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# SP CH 795Z

NCS

GPA 3.54

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This 237 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 795Z at North Carolina State University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/224002/ch-795z-north-carolina-state-university in Chemistry at North Carolina State University.

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Date Created: 10/15/15

The Variation Method The variational theorem Application to LCAOMO theory The He atom The variational theorem The variation method allows us to obtain an approximation to the ground state energy of the system without solving the Schrodinger equation The variation method is based on the following theorem Given a system with hamiltonian operator H then if I is any normalized wellbehaved function that satisfies the boundary conditions it is true that pHpd C 2 EO where E0 is the true value of the lowest energy eigenvalue of H This important theorem allows us to calculate an upper bound for the ground state energy Practical significance The variation method serves as the basis for all methods that use combinations of hydrogenlike orbitals to solve for the eigenfunctions wave functions and eigenvalues energies of atoms and molecules The radial part of the hydrogenlike wave functions is modified by a variational parameter which is minimized The theorem allows us to set the derivative with respect to any parameter 0c equal to zero to find the value of that parameter that minimizes the energy a gtxlt a p Hpd t 0 We can be sure that the energy calculated in this way will be greater than the true energy an upper bound Application of the Variational Method The linear combination of atomic orbitals LCAO Calculations of the energy and properties of molecules requires hydrogenlike wave functions on each of the nuclei The HartreeFock method begins with assumption that molecular orbitals can be formed as a linear combination of atomic orbitals N bi 211 Crux The basis functions x are hydrogenlike atomic orbitals that have been optimized by a variational procedure The HF procedure is a variational procedure to minimize the coefficients Cm Note that we use the index m for atomic orbitals and i orj for molecular orbitals Common types of atomic orbitals Slatertype orbitals STOs x 0C 6 The STOs are like hydrogen atom wave functions The problem with STOs arises in multicenter integrals The Coulomb and exchange integrals involve electrons on different nuclei and so the distance r has a different origin Gaussiantype orbitals GTOs W2 XHOC 9 Gaussian orbitals can be used to mimic the shape of exponentials ie the form of the solutions for the hydrogen atom Multicenter Gaussian integrals can be solved analytically STOs vs GTOs GTOs are mathematically easy to work with Normalized Amplitude r atomic units STOs vs GTOs GTOs are mathematically easy to work with but the shape of a Gaussian is not that similar to that of an exponential Normalized Amplitude r atomic units STOs vs GTOs Therefore linear combinations of Gaussians are used to imitate the shape of an exponential Shown is a representation of the 3Gaussian model of a STO 10 l exp r Sum STO3G Gaussian components O2exp 016 r2 04exp067 r2 O3exp25 r2 Normalized Amplitude r atomic units Doublezeta basis sets Since the remaining atoms have a different exponential dependence than hydrogen it is often convenient to include more parameters Car r X Dude pre b The second exponential is a diffuse function It accounts for properties of a valence electrons involved in bonding When GTOs are used there are always multiple Gaussians required because the shape of Gaussians must be matched as closely as possible to that of exponentials In a doublezeta basis there may be up to 3 Gaussians used to represent the first exponent Q8 and 1 for the second exponent Qb In a socalled 631G basis set in the GAUSSIAN program there are 6 Gaussians for core electrons and then 3 for Q8 and 1 for Qb The variation procedure applied to the HF wave functions The HF procedure uses the variational method to obtain the value of parameters that minimizes the energy SE 8lt lHlltgt gt 8 IH d1 0 subject to the constraint the wave functions remain rth l o ogona lt i jgt I bl de 51 The minimization of an equation subject to a constraint is carried out using the method of LaGrange undetermined multipliers 8lt IHIltIgt gt EltltIgtIltIgt gt 1 0 Note that the multiplier is the energy E and it will be determined during the procedure The variational method in HF The variation can also be viewed as a variation of the function by an infinitesimal amount I ltlgt39 84v The energy then becomes Eltlgt 54gt ltltlgt 5 Hltlgt 5 gt ltltlgtHltlgtgt ltltlgt 5 Hlltlgtgt ltltlgtHltlgt 5 gt E SE In the variation method we are looking for the wave function fthat will minimize the energy Here the condition is that the variation in the energy SE O as indicated on the previous slide This condition assures that E is stationary A stationary point is usually a minimum Introduction of the basis At this point we can substitute our trial wave function into the Fock equations E ltiIIilt gt g C CvltxMiHixv gt The Lagrangian becomes L ltltgtiHiltlgt gt E lt ltlgtiltlgt gt 1 CZCvltxiHixv CZCVltXMiXV gt 1 The first variation in the Lagrangian is set equal to zero 5L SCZCVltXMHXV gt CZ5CVltXMHXV gt 5C3CVltX m gt CZSCVltX m gt 0 Expansion in terms of coefficients Since both E and L are real we can collect terms and exchange indices to obtain N Z 5CZ HKHCW EiSKHCW K 123N 1 z HKHltxKiHixMgt xinudr SW ltxKixM gt xixudr In other words H11 EZSHC11 H12 IiiSum Hm EISWCM 0 H21 E1821Cu H22 IiiSum Hm EisgNCNi 0 Hm EZSMCU Hm EiSN2C2i HNN EZSNNCM 0 The secular determinant There are N equations and N 1 unknown variables C1iCziC3i CM and E In order for the equations to have meaningful nonzero solutions they must comprise a secular determinant ray 55H f Z ESU y ESE H th ESM E f ESH EQy ESB H u4da HQ Eamp1 Hg Eamp2 Hg Eamp3 H 0 which leads to N eigenvalues Ei i 123N The solutions for the bi C1ix1 02in CNixN under the constraint N N chcs1 u1v1 W W V Matrix representation The fock hamiltonian is an effective oneelectron hamiltonian He iZEi i He Efr AOS X19 X29 X39 399XN N Mos 4 Z CHIx1 1 The matrix representation for the overlap and interaction energies is Matrix representation Thus HCi EiSCi Define the matrices of coefficients and eigenvalues as C11C12C13C1N E110 o 0 C2 C21C22C23C2N E 0 E220 0 Then the matrix form is HC SCE This system of equations is diagonalized if detH ES 0 However this is possible only in the MO basis A0 baSiS X17 X27 X37 39gtXN baSiS 117 127 137 gt N S W ltXMXVgt Not orthogonal 51 lt i jgt Orthogonal HMV ltXHHe xVgt Not diagonal Eloy lt iHe Jgt Diagonal Eigenvalues are energies and eigenvectors are MOs The procedure is carried out until a selfconsistent minimum energy is found The resulting selfconsistent field SCF energy is the HartreeFock procedure The values for the coefficients that give the SCF energies for the electronic states give the MOs As shown previously we can think of these as linear combinations of atomic orbitals The linear combinations have distinctive shapes such as those shown for the MOs of N2 in the following Summary of methods 1 Determine the optimum atomic orbitals This is done by a variational procedure for each atom The exponent of the STD or GTO is optimized More than one STO can be used per atomic orbital eg doublezeta basis The GTO requires parameterization of multiple Gaussian functions 2 Form linear combinations of the atomic orbitals at the positions indicated by a molecular geometry Note that this is initially just a guess and that the geometry which gives the lowest energy must be calculated in a number of tries cycles of the HF procedure 3 Perform the HF procedure repeatedly until a selfconsistent solution of the equations for the coefficients is obtained This is the selfconsistent field SCF method Example of the Variation Method Screening of Electrons in He The hamiltonian for H The electronic hamiltonian for the hydrogen atom consists of a kinetic energy term for the electron and the Coulomb attraction of the electron and proton nucleus 1 2 n V2 Z62 A Of course the nuclear charge of hydrogen is Z 1 so the Z is included for completeness We know that the solutions Of the Schrodinger equation HLP ELF gives energy levels 2 n is the principal quantum number En 267 a0 is the Bohr radius 0 The hamiltonian for He For helium the same kinetic energy and Coulomb attraction terms are present but there is also a Coulomb repulsion between the two electrons that must be included 62 12 n 2 r Jia 22 2m 2 6 V1 Ze2 V2 Because of the Coulomb repulsion there is no exact solution for He To solve the problem we use two 1s orbitals from the solution for hydrogen and then apply the variational method The4eNavefunc on The hydrogen 1s wave functions for electrons 1 and 2 are 32 32 ra Z ra radio ems am The aufbau approach for atoms assumes that the total wave function for a manyelectron atom is just a product of one electron wave functions In the present case m Note that the hydrogen wave functions are normalized so mm mm1 Variational approach for the He atom The He wave function used for the variation method is a product of two hydrogen 1s orbitals However instead of the nuclear charge Z we use a variational parameter Q C 3 p e eraOe Qrzao 5 0 Q has a physical interpretation Since one electron tends to screen the other from the nucleus each electron is subject to a nuclear charge that is less than Z The hamiltonian is Evaluation of the integrals If we consider only the part of the hamiltonian in parentheses We have the solution to a hydrogen atom with two electrons in the ls orbital 2 62 12 62 2 I l7fC 1 73Cpg220p 2m 7 1 2m 72 where the right hand side is twice the energy of a 13 electron Using this result we have 2e2 2 9pdr 2 9pdr 2 9pdr IpHpdT Ca O ppd tC ZeTi C ZkITte T The integrals have the following values ltpltpdr ltpltpdr C 2 9de 5C 62 gtllt Immdr 1 71 I 72 a0 6 712 8610 Evaluation of the variational parameter Q We have pHpdT Q2 22g QZ We now vary Q to minimize the variational integral f pHpdt Q2 22g QZ 2g 2z0 i C Z 16 The variational energy is t 2 25e2 ize2 PH P Z8 256ao 21600 The variational energy comparison with experiment The experimental ionization energy of He is 245 eV Our first guess would be to calculate the energy of the 13 Electron in He using the hydrogen energy level with a nuclear charge Z 2 E Ze2a0 This gives 2136 eV 272 eV Using the value obtained by the variational method we have E 2716e2a0 2716136 eV 2295 eV The value is much closer to the true value In accord with the variational theorem the true ground state energy is less than that given by variational method Summary The hydrogen atom is the only atom with an exact solution Hydrogen wave functions are used as the approximation for atomic wave functions in multielectron atoms The variational principle states that any wave function we choose that satisfies the Schrodinger equation will give an energy greater than the true energy of the system The variation method provides a general prescription for improving on any wave function with a parameter by minimizing that function with respect to the parameter The connection to experiment Fermi Golden Rule Transition probability Einstein coefficients Relation to Emission Relation to Absorption The Fermi Golden Rule Starting with the definition of energy density uv 80E2 f p vdv and the Fermi Golden Rule P 1 H12 E0 2 4h2 5V V12 8V V12 we can obtain the transition probability per unit time 0312 In the FGR expression LL12 is the transition moment of the molecule and E0 is the electric field vector of the light interacting with the molecule The delta functions guarantee that the energy the incident light must equal that of the transition n21 2 The transition probability The transition probability per unit time as a function of the incident energy density is H12 39 3 29V12 12 t 28th where e hat is a unit vector along the direction of the electric field of the light What happened to the delta function We have used the property of a delta function that the integral over dx x0 is equal to the function evaluated at x0 I fX8X X0dX fX0 The energy density is typically polarized along one direction x y or 2 Therefore the transition probability per unit time is 13 as large as the above derivation Orientation average of 00326 The orientation average of 00329 is 13 TE 27 ltc0529gt if f COS2GSin9d9d 4n 0 0 ltc0529gt COS2GSin9d9 0 1 1 1 3 3 2 112 1x ltcosegt 2Lde 2 Relation to Absorption and Emission The energy density is typically polarized along one direction x y or 2 Therefore the transition probability per unit time is 13 as large as the above derivation H1229V12 680h2 We can compare this expression with Einstein rate W12 B12pV12 680h2 Based on the relationship between A21 and B12 we have A 16n3lu122n3v3 21 380703 12 Thus 12 Photon density of states In order to derive the emission probability we will need the photon density of states In a box of length L boundary conditions require that exp ikx exp ikx L The number of photon states dN in an infinitesimal volume is 3 dN L dkxdk dkz V kzdk sinededq 27 y 2 TE 3 or using the definition d9 sinededcl we can write dN V cozdcon 2703 30 CIN V 032 CIN V 032 dN dE deE and CIQ dE 2703 n dE 2703 n Understanding the emission rate constant We have used cn for the speed of light in a medium with index of refraction n This derivation assumes that all of the radiation along a given direction eg xpolarized is collected In reality we must account for the experimental geometry The collection optic has a finite size and is at a fixed distance from the sample To take this properly into account we use a photon density of states as a function of the solid angle For emission we can make a substitution similar to the above UV N SOE This is valid because the energy density is the number of photons N per unit volume V times the energy per photon 2th 80V So E3 Emission rate constant For emission we take the case where the energy of final state is lower than the energy of the initial state W 2nu12322Nho 12 4h 80 Here we have made use of the nonintuitive property of a delta function 5a 0021 h6E E21 We make the association between the delta function and the photon density of states 5E E21 5E E12 gt 5E E21 dE This is like saying that the delta function represents the number of states per unit energy Emission rate constant We express the emission rate as a differential rate per unit solid angle 2RH12322Nh V 02 dll12 4h 3 d9 80V 2750 n d Nw2u12 e 2 d9 227520380h Again as above we can assume that radiation is being detected along one polarization and we divide the above expression by 3 dVV12 Nm2H122 d9 627c20380h Emission rate constant Furthermore we can substitute Cv for c as above and 0 27w d9 380703 Which is the same as the expression above if we integrate over d9 7 2n dof sinedef d 47c 0 0 2 H12 Since there are two polarizations of light there is an additional factor of two required With this factor the expression is identical to the expression for A21 given above N167c3n3v2 W12 38 hC3 IH122NA21 o Emission rate constant Note that as above the rate of upward transitions is equal to the rate of downward transitions at equilibrium thus W21 W12 The number of molecules N represents the number in the particular state and can N1 or N2 depending on the context To summarize the point here is that we can determine the emission a function of the solid angle of the collection optics as shown below Large solid angle dQ Small solid angle dQ Sample Fluorescence Lifetime and Yield These expressions show that the rate of absorption and spontaneous emission are proportional to the square of the transition dipole moment Since A21 is proportional to 11rad we see that the radiative lifetime gets shorter the larger the transition moment Thus we have an experimental connection between the of measured rate of decay of an excited state and the transition moment However this is not the best way to estimate 2 the transition moment First it must be realized that the observed fluorescence lifetime may not correspond kWhradiative to the intrinsic radiative lifetime v V kradiative Fluorescence Lifetime and Yield If there are both nonradiative and radiative pathways to the ground state the observed rate of the fluorescence decay is kfluorescence kobserved knonradiative kradiative or 1Tfluorescence 1l tobserved 1Tnonradiative 1l tradiative The intrinsic fluorescence lifetime can be determined from a measurement of both the decay kinetics and the fluorescence quantum yield The emission yield is 1 krad 1lfLrad f krad knonrad 1lfLrad 1T nonrad 1 Crad If 1Tf Trad The absolute fluorescence quantum yield is still not easy to measure The amount of fluorescence collected depends on the solid angle of the collection optics in a fluorometer Fluorescence Lifetime and Yield The relative yield of the known compared to the standard dye can be used to determine the absolute quantum yield There are additional complications if other states are present We have thus far only discussed a two level system There are several other processes that can compete with radiative decay including 1 lntersystem crossing eg conversion from a singlet to triplet state 2 Electron transfer or energy transfer to a neighboring molecule Integrated Absorption Band The transition moment can also be estimated from the integrated absorption and emission spectra To see this we start with Beer39s law I low EM The concentration is C and the extinction coefficient is l in units L mol391 cm391 The factor of 2303 is the conversion from base e to base 10 Usually we see Beer39s law as dl thw12dx If there are N molecules per cm3 then we can also express the change in intensity as 23036VC W W Integrated Absorption Band The above way of writing the change in intensity considers the photons of energy hn absorbed or emitted by N molecules each with a probability W12 We can write 23036VC W 2 th It is convenient to replace l by cun and the molar concentration C by 1000NNA where N is the number of molecules per cm3 and NA is Avagadro39s number 2303Cevu or W1223030f elvlplvldv W 12 N Ahvn using the definition u f pVdV Integrated Absorption Band This expression can be related the B12 coefficient A band Finally this expression can be related to the square of the transition moment 2 680h223030f EmaV band 12 This expression shows that the magnitude of the square of the transition moment can be related to the integrated band regardless of the band shape This suggests that there is a shape factor that does not change the overall strength of the absorption Review of Quantum Mechanics Postulates of quantum mechanics The Wavefunction Averaging Postulates of quantum mechanics are assumptions found to be consistent with observation The first postulate states that the state of a system can be represented by a wavefunction quotPql q2 q3n t The q are coordinates of the particles in the system and t is time The wavefunction can also be timeindependent or stationary wq1 ql q3n Postulate 2 The probability of finding a particle in a region of space is given by Pa r V Pdt Postulate 2 Assumptions 1 KP is real quotP is Hermitian 2 The wavefunction is normalized 3 We integrate over all relevant space Normalization is needed so that probabilities are meaningful Normalization means that the integral of the square of the wavefunction probability density over all space is equal to one P Pd t 1 space The significance of this equation is that the probability of finding the particle somewhere in the universe is one Normalization is needed so that probabilities are meaningful Normalization means that the integral of the square of the wavefunction probability density over all space is equal to one lt P Pgt1 The braket representation is equivalent to writing the integral over all space The bra signifies the complex conjugate The wavefunctions form an orthonormal set Orthogonal wavefunctions have zero overlap f krfkrjdx 0 all space Normalized wavefunctions integrate to l f Tfkl dr 1 all space The wavefunctions form an orthonormal set Orthogonal wavefunctions have zero overlap Hngto Normalized wavefunctions integrate to l MEgt1 Postulate 3 Every physical observable is associated with a linear Hermitian operator Observables are energy momentum position dipole moment etc operator 13 gt observable P The fact that the operator is Herrnitian ensures that the observable will be real Postulate 4 The average value of a physical property can be calculated by PTLPdI ltP Normalization Postulate 4 The calculation of a physical observable can be written as an eigenvalue equation PrepsI This is an operator equation that returns the same Wavefunction multiplied by the constant P P is an eigenvalue An eigenvalue is a number The form of the operators is Position vii q Momentum P ifig 661 Time t t Energy P7 i722 6t Stationary State Wave Equation Quantum Mechanical Description Hamiltonian Eigenvalue Energy Operator Energy value l HO P EO P Wave function The Hamiltonian and Wavefunction are timeindependent The wavefunction is composed of electronic and nuclear parts LP Welectronicx nuclear 1 f l Total Electronic Nuclear The wavefunction represents the probability amplitude of electrons and nuclei The wave equation can be separated into electronic and nuclear parts Hamiltonian Eigenvalue Energy Operamr Energy value Helecw E elecql A Hmch Emclx Wavefunctions The Hamiltonian contains both kinetic and potential energy terms Kinetic Energy Potential Energy l l l l Analysis of porphine molecular orbitals and spectra using group theory Assignment of the point group Four vertical mirror planes 69V c d Assignment of the point group Two 90 rotations Four 1800 rotations n Four 180 rotations Four 180 rotations Four 180 rotations Character table for D 4h point group E 202 2Cquot2 i 254 ch 26V 26d S quadratic A12 1 1 1 1 1 1 1 1 x2y2z2 A22 1 1 1 1 1 1 1 1 R2 B12 1 1 1 1 1 1 1 1 xz y2 132g 1 1 1 1 1 1 1 xy E2 2 O O 2 O 2 O O RX RV XZ yz A111 1 1 1 1 1 1 1 1 A2 1 1 1 1 1 1 1 1 z B111 1 1 1 1 1 1 1 1 B2 1 1 1 1 1 1 Eu 2 0 0 2 0 2 0 0 x y Linear combination of basis porbitals Method for determination of irreducible representations of molecular orbitals Determine how the orbitals of the nsystem transform as a result of each symmetry operation If a particular porbital moves its contribution is zero If the orbital is not moved by a symmetry operation it contributes either 1 same phase or 1 opposite phase The character is determined for each symmetry operation and then decomposed into irreducible representations Initial basis H The reducible representation E 204 2 202 202 i284 ch 20V 20d 160 04 4 0016 4 4 The A19 irreducible representation E 204 2 202 202 1234 ch 20V 20d 16 o o 4 4 o o 16 4 4 1 1 1 1 1 1 1 1 1 1 16 o o 4 4 o o 16 4 4 Summing all of the contributions we find FA1g O The A29 irreducible representation E 204 2 202 202 1234 ch 20V 20d 160044001644 1 1 1 1 1 1 1 1 1 1 160044001644 Summing all of the contributions we find FAzg O The B1g irreducible representation E 20V 2 202 202 1234 ch 20V 20d 16 o o 4 4 o o 16 4 4 1 1 1 1 1 1 1 1 1 1 16 o o 4 4 o o 16 4 4 Summing all of the contributions we find FB1g O The B2g irreducible representation E 20V 2 202 202 1234 ch 20V 20d 16 o o 4 4 o o 16 4 4 1 1 1 1 1 1 1 1 1 1 16 o o 4 4 o o 16 4 4 Summing all of the contributions we find FBzg O The E9 irreducible representation E 20V 2 202 202 i284 oh 20V 20d 16 o o 4 4 o o 16 4 4 2 o 2 o o 2 o 2 o o 32 o o o o o o 32 o o Summing all of the contributions we find FEg 6416 4 The A1U irreducible representation E 204 2 202 202 1234 ch 20V 20d 160044001644 1111 1 1 1 1 11 160 04 4 oo 16 4 4 Summing all of the contributions we find FA1u 1616 1 The A2U irreducible representation E 204 2 20 2 26 2 i284 6h 26V 26d 16 o o 4 4 o o 16 4 4 1 1 1 1 1 1 1 1 1 1 16 o o 4 4 o o 16 4 4 Summing all of the contributions we find FA2u 4816 3 The B1U irreducible representation E 204 2 20 2 26 2 i284 6h 26V 26d 16 o o 4 4 o o 16 4 4 1 1 1 1 1 1 1 1 1 1 16 o o 4 4 o o 16 4 4 Summing all of the contributions we find FB1u 3216 2 The B2U irreducible representation E 204 2 20 2 26 2 i284 6h 26V 26d 16 o o 4 4 o o 16 4 4 1 1 1 1 1 1 1 1 1 1 16 o o 4 4 o o 16 4 4 Summing all of the contributions we find FBZU 3216 2 The Eu irreducible representation E 204 2 202 202 i284 oh 20V 20d 16 o o 4 4 o o 16 4 4 2 o 2 o o 2 o 2 o o 32 o o o o o o 32 o o Summing all of the contributions we find FEu O The decomposition Summing all of the contributions we find F A1U 3A2U ZB1U ZB2U 4Eg Note that the total dimension is the same as the original number of orbitals Le 16 Since Eg is doublydegenerate each Eg counts as two The molecular orbitals can be constructed using the method of projection operators Unique basis 4 1 C4 rotation Clockwise 02 rotation C4 rotation 16 Counter clockwise ltlgt1 C 2 rotation 1 16 C 2 rotation 2 C 2 rotation 1 2 C 2 rotation 2 ch reflection 2 S4 rotation Clockwise i inversion S4 rotation 16 Counter clockwise csquot reflection 16 GquotV reflection c d reflection c d reflection b2 PE ltlgt1ltlgt2ltlgtsltlgt4 PC 4 15 16 I 18 PCz 19 110 111 112 PC 4 113 114 115 16 PC 12 11 414 15 16 PC 22 be 17 18 19 PCquot12 Fr Fr ltlgt4 ltlgt5 PCquot22 1o 111 12 113 Pi 9 10 11 12 PS 4 5 6 7 8 P5h 1 2 3 4 PS 4 13 14 15 16 PG v 11 114 115 16 PGquotV be 17 be 19 PG d 12 13 14 15 PG d 110 111 112 113 4 C 44444444 I I I I I I I I 44444444 PE ltlgt1ltlgt2ltlgtsltlgt4 PC 4 15 16 I 18 PCz 19 110 111 112 PC 4 113 114 115 16 PC 12 11 414 15 16 PC 22 b6 I b8 19 PC 12 2 3 4 15 PCquot22 1o 111 12 113 Pi 9 10 11 12 PS 4 5 6 7 8 P5h 1 2 3 4 PS 4 13 14 15 16 PG v 11 114 115 16 PG v b6 I b8 19 PG d 12 13 14 15 PG d 110 111 112 113 gt N C I I I I I I I 44444444 PE ltlgt1ltlgt2ltlgtsltlgt4 PC 4 15 16 I 18 PCz 19 110 111 112 PC 4 113 114 115 16 PC 12 11 414 15 16 PC 22 b6 I b8 19 PC 12 2 3 4 15 PCquot22 1o 111 12 113 Pi 9 10 11 12 PS 4 5 6 7 8 P5h 1 2 3 4 PS 4 13 14 15 16 PG v 11 114 115 16 PG v b6 I b8 19 PG d 12 13 14 15 PG d 110 111 112 113 gt LP A 2u U llllllw AA A AAA A A PE ltlgt1ltlgt2ltlgtsltlgt4 PC 4 15 16 I 18 PCz 19 110 111 112 PC 4 113 114 115 16 PC 12 11 414 15 16 PC 22 b6 I b8 19 PC 12 2 3 4 15 PCquot22 1o 111 12 113 Pi 9 10 11 12 PS 4 5 6 7 8 P5h 1 2 3 4 PS 4 13 14 15 16 PG v 11 114 115 16 PG v b6 I b8 19 PG d 12 13 14 15 PG d 110 111 112 113 gt LIJB1u PE ltlgt1ltlgt2ltlgtsltlgt4 PC 4 15 16 I 18 PCz 19 110 111 112 PC 4 113 114 115 16 PC 12 11 414 15 16 PC 22 b6 I b8 19 PC 12 2 3 4 15 PCquot22 1o 111 12 113 Pi 9 10 11 12 PS 4 5 6 7 8 P5h 1 2 3 4 PS 4 13 14 15 16 PG v 11 114 115 16 PG v b6 I b8 19 PG d 12 13 14 15 PG d 110 111 112 113 gt L11B2u PE ltlgt1ltlgt2ltlgtsltlgt4 PC 4 15 16 I 18 PCz 19 110 111 112 PC 4 113 114 115 16 PC 12 11 414 15 16 PC 22 b6 I b8 19 PC 12 2 3 4 15 PCquot22 1o 111 12 113 Pi 9 10 11 12 PS 4 5 6 7 8 P5h 1 2 3 4 PS 4 13 14 15 16 PG v 11 114 115 16 PG v b6 I b8 19 PG d 12 13 14 15 PG d 110 111 112 113 gt L11Eg LIJIA2u 41 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 No nodes LPB1u 2 3 4 6 7 8 1o 111 12 14 15 16 LPB1u 2 3 4 6 7 8 1o 111 12 14 15 16 2 nodes L11Bzu 41 39 15 19 39 113 L11Bzu 41 39 15 19 39 113 2 nodes LPEg 41 2 3 439 939 1039 1139 12 LPEg 41 2 3 439 939 1039 1139 12 1 node LPEg 413 14 115 1639 539 639 1739 18 LPEg 413 14 115 1639 539 639 1739 18 1 node PA2u 41 12 13 4 15 b6 I 18 9 1o 111 12 13 14 15 116 LPB1u 2 3 439 639 739 8 1o 111 1239 1439 1539 16 LP82u 41 39 15 939 113 LPEg 41 2 3 439 939 1039 1139 12 Second unique basis 4 1 C4 rotation Clockwise 02 rotation C4 rotation 16 Counter clockwise ltlgt1 C 2 rotation 1 16 C 2 rotation 2 C 2 rotation 1 2 C 2 rotation 2 PE 11 239 339 4 PC 4 15 16 39 I 39 18 PCz 19 1o39 111 39 112 PC 4 113 114 39 115 39 16 PC 12 11 114 15 16 PC 22 16 7 8 19 PCquot12 12 3 4 15 PCquot22 110 111 12 113 4 C 44444444 PE 11 239 339 4 PC 4 15 16 39 I 39 18 PCz 19 1o39 111 39 112 PC 4 113 114 39 115 39 16 PC 12 11 114 15 16 PC 22 b6 4 8 19 PC 12 12 3 4 15 PCquot22 110 111 12 113 gt WA J gt I I I I 4444 WE 11 239 339 4 PC 4 15 16 39 I 39 18 PCz 19 1o39 111 39 112 PC 4 113 114 39 115 39 16 PC 12 11 114 15 16 PC 22 b6 4 8 19 PC 12 12 3 4 15 PCquot22 110 111 12 113 gt W2 J IIAIAIAw AA A A U PE 11 239 339 4 PC 4 15 16 39 I 39 18 PCz 19 1o39 111 39 112 PC 4 113 114 39 115 39 16 PC 12 11 114 15 16 PC 22 b6 4 8 19 PC 12 12 3 4 15 PCquot22 110 111 12 113 gt PB1u J I Nw U PE 11 239 339 4 PC 4 15 16 39 I 39 18 PCz 19 1o39 111 39 112 PC 4 113 114 39 115 39 16 PC 12 11 114 15 16 PC 22 b6 4 8 19 PC 12 12 3 4 15 PCquot22 110 111 12 113 gt mg J OOOOOI39vONlTI PE 11 239 339 4 PC 4 15 16 39 I 39 18 PCz 19 1o39 111 39 112 PC 4 113 114 39 115 39 16 PC 12 11 114 15 16 PC 22 b6 4 8 19 PC 12 12 3 4 15 PCquot22 110 111 12 113 gt mg J LPA1u 42 39 4 639 18 1039 112 1439 116 LPA1u 42 39 4 639 18 1039 112 1439 116 4 nodes LI1mm 41 3 5 7 9 111 13 115 LI1mm 41 3 5 7 9 111 13 115 LI103m 3 17 111 115 LP0320 11 2 4 5 6 8 9 10 12 13 14 116 L11059 41 2 3 4 9 10 11 12 L11059 41 2 3 4 9 10 11 12 W 3 nodes L11059 45 6 7 8 13 14 15 16 L11059 45 6 7 8 13 14 15 16 Y 3 nodes Third unique basis 4 1 C4 rotation Clockwise 02 rotation C4 rotation 16 Counter clockwise C 2 rotation 2 C 2 rotation 1 2 C 2 rotation 2 PE 139 239 3 4 PC 4 15 39 b6 39 I 18 PCz 939 1o39 111 112 PC 4 113 39 114 39 115 16 PC 12 11 114 115 16 PC 22 16 quot39 17 quot39 18 19 PCquot12 12 quot39 13 quot39 14 15 PCquot22 110 111 12 113 44444444 4 C PE 139 239 3 4 PC 4 15 39 b6 39 I 18 PCz 939 1o39 11 12 PC 4 13 39 14 39 15 16 PC 12 1 14 15 16 PC 22 b6 I 8 19 PC 12 12 13 4 15 PCquot22 1o 11 12 13 J gt WA This wavefunction is mathematically identical to that found with basis number 2 gt I I I I 4444 WE 139 239 3 4 PC 4 15 39 b6 39 I 18 PCz 939 1o39 11 12 PC 4 13 39 14 39 15 16 PC 12 1 14 15 16 PC 22 b6 I b8 19 PC 12 12 13 4 15 PCquot22 1o 11 12 13 gt W2 This wavefunction is mathematically identical to that found with basis number 2 lw U PE 139 239 3 4 PC 4 15 39 b6 39 I 18 PCz 939 1o39 11 12 PC 4 13 39 14 39 15 16 PC 12 1 14 15 16 PC 22 b6 I 8 19 PC 12 12 13 4 15 PCquot22 1o 11 12 13 gt PB1u This wavefunction is mathematically identical to that found with basis number 2 w N U PE 139 239 3 4 PC 4 15 39 b6 39 I 18 PCz 939 1o39 11 12 PC 4 13 39 14 39 15 16 PC 12 1 14 15 16 PC 22 b6 I b8 19 PC 12 12 13 4 15 PCquot22 1o 11 12 13 gt mg This wavefunction is mathematically identical to that found with basis number 2 OOOOOI39vONlTI PE 139 239 3 4 PC 4 15 39 b6 39 I 18 PCz 939 1039 11 12 PC 4 113 39 114 39 115 16 PC 12 11 114 115 16 PC 22 b6 I b8 19 PC 12 12 13 14 15 gt mg PCquot22 110 111 12 113 J L11059 41 2 3 4 9 10 11 12 L11059 5 6 7 be 113 14 15 16 Fourth unique basis 4 1 C4 rotation Clockwise 02 rotation C4 rotation 16 Counter clockwise C 2 rotation 2 C 2 rotation 1 2 C 2 rotation 2 PE 1 2 3 4 PC 4 15 b6 I 18 PCz 9 110 111 112 PC 4 13 114 115 16 PC 12 11 14 115 16 PC 22 16 I b8 19 PCquot12 12 13 14 15 PCquot22 1o 111 12 113 4 C 44444444 PE 1 2 Ike 14 PC 4 15 b6 I 18 PCz 9 110 111 112 PC 4 13 114 115 16 PC 12 11 14 115 16 PC 22 b6 I b8 19 PC 12 12 13 14 15 PCquot22 1o 111 12 113 J gt I I I I 4444 WE 1 2 Ike 14 PC 4 15 b6 I 18 PCz 9 110 111 112 PC 4 13 114 115 16 PC 12 11 14 115 16 PC 22 b6 I b8 19 PC 12 12 13 14 15 PCquot22 1o 111 12 113 gt W2 J IIAIAIAw AA A A U PE 1 2 Ike 14 PC 4 15 b6 I 18 PCz 9 110 111 112 PC 4 13 114 115 16 PC 12 11 14 115 16 PC 22 b6 I b8 19 PC 12 12 13 14 15 PCquot22 1o 111 12 113 gt PB1u J I Nw U PE 1 2 Ike 14 PC 4 15 b6 I 18 PCz 9 110 111 112 PC 4 13 114 115 16 PC 12 11 14 115 16 PC 22 b6 I b8 19 PC 12 12 13 14 15 PCquot22 1o 111 12 113 gt mg J OOOOOI39vONlTI PE 1 2 Ike 14 PC 4 15 b6 I 18 PCz 9 110 111 112 PC 4 13 114 115 16 PC 12 11 14 115 16 PC 22 b6 I b8 19 PC 12 12 13 14 15 PCquot22 1o 111 12 113 gt mg J LIJIA2u 139 2 339 4 539 6 739 8 939 10 111 39 12 1339 14 1539 16 LIJIA2u 139 2 339 4 539 6 739 8 939 10 111 39 12 1339 14 1539 16 8 nodes 3 x V LIJB1u 2 3 4 6 7 Pa 1104quot bu 112 14 15 116 L11I32u 1 2 3 4 9 10 11 12 L11059 41 2 3 4 9 10 11 12 L11059 5 6 7 8 113 14 15 16 Determining the excited state configuration The excited state configuration is determined by the direct product of the irreducible representations of each of the molecular orbitals By definition the configuration of the ground state is A19 Why To obtain the excited state configuration shown in the previous slide we use the product of A1 uEg which is obtained by multiplying the characters for each will give the symmetry of the excited state configuration In the case of porphyrins and many other aromatic molecules there are two excited state configurations that can mix by configuration interaction The requirement for mixing is that the excited state configurations must have the same symmetry The direct product of A1U Eg 20V 2 20 2 20 2 1234 ch 20V 20d 1111 1 1 1 1 1 E 1 2020020200 2020020200 The direct product of A2U Eg 20V 2 20 2 20 2 1234 ch 20V 20d 111 1 1 1 111 E 1 2020020200 2020020200 The direct product of Eu Eu E 2cv c2 202 202 i284 oh 20V 20d 2 O 2 O O 2 O 2 O O 2 O 2 O O 2 O 2 O O 4 O 4 O O 4 O 4 O O The decomposition of the product E 2cv c2 202 202 i284 oh 20V 20d 1 1 1 1 1 1 1 1 1 1 4 O 4 O O 4 O 4 O O 4 O 4 O O 4 O 4 O O FA1g16161 The decomposition of the product E 20V 2 20 2 2C3 2 1234 ch 20V 20d A291 1 1 1 1 1 1 1 1 1 r 4 o 4 o o 4 o 4 o o 4 o 4 o o 4 o 4 o o FAzg16161 E 20V 2 20 2 2C3 2 1234 ch 20V 20d B191 1 1 1 1 1 1 1 1 1 r 4 o 4 o o 4 o 4 o o 4 o 4 o o 4 o 4 o o FB1g 1616 1 The decomposition of the product E 20V 2 20 2 20 2 i284 oh 20V 20d 1 1 1 1 1 1 1 1 1 1 4 o 4 o o 4 o 4 o o 4 o 4 o o 4 o 4 o o FBzg16161 E 20V 2 20 2 20 2 i284 oh 20V 20d 2 o 2 o o 2 o 2 o o 4 o 4 o o 4 o 4 o o 8 o 8 o o 8 o 8 o o rEgo The significance of excited state mixing We have determined that the irreducible representations That promote mixing of the direct product EUEu is given by r A1g Azg B1g B2g Thus these are the symmetries of the vibrational normal modes that give rise to FranckCondon activity and vibronic coupling Another way to look at this situation is to examine which optical transitions are allowed In the character table you can see that xy transform as Eu and z transforms as A2 These irreducible representations correspond to the polar ization of electromagnetic radiation that interacts with a molecule Using symmetry to predict allowed transitions We have seen that the transition moment is proportional to the integral Mge LPgeGLPed C o X y z The symmetry requirement for an allowed transition is that the direct product of the ground state excited state and the polarization of radiation should contain the totally symmetry irreducible representation In the case of porphine we can make the following lden ca ons LP AWAZU L11e 59 ml Eu 2 A 2u Using symmetry to predict allowed transitions For example for xy polarized light we have the following Product of irreducible representations AZUEUEQ or A 1UEUEQ The product of ground state wave function excited state wave function and electromagnetic polarization is known as the triple product In this case there are two triple products since there are two possible excited state configurations Note that we have already worked out the direct product of these and they are rA1gAzg B1g B2g Irreducible representations of the Fe dorbitals s o dX2y2 B1g 4 dxz yz E9 96 ft dxy 829 Low spin FeII Irreducible representations of the Fe dorbitals S 86 44 2 L dX2y2 B1g L dZ2 dXZ2dyz E 9 1 I dXY Bzg High spin FeII A more realistic energy level ordering is observed due to Blg distortion of the heme 50 52 dX2 2 dX2y2 B1g i dz 4 dz 88tt it E9 88 L ft dxy c 3 F Low spin FeII High spin FeII Review of Electrostatics The Coulombic force on charge due to charge i is T 1 qui s Wm The Coulombic force is additive The combined force is a superposition The force on charge k due to a number of charges with the index is The constant an is the permittivity ofvacuum ln MKS units the value is an 8854 x1012 C2 N391 m39 In the cgsesu unit system the permittivity of tree space is 147 and the constant 147rcn does not appear in the Coulomb force Electric Field The electric eld is is the force per unit charge The most precise statement is that it is the force per unit charge in the limit that the charge is in nitesimally small When applied to the Coulomb force the electric eld becomes E 1 M k 4180 qr Electrostatic Potential The electric eld is the negative gradient of the scalar potential 7 The potential at a distance rfrom a charge is 7 47r2nr The electric eld represents the force per unit charge The potential is the work per unit charge W12 72 71 In MKS units the potential has units owahere 1 V 1 JC Potential and Field due to a Dipole The potential due to a dipole is If 7 r 4n8ur3 The assumption in this equation is that the distance between the charge and dipole r is large relative to the separation of charges in the dipole d r d The electric eld due to a dipole is Using the expression W u E we can calculate the interaction energy oftwo dipoles 31 mi 7 r5 Example Effect of Dipole Orientation Consider two dipoles which have the orientations below that we can call aligne and headtotail Aligned 111112 2111 u r 112 r u r W 41124112an Headtotail 111112 u2 111 r 0 illz39r 0 W 1124712an Electric moments The potential due to discrete charge distribution is 71 R L M 47mm R rk If the distance R of the test charge is large relative to the distances between the charges then the expansion 1 1 1 1 1 m rl7 rgrkl7l1 can be made V Electric moments The potential is then given by q 1 1 1 4nsn R i 11 79 V This is the multipole expansion The terms are the charge also called the monopole q the dipole u the quadrupole G and higher orderterms q 19de Ju qfi I 9737d7 mg q7r7fplt7gt7raf q is a scalar u is a vector a rst rank tensor 6 is a matrix a second rank tensor Interaction of electric moments with the electric field The interaction ofa collection of charges subjected to an electric eld is given b W q pE 7E The picture is that of a charge interacting with the potential the dipole interacting with a eld the quadrupole interacting with the eld gradient etc An electric eld can exert a force FZ mar or a torque TZ r x mm on a collection of charges Polarizability In the presence ofan externally applied electric eld the eipole moment ofthe molecule can also be expressed as an expansion in terms ofmoments39 u uo aE pEE The leading term in this expansion is the permanent dipole moment u The polarizability is a tensor whose components can be described as a follows a 6m xv aEy D Where the 0 subscript refers to the fact that the derivative is Evalua eld The Btensoris called the hyper polarizability and is third ranked tensor Polarizability as second rank tensor The dipole moment components each can depend on as many as three different polarizability components as described by the matrIx lix 0w 0w mxz Ex y aw 0 DWI Ev liz my DLzy 0 22 E z lfa molecule has a center of symmetry eg CCIA then The polarizability is a scalar ie the induced dipole moment ls always in the direction of the applied eld However for noncentrosymmetric molecules components can be induced in other directions The directions are often determined by the directions of chemical bonds which may not be aligned with the eld This is the signi cance of the tensor Properties of the polarizability tensor Like the quadrupole moment the polarizability can be made diagonal in the principle axes ofthe molecule In the laboratory frame of reference the polarizability depends on the orientation ofthe molecu e The average polarizability is independent of orientation It is given byt e Trace which is written Tr 0 Tr a xXX 1W an The polarizability increases with the number of electrons in The molecule orwith the volume of the charge distribution Classically for a molecule of radius a o 4n ua3 Lorentz classical polarizability The Lorentz model is given here An electron in one dimension is subject to an electric eld that results in a displacement from the center of charge For a displacement in the x direction this is the polarization ofthe electric eld ie the electric vector of radiation the induced moment is given by Hm eX IE The restoring force is F kx mmuzx The force is balanced by the force due to the applied electric eld F eE The balance offorces is moaDZX eE Lorentz polarizability Thusxis X eE moan2 and the dipole moment is e2 lima WE The classical Lorentz polarizability is a 6 quot703m The expression can be generalized by letting there be N electrons on a molecule Each electron has an intrinsic harmonic oscillatorfrequency The 39action of the electrons in each ofthej modes is The polarizability for all ofthe electrons is e2 f u E 2 Frequency dependence of polarizability The quantity is the oscillator strength The above treatment is for a static eld In the case where Et is a sinusoidal eld the equation resembles that ofa driven harmonic oscillator Jx 2 F m eEt mwEl xt th In addition to the electric force and the harmonic restoring force we have added a frictional force The time dependent electric eld must have the form Et Enxexpiky 030 Therefore we try a solution of the form xt xnexpiky wt imxuexpiky oat df 2 g anX exp i ky oaf dtz u Frequency dependence of polarizability The equation becomes w2xt at wuzxa xt rxt we have e2 f ale wLwL We can also de ne the polarizability o extEt so the polarizability has real and imaginary parts These are obtained by multiplying both numerator and denominator by of w iw1quotlm Frequency dependence of polarizability The complex polarizability is um12 z 0 j 2 03 m Clearly the polarizability consists of real and imaginary parts We can write it s w w I39m1m Kramer s Kronig Relations The real and imaginary parts are related to one another through the KramersKronig relations 2 Nso sds i W 2w No sds T W Oscillator Strength The oscillator strength is a classical formalism The oscillator strength U is proportional to the intensity of a transition i 9 j It is a number less than one and in fact the sum of all of the oscillator strengths in the molecule equals one Thus U Comparison of the quantdm treatment with the Lorentz formulation gives 2 f 2mium 03m l 3e2i7 The factor of 13 arises because only one polarization of electromagnetic radiation is considered here Raman Spectroscopy Experlrnental eensiueratiens KramerseH elsenbergeDlrac as a sumuverestates furrnallsrn Tlrneecurrelatur Practical ealeulatiun er absurptlun and Rarnan eruss semiuns NC State Un vers ty Resonance Raman spectrum is obtained by a laser light scattering experiment Detector p Lens gt 4 Inelastic light scattering produces a frequency shift T r is exchange of energy between the vibrations ofthe molecule and the incident phot Raman scattering Resonance Raman is a two photon process Incident photon from a laser hv Scattered photon has an energy shift Elergy l i The difference is 7 7 because the molecule is left in an excited vibrational state z u 2 1 Nuclear Disptauement 72 u 2 4 Nuclear Dlsplacemenl pl urElS h Wn KramerHeisenbergDirac n scatteran depends on the rneleeular pularlzablllty nelu drlvesthe system inte Excite sta e ngt a The eleetrie and then from lngt it returns tn the tin l state lrgt i Z ltilulmltnlulrgt ltiliilngtltnli lrgt r hii mnmlf 39 mnemmeirquot anytvvu er the Cartesian dlfferences an include a Two electrieneld interaetiuns are required in the abuve lEI a d a nergy al euntributiuns e i urz Thee buth ElE aniE and yibratiun hi or Muir an Eunsider buth nuneresunant and resunant Wavepacket dynamics Ra man litgt HW lD lt itgt Oi Oi Nuclear Displacement Tlme We eases fur KHD Energy Fl39 of ltf itgt yields the REP J Trmgl39ixlarwirwuremia lt0l0gtlt0l1gt ltL0l1gtlt1l1gt l 39i i aquot a Ahsnlmm my my Nuclear Displacement Calculating Resonant Raman Cross Sections F39erfurrn an uptical absorption calculatiun ldentify reieyant states 2 9 four urbital mudel rn a vibrational frequency calculatiun F39ruyect alung eigenvecturs ter eaen nurrnal made F39lut potential energy surfaces and determine displacements Use dimEnSiDnlESS deltas in lime eurreiaturtu calculate an NC State University We can construct molecular orbitals of benzene using the six electrons in p orbitals l H cEKH r i n Benzene Structure Electronic Energy Levels An electronic wavefunction corresponds to each energy level NODES Benzene transitions 20 I 22 21 39 Relevant states furfuur urbital mudel All are predicted te naye ruu niytne same intensity and energy by DM en This is at Wman Observed Weak band ZED nrn and intense band lEE nrn TDDFT calculation two own no Predictiuri isfurunly a strong band att7t nrn n anrntro o Tnts ts o ly rntne bsewed oanot ifquot Tne reason tne Weak oano ts nottouno tstnat no Vibrunl eouoitng ts tneiuoeo V Awll l ern1 Ewl a cml Potential energy surfaces Unsntfteo TIMETHERM TlMETHERM ts a orograrn tnat eaieuiates ootn tne aosorotton soeetrurn and Raman Excitatiun Profile REP ustn tne ttrne eorreiator Tnts ts a firstrprinciples eaieuiatton ano tt ts Exact Tne assurnptton ts tnattne potenttai Energy surfaces are narrnonte Tne tnouts are tne oararneters neeoeo tor tne ttrne eorreiator Mooei Foreacn Raman amive rno e s Electrunrplwunun eoupitng constant 5 AZz otrnenstoniess otsoiaeernent ootatneo trorn F39ES o treguenoy tn ernt Overallthere ean be a Gaussian and Lurentzian oroaoentng r Lurentzian oroaoentng tn Eml o Gaussian oroaoentng tn ern 1 Calculating the dimensionless 39 placement Tne utsoiaeernentottne ootenttai Energy surtaee ts tn atornte units in urdertu urtverttu dimensionless units the rnust be norrnaitz asfulluvvs AgE Energy l 72 o z 4 Nuclear otsptacetnem Benzene example Benzene ts a rnoieeuie of ntgn symmetryDen As a result Only A1v rnoue rnan active Huvveven E2v rnoues c be vibrunically eouoieo Tnts eornoiteates tne tnteroretatton uleMETHERM tnout in tneoryTiMETHERM only Wurks tor Francerundun active modes t e A symmet Huvveven tne oouoieweii ot tne E v es ean oe tnteroreteu as a otsoiaeernent anu tttnts ts uone tnen better agreement wttn Experiment ts found t Art Example ts pruvided on tne nextslide Calculated Benzene Absorption and Raman Spectra roauentng ts aoueu artificially based on Experimental Line o soeetrai Wtotns Paracresol example Paracresul is a model furtne amlnu acid tyrosine The luvver symmetry er paracresul implies meme issue er vibruni eeupiing is greatly reduced in the fulluvving We Examine a survey erme nurrnal melee ufparacresul lri order m understand some er the issues that arise when the putantial Energy surface is calculated using the metnuds Eveluped rm Estimating the dimensionless displacement Paracresol transitions 29 31 Vibrational Modes of paracresol A Model for Tyrosine Mode 15 844 cm391 Vibrational Modes of paracresol A Model forTyrosine W Mode 22 1159 cm391 Vibrational Modes of paracresol A Model for Tyrosine fe 39l Enmgy M m Mode 231174 cm391 ammu emu Vibrational Modes of paracresol A Model forTyrosine Vibrational Modes of Paracresol A Model for Tyrosine Mode 33 1600 cmquot E Vibrational Modes of Paracresol A Model for Tyrosine Mode 34 1635 cmquot TIMETHERM input file for para cresol 9 llKlEIEIEI l l zuunun 46U7EIEIEI mm mm Example contd DD Designate tlne rnuue fur which tlne excitauun a pm lewlllbe calculateu fulluwedby tlne change in quanta fur tlne acuve rnuue 31 uu susluun 25m mm 61 M 1m In M m m Enter LEIWhEn dunefureach REF llSBEIEIEIEI nusz zunnnu lmanuu mm mm 5 lZlZEIEIEIEI I113 zunnnu uu l EIEIIEIEIEI mm mm 7 1634mm I123 znunun 9 uu 91 Example contd Example contd crez abs I I1 cunvergence criteria 3 Output les fur alusurpuun n 23 transiuun rnurnent A mam ram and Raman excitatiun pru le l 33 sulventinuex ufre 39acuun U m number ufurn mam mm H 4 ax ume m ps n May savetlne urne currelatnr 35857 M transmun energy ems mm m results furtliereal anu irnannary znuu numberufenergy nuints U parts Hyes Ann zunun energy range aruunnl EUrEI mam ram LurentzianGau sian bruademng l s HUD humugeneuus Lurentzian crnrl n umugeneuus Gaussian crnrl znn mhumugeneuus Gaussian std ueViauun n ncrles rarn n Ranian Excitation Pro les Ext MndzlS th222 th224 Mum m r l lu2232 mums mm mm arku mm lqu A mm mm 035nm Him mum la 9 5mm HIng nuazss mum in 3 5mm HEI ZZ mm um am First column is excitation frequency preceded by cross sections for each active mode Calculated Raman excitation pro le Raman Cross Section ma E EID15 3 40 Wavenumbers MEI cm Elm umu Resonant Raman Spectrum for Tyrosine PBE Z nm H Raman Crass Seminn MU Ramaquot Shi cmquot Resonance Raman Spectrum for Paracresol at Variant Excitations F aman Cross Section MD 72m nm 7218 nm Ramansmmemquot Experimental vs Calculated ween imam ThmsUVmem1KmmJ m w my mam mamMew mm m 2 mmquot EWWWSJWX 63 is m m m gmmpvammmmmmi Fem mm Perturbation Theory We will be interested in the effect on molecules of weak interactions such as electric or magnetic fields Moreover Almost any process that is considered forbidden in quantum mechanics can be observed due to some higher order effect In all of these cases we treat the effect of a small perturbation on the system using Perturbation Theory The hamiltonian HlO is modified by a perturbative term H H Hlt gt m We assume that we know the zeroorder O eigenfunctions and energies O O O O Perturbed Energy and Wave Function The Iamda A is just a number It can be used to turn on the Perturbation If A 0 then the perturbation is not present If A 1 then the perturbation is present This method is convenient so we can keep track of any changes We can Also consider various orders or perturbation first order A second order A2 etc T L11 ALP AZLsz We say the L111 is first order correction to the wave function Similarly the energies can be written E E O AE AZE Z Convergence requires the EU ltlt E0 etc Perturbed Energy and Wave Function If the states are not degenerate then we find that E ltLPK gtHILPK gtgt 5 Hi ltLP0HIILPI0gt 0 H 1 u 0 LP Ego Ego LP Ere Ereij I j I j The perturbed wave function is composed of a linear combination of terms derived from the zeroorder wave function LPin Z CjiLPer j I The prerequisite for the existence of a perturbed wave function is that the matrix elements ltLPjH LPigt are not zero E C m I m Asmiii wv C E E C C E Asi wv m I 8T M A iire wv C E u AEEAOVEV C E C C E E C E Asi wv m I Asisvmgvsvm A6151 59v 2 E C C E C C E C E Aeia vavm Aavia vem u Aeiavlro v Aaviklrgmgv 3AOWMK SWVFMMK swigamm H CWVFBIA SWVHEI Jammie a amp ammgxcmma ammv n a amp 63431 QIV 3 Namp 3 amp 6 H e 95 rHE E 2 CE E CE 2 A 6126 v n o 99 air 0 now u 6 The Anharmonic Oscillator 2 2 n0 K 2 ZHanxv 20 xv vav 2 2 lt0gtn6 52 H 2MQZ2Q H1Q3Q4 xggte 4e wz2 xaogte 4mm 59 dek gtoQ3Q4 we 00 E81 1zf dQe OLQ2Q3 Q4 2i 2 doeWZQ4 lt4gt 2 OL2 1 01200 2 XQ2B3 Y 4 E12 IdQO Qe T mew i 32a2 25 20 Energy units of hv 05 00 Zeroth Order Energies E1 32hv V E0 12hv I I I 1 0 1 Nuclear Coordinate normalized units First Order Corrections 25 20 y O1oc E 15 AE1 5y320c2 i 10 m 05 AEO y320c2 00 V 2 1 O 1 2 Nuclear Coordinate normalized units In this example the anharmonic correction is a little more than 1 Example Normal modes of a linear triatomic ABA 9X 1 Consider only motion along X for this example 2 Assume that the force constant for the A B bond is k V kx2 x12 kx3 x22 The coupled equations have the form mfg kx1 x2 O m3kx2 xl kx2 x30 mfg kx3 x20 In massweighted Cartesian coordinates 2 m6n1k n1 m 20 6 m7 m 62112 n2 n1 n2 m 0 m Barz chm B Jm A kJm B Jm A 6 n3 m m A 6Z2 kbm A m 0 Substituting in the values k all a33m a 22k 22 m3 k a12a23 mAmB We obtain the equations 62 611 ka11n1 6112112 O 62 612 k6122112 6112111 k6122112 6112113 2 O 62 613 ka33n3 6112112 O These equations can be Wiitten in matrix foim as all t a12 O det a12 6122 t a12 O 0 a12 all k Note that we have assumed that am 0 for simplicity The detenninant can be expanded as 6111 A and 12k 6112le all may a 2a122au a One solution is V Z all m A The second solution is obtained by expansion of the terms inside the brackets k2 6111 6 22L aliazz 26122 Z O 2 2 9L 6111 6 22 3 a11 6 22 461116122 8612 2 This can be simplified as follows a a i a a 28a 2 A ll 22 ll 22 12 k 2k k 212 k2 m Am Bi 8mAmB A 2 k 2k k 2k Wm Blilm AW sz k 2k im3 i 0 mA The eigenvectors are the matrix terms that transform from mass Weighted Cartesian space to normal coordinate space The eigenvectors can be found by substituting in each of the eigenvalues into the equation 0 ail 9L 6 12 O 111 o 6 12 6122 L 6 12 112 O O 6 12 ail 9L T For example for 9 all we have 0 an 0 n 6 12 6122 6111 6 12 11g 0 0 an 0 112 Thus 012n20 a 1120 01211 6122 a11ng 012113 O 611211 6112113 2 O n 112 Finally we use normalization to determine the magnitude of the coef cients 1102 1192 1192 1 2n21 o l 111 The eigenvector corresponding to this normal mode is 2 L 10 E The normal mode corresponds to a symmetric stretch A The normal coordinate is Q L 1 m Tl 3 2 x1 x3 The central atom B does not move in this normal mode This illustrates the point that in the normal mode description the center of mass does not move We can repeat the calculation using the roots M and h The eigenvectors in these cases are Where M 2mA mB In 12 the negative sign on the central atom B indicates that the displacement is opposite to the end atoms A Thus this mode is an asymmetric stretch The third eigenvector represents translation Translation has not been separated out since we are working in Cartesian coordinates Polarizability of the hydrogen atom In order to calculate the polarizability of the hydrogen atom we need the transition moments for the allowed transitions In the presence of a Zpolarized applied electric field we have the induced dipole moment ltnmZ1OOgt2 E1 En Since u 0ch we have for the polarizability lt nmZ100 gt2 E1 En obtained from CohenTannoudji page 1280 It appears from the selection rules that An any value AI i1 and Am 0 ltLPoqZILP0gt ZQZEOZN3 or 2q2 Polarizability of the hydrogen atom We have the following matrix elements 00 1 r r2a 1 1 32 ra 2 In 2 271 lt21OZ100gt 0 0 dr COS GSnedG d lo 32nag2e rnao e r O l 0 1 making the substitution x cose and collecting constants we have 27 1 00 32 4 I1 e aOr dr X2dX V 327563o 1 Let u 3r2a0 then r 2a03u and dr 2a03du so lt210Z100gt 1 lt210Z100gt e uu4du X2dX 1 64 are 243J3 2 00 Polarizability of the hydrogen atom Evaluating the integrals we obtain 4 2 lt21 OZ1 OOgt a4 O744a quot 243J3 203 The energy levels are given by i72 2iia2n2 The energy difference in the denominator is hz 1 E Equot 2Ma n2 1 Polarizability of the hydrogen atom The polarizability can be expressed 2pq2a022 lt nmZ1OOgt2 0 2 n n l 1 I7 Thus for example for the 210 state we have 2 4 2 OL210 2Hquot 0741 A3 1 The constant in front of the polarizability expression is 2 4 2Mq2ao4 29109 gtlt 103931kg162 X 10 529 x 10 11m 72 1054 X 103934Js2 Polarizability of the hydrogen atom The units work out as follows 2 4 ZWZI fo 337 x 1041 sz2J 2quao4 337 x 10 sz2J eohz 8854 x 10 1ZCZm 1J 380 gtlt1O3O m3 380 A3 Polarizability of the hydrogen atom In orderto calculate the polarizability ofthe hydrogen atom e e transition moments ort e allowed transitions In the presence ofa Zpolarized applied electric eld we have the induced dipole moment ltnlmZ100gt2 ltWuq2wugt MED n Since it QED we have forthe polarizability lt n lm Z 1 0 0 gt2 LMZ E1 n obtained from CohenTannoudji page 1280 It appears 39om the selection rules that An any value AI i1 nd Am 0 Polarizability of the hydrogen atom We have the following matrix elements NTLWJTLMq fz I lt2 1 0izi1 0 0gt lmh wy ur GD e Urdrucos Sweden m making the substitution x cose and collecting constants M1 Let u 3r2an then r 2aD3u and dr 2aD3du so 2 m i lt21021oogt quot e 3 ua r drl xzdx 64 w 1 lt21oz1oogt d1 2d HIIH 243Tzauueuu71xx Polarizability of the hydrogen atom Evaluating the integrals we obtain 4 2 lt2102100gt a 4 0744a l l I I i l I 3 El The energylevels are given by h2 En ZWSHZ The energy difference in the denominator is h21 El En 2mgan 1 Polarizability of the hydrogen atom The polarizability can be expressed 2 2 2 lt gt2 a 728D Z nmZ100 n n 1 I7 Thus for example for the 210 state we have 2 4 2 mm W ow 38A30738 280 A3 1 7 4 The constant in front of the polarizability expression is 1054x10 3 Js2 Polarizability of the hydrogen atom The units work out as follows 2 A zl Z f 337 x 10M CZmZIJ ZWZauA 337 x 10 czm2J gun 8854 x 10 Zczm4J 380 gtlt10393D m3 380 A3 Dielectric Polarization We divide matter into two categories conductors and insulators Free charges in a conductor will respond to exactly cancel an applied eld The charges in an insulator will respond to an applied eld in such a way as to partially cancel an applied electric eld The situation in an insulator is more complicated however since a molecule in the insulator will also experience a eld due to the response ofthe insulator There is a reaction eld due to the response ofthe medium to charges on the molecule and there is a local eld due to polarization ofthe solvent in the applied eld he e issues are important for relating the molecular polarizability to the bulk polarization Here we shall demonstrate the role of the dielectric constant also called the relative permittivity as a factor that relates the polarization ofan insulatorto an applied electric eld The Electric Field in a Capacitor The experimental geometry that is most convenient for the purpose of demonstrating the dielectric constant is the parallel plate capacitor We compare a capacitor with vacuum between the charged plates to one with a dielectric In the case of vacuum the eld is G EU 8 D 32 Notice that the way the eld is de ned it is independent of what is placed in the capacitor Field calculations are simple Just use the voltage and divide by the separation distance Common units of eld are Vcm orVm Illustration of the parallel plate capacitor The surface charge density is o qA The vacuum permittivity is an The potential is 1 and the distance between the plates is d The unit vectorz is normal to the capacitor plates These features are illustrated below charged plate z gmunded plate Definition of the Dielectric Constant If we now place a dielectric medium between the plates at constant voltage we have EU 22 Note that the eld is unchanged It is still the voltage divided by the distance However the surface charge dens39ty required to attain this eld is different 039 instead ofo because the medium has a permittivity a Note that since the surface charge density increases when the dielectric is present the capacitance also increases The relationship between the dielectric constant and the permittivity of vacuum is a a D The relative permittivity a is commonly called the dielectric constant The dielectric constant is greater than 1 and can be as large 1110 for formamide see Table 31 in Mol Spectroscopy by McHale Dielectric Polarization The larger the dipole moment the greater the tendency ofthe solvent to respond to an applied eld by reorientation of the microscopic dipoles However inspection ofTable 31 shows that there are exceptions and that liquid structure and collective dynamics also play a role To see the connection between the dielectric constant and the polarization we perform an experiment We charge up the capacitor in vacuum The eld39s E o a insulating medium with dielectric constant a leaving the charges constant Now the eld39s E 02 clean The eld is reduced The difference between E and EEl is due to the polarization of the medium P Dielectric Polarization We call E the macroscopic eld The polarization is proportional to this eld P 848 1E SEer where 15 is the electric susceptibility One main goal of studies of dielectric polarization is to relate macroscopic properties such as the dielectric constant to microscopic properties such as the polarizability Nonpolar gas phase molecules The relationship between polarizability and susceptibility is simple for nonpolar molecules in the gas phase where intermolecular interactions can be ignored The polarization be immediately expressed in terms of both electric susceptibility macroscopic and polarizability microscopic P MOLE XSSDE V We can see that Na X 7 e Ven and since 2 1 1E Na 8 1 Van Furthermore since 2 n2 we have 1 1 No 2Vsu Nonpolar molecules in condensed phase Interactions between nonpolar molecules cannot be neglected in condensed phases The treatment considers a local eld F inside the dielectric and its relation to an applied eld E The eld inside this uniformly polarized sphere behaves as if it were due to a dipole given by 3 476 P 1 3 Since P is the polarization per unit volume and 4na33 is the volume ofthe sphere we see that u is the induced dipole moment or polarization these are equivalent The local eld is the macroscopic eld E minus the contribution ofthe due to the matterin the sphere P F E Em E 380 Lorentz local field Since P 242 1E the Lorentz local eld is g 2E Since a 1 for vacuum and 2 gt1 for all dielectric media it is apparent that the local eld is always larger than the applied eld This simple consequence ofthe theory ofdielectric polarization causes confusion We usually thInk ofthe dielectric constant as providing a screening of the applied eld Therefore we might be inclined to think of a local eld as smaller than the applied eld However this na39I39ve view ignores the role ofthe polarization of the dielectric itself Inside the sphere we have carved out ofthe dielectric we observe the macroscopic applied eld plus the eld due to the polarization ofthe edium The sum ofthese two contributions leads to a eld that is always largerthan the applied electric eld The ClausiusMosotti Equation The polarization is the number density times the polarizability times the local eld P V fF g 2E sus 1E We eliminate E to obtain the ClausiusMossotti equation This equation connects the macroscopic dielectric constant a to the microscopic polarizability Since a n2 we can replace these to obtain the LorentzLorentz equation LorentzLorentz Equation Again here the equation connects the index of refraction macroscopic property to the polarizability microscopic o e number density NN can be replaced by the bulk density p gmcma through a NAP V M where NA is Avagadro39s number and M is the molar mass hus the LorentzLorentz equation that connects the index of refraction with the polarizability is n2 1 NAP a n2 2 3M8 Polar molecules The polarization we have discussed up to now is the electronic polarization lfa collection of nonpolar molecules is subjected to an ap lied electric eld the polarization is induced only in their electron distribution Howeve 39 molecules in the collection posses a permanent ground state ipole moment these molecules will tend to reorient in the applied eld The alignment of the dipoles will be disrupted by thermal motion that tends to randomize the orientation ofthe dipoles The nuclear polarization will then be an equilibrium or ensemble average ofdipoles aligned in the eld Nu T Pu Equilibrium averaged dipole moment The angle brackets indicate the equilibrium average If the permanent dipole moment is u then the interaction with the eld is W DxF u Fcose where e is the angle between the dipole and the eld direction Thus the average dipole oment is ltHgt H0ltcos egt The average indicated is an average over a Boltzmann distribution W Here cos Oex sin OdO I p kT n W ex i sIn OdO I p M ltcos Ogt Equilibrium averaged dipole moment Substituting in forthe interaction energy W we nd uDF cos O M sln OdO cos Oexp ltcos egt n uDF cos O 1 exp 7 sln OdO We make the substitutions HUF kT The integral is u X 00 6 The Langevin Function And the integral becomes 1 u ru xex ux dx w w I l l 2 2 cosO 1 8 e L expuxdx u u eu ew e e e e The Jnction cothu 1u is knovm as the Langevin Jnction lt approac es u or u ltlt an 1 n u is large The limit forlarge u is easy to see The limit for small u requires carrying out a Taylor39s series expansion of the function to many higher order terms coth u 1y The Langevin Function Fortypical elds employed unFkT ltlt 1 You can convince yourself ofthis using the following handy conversion factors u F 68 x 10395 cm39lDVcm k 0 697 cm39lK liquid water pus 24 D in a 10000 Vcm eld we have W 04 cm39l Here u FkT is ofthe order of11000 Thus we can express the orientational polarization as P Nu0 cos 6 NHOZ F 0 V T The total polarization is the sum of the electronic and orientational polarization terms 2 Pm P P Vla gngF The Debye Equation Following the same protocol used above to derive the ClausiusMossotti equation we obtain the Debye equation forthe molar polarization 2 M81 NA 0c 390 p s 2 380 3kT This equation works reasonable well for some organics however it fails for water The reason for the failure ofthe Debye m 39 F PM o e Is that the Lorentz local eld correction begins with a cavity large compared to molecular dimension and thus ignores local interactions of solvent dipoles The Local Field Problem The local eld problem is one of the most vexing problems of condensed phase electrostatics Following Lorentz there are two models the Onsager model and the Kirkwood model that attempt to account forthe local interactions ofsolvent molecules in an applied electric eld The approaches discussed here are all continuum approaches in that there is a cavity and outside that cavity the medium is treated as a continuum dielectric with dielectric constant 2 The models differ in how they de ne the cavity As stated above Lorentz model a umes a large cavity a is much largerthan the molecule size The Onsager model focuses on the creation of a cavity around a single molecule of interest a is equal to the molecule size The Kirkwood model includes a cluster around the molecule to account for local structure The Onsager Model The Debye model assumes that the dipole um is not affected by the solvation shell Yet consider water which has a phase dipole moment of186 D and in condensed phase has a dipole moment in the range 23 24 D The neighboring water molecules ave a large effect inducing a dipole moment more than 25 largerthan the gas phase dipole moment The dipole moment In is the sum ofthe permanent and induced arts m um ocF The local eld F has two contributions the cavity eld G and the reaction eld R F G R The cavity eld is given the spherical cavity approximation in terms ofthe applied eld 3 fE 28 1 The Onsager Model Notice that the cavity eld is always greater than one This is exactly analogous to the Lorentz local eld However the Lorentz local eld increases without bound as 2 increases The Onsager cavity eld increases from 1 to approaches no The reaction eld is proportional to the dipole moment of the molecule in the cavity Ply 1 m R 22 12n332n gm The reaction eld is always parallel to the permanent dipole moment Only the cavity eld can exert a tor ue on t e dipole and cause it to align in the applied eld By separating these two effects the Onsager model improves upon the Debye equation The Onsager Model The Onsager reaction eld is also an important relation for understanding the effect of solvents on the absorption and to obtain an estimate ofthe reorientational dynamics of solvents Frequency dependent dielectric function Viewed from a microscopic perspective we know that the molecular polarizability is frequency dependent Electronic polarizability is present in all molecules and has a response time that is rapid gt 10M s39l The high frequency response can follow the undulations ofelectromagnetic radiation in the visible region and hence this response gives rise to refraction of light This contribution is the high frequency or optical dielectric constant am There is a nuclear polarizability in lecules due to theirtendencyto align in an applied 1 2 They also contribute to the low frequency or static dielectric constant an The static dielectric constant is not really static but rather is due to changes in the electrical response due to dipolar reorientation Complex dielectric function We shall dissect the relative permittivity a into real and imaginary parts 803 803 84m These two contributions represent the inphase c39 and outofphase 2quot components ofthe frequency res onse of the medium The inphase component results in dispersion Physically this means refraction ofthe electromagnetic radiation as it passes through the medium The outofphase ent gives rise to absorption Absorption occurs in the visible electronic state transitions in 39ared vibrational transitions and microwave rotational transitions The real and imaginary parts ofthe frequency dependence dielectric response are related to one another by KramersKronig relations KramersKronig Relations asst My 9 n measuring dispersion For example di use re ectance spectra from crystals can be transformed into absorption ectra The refractive index can also be represented as a complex quantity Noa noa 0a Relationship of index of refraction The high frequency part ofthe dielectric response is equal to the square ofthe index of refraction 24w Nw2 Equating real and imaginary parts leads to 8103 quot303 Kzlm 2103 2nmKm The real part of the index of refraction n w is the factor by which the speed oflight is reduced as it traverses a medium The ima ina art of the index of refraction Kw is an absorption coef cient To understand the effects of these two terms consider an electric eld E Re Enexp ikx i 01 Absorption The wavevector in vacuum is k 2 7 27rN 7 Considering both the real and imaginary parts of the index of refraction we have E Re Enexp 27 1 wtexp 272a The exponentially decaying term represents the attenuation of radiation as it passes through an absorptive medium Since the intensity is proportional to the square of the amplitude of eld I Inexp 7x Inexp and in a dielectric medium it is k the electnc The absorption coefficient The absorption coef cient y is The absorption coef cient can be related to the molar absorptivity E units of L mol39l cm39l by comparing Beer39s law to the above expression Inexp 2303 6 OX ID1039ECX In the above expression x is the pathlength and C is the concentration Using these relations we can establish the connection between the imaginary part of the dielectric constant and the molar absorptivity Relationship to extinction coefficient The imaginary part of the dielectric function is related to The extinction coef cient as fo 0 39 a 23036NCI7 mNA where N is the number ofabsorbing molecules per cm3 This can also be expressed as the number of moles per L 2303600n 39 T Using the relation c wk and k 27d this expression can also be recast as 23036 021m A An example of matrix diagonalization Rotation of a quadrupole tensor We refer to the example of a quadrupole tensor A quadrupole moment is represented by a second rank tensor which is a matrix For charges at the following positions 1 at xy 11 and 11 1 at xy 11 and 11 The elements of the quadrupole tensor are obtained from Ga 2 qtriri The elements of the quadrupole tensor are xx 111111111111 0 9W 111111111111 0 Xy 111111111111 4 yX 111111111111 4 23 By inspection we can see that a 754 rotation of the coordinate system will diagonalize the matrix C 1 1 This is shown as a rotation of the charges for simplicity In the new coordinate system we have 1 at Xy 0120 and x20 1 at Xy 042 and 042 9 2 61m The elements of the quadrupole tensor are XX lV2V2l V2 V2l00 l00 4 yy l00l00lV2V2lV2V2 4 Xy lV20lV20l0V2l0 V2 0 yx l0V2l0V2lV20lV20 0 32 This problem is that of diagonalization of a matriX We began with a nondiagonal matriX and ended up with a diagonal matrix We did this using a transformation of coordinate system In this case the transformation was a rotation The general case of a clockwise rotation of a vector by an angle 9 is cosG sine x1 2 x2 sin6 cosG yl yz In this case we know that the rotation occurs through an angle 9 7t4 45 Thus cosG sinG l2 The rotation matrix is l l a 5 x1 M L L y1 y2 J to rotate the vector x1y1 by 45 We seem to know the answer We use this example to demonstrate a general method of matrix diagonalization and to show that this method also generates a transformation matrix The elements of the diagonalized matrix are called eigenvalues and the rows of the transformation matrix are called eigenvectors How do we obtain these eigenvectors and eigenvalues We wish to diagonalize the matrix 6 We require that 9 XIcT 0 CT is a column vector This is the same thing as a n x 1 column matrix We can write this as xx xyl0 c 0 yx G O L c We have substituted in one possible solution that we call 9o There is a second solution that we call 7L 9 xy x 0 610 oyx oyy 0 x c The solution to the above set of equations will exist only if the determinant of the matrix 9 M vanishes Using the explicit values above we have the determinant am x 216 x det k2 16 O yx GM 7L 4 1 Which results in the solution ki4 We see that the eigenvalue matrix is A 2 h 0 2 4 0 0 7t 0 4 This eigenvalue matrix A is exactly the same as the diagonalized matrix that we obtained above using geometric means To obtain the eigenvectors we substitute in each eigenvalue and solve for the coef cients We must use a normalization condition as a constraint 0120221 First we substitute in 9d Using the normalization condition Then we substitute in 7L Again the normalization condition is applied 0102 62 2CI2 1 Finally we construct a matrix of coefficients L L C 2 cr c 2 r J C 35 i L af af This is the transformation matrix It is equal to the rotation matrix that we discussed above The form of the transformation is o ACT 0 CT ACT CTA C lA CEDC 1 CC IA A This procedure defines the similarity transform CBC 1 A We have used the fact that C is a unitary matrix and therefore its inverse is equal to its transpose C C T The similarity transform is simply a rotation in the present case We have transformed the coordinate system into one which is rotated by 450 with respect to the charges in the quadrupole Thus we can show that 1 0 4 li li 4 O 4 O Ij lj lj LE i La 16 1E 1 w C Time dependence in quantum mechanics The Schrodinger picture timedependent wave functions The timedependent Schrodinger equation is 8w HM In The formal solution is thh iEth W3 e Ws0 e Ws0 provided ws is an eigenfunction of H The subscript 8 refers to the Schrodinger equation The timeindependent Schrodinger equation is a special case HwEtJ We can treat w as an eigenfunction of the timeindependent equation and then the factor e39iEt h is an arbitrary phase factor The Schrodinger Representation To proceed we introduce braket notation We can express the timedependent wave function as IWstgt 9 thhIWsOgt ltWst ltWs0lethh The expectation value at time t is lt AU gt lt WstAsWst gt lt wSOIethhA Se thhlws0 gt where AS represents the operator in the Schrodinger representation The Heisenberg Representation The above expectation value suggests that we may define the operator to be timedependent instead of the wave function This is known as the Heisenberg representation The connection is given by A eHthA eHth H S We may obtain an equation of motion by taking the timederivative of both sides of the above expectation value ltAtgt lt ws0le HthAsariMO gt lt WSOIeIHz hA Se thhlws0 gt lt WSOIethhe thhlws0 gt dt lt ws0e H A 3deC i Hth180 gt The Heisenberg Representation We have applied the product rule to the three quantities that depend on the time Since de Ht ethh dt n we can write the above expression as d i h i h a At E lt ws0eH HAse Hquot wso gt dAS thh lt WsOIeIHthwe IWsO gt lt ws0eIHthASHe thhws0 gt ltw80ethhe thhlws0 gt lt WSOethhH Ase thhws0 gt The Heisenberg Representation Ifthe expression is evaluated at t 0 then it can be written as d dAs i Elt At gt10 lt dt gt 5 lt HAS gt where the angle brackets represent averages with respect to ulSO Ifthe operator AS does not depend on time this is usually the case for a Schrodinger operator then the equation becomes we ltHAsigt Timedependent perturbation theory We treat the hamiltonian as a zeroorder part that does not depend on time HO and a perturbation H39 that does The Schrodinger equation gen takes the form r ll H0 H y We assume that the zeroorder eigenfunctions and eigenvalues are known Defining Vn0 a ngt as eigenfunctions of H Hngt Engt we can use these zeroorder solutions as a basis for expanding the perturbed wave function Iwtgt CnltlequotEquot Ingt The coefficients cnt and phase factor eiEVh carry the time dependence The coefficients give the timedependent probability amplitudes for the zero order states n Timedependent perturbation theory We substitute this firstorder expression into the Schrodinger equation to obtain Zn cntequot39En H0 H39ngt in cntequot5n ngt The sum over n runs over an infinite number of eigenstates If we pick one of these states call it m and multiply both sides by the complex conjugate of the wave function of state m and then integrate over all space we find cntequotEn hltmHOngt ltmH ngt 39 6 i h In ECnte En ltmngt The eigenfunctions are orthonormal ltnmgt Snm so out of the infinite sum on the righthand side only the term in n m survives Timedependent perturbation theory Since ltmHOngt Em5nm we have Cmte 1EmtnEm Zn Cnte 1Entn ltmHIngt acmt E quot IEmth quot m IEmtn m at e m h cmte Note that there are terms on each side that cancel and thus we can write a set of coupled linear equations for the coefficients 39 dcmm Cntequotmquotmthnt dt where some standard definitions have been used ant E ltmH39ngt En Em 0 E nm n Timedependent perturbation theory The solution thus far is exact However because the coefficients comprise a set of linear coupled equations the solution is still not practical We can obtain a perturbation theory solution by assuming that all of the coefficients on the right hand side are equal to their values at t O cnt z cn0 5m This eliminates all of the terms in the summation on the right hand side except one We assume that the wave function starts out in an initial state i and thus at time zero all of the population is in l ci0 1 We wish to know the time dependence of the coefficient for a final state f We can integrate directly to find this dcf t dt cit if dt39er v lt39 eim tv a Timedependent perturbation theory The coefficient gives a probability amplitude The probability is the square of the wave function LIMP LP2 if LP is real Thus the probability for observing population in state fat time t is 2 Pft 712 t dt39eiw t39v a39

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