- #1
jameson2
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Firstly, I have been able to find almost nothing on this kind of question in textbooks or online anywhere. Most places (including my lecture notes) give at most the definition of the operator and that's all. One page if you're lucky out of a whole book. I'd kill for some examples, if you could direct me anywhere it'd be great. Here's my attempt anyway...
(i)Consider a diatomic molecule, of atoms A and B. It's electronic structure can be given by the 1s orbitals of A and B, |A> and |B> (assumed orthonormal). Calculate the associated density matrix expanded over the basis set |A>, |B> as a function of number of electrons N=1,2,3,4. Use the nearest neighbour approximation with on site energies [tex] \epsilon_A=-4eV , \epsilon_B=4eV [/tex] and hopping parameter [tex] \gamma=-3eV [/tex]
(ii) By using ρ from part 1 evaluate the total energy for N=1,2,3,4.
(i) [tex] \rho=\sum_{j}^{occupied} |\psi_j><\psi_j| [/tex]
(ii)[tex] \begin{bmatrix} \epsilon_A & \gamma \\ \gamma & \epsilon_B \end{bmatrix}\left[ \begin{array}{c} \psi_A \\ \psi_B \end{array} \right]= E\left[ \begin{array}{c} \psi_A \\ \psi_B \end{array} \right] [/tex]
(iii)[tex] L=Tr(L\rho) [/tex] for an operator L.
(i) I don't know how to go about this, mainly the fact that the question brings in the number of electrons is confusing me. Here's what I've reasoned so far: Since it's a molecule, all the electrons are in the bonding state. Solving the matrix equation to find the bonding state, I get bonding energy of -5eV which leads to a bonding state of [tex] \frac{1}{\sqrt{10}} (3,1) [/tex]. Applying the formula for the density matrix for N electrons in the bonding state, I get [tex] \rho= N\frac{1}{10}\begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix} [/tex]
(ii) Using the formula above, I guess the answer comes from [tex] E=Tr(H\rho) [/tex], but I need the density matrix first. Assuming I have the right density matrix, plugging in the Hamiltonian [tex] \begin{bmatrix} -4 & -3 \\ -3 & 4 \end{bmatrix} [/tex] I get the energy to be -5N for N=1,2,3,4.
Thanks a lot for any help, or pointers towards somewhere that can help me. As I said, I'm having trouble finding info on this topic, you're my last chance!
Homework Statement
(i)Consider a diatomic molecule, of atoms A and B. It's electronic structure can be given by the 1s orbitals of A and B, |A> and |B> (assumed orthonormal). Calculate the associated density matrix expanded over the basis set |A>, |B> as a function of number of electrons N=1,2,3,4. Use the nearest neighbour approximation with on site energies [tex] \epsilon_A=-4eV , \epsilon_B=4eV [/tex] and hopping parameter [tex] \gamma=-3eV [/tex]
(ii) By using ρ from part 1 evaluate the total energy for N=1,2,3,4.
Homework Equations
(i) [tex] \rho=\sum_{j}^{occupied} |\psi_j><\psi_j| [/tex]
(ii)[tex] \begin{bmatrix} \epsilon_A & \gamma \\ \gamma & \epsilon_B \end{bmatrix}\left[ \begin{array}{c} \psi_A \\ \psi_B \end{array} \right]= E\left[ \begin{array}{c} \psi_A \\ \psi_B \end{array} \right] [/tex]
(iii)[tex] L=Tr(L\rho) [/tex] for an operator L.
The Attempt at a Solution
(i) I don't know how to go about this, mainly the fact that the question brings in the number of electrons is confusing me. Here's what I've reasoned so far: Since it's a molecule, all the electrons are in the bonding state. Solving the matrix equation to find the bonding state, I get bonding energy of -5eV which leads to a bonding state of [tex] \frac{1}{\sqrt{10}} (3,1) [/tex]. Applying the formula for the density matrix for N electrons in the bonding state, I get [tex] \rho= N\frac{1}{10}\begin{bmatrix} 9 & 3 \\ 3 & 1 \end{bmatrix} [/tex]
(ii) Using the formula above, I guess the answer comes from [tex] E=Tr(H\rho) [/tex], but I need the density matrix first. Assuming I have the right density matrix, plugging in the Hamiltonian [tex] \begin{bmatrix} -4 & -3 \\ -3 & 4 \end{bmatrix} [/tex] I get the energy to be -5N for N=1,2,3,4.
Thanks a lot for any help, or pointers towards somewhere that can help me. As I said, I'm having trouble finding info on this topic, you're my last chance!