Computational Physics - M. Jensen Episode 2 Part 4 ppsx

12.3. SIMULATION OF MOLECULAR SYSTEMS 229 protons. In Fig. 12.5 we show a plot of the potential energy (12.77) Here we have fixed og , being 2 and 8 Bohr radii, respectively. Note that in the region between (units are in this figure, with ) and the electron can tunnel through the potential barrier. Recall that og correspond to the positions of the two protons. We note also that if is increased, the potential becomes less attractive. This has consequences for the binding energy of the molecule. The binding energy decreases as the distance increases. Since the potential is symmetric with nm eV nm [eV] 86 42 0 -2-4 -6-8 0 -10 -20 -30 -40 -50 -60 Figure 12.5: Plot of for =0.1 and 0.4 nm. Units along the -axis are . The straight line is the binding energy of the hydrogen atom, eV. respect to the interchange of and it means that the probability for the electron to move from one proton to the other must be equal in both directions. We can say that the electron shares it’s time between both protons. With this caveat, we can now construct a model for simulating this molecule. Since we have only one elctron, we could assume that in the limit , i.e., when the distance between the two protons is large, the electron is essentially bound to only one of the protons. This should correspond to a hydrogen atom. As a trial wave function, we could therefore use the electronic wave function for the ground state of hydrogen, namely (12.78) 230 CHAPTER 12. QUANTUM MONTE CARLO METHODS Since we do not know exactly where the electron is, we have to allow for the possibility that the electron can be coupled to one of the two protons. This form includes the ’cusp’-condition discussed in the previous section. We define thence two hydrogen wave functions (12.79) and (12.80) Based on these two wave functions, which represent where the electron can be, we attempt at the following linear combination (12.81) with a constant. 12.3.2 Physics project: the H molecule in preparation 12.4 Many-body systems 12.4.1 Liquid He Liquid He is an example of a so-called extended system, with an infinite number of particles. The density of the system varies from dilute to extremely dense. It is fairly obvious that we cannot attempt a simulation with such degrees of freedom. There are however ways to circum- vent this problem. The usual way of dealing with such systems, using concepts from statistical Physics, consists in representing the system in a simulation cell with e.g., periodic boundary conditions, as we did for the Ising model. If the cell has length , the density of the system is determined by putting a given number of particles in a simulation cell with volume . The density becomes then . In general, when dealing with such systems of many interacting particles, the interaction it- self is not known analytically. Rather, we will have to rely on parametrizations based on e.g., scattering experiments in order to determine a parametrization of the potential energy. The in- teraction between atoms and/or molecules can be either repulsive or attractive, depending on the distance between two atoms or molecules. One can approximate this interaction as (12.82) where are some integers and constans with dimension energy and length, and with units in e.g., eVnm. The constants and the integers are determined by the constraints 12.4. MANY-BODY SYSTEMS 231 [nm] 0.50.450.40.350.30.25 0.007 0.006 0.005 0.004 0.003 0.002 0.001 0 -0.001 Figure 12.6: Plot for the Van der Waals interaction between helium atoms. The equilibrium position is nm. that we wish to reproduce both scattering data and the binding energy of say a given molecule. It is thus an example of a parametrized interaction, and does not enjoy the status of being a fundamental interaction such as the Coulomb interaction does. A well-known parametrization is the so-called Lennard-Jones potential (12.83) where eV and nm for helium atoms. Fig. 12.6 displays this interaction model. The interaction is both attractive and repulsive and exhibits a minimum at . The reason why we have repulsion at small distances is that the electrons in two different helium atoms start repelling each other. In addition, the Pauli exclusion principle forbids two electrons to have the same set of quantum numbers. Let us now assume that we have a simple trial wave function of the form (12.84) where we assume that the correlation function can be written as (12.85) with being the only variational parameter. Can we fix the value of using the ’cusp’-conditions discussed in connection with the helium atom? We see from the form of the potential, that it 232 CHAPTER 12. QUANTUM MONTE CARLO METHODS diverges at small interparticle distances. Since the energy is finite, it means that the kinetic energy term has to cancel this divergence at small . Let us assume that electrons and are very close to each other. For the sake of convenience, we replace . At small we require then that (12.86) In the limit we have (12.87) resulting in and thus (12.88) with (12.89) as trial wave function. We can rewrite the above equation as (12.90) with For this variational wave function, the analytical expression for the local energy is rather simple. The tricky part comes again from the kinetic energy given by (12.91) It is possible to show, after some tedious algebra, that (12.92) In actual calculations employing e.g., the Metropolis algorithm, all moves are recast into the chosen simulation cell with periodic boundary conditions. To carry out consistently the Metropolis moves, it has to be assumed that the correlation function has a range shorter than . Then, to decide if a move of a single particle is accepted or not, only the set of particles contained in a sphere of radius centered at the referred particle have to be considered. 12.4.2 Bose-Einstein condensation in preparation 12.4. MANY-BODY SYSTEMS 233 12.4.3 Quantum dots in preparation 12.4.4 Multi-electron atoms in preparation Chapter 13 Eigensystems 13.1 Introduction In this chapter we discuss methods which are useful in solving eigenvalue problems in physics. 13.2 Eigenvalue problems Let us consider the matrix A of dimension n. The eigenvalues of A is defined through the matrix equation (13.1) where are the eigenvalues and the corresponding eigenvectors. This is equivalent to a set of equations with unknowns W can rewrite eq (13.1) as with being the unity matrix. This equation provides a solution to the problem if and only if the determinant is zero, namely which in turn means that the determinant is a polynomial of degree in and in general we will have distinct zeros, viz., 235 236 CHAPTER 13. EIGENSYSTEMS Procedures based on these ideas con be used if only a small fraction of all eigenvalues and eigenvectors are required, but the standard approach to solve eq. (13.1) is to perform a given number of similarity transformations so as to render the original matrix in: 1) a diagonal form or: 2) as a tri-diagonal matrix which then can be be diagonalized by computational very effective procedures. The first method leads us to e.g., Jacobi’s method whereas the second one is e.g., given by Householder’s algorithm for tri-diagonal transformations. We will discuss both methods below. 13.2.1 Similarity transformations In the present discussion we assume that our matrix is real and symmetric, although it is rather straightforward to extend it to the case of a hermitian matrix. The matrix has eigenvalues (distinct or not). Let be the diagonal matrix with the eigenvalues on the diagonal (13.2) The algorithm behind all current methods for obtaning eigenvalues is to perform a series of similarity transformations on the original matrix to reduce it either into a diagonal form as above or into a tri-diagonal form. We say that a matrix is a similarity transform of if where (13.3) The importance of a similarity transformation lies in the fact that the resulting matrix has the same eigenvalues, but the eigenvectors are in general different. To prove this, suppose that (13.4) Multiply the first equation on the left by and insert between and . Then we get (13.5) which is the same as (13.6) Thus is an eigenvalue of as well, but with eigenvector . Now the basic philosophy is to either apply subsequent similarity transformations so that (13.7) or apply subsequent similarity transformations so that A becomes tri-diagonal. Thereafter, techniques for obtaining eigenvalues from tri-diagonal matrices can be used. Let us look at the first method, better known as Jacobi’s method. 13.2. EIGENVALUE PROBLEMS 237 13.2.2 Jacobi’s method Consider a ( ) orthogonal transformation matrix (13.8) with property . It performs a plane rotation around an angle in the Euclidean dimensional space. It means that its matrix elements different from zero are given by (13.9) A similarity transformation (13.10) results in (13.11) The angle is arbitrary. Now the recipe is to choose so that all non-diagonal matrix elements become zero which gives (13.12) If the denominator is zero, we can choose . Having defined through , we do not need to evaluate the other trigonometric functions, we can simply use relations like e.g., (13.13) and (13.14) The algorithm is then quite simple. We perform a number of iterations untill the sum over the squared non-diagonal matrix elements are less than a prefixed test (ideally equal zero). The algorithm is more or less foolproof for all real symmetric matrices, but becomes much slower than methods based on tri-diagonalization for large matrices. We do therefore not recommend the use of this method for large scale problems. The philosophy however, performing a series of similarity transformations pertains to all current models for matrix diagonalization. 238 CHAPTER 13. EIGENSYSTEMS 13.2.3 Diagonalization through the Householder’s method for tri-diagonalization In this case the energy diagonalization is performed in two steps: First, the matrix is transformed into a tri-diagonal form by the Householder similarity transformation and second, the tri-diagonal matrix is then diagonalized. The reason for this two-step process is that diagonalising a tri- diagonal matrix is computational much faster then the corresponding diagonalization of a general symmetric matrix. Let us discuss the two steps in more detail. The Householder’s method for tri-diagonalization The first step consists in finding an orthogonal matrix which is the product of orthog- onal matrices (13.15) each of which successively transforms one row and one column of into the required tri- diagonal form. Only transformations are required, since the last two elements are al- ready in tri-diagonal form. In order to determine each let us see what happens after the first multiplication, namely, (13.16) where the primed quantities represent a matrix of dimension which will subsequentely be transformed by . The factor is a possibly non-vanishing element. The next transformation produced by has the same effect as but now on the submatirx only (13.17) Note that the effective size of the matrix on which we apply the transformation reduces for every new step. In the previous Jacobi method each similarity transformation is performed on the full size of the original matrix. After a series of such transformations, we end with a set of diagonal matrix elements (13.18) and off-diagonal matrix elements (13.19) [...]... points ặ with ệẹ ề ẵẳ and ệẹ ĩ ẵẳ ặ ẳ 50 9.898985E-01 100 9.9 748 93E-01 20 0 9.993715E-01 40 0 9.99 846 4E-01 1000 1.000053E+00 ẵ 2. 949 052E+00 2. 98 744 2E+00 2. 996864E+00 2. 99 921 9E+00 2. 999917E+00 ắ 4. 86 622 3E+00 4. 96 727 7E+00 4. 991877E+00 4. 997976E+00 4. 999 723 E+00 6.739916E+00 6.936913E+00 6.9 843 35E+00 6.996094E+00 6.999353E+00 8.56 844 2E+00 8.89 628 2E+00 8.9 743 01E+00 8.993599E+00 8.999016E+00 The agreement... ễ ẵ  ắ (13. 42 ) we can write the latter equation as a ẳ ẵ ẵ ắ ẵ ắ ắ ặìỉ ễ ẵ  ặìỉ ễ ẵ ẵ (13 .43 ) or if we wish to be more detailed, we can write the tri-diagonal matrix as ẳ  ãẵẻẵ ắ ẵắ ẳẵ ẳ ắ ắ ã ẻắ ắ ẳ ẳ ẵắ ắắ ã ẻ ẵắ ẳ ẳ ẳ ẳ ẳ ắắ ắắ ã ẻặìỉ ễắ ẵắ ẵ ẳ ẳ ẳ ẵắ ắắ ã ẻặ ẵ ìỉ ễ (13 .44 ) 13.3 SCHRệDINGERS EQUATION (SE) THROUGH DIAGONALIZATION ẵ 24 3 ẵ This is a matrix problem with a tri-diagonal matrix... constant è ẩ ắ è ắ ề ắ ắ ẵ Ư Nowwe can rewrite Eq (13 .25 ) as ắè à (13 .26 ) (13 .27 ) 24 0 CHAPTER 13 EIGENSYSTEMS and taking the scalar product of this equation with itself and obtain ắè àắ ắ Ư which nally determines ắẵ à (13 .28 ) ắè à (13 .29 ) In solving Eq (13 .28 ) great care has to be exercised so as to choose those values which make the right-hand largest in order to avoid loss of numerical precision... (13 .22 ) ẵ where is the ề unity matrix and is an ề column vector with norm è (inner product Note that è is an outer product giving a awith dimension ( ề Â ề ) Each matrix element of ẩ then reads ẩ ặ (13 .23 ) ẵ ẵà ẵà ắ ẵ where and range from to ề Applying the transformation ẫẵ results in ẵẵ ẫè ẫẵ ẵ where è ắẵ ẵ ĂĂĂ ẳ (13 . 24 ) à ẳ ẳ ĂĂĂ è ềẵ and P must satisfy (ẩ ẩ ẵẳẳ ẩ ẩàè ắè à Then (13 .25 )... transformations till we have a tri-diagonal matrix suitable for obtaining the eigenvalues Diagonalization of a tri-diagonal matrix The matrix is now transformed into tri-diagonal form and the last step is to transform it into a diagonal matrix giving the eigenvalues on the diagonal The programs which performs these transformations are matrix tri-diagonal matrix diagonal matrix void trd2(double ÊÊa, int n, double... potential Let us see how this where recipe may lead to a matrix reformulation of the SE Dene rst the diagonal matrix element ắ ãẻ ắ (13 .40 ) ẵắ (13 .41 ) and the non-diagonal matrix element In this case the non-diagonal matrix elements are given by a mere constant All non-diagonal matrix elements are equal With these denitions the SE takes the following form ã ẵ ẵ ã ãẵ ãẵ ẵ where is unknown Since we... needed in this problem The nucleon-nucleon interaction has a nite and small range, typically of some few fm1 We will in this exercise set fmẵ It is then proportional to the mass of the pion The pion is the lightest meson, and sets therefore the range of the nucleon-nucleon interaction For low-energy problems we can describe the nucleon-nucleon interaction through meson-exchange models, and the pion...13 .2 EIGENVALUE PROBLEMS 23 9 The resulting matrix reads ẳ ẵẵ ẵ ẳ ẫè ẫ ẳ ẵ ẳ ắắ ắ ẳ ẳ ắ ẳẳ ẳ ẳ ẳ ềẵà ềắ ềẵ ẳ ẳ  ẳ ẳ ẳ ẵ (13 .20 ) ềẵ ềẵà ềẵ Now it remains to nd a recipe for determining the transformation ẫề all of which has basicly the same form, but operating on a lower dimensional matrix We illustrate the method for ẫẵ which we assume takes the form ẵ ẳè ẳ ẩ ẫẵ (13 .21 ) ẳẳ ẵà ẵà ẵà... can be stored in RAM The obvious question which then arises is whether this scheme is nothing but a mere example of matrix diagonalization, with few practical applications of interest  24 8 CHAPTER 13 EIGENSYSTEMS 13 .4 Physics projects: Bound states in momentum space In this problem we will solve the Schrửdinger equation (SE) in momentum space for the deuteron MeV The ground state is given by The deuteron... to the spatial part, since é To obtain a totally anti-symmetric wave function we need to introduce another quantum number, namely isospin The deuteron has isospin è , which gives a nal wave function which is anti-symmetric We are going to use a simplied model for the interaction between the neutron and the proton We will assume that it goes like ẳ ẵ ắ ắắ ẵ ẳ ẳ ẻ ệà ẳ ẻẳ ĩễ ệà (13 .46 ) ệ where has . 6.739916E+00 8.56 844 2E+00 100 9.9 748 93E-01 2. 98 744 2E+00 4. 96 727 7E+00 6.936913E+00 8.89 628 2E+00 20 0 9.993715E-01 2. 996864E+00 4. 991877E+00 6.9 843 35E+00 8.9 743 01E+00 40 0 9.99 846 4E-01 2. 99 921 9E+00 4. 997976E+00. with nm eV nm [eV] 86 42 0 -2 -4 -6 -8 0 -1 0 -2 0 -3 0 -4 0 -5 0 -6 0 Figure 12. 5: Plot of for =0.1 and 0 .4 nm. Units along the -axis are . The straight line is the binding energy of the hydrogen atom, eV. respect. preparation 12. 4. MANY-BODY SYSTEMS 23 3 12. 4. 3 Quantum dots in preparation 12. 4. 4 Multi-electron atoms in preparation Chapter 13 Eigensystems 13.1 Introduction In this chapter we discuss methods