:53 Let s apply this to another eample, namely the dimerized solid. a mn- mn t mn+ Again the general h() can be written as: [ ( )] i t [ ] ( d m d n h H ) nm e m h() for the dimerized solid can be written as: n, t n, m n m n m n+ m, m, ε t t ε t [ ] ia ia h ( ) + e + e [ h() ] ε Δ * ( ) Δ( ) ε ( ia t + t e ) Δ

84 Dispersion Relation for D Dimerized Solid 33:7 Net we want to find the eigenvalues. t ia ε t + [ ] te h( ) + ia + te igenvalues are: ε ε ± Δ Δ t + t e ia t t t t Δ Δ t + t t + t ε π π a

85 ample: D Square Solid 35: y Unit Cell Tae as shown origin ˆ + y yˆ n a ε n m t a and set the origin 3 d d d n 4 n Want to evaluate: valuating Hnn results in ε (self-energy) and evaluating Hnm results in a value that we call t. So we have 5 terms in our summation: h( ) Thus, after adding all of them we get: ε + ε + + i ( ) ( dm dn H e ) d h( ) ε + t( cos a + cos b) te h 4 te i m m i ( a ˆ ) te i ( ya ˆ ) + nm ( d d ) te m + n i ( te ya ˆ ) i ( a ˆ ) y

86 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University.8.4 Lecture : Reciprocal Lattice Ref. Chapter 5. Networ for Computational Nanotechnology

87 Dispersion Relation Quantized States :35 Given any periodic structure we ve discussed how to calculate the - relationship. For eample consider a D dimerized solid with two orbitals per unit cell a 3 3 N N LNa The number of allowed values of can be found as: π / a π / L Notice that the spacing between the states comes from the imposition of periodic boundary conditions. To see this consider: L a 3 n- n n+ N N N π a π L π + a e e ina ina i( e e n+ N ) a iπ (integer) Na πν ν : integer π ν Na

88 Brillouin Zone 8:6 To see why the values are bounded between pi/a and pi/a consider the simple dispersion relation of D solid: ε +t cosa N π a π L The two states give us the same wavefunction π + a The point is that if you tae any value of within the range and add pi/a to it, you will not get a new independent wavefunction. To see this consider the solution () ψ n ψ n e ina () ψ ψ ψ e ina i n e e π ina You can clearly see that and are the same. This is why we do not need to consider any values outside of the range pi/a pi/a. The point is that corresponding to any point outside the range there is a point within the range which is an integer multiple of pi/a from it. It is the same story for all of them. Add this amount to and you will get the same answer. This symmetric interval around states that gives us a complete set of values is called the first Brillouin zone.

89 Reciprocal Lattice in D 4:58 Any point in the real space can be written as: Where the general solution is: i r To construct the reciprocal lattice we need to find a vector K such that: i( + K ) r i r e the y a a ψ ( r ) ψ e e ik r e r Real Space Structure ma ˆ + n ˆ ya K r a (π )υ a a a Reciprocal Lattice First Brillouin zone K MA Any point in the reciprocal lattice can be written as: A To find K the general procedure is to find the vectors A and A that satisfy: A a A π A ˆ( π / a) A a A A yˆ( π / a) K r π/a π/a Γ N We can now chec to see if we have right answer: π ( Mm + Nn y π/ a + π π ( m a ˆ + nyˆ a) ( ˆ + yˆ ) a a )

90 - Diagram and Conduction 36:9 F Conduction Band Conduction valleys G We are studying this course to understand the electrical properties of semiconductors. Looing at this - diagram, you can see that the chemical potential lies in the gap between the bands. Ordinary there is no conduction. But if we move the levels relative to f via applying a voltage we get conduction. Valence Band π / a + π / a

91 FCC Lattice to BCC Reciprocal Lattice 4:4 FCC in Real Space BCC in Reciprocal Space Brillouin Zone in Reciprocal Lattice μ L Γ X

92 Review of Reciprocal Lattice In the last class we learned how to construct the reciprocal lattice. For D w have: Real-Space: a R ˆ a In general for periodic structures we can write 3 basis vectors such that any point in the lattice can be written as a linear combination of them with the condition that the coefficients must be integers. R -Space: - /a m a..4 BZ / a K ˆ / n a p a a 3 :5 Similarly any point in the reciprocal lattice can be written as: K MA NA P A 3 How are the vectors A related to vectors a? The defining condition is: A j a i ij The significance of reciprocal lattice vectors A is that points in space which are apart from each other by an integer multiple of Ai s, give is the same wavefunction solution. ij ij for for i i j j

93 Graphene Graphene is made up of carbon atoms bonded in a heagonal D plane. Graphite is 3D structure that is made up of wealy coupled Graphene sheets. This is of particular importance because carbon nanotubes are made up of a Graphene sheet that is rolled up lie cylinder. Carbon nanotubes themselves are of interest because people believe they can mae all inds of Nano devices with them...4 6:5

94 Reciprocal Lattice in 3D 8:3 Semiconductors of interest to us have what is called a diamond structure. The diamond structure is composed of to interpenetrating FCC lattices the following way: Imagine two FCC lattices such that each atom of each lattice is on top of the corresponding atom of the other lattice. You should only be seeing FCC lattice as of now. Then fi one lattice and move the other one in the direction of the body diagonal of the fied one by ¼ of the body diagonal. Now you ve yourself a diamond lattice. If the two FCC lattices are made up of two different types of atoms, the structure is then called a Zinchblend lattice. To visualize the reciprocal lattice focus only on one FCC lattice in the diamond structure. FCC in Real Space BCC in Reciprocal Space Brillouin Zone in Reciprocal Lattice..4

95 - Diagrams for 3D Reciprocal Lattices Since the reciprocal space is now 3 dimensional, to draw the - diagram we have choose particular directions and draw - diagram along those directions: L X..4 :6 Some useful information: The top of the valence band usually occurs at the Gamma point (). The bottom of conduction band however does not always lie at. For eample consider Silicon: If both conduction band minimum and the valence band maimum lie at the same value of, the material is called a direction bandgap semiconductor. Other wise the material is indirect lie Si.

96 Parabolic Approimation Usually, it is necessary to derive an epression for (, y, z ) about the conduction points of a bul solid For silicon, use the parabolic approimation c m * ( m y * z ) where m* is the effective mass. For nanotubes we can derive a similar parabolic epression via a Taylor series epansion that approimates the subbands near the conduction valleys..4 7:8 Silicon Parabolic Conduction Band Approimation Approimation y

97 - Relation for Graphene Let s get bac to Graphene. First identify the basic unit cell Basic Unit Cell The lattice structure only repeats in pairs of!..4 :5 Remember the general result of principle of bandstructure: h i d m d n h H nm e m To write h() consider one unit cell an its nearest neighbors. Figure shows that there will be 5 terms in the summation for h(). a b a a

98 Graphene - Diagram Remember the general result of principle of bandstructure: h i d m d n h H e m nm To write h() consider one unit cell an its nearest neighbors. Figure shows that there will be 5 terms in the summation for h(). a b a a..4 3:45 Writing the summation terms and adding them up we get: Where h t h ( The eigenvalues of this matri are given by: ) h h * i a i a e e h ( ) Conduction Point { filled states Conduction Point

99 Magnitude of h() 9:5 Net we lie to locate the conduction points in the dimensional space: h t a a a i a i a e e y Unit Cell a a ˆ by ˆ a a ˆ by ˆ ˆ y y ˆ i a y b i a y b t e e 3 3 a a ˆ ˆ 3 3 a a a e i cos b t y y ˆ y ˆ To find the conduction points we need to set h(). So we need to find h() :..4 so, h h ( ) h t h * t 4 cos 4 cos a cos b 4 cos b a cos y b y 4 cos y b y

100 Conduction Valleys Now let ( ) t 4 cos a cos y b 4 cos b h y Let a and investigate h() as a function of y. cos b for h t y b to get h () y Let api and investigate h() as a function of y. cos h t y b for 3 y b to get h () 3..4 Conduction Valley Conduction Valley (, /3b) y (/a,/3 b) (/a,-/3b) 38:35

101 Two Full Valleys The si Brillouin valleys really only give independent valleys, e.g. in each group of 3 that are in the picture two of the valleys are away form the other by a reciprocal lattice unit vector; hence represent the same state. One can thin that each corner in the st Brillouin zone contributes /3 rd./3 6 (left figure). Alternatively we can translate two of the corners in each group to get the full valleys on the right. 3 Dispersion relation along y. h t cos y b for..4 y 3 3 Translating two of the corners in each group of 3 Conduction Valley 43:45 y h ( ) Conduction +3 t Valley -3 t y

102 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University..4 Lecture : Graphene Bandstructure Ref. Chapter 6. Networ for Computational Nanotechnology

103 Review of Reciprocal Lattice :5 In the last class we learned how to construct the reciprocal lattice. For D w have: Real-Space: -Space: - π/a a BZ π/ a R In general for periodic structures we can write 3 basis vectors such that any point in the lattice can be written as a linear combination of them with the condition that the coefficients must be integers. R..4 m a + n a + p a 3 ˆ a K ˆ π / a Similarly any point in the reciprocal lattice can be written as: K MA + NA + PA3 How are the vectors A related to vectors a? The defining condition is: A j a i πδ ij The significance of reciprocal lattice vectors A is that points in space which are apart from each other by an integer multiple of Ai s, give is the same wavefunction solution. δ ij ij for i δ for i j j

104 Graphene 6:5 Graphene is made up of carbon atoms bonded in a heagonal D plane. Graphite is 3D structure that is made up of wealy coupled Graphene sheets. This is of particular importance because carbon nanotubes are made up of a Graphene sheet that is rolled up lie cylinder. Carbon nanotubes themselves are of interest because people believe they can mae all inds of Nano devices with them...4

105 Reciprocal Lattice in 3D 8:3 Semiconductors of interest to us have what is called a diamond structure. The diamond structure is composed of to interpenetrating FCC lattices the following way: Imagine two FCC lattices such that each atom of each lattice is on top of the corresponding atom of the other lattice. You should only be seeing FCC lattice as of now. Then fi one lattice and move the other one in the direction of the body diagonal of the fied one by ¼ of the body diagonal. Now you ve yourself a diamond lattice. If the two FCC lattices are made up of two different types of atoms, the structure is then called a Zinchblend lattice. To visualize the reciprocal lattice focus only on one FCC lattice in the diamond structure. FCC in Real Space BCC in Reciprocal Space Brillouin Zone in Reciprocal Lattice..4

106 - Diagrams for 3D Reciprocal Lattices :6 Since the reciprocal space is now 3 dimensional, to draw the - diagram we have choose particular directions and draw - diagram along those directions: Some useful information: The top of the valence band usually occurs at the Gamma point (). The bottom of conduction band however does not always lie at. For eample consider Silicon: L Γ X..4 If both conduction band minimum and the valence band maimum lie at the same value of, the material is called a direction bandgap semiconductor. Other wise the material is indirect lie Si.

107 Parabolic Approimation 7:8 Usually, it is necessary to derive an epression for (, y, z ) about the conduction points of a bul solid Silicon Parabolic Conduction Band Approimation For silicon, use the parabolic approimation c + m* ( + y m* + z ) Approimation where m* is the effective mass. For nanotubes we can derive a similar parabolic epression via a Taylor series epansion that approimates the subbands near the conduction valleys y..4

108 - Relation for Graphene :5 Let s get bac to Graphene. First identify the basic unit cell The lattice structure only repeats in pairs of! Basic Unit Cell Remember the general result of principle of bandstructure: { φ } [ ( )]{ } h φ [ ( )] i [ ] ( d m d n h H ) nm e m To write h() consider one unit cell an its nearest neighbors. Figure shows that there will be 5 terms in the summation for h(). a a b a..4

109 Graphene - Diagram 3:45 Remember the general result of principle of bandstructure: { φ } h( ) To write h() consider one unit cell an its nearest neighbors. Figure shows that there will be 5 terms in the summation for h()...4 [ ]{ } φ [ ( )] i [ ] ( d m d n h H e ) m nm a a a b Writing the summation terms and adding them up we get: Where h t ε h( ) h * The eigenvalues of this matri are given by: ε h ε ( ) i a i a + e + e ( ) ε ± h ( ) Conduction Point { filled states Conduction Point

110 ..4 Magnitude of h() 9:5 Net we lie to locate the conduction points in the dimensional space: To find the conduction points we need to set h(). So we need to find h() : Unit Cell a a y y a a by a a y a a by a a ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ a ( ) ( ) ( ) b a i b a i y y e e t ( ) a i a i e e t h + + y y ˆ ˆ + ( ) b e t y a i cos + ( ) b b a t h b b a t h h h y y y y * 4 cos cos 4 cos ) ( so, 4 cos cos 4 cos

111 Conduction Valleys 38:35 Now let h ( ) t + 4 cos a cos yb + 4 cos yb Let a and investigate h() as a function of y. ( + cos b) for h t y π y b to get h () 3 Let api and investigate h() as a function of y. h ( cos b) for t π y y b to get h () π 3..4 Conduction Valley Conduction Valley (, π/3b) y (π/a,π/3 b) (π/a,-π/3b)

112 Two Full Valleys 43:45 The si Brillouin valleys really only give independent valleys, e.g. in each group of 3 that are in the picture two of the valleys are away form the other by a reciprocal lattice unit vector; hence represent the same state. One can thin that each corner in the st Brillouin zone contributes /3 rd./3 6 (left figure). Alternatively we can translate two of the corners in each group to get the full valleys on the right. Dispersion relation along y. h t + cos yb for..4 3 ( ) y 3 ε 3 Translating two of the corners in each group of 3 ε ± Conduction Valley h ( ) ε+3 t y Conduction Valley ε-3 t y

113 Graphite Review In the last couple of sessions we ve been taling about graphite. By rolling up a sheet of graphite we get a carbon nanotube which is of great interest because of its possible applications for maing of transistors, electronic sensors, etc. Let s review what we ve learned about graphene: a a y Unit Cell a a a ˆ a ˆ by ˆ by ˆ 3 3 h h a a ˆ ˆ 3 3 * h a a y ˆ y ˆ :3 Dispersion relation: h cos b for t y Conduction Valley h ( ) +3 t Conduction Valley y -3 t Conduction Valley (, /3b) (/a,/b) Conduction Valley y

114 Graphite Review 9:5 Recall, the si Brillouin valleys really only give independent valleys, e.g. in each group of 3 that are in the picture two of the valleys are away form the other by a reciprocal lattice unit vector; hence represent the same state. One can thin that each corner in the st Brillouin zone contributes /3 rd./3 6 (Left). Alternatively we can translate two of the corners in each group to get the full valleys on the right. Points and are separated by /a, so they really are one and the same point. Graphically, we visualize and as ½ points in -space 3 y Recall:D 3 h ( ) t y 3 Translating two of the corners in each group of 3 a e i cos b a y Point Point a

115 Carbon Nanotube Boundary Conditions : For a sheet of material or a D surface, the correct boundary conditions are the infinite potentials at the edges but for mathematical convenience people usually use periodic boundary conditions. What saves us is that if one is interested in the bul properties at the center of the sheet will not depend what s happening at the edges. But for nanostructures this cannot be done because what s happening at the edges can effect the middle of the solid so Periodic Boundary Conditions (PBC) cannot be used. Carbon nanotubes are a case where PBC is real because we roll up Graphene and in doing so boundary points at one edge overlap with the boundary points at the other edge; hence the periodicity. PBC : L The question is how to impose PBC in a two dimensional case. L

116 Rolling Up Graphite a a y Unit Cell a a by ˆ a a ˆ by ˆ ˆ and It is clear that Graphene can be rolled up into a nanotube in various ways depending on how one rolls it up. So how can we classify or define the nanotubes? The idea is that in the process of rolling up Graphene, a unit cell gets superimposed on another. One uses the vector that connects the two unit cells in Graphene to describe the corresponding nanotube. This vector is called the circumferential vector and can be written as a combination of vectors a and a with c ma n a integer factors: Type of nanotube is described by : (m, n) 4:4 An initial simple eample is to roll in the direction. For instance, create a tube described as: (,) We use the circumferential vector c a ˆ c am ˆ to denote the direction and length. Here, but in general wheremis an integer. This will result in RAL periodic B.Cs because each point on graphite will coincide with a similar one after being rolled up. (Notice that the magnitude of gives you the circumference of a cross section of the tube hence the name circumferential vector.) c

117 Zigzag/Armchair For the most part only two types of folding are of interest. The first, a fold in the direction resulting in the ŷ zig-zag nano-tube Zigzag (m, ) Armchair ˆ since the circumferential edge loos lie a zig-zag. The second, a fold in the direction resulting in the armchair nano-tube ( m, m ) 8:9

118 PBC Subbands Discrete nergy Levels :5 ˆ ˆ c The periodic B.C along the circumference requires that ( is an integer) c ma ˆ ( ) ˆ na a m n yb ( m n ) y y c a ( m n ) b ( m n ) y v v Subbands in Direction c ma n a Commonly, the carbon nano-tube - diagrams, about the conduction points, given by this model loo lie: y y

119 Carbon Nanotube: lectrical Conduction 8: A nano-tube will only conduct if one of its subbands passes through the si corners of the Brillouin Zone. In particular consider the corner mared by the red dot: Conduction Valley h ) t ( 3 b b h ( ) Conduction Valley i a e cos b y y Conduction Valley Conduction Valley (, /3b) y (/a,/b ), / 3 b : coordinate of the corner quation of the line: c a ( m n ) y b ( m n ) ( m n ) A nanotube is described by the pair of integers (m, n) m 3 n : integer Condition for Conduction

120 Taylor Approimation Overview 39:3 Net let s try to simplify the epression for h(). The following simplification is based on Taylor series epansion and it is only valid close to the conduction valleys. Conduction Valley (, /3b) (/a,/b) Conduction Valley Conduction Valley y Conduction Valley The approimation will result in these tangent y lines close to the conduction valleys

121 Taylor Approimation 4:4 To do the approimation consider the - relation: h ( ) b t 4cos a b which after simplification can cos y 4cos y be written as ta y y 3a where y, a 3 b So, to approimate the energy epression we Taylor epand h about the conduction valleys (, y ) (, ()/3b) h i 3 at where y h 3 a (, y t 3 i a t 3 i y b ) 3 b y 3 b 3 b h y (, y 3, b )

122 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University..4 Lecture : Carbon Nanotubes Ref. Chapter 6. Networ for Computational Nanotechnology

123 Graphite Review :3 In the last couple of sessions we ve been taling about graphite. By rolling up a sheet of graphite we get a carbon nanotube which is of great interest because of its possible applications for maing of transistors, electronic sensors, etc. Let s review what we ve learned about graphene: a a a a aˆ + byˆ aˆ byˆ [ h( )] 3 3 ε h a a ˆ + ˆ 3 3 * h ( ) ( ) ε y a a Unit Cell yˆ yˆ Dispersion relation: h ε ( + cos b) for t y Conduction Valley Conduction Valley Conduction Valley ε ± ε+3 t ε-3 t y h ( ) Conduction Valley (, π/3b) y (π/a,π/b)

124 Graphite Review 9:5 Recall, the si Brillouin valleys really only give independent valleys, e.g. in each group of 3 that are in the picture two of the valleys are away form the other by a reciprocal lattice unit vector; hence represent the same state. One can thin that each corner in the st Brillouin zone contributes /3 rd./3 6 (Left). Alternatively we can translate two of the corners in each group to get the full valleys on the right. Points and are separated by π/a, so they really are one and the same point. Graphically, we visualize and as ½ points in -space 3 y 3 3 Recall:D h ( ) t + Translating two of the corners in each group of 3 ( i a e cos b) π a y Point Point y + π a

125 Carbon Nanotube Boundary Conditions : For a sheet of material or a D surface, the correct boundary conditions are the infinite potentials at the edges but for mathematical convenience people usually use periodic boundary conditions. What saves us is that if one is interested in the bul properties at the center of the sheet will not depend what s happening at the edges. But for nanostructures this cannot be done because what s happening at the edges can effect the middle of the solid so Periodic Boundary Conditions (PBC) cannot be used. Carbon nanotubes are a case where PBC is real because we roll up Graphene and in doing so boundary points at one edge overlap with the boundary points at the other edge; hence the periodicity. PBC : L πν The question is how to impose PBC in a two dimensional case. L

126 Rolling Up Graphite 4:4 a a a aˆ + byˆ and a Unit Cell It is clear that Graphene can be rolled up into a nanotube in various ways depending on how one rolls it up. So how can we classify or define the nanotubes? The idea is that in the process of rolling up Graphene, a unit cell gets superimposed on another. One uses the vector that connects the two unit cells in Graphene to describe the corresponding nanotube. This vector is called the circumferential vector and can be written as a combination of vectors a and a with integer factors: c ma + na Type of nanotube is described by : (m, n) y aˆ byˆ An initial simple eample is to roll in the direction. For instance, create a tube described as: (,) We use the circumferential vector c to denote the direction and length. Here, but in general c aˆ c amˆ where m is an integer. This will result in RAL periodic B.Cs because each point on graphite will coincide with a similar one after being rolled up. (Notice that the magnitude of gives you the circumference of a cross section of the tube hence the name circumferential vector.)

127 Zigzag/Armchair 8:9 Armchair For the most part only two types of folding are of interest. The first, a fold in the ŷ direction resulting in the zig-zag nano-tube Zigzag (m,) since the circumferential edge loos lie a zig-zag. The second, a fold in the ˆ direction resulting in the armchair nano-tube ( m, m )

128 PBC Subbands Discrete nergy Levels :5 c πv The periodic B.C along the circumference requires that c ma ˆ ( ) ˆ + na a m + n + yb( m n c ˆ + yˆ y a( m + n) + b( m n) y ) v ( is an integer) πν Subbands in Direction c ma + n a Commonly, the carbon nano-tube - diagrams, about the conduction points, given by this model loo lie: y y

129 Carbon Nanotube: lectrical Conduction 8: A nano-tube will only conduct if one of its subbands passes through the si corners of the Brillouin Zone. In particular consider the corner mared by the red dot: ε Conduction Valley h ( ) t π 3b ε ± h ( ) Conduction Valley ( i a + e cos b) quation of the line: b ( m n ) y πν y c A nanotube is described by the pair of integers (m, n) m Conduction Valley Conduction Valley (, π/3b) y (π/a,π/b ) (,π / 3b) : coordinate of the corner 3 a( m + n) + b( m n) n y ν : integer πν Condition for Conduction

130 Taylor Approimation Overview 39:3 Net let s try to simplify the epression for h(). The following simplification is based on Taylor series epansion and it is only valid close to the conduction valleys. Conduction Valley (, π/3b) (π/a,π/b) ε Conduction Valley Conduction Valley y Conduction Valley y The approimation will result in these tangent lines close to the conduction valleys

131 Taylor Approimation 4:4 To do the approimation consider the - relation: which after simplification can be written as ± ) ( h ε b b a t y y 4cos cos 4cos + + ± ε y ta β ε + ± ( ) 3a a, 3 where ± b y β y π So, to approimate the energy epression we Taylor epand h about the conduction valleys (, y ) (, ±(π)/3b) ( ) ( ) ( ) ( ) b i t a i b t a at i h b h h y y y y b y y b 3 where, ) 3 (, ) 3 (, π β β π π π π ± ± ± ± ± + ± ±

132 Fundamentals of Nanoelectronics Prof. Supriyo Datta C 453 Purdue University.5.4 Lecture 3: Subbands Ref. Chapter 6. Networ for Computational Nanotechnology

133 Review :5 a a h [ ( )] h * h ( ) i. a i. a h ( ) t + e + e, ε ± a a h aˆ + byˆ aˆ byˆ ( ) Conduction Valley Conduction Valley (, π/3b) y (π/a,π/3 b) a t( + e cos ) iat + at ε Unit Cell Conduction Valley π b i y y 3 Conduction Valley y

134 Taylor pansion 4:3 Recall, given ε ± h we Taylor epand h about the conduction points panding and solving ( ) ( ) ) 3 (, ) 3 (, 3 b y y b h b h h π π π + Finally we get The above approimation describes quite well the behavior of semiconducting and conducting nanotubes. This, of course, is true because all the action (electrical and optical effects) tend to occur at or close to the conduction point. ( ) ( ) b ta i h y y y 3 where π β β + ( ) ( ) ( ) ta a t tb tb h ita a t i h b tbe h b e ia t h b y b y a i y y a i sin, cos 3, 3, π π ( ) + + b at iat e t h y y a i 3 cos ) ( π

135 Taylor Approimation 9: ε h( ) Conduction Valley t ( a ) + e cos iat + at Conduction Valley y π b i y y 3 3ta ε ± + y π 3b This approimation wors quit well and will result in these tangent lines close to the conduction valleys. This, of course, is true because all the action (electrical and optical effects) tend to occur at or close to the conduction point.

136 Folding Graphene / Zigzag Nanotube :3 The periodic boundary conditions along the circumference requires that c πν where v is an integer. c ma ˆ ( ) ˆ + na a m + n + yb( m n) c ˆ + yˆ y πν Zigzag nanotube ŷ A fold in the direction has the circumferential vector c mb yˆ, where m is an integer, and the resulting subbands loo something lie y ( bm) πυ πυ y bm y y

137 Zigzag Nanotube: - Plot 8:5 y A general energy epression for a ŷ fold, zigzag, nanotube is: v ( where ) ε ± at v π 3b A nanotube will only conduct if one of its subbands pass through the si corners of the Brillouin Zone. The condition for conduction is πv π mb 3b or v m 3 Therefore only if m is a multiple of 3 then conduction will in a zigzag nanotube. 3v m + v Semi-Conducting Nanotube Conducting Nanotube m 3υ m 3υ v (m)/3 + v (m)/3 + v (m)/3, π 3b Gap

138 Semi-Conducting Gap 5:5 The minimum is then: Semi-Conducting Nanotube m 3υ Gap a π at + where a 3b m 3 π 3v v ( ) ε ± at + How can we estimate the semi-conducting band gap? This can be done by finding the smallest value of the second term in the equation above: at + 3b m π 3v m 3b m is minimized if: 3 ν m The quantity above represents half of the band gap shown in the figure given that. 3a t 4 g 3 d d: diameter of nanotube a d g Let s put some numbers in to find g for a carbon nanotube with diameter: d8 Å. ( m)(.5ev ).4 g. 5eV t ( 8 m)

139 Confinement Subbands 33:53 a a y πυ y bm Unit Cell As long as we have a sheet of a material, the confinement is only in dimension. Then all values on the, y plane are allowed. The - plot will then be 3 dimensional because at each point in the plane, there is a corresponding energy value. The imposition of periodic boundary conditions by folding a sheet of graphite to a carbon nanotube -for eample a zigzag nanotubemaes only certain values of y acceptable. These are the horizontal lines shown in the figure. Increasing the diameter of nanotube will mae the lines closer to each other and decreasing the diameter will mae the lines further apart. In the limit of very big diameters, we get the same thing as a sheet of graphite: the acceptable lines get really close to each other and it starts to loo lie a bul piece of graphite sheet namely: Graphene.

140 Dimensional Confinement 4: How do we loo at this process of dimensional confinement in a general manner? Where does the carbon nano-tube fit in? Consider the well nown bul solid Assuming periodic boundary conditions for L z, we get z (vπ) / L z and our - function is v (, y ) (, y, z (πv)/l z ) (where v is an integer) Similarly, constraining along the y direction results in a quantum wire L z L y Bul Solid without any constraints, very long, wide and deep. It has a general - behavior epressed by the function (, y, z ) If we constrain the bul solid in one direction, say z, to a comparably short length, L z, we get what is nown as a quantum well... Quantum Wire For the quantum wire we have y (v π)/l y and the - function ( ) v π vπ v, v, y, z Ly Lz Note: a carbon nanotube is really, in the general sense, a form of quantum wire!

141 More Dimensional Confinement 46:36 Finally, confinement in the -direction as well leads to a quantum dot L z L The quantum dot includes (v π)/l such that the - behavior is given by v π v π vπ v, v, v, y, z L Ly Lz The quantum energy levels are discretized in the same way as those of an atom and so quantum dots are often referred to as artificial atoms L y Quantum Dot One important question, when do constraints begin to lead to quantization of the bandstructure? ssentially, quantization depends on the thermal energy B T. Because the thermal energy tends to smooth out the difference between energy levels, the discretization corresponding to /y/z (qπ)/l /y/z must be less than or comparable to B T to eperimentally (and hence physically) matter. Note: this is often the motivation for conducting eperiments at very low temperatures