1.5 Eigenvalue Problems

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1 5 Eigenvalue Problems The eigenvalue problem, for matrices, reads: Given a matrix A 2 IR n n,find some/all of the set of vectors {v i } n i= and numbers { i} n i= such that: A v i = i v i (45) In general, for a vector y, the linear operation (matrix-vector multiplication) A y can be thought of in terms of rotations and stretches of y The eigenvectors, v i,arethespecialvectorsforwhicha v i is parallel to v i, only the length has changed There are continuous versions of the eigenvalue problem the time-independent form of Schrödinger s equation, for a particle of mass m moving in the presence of a potential energy V (x) (in one spatial dimension) is apple d 2 + V (x) (x) =E (x), (46) dx2 where we have some sort of boundary condition, like (±)! 0 We solve this equation for both (x) and E the left-hand side of the equation represents a linear di erential operator acting on (x), and if we can satisfy the equation, the e ect of that linear operator is to scale (x) by E this is a continuous eigenvalue problem, with (x) the eigenfunction and E the eigenvalue What we will do is replace the di erential operators with finite di erence approximations, turning the continuous eigenvalue problem presented by Schrödinger s equation into a discretized matrix eigenvalue problem of the form (45) Introduce a grid x j = j x with x fixed and j =0,,N+ (the left-hand boundary is at x =0with j =0, the right-hand boundary will be at x f,with j = N + this means we can again take x = x f /(N + )) Then, letting j (x j ), and recalling the derivative approximation in (32), the projection of Schrödinger s equation onto the grid is: and this holds for j =2! N wave function vanishes (so that j =: j+ 2 j + j x 2 + V (x j ) j E j, (47) We ll assume that at the boundaries, the! gets remapped to 0! x f ), then for 2 2 x 2 + V (x ) = E, (48) 2

2 and for j = N, 2 N + N x 2 + V (x N ) N = E N, (49) and once again, we can box up the approximation in a matrix-vector multiplication let j be the approximation to j that is our target, and define the vector form: 0 = B 2 n A (50) The matrix that encapsulates (47) and the boundary points is, letting p x 2, 0 2 p + V (x ) p 0 0 p 2 p + V (x 2 ) p 0 H = 0 p 2 p + V (x 3 ) p 0 B 0 0 A 0 0 p 2 p + V (x N ) (5) This is again tridiagonal, and can be made relatively easily using your function(s) from the previous section Now we have turned the continuous eigenvalue problem into a discrete one we want and Ē that solve H = Ē, (52) a matrix eigenvalue problem If you can construct the matrix H, then you can use the built-in command Eigensystem in Mathematica to get the eigenvalues (the set of energies) and eigenvectors (the associated wave functions) of the matrix 5 Example For a particle in a box of length a, wehave: d 2 (x) =E (x), (53) dx2 22

3 with (0) = (a) =0 Taking x = x 0 X, we can render the equation dimensionless, letting Ẽ x2 0 E be the dimensionless energy, we have d 2 (X) dx 2 = Ẽ (X), (54) with (0) = 0 and (a/x 0 )=0, suggesting we take x 0 = a The full solution to this problem is, of course: (X) =A sin(pẽx) (55) for constant A, which will be set by normalization, and with pẽ = n for integer n (to get () = 0), so that in these units, Ẽ = n 2 2, or, going back to the original units in the problem: E = ~2 a Ẽ = ~2 n a 2 If we wanted to solve this problem numerically, we would approximate the derivative in (54) with a finite di erence The dimensionless spatial coordinate X runs from 0!, soletx j = j X with X =/(N + ) for N the number of grid points Let j = (X j ) be the unknown values associated with on the spatial grid Then using our familiar finite di erence approximation to the second derivative, (54) becomes: j+ 2 j + j X 2 = Ẽ j (56) for j =2! N At j =and j = N we need to enforce our boundary conditions: 0 =0and N+ =0, so those two special cases have: If we define the vector 2 2 X 2 = Ẽ 2 N + N X 2 = Ẽ N 0 = B then the discretized problem can be written as 2 N (57) A, (58) H = Ẽ (59) 23

4 with H = X 2 0 B (60) A ( similar to (36)) In Algorithm 7, we see the pseudocode that generates the matrix H given X f (here, one) and the size of the grid, N Algorithm 7 Pbox(X f,n) X X f /(N + ) H N N table of zeros H 2/ X 2 H 2 / X 2 for j =2! N do H jj / X 2 H jj 2/ X 2 H jj+ / X 2 end for H NN / X 2 H NN 2/ X 2 return H 52 Hydrogen To get the spectrum of hydrogen, we just need to put in the appropriate potential energy for an electron and a proton interacting electrostatically, we have V (x) = e x (6) where e is the charge of the electron We can continue to work on the half - line, letting x =0!, and we want to solve d 2 (x) e 2 dx x (x) =E (x) (62) with (0) = 0 and () =0 24

5 Let x = x 0 X for dimensionless X and where x 0 has dimension of length Then (62) can be written: d 2 (X) dx x 0 e X (X) =x2 0 E (X), (63) and we ll clean up the potential term by requiring that mx 0 e 2 =, (64) 4 0 which serves to define x 0 Letting Ẽ = x2 0 E (so that Ẽ is dimensionless), the dimensionless form of (62) reads: d 2 (X) dx 2 2 X (X) =Ẽ (X) (65) When we discretize in position, we cannot go all the way out to spatial infinity so we pick a large value of X (since X is dimensionless, large means X ) Let X f be the endpoint of our grid, then X = X f /(N + ) The discretized form of (65) is (for j = (X j ),anunknownvaluefor evaluated at X j = j X) j+ 2 j + j X 2 2 X j j = Ẽ j (66) and this is for j =2! N boundary condition) to get forj =,weusethefactthat 0 =0(a 2 2 X 2 2 X = Ẽ, (67) and similarly for j = N, with N+ =0(the boundary at infinity ), we have 2 N + N X 2 2 X N N = Ẽ N (68) Problem 5-2 Find the value of x 0 from (64), this is a characteristic length for hydrogen For ~ this value of x 0, find the energy scale 2 that you will use in converting the x 2 0 dimensionless Ẽ back to E What is the value of this energy in electron-volts? 25

6 Problem 5- Indicate the non-zero entries of the matrix associated with the dimensionless hydrogen problem ie take (66), (67), (68) and write the entries of the matrix H appearing in the discrete eigenvalue problem: H = Ẽ where 0 = B 2 N A (69) Problem 50 The finite di erence (56) can be solved analytically Take the ansatz: j = Ae i nj X,insertin j+ 2 j + j X 2 = Ẽn j, (70) and find Ẽn takethelimitas that you recover n 2 2 X! 0 (using you-know-what!) and show Problem 5 Generate the appropriate matrix H for the particle-in-a-box problem from Algorithm 7 using X f =and N = 00 Find the eigenvalues using Eigenvalues Sort your list (using Sort) so that you get the values from smallest to largest what is the di erence between the first two smallest eigenvalues and their analytic values from your solution to Problem 50? Problem 52 Modify your matrix from the previous problem so that its entries come from your solution to Problem 5- ie make a matrix that reflects the content of (66), (67), and (68) Using X f = 30 and N = 0000, buildthematrixandfind its eigenvalues using Eigenvalues List the first four (sorted) eigenvalues (they should be negative) using evals[[;;4]] (if you called your sorted list evals) Guess the form of these eigenvalues if you associate the lowest one with n =, the second with n = 2 and so on These are dimensionless eigenvalues 26

7 restore dimensions of energy using your result from Problem 5-2, and express the spectrum of hydrogen, E n =? in ev 27

Physics 202 Laboratory 5. Linear Algebra 1. Laboratory 5. Physics 202 Laboratory

Physics 202 Laboratory 5. Linear Algebra 1. Laboratory 5. Physics 202 Laboratory Physics 202 Laboratory 5 Linear Algebra Laboratory 5 Physics 202 Laboratory We close our whirlwind tour of numerical methods by advertising some elements of (numerical) linear algebra. There are three