MIT OpenCourseWare http://ocw.mit.edu 3.23 Electrical, Optical, and Magnetic Properties of MaterialsFall 2007For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.Homework # 2 September 19, 2007 Homework is due on Wednesday September 26th, 5pm 1 The power of spinach MIT researchers have been able to use the Photosystem 1 in spinach chloroplasts to convert sunlight to energy into a solid-state electronic spinach sandwich de-vice that may one day power laptops and cell phones. What would be the size of the confining region for the electron responsible for the green color of spinach ? (Hint: To study this problem, assume that the electronic transition is from the ground state to the first excited state. You can think at Photosystem 1 as being a 1-d infinite well) 2 Molecular orbitals and the Linear Combinaition of Atomic Orbitals method (LCAO) Background on LCAO In this problem, we will be interested in character-izing the two lowest lying molecular orbitals of an electron in the Hydrogen molecule H2. To find those molecular orbitals, we will use a well-known tech-nique in Quantum Chemistry called the Linear Combinaition of Atomic Orbitals (LCAO). The spirit of the technique is the following: Imagine that you take two hydrogen atoms far from each other. In that cas e each atom has one elec-tron, and the wavefunction of each electron is des cribed by a 1s atomic orbital, φ1s = 1 e− ar 0 , where a0 is the Bohr radius. When one then reduces the √π a30 distance between the two hydrogen atoms up to atomic dimensions, the ground state wavefunction of each electron is no longer described by a 1s orbital because an electron localized around one of the proton can hop to the second proton of the H2 molecule. So what is the wavefunction then? The LCAO technique ap-plied to the c ase of the H2 molecule consists in writing the total wavefunction of an electron as a linear combinaison of two 1s orbitals, one centered around each proton. The new wavefunction is then called a molecular orbital because it spreads over the entire molecule and is not localized around a particular atom. Hence the molecular orbital looks like this: ψ = c1φ1s(1) + c2φ1s(2) 1 Nicolas Poilvert & Nicola Marzari� � � � � � � � � where (1) and (2) stand for proton 1 and 2. We can compactly represent this linear combinaison as a 2-dimensional vector in the basis of φ1s(1) and φ1s(2) as: c1ψ = c2 Expressing the Hamiltonian of an electron Let’s call Hˆthe hamiltonian of a single electron in an H2 molecule. Since we expressed the electronic wave-function as a linear combinaition of φ1s(1) and φ1s(2) only, we need to find the hamiltonian matrix in the basis formed by those two atomic orbitals if we want to be able to find the eigenenergies (i.e the molecular energies) and the corresponding eigenfunctions (i.e the molecular orbitals). So we see that instead of expressing the hamiltonian in an infinit e and complete basis, we chose a two-dimensional basis. This is clearly an approximation but if the two basis functions are carrefully chosen, then the eigenfunctions obtained by diagonaliz-ing the hamiltonian matrix in this 2-dimensional basis can be very accurate in describing the ”exact” molecular orbitals. Let’s define the matrix elements of the hamiltonian as: E0 = � (1)|Hˆ|φ1s(1) � = � 1sHφ1sφ1sφ∗ (1) ˆ(1) which by symmetry is equal to: E0 = � (2)|Hˆ|φ1s(2) � = � 1sHφ1sφ1sφ∗ (2) ˆ(2) and V = � 1sHφ1s(2) = � φ∗ (2) ˆ(1)φ∗ (1) ˆ1sHφ1sThe off-diagonal elements shown above are in general complex conjuguate to each other, because the hamiltonian matrix must be hermitian, i.e Hˆ† = Hˆ. Using E0 and V we find that the hamiltonian matrix expressed in the two-dimensional basis of the φ1s(1) and φ1s(2) atomic orbitals is: ˆE0 V H = V E0 The atomic orbitals φ1s(1) and φ1s(2) are real and normalized wavefunctions -i.e |φ1s(1)|2 = φ1s(1)2 = 1 and the same for φ1s(2)- but they are not orthogonal because φ1s(1)φ1s(2) = S =� 0. In order to find the eigenergies we will have to equate the determinant of the following matrix to zero (This is a generalization of the determinantal eigenvalue equation for non-orthogonal basis): E0 − λ V − λS V − λS E0 − λ 1. Find the eigenenergies (λ+ and λ ) of an electron in an H2 molecule by −solving the determinantal equation above for λ. λ+ will be the smallest eigenenergy (knowing that S is positive and V is negative) and λ the −largest. 2� � � � �� � 3 2. For each of the two possible eigenenergies λ+ and λ−, find the corre-sponding normalized eigenvectors by solving the Schrodinger equation in matrix form (This is a generalization of the matrix eigenvalue equation for non-orthogonal basis): � �� � E0 − λ± V − λ±S c1 = 0 V − λ±S E0 − λ± c2 3. From the calculations carried out in the previous questions, we see that an electron has two possible states in the H2 molecule. Because of the spin degeneracy, the ground state of the H2 molecule consists of the two electrons accupying the lowest energy state with opposite spins. Then what is the total energy of the H2 molecule? (in its ground state) 4. If hˆis the hamiltonian of an electron in the hydrogen atom, then φ1s is an eigenfunction (the ground state) of hˆfor the eigenvalue �0 (the ground state energy). Then what is the total energy of two isolated hydrogen atoms? (in their ground state) 5. If one considers that �0 is equal to E0 (this is almost true for the H2 molecule), can you demonstrate that the formation of a H2 molecule is favored energetically with respect to two isolated atoms? The spin of the electron During the 1920’s, physicists realized that in order for them to be able to explain all the experimental results accumulated over the previous 20 years, they needed to give to the electron an intrinsic property, the spin. In Quantum mechanics, the spin of the electron is described by a 2 dimensional complex vector like: a |ψ� = b where a and b are complex numbers. In this exercise, we will consider that the electron spin is described by the following quantum state: 1|ψ� = − √√212 Since the spin is a