Lecture 5 Bonds to Bands
Lecture 5 Bonds to Bands
Lecture 5 Bonds to Bands
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<strong>Bonds</strong> <strong>to</strong> <strong>Bands</strong>• Determinant becomes:HH( HABAAAA(1 − S− E− ES− E)(H2AB) EAB2BBHABH+ (2H− ESBB− E− E)− ( HABAB− 2HAB= 0AA− ESAB) E + ( H2) = 02AA− H2AB) = 0• Solve quadratic eqs and get 2 solutions:HE =HE =AAAA− HABS(1 − S− HABS(1 − SHAA+ HE1=1+SABABABHAA− HE2=1−SABABABABAB+ HAB− H)(1 + S )− HAB+ H)(1 + S )( + )(-)ABABAAAASSABAB+ sol' n− sol' n<strong>Lecture</strong> 5Energy diagram:ψ1=1( φISA2−φISB)ψ2=1( φISA2+ φISB)βαE 2E 17Molecular system H 2+• Most simple molecular system H 2+(molecular ion with one electron)• For trial functions choose 2 simple 1s a<strong>to</strong>mic orbitals<strong>Lecture</strong> 584
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