ChemSiddiqui Posted August 4, 2011 Share Posted August 4, 2011 Hello everyone, I am trying to teach myself theoretical chemsitry and have done well so far. I am using this book called Modern Quantum Chemistry by Szabo. I was reading about Couloumb and Exchange opertaor and I have a question. I understand why Coloumb and exchange operator integrals arise, due to electrostatic potential and antisymmetry of the slater determinant. Then there was this exercise which made me pause. It asks to prove that the coulomb operator denoted by Jii and the exchange operator denoted by Kii are equal. Now in Chemists notation it the couloub integral Jab is given as Jij = (ii|jj) = <ij|ij> and Kab is defined as Kii = (ij|ji) = <ij|ji>. Now the following to prove is; 1).Jii = Kii My attempt at it is Jii = (ii|ii) and also Kii = (ii|ii) so they are equal 2). Jij*= Jij My attempt at solution: Jij = (ii|jj) = (ij|ij) = Jij* other to prove are: 3). Kij* = Kij 4). Jij = Jji 5). Kij = Kji I dont want to consider (3-5) yet unless I know I am on the right direction on proving them and so I ask if you think the above solutions are reasonable? any suggestion you can give me to maybe do these sort of proofs? Any help will be most appreciated. Thanks. Link to comment Share on other sites More sharing options...

mississippichem Posted August 4, 2011 Share Posted August 4, 2011 (edited) Hello everyone, I am trying to teach myself theoretical chemsitry and have done well so far. I am using this book called Modern Quantum Chemistry by Szabo. I was reading about Couloumb and Exchange opertaor and I have a question. I understand why Coloumb and exchange operator integrals arise, due to electrostatic potential and antisymmetry of the slater determinant. Then there was this exercise which made me pause. It asks to prove that the coulomb operator denoted by Jii and the exchange operator denoted by Kii are equal. Now in Chemists notation it the couloub integral Jab is given as Jij = (ii|jj) = <ij|ij> and Kab is defined as Kii = (ij|ji) = <ij|ji>. Now the following to prove is; 1).Jii = Kii My attempt at it is Jii = (ii|ii) and also Kii = (ii|ii) so they are equal I'm not sure I understand this bottom part here. Did you mean [math] \hat{J}_{ii} = \langle ii | ii \rangle [/math] ? and the same for [math] \hat{K}_{ii} [/math]? I don't really follow here. Explain further, Why don't you try to show these operators acting on some simple [math] \phi [/math] function and see if they give the same eigenvalues? Show how the exchange operator makes [math] \phi_{i} [/math] exchange with [math] \phi_{j} [/math]. I think this will give you a hint. Or if you try to write them down in full Dirac notation like: [math] \langle \phi_{i} \phi_{j} | \hat{O} | \phi_{i} \phi_{j} \rangle [/math], then I think you will be able to see how they are the same by mere inspection or by their matrix forms. I tend to be comfortable with the matrix formulations of Hartree-Fock stuff, I've told I'm strange for that though so... Also, I know you can write down the antisymmetrization operator, [math] \hat{A} [/math], in terms of the exchange operator acting on the Hartree product. See if you can do the same for the Coloumb operator. Edited August 4, 2011 by mississippichem Link to comment Share on other sites More sharing options...

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