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Problem set 6 on using the DVR approach to vibrational problems,

a discussion of multimode problems, and non-Born-Oppenheimer Vibronic Coupling effects.

Due Date: Wednesday, November 11 In this problem set we will cover 3 topics. In part A you will use the DVR approach to solve the energy levels in a one-dimensional potential energy surface. In part B, we will look at some problems involving multiple normal modes. The main purpose is to get you familiar with coding such problems. In part C we will go beyond the Born-Oppenheimer approximation and discuss vibronic coupling. Your first problem will be using 1 normal mode, in the second we generalize to two normal modes. This model problem is rich enough to address many of the issues. I suggest that you send your python, Matlab, Mathcad or Maple files to me. If I ask you to do the problem for several values of computational parameters, there is no need for me to see all the results. I would like to see a discussion or conclusions from your similations. Good luck! A. Solving for the vibrational energy levels of a 1d potential energy curve using the DVR approach. 1. Testing the approach on a known problem: The displaced harmonic oscillator.

As the potential we will use 21( ) ( )2

V q q aω= − . The energy levels are expected to be

equidistant, if we use a large enough basis. Take 0.1 , 0.2eV aω = = . Take on the order of 50 basis functions (M=50).

a) Calculate the matrix †1 ˆ ˆ2

n q m n b b m= +

b) Diagonalize this matrix and collect the eigenvalues and eigenvectors. ,i niCλ . This defines the DVR basis. c) Calculate the potential matrix in the DVR basis ˆ ( )i j j ijV Vλ λ λ δ= (diagonal matrix), and transform to the original H.O. basis set, using the eigenvectors C. Let us call this back-transformed matrix the potential matrix V. You can test it is OK, by diagonalizing the matrix once more. You should get the same eigenvalues.

d) Calculate the kinetic energy operator 2 † 21 ˆ ˆˆ ( )4

T n p m n b b mω= = − − -> T

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e) Diagonalize the full Hamiltonian T+V and check the energy levels as a function of the # of basis functions. Compare to the exact solution which you considered before for this problem. f) Construct the kinetic energy operator in an alternative way:

† 1ˆ ˆ( ) ,4nmD n b b m T ω= − = − DD

Compare the matrices obtained under d) and f). Can you spot the difference? g) Calculate the eigenvalues using the Hamiltonian using the matrix T as defined under f). Which approach e) or g) is preferred? Does it make a difference for larger M? I do not know the answer to this question. Let me know, I am curious! 2. Using DVR to calculate energy levels of a Morse potential We use precisely the same approach as in question 1), but now the potential is a Morse

potential 2 2 1( ) (1 exp( )) ; 0.1 ; 100 10 ,2

V q D q eV D eV Dα ω ω α ω= − − = = = = .

a) Draw the Morse potential, and also the harmonic approximation 21( )2qω to the

potential. b) Using the DVR procedure to calculate the energy levels of this Morse potential. See if you can find the analytical result searching the literature (Google) and compare. 3. Calculating FC factors between two Morse potentials in ground and excited states. Graduate students only: The ground state is given by the potential in problem 2. The excited state is described by a displaced Morse potential of the form

20

20

( ) (1 exp( ( )) ;10.1 ; 20 2 , , 5 , 0.22

x x

x

V q E D q q

eV D eV D E eV q

α

ω ω α ω

= + − − −

= = = = = =

Calculate the energy levels in ground and excited states and calculate both the absorption spectrum and the emission spectrum, using the Franck-Condon overlap integrals. Do you observe the mirror symmetry we saw before for a displaced oscillator (I do not know the answer. You tell me). 4. Further suggestion (optional, for further enjoyment). In the previous problem set 3, question 5, you set up the problem to do a quartic potential using the second quantized approach. This problem is very easy to solve using the DVR setup. Just supply the quartic formula for the potential. You can use this to compare the two approaches. I expect them to converge to the same answer for sufficiently large basis sets. You can use both forms of the kinetic energy (as under 1), to see which converges fastest in the DVR context. I assume the finite basis product representation is best. It would be of real interest to compare the two DVR approaches to the exact finite basis solution to the problem, in order to decide which is best, in particular also for small sizes of the basis set. This is often what we need to use in practice.

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B. vibrational problems in two-dimensions. 5. Doubly displaced Harmonic oscillator: a 2d Franck-Condon spectrum Consider the excited state Hamiltonian: † † † †

1 1 1 1 1 1 1 2 2 2 2 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( )H E b b b b b b b bω λ ω λ= + + + + + + ,

with 1 1 1 2 25, 0.1, 0.1, 0.15, 0.2E ω λ ω λ= = = = = The basis states we use do now have 2 quantum numbers, 1 2,n n , and the operators act according to

†1 1 2 1 1 2 1 1 2 1 1 2

†2 1 2 1 1 2 2 1 2 2 1 2

ˆ ˆ, 1, , , 1 1,ˆ ˆ, , 1 , , 1 , 1

b n n n n n b n n n n n

b n n n n n b n n n n n

= − = + +

= − = + +

You can set up the matrix 1 2 1 2, , , , 0.. 1i i im m H n n n m M= − . Here the total number of basis states for each mode can in principle be different. In order to do this we need to think of the two indices as one compound index. An easy way to do this would be to define 1 2 1 2 2 1 2 2 1 2, , , * ; *ijH m m H n n i m m M j n n M= = + = + Set up the matrix, diagonalize it to obtain the eigenvalues and eigenvectors. If you assume that the groundstate vibrational wave function is 0,0 , you can calculate the Franck-Condon factors as the first component of each eigenvector. Plot the spectrum you obtain in this way. The energies all have the form 1 1 2xE n nω ω+ +% % . By varying the iλ (one small or zero, one large), you can see how the spectrum can be thought to be built up from the individual modes. Check if your calculations are converged with M. 6. Mixing of normal modes: ‘Duschinsky’ rotation. Graduate students only. Consider the following 2d potential for an excited state

( ) 11

1 2 1 222

1 2

1( , ) , ,2

5, 0.1, 0.02, 0.15, 0.01, 0.03 ( )

x

x

qV q q E q q

q

E all eV

ω α γγ ω β

ω α ω β γ

⎛ ⎞+ ⎛ ⎞= + ⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠

= = = = = − =

Set up the Hamiltonian in the Harmonic oscillator basis and diagonalize. Also calculate the absorption spectrum, assuming the vibrational groundstate is 0,0 , and the vibrational frequencies are 1 2,ω ω respectively. Also diagonalize the force constant matrix and obtain the normal modes in the excited state, as well as the vibrational frequencies in the excited state. Verify that the energy levels you obtain follow the equidistant Harmonic oscillator levels, but for the 2d H.O. The fact that the normal modes are different for ground and excited states is called the Duschinsky rotation effect. Here you calculated how this is reflected in the absorption spectrum. I did not include a linear displacements, only pure rotation, and so few levels are expected to have significant oscillator strength.

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C. Vibronic coupling. In this third set of problems we will consider coupled electronic states, and build up gradually to the multimode-multistate problem. 7. Two electronic states, 1 normal mode: Comparison of approximate Franck-Condon vs. Vibronic Coupling. Consider the vibronic Hamiltonian which describes two close-lying excited states and has

the form using dimensionless reduced normal mode coordinate †1 ˆ ˆ( )2

q b b= + :

†

1 11 12

†21 221

ˆ ˆ ˆ ˆ( )ˆˆ ˆ ˆ ˆ( )

E b b q H HH

H Hq E b b

ω µ

µ ω

⎡ ⎤ ⎡ ⎤+ Δ += =⎢ ⎥ ⎢ ⎥

− Δ +⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

The potential matrix in the first (delocalized) diabatic basis is hence given by

21

21

1( )2

1( )2

E q q

q E q

ω µ

µ ω

⎡ ⎤+Δ +⎢ ⎥⎢ ⎥⎢ ⎥−Δ +⎢ ⎥⎣ ⎦

where I have assumed for simplicity that the harmonic oscillators for states 1 and 2 have the same frequency, 1 0.1ω = eV.

a) Show that by the similarity transform 1 111 12⎛ ⎞⎜ ⎟−⎝ ⎠

this potential transforms to

21

21

12

12

E q q

E q q

ω µ

ω µ

⎡ ⎤+ + Δ⎢ ⎥⎢ ⎥⎢ ⎥Δ + −⎢ ⎥⎣ ⎦

This is the second, localized, basis that we will use. In the vibronic coupling treatment of this problem the same results will be obtained, independent of the diabatic basis used. Let us take 5E eV= . The parameter Δ in the problem could be varied, as in the lecture notes for the O=C=O core-hole example: 0.01, 0.1,0.25,0.5,1.0 eVΔ = , while keeping

0.3µ = . Alternatively, we will explore the case 0.25Δ = , while the parameter 0.1,0.15,0.3,0.5µ = varies. The basis states are now characterized as

, , 0.. 1, 1,2n a n M a= − = and we need to assemble the Hamiltonian with matrix

elements , ,m b H n a . The Hamiltonian would have 4 MxM blocks, charactererized by the electronic labels a,b. As before a matrix has only two labels, and hence we have to identify

, , , *( 1), *( 1)ijH m b H n a i n M a j m M b= = + − = + −

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To calculate a spectrum we have to assume transition moments between ground and excited states. Let us assume that in the localized basis, the transition dipoles are equal (both unity). This means that in the delocalized basis the transition dipoles are

1 ( ) 2 ( )21 ( ) 0 (sec )2

g L R

u L R

x x x first state

x x x ond state

= + =

= − =

Carry out the following calculations, for each value of µ : b) Plot the potential energy surfaces by diagonalizing the matrix as a function of q. In the same plot also plot the diagonal potentials from the localized model. You should clearly see the effect of Δ . c) Construct the vibronic Hamiltonian and diagonalize to obtain the eigenvectors and eigenstates. You should use both forms of the Vibronic Hamiltonian. Doing it both ways allows you to check your results, and verify the statement that the result is the same. d) Calculate overlap matrix elements 1 0,1 2 0,2xC x Cλ λ+ for each eigenvector λ and calculate the corresponding intensity. Plot the spectrum. Here you should use different transition moments for the localized and delocalized transition moments, as discussed above. The spectra from both full vibronic approaches should be exactly the same! e) You can calculate approximate FC spectra by taking the localized diabatic potential energy surfaces and neglecting the off-diagonal coupling. As an aside: a more accurate approach would be to diagonalize the full potential energy matrix (symbolically) and obtain a quadratic surface for the upper and lower potential that can be used in a FC calculation. You could also use the DVR approach, calculating the exact potential eigenvalues in the selected DVR points, obtained from diagonalizing the operator q. This would be the most accurate FC approach. Also the transition dipoles would be functions of molecular geometry, in principle. 8) Vibronic Coupling involving two normal modes. Consider the vibronic Hamiltonian of the form using dimensionless reduced normal mode coordinates 1q and 2q , which describes two close-lying excited states.

1 1 1 2 11 12

2 2 2 1 21 22

ˆ ˆ. .ˆˆ ˆ. .

E h o q q H HH

q E h o q H H

λ µµ λ

⎡ ⎤+ +⎡ ⎤= = ⎢ ⎥⎢ ⎥+ +⎣ ⎦ ⎢ ⎥⎣ ⎦

Assume the harmonic oscillators (h.o.) for normal modes 1 and 2 have the same

frequency 1 2 0.1ω ω= = eV, i.e. 2 2

2 2 † †1 1 2 1 1 1 1 2 22 2

1 2

1 1 1 1. . ( )2 2 2 2

h o q q b b b bq q

ω ω ω∂ ∂= − + − + = +∂ ∂

.

The potential for the initial ground state is assumed to be a harmonic oscillator with these same frequencies and electronic energy 0 0.0E = . Further assume 1 23.0, 4.0E E= = , both in eV. Moreover, we assume the off-diagonal coupling constant 0.3µ = eV in all

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calculations. We will assume a basis set characterized as 1 2, ,n n a , where 1 10.. 1n M= − denotes the number of quanta in mode 1, 2 20.. 1n M= − indicates the number of quanta in mode 2, while a labels the diabatic state. The Hamiltonian matrix-elements can then be evaluated as 1 2 1 2

ˆ, , , ,m m b H n n a . A vibronic eigenstate can then be described as

1, 2

1 2

1 2 , ;, ,

, , n n an n a

n n a cν νΨ = ∑ . As always, we wish to label the vibronic basis states with

one label, so we can define a matrix. Hence define 1 2 1 1 2 1 2 1 1 2 2, , , ( 1), 0... 1, 0... 1, 1,2K n n a K n M n M M a n M n M a= = + + − = − − = . The

total dimension of the space is 1 22M M then. This allows you to define the matrix KLH which can be diagonalized to find vibronic eigenstates. To evaluate the matrix elements you want to write the Hamiltonian in second quantization first, using the conventional definitions for the annihilation and creation operators. We will consider the problem set for four different values of 1 2,λ λ , namely:

1 2

) 0.2 0.2) 0.2 0.2) 0.2 0.0) 0.0 0.2) 0.0 0.0

iiiiiiivv

λ λ

−

Questions (for each set of parameters). It is best to fully work the problem for one set of parameters, and see if it works for the next set of parameters. Bless Mathcad!): a) Plot the potential energy surfaces in the two-dimensional 1 2,q q space. This means diagonalizing the potential on a grid of points and plotting the corresponding energies. You might also plot slices of the potential as a function of 1q for different values of 2q . You will see that the shape of the potential along 1q depends on the value for 2q . This potential is clearly not a sum of independent harmonic oscillators, and is a major reason why the vibronic approach is more accurate than Franck-Condon using the harmonic approximation. b) Set up the vibronic Hamiltonian in second quantization and solve the full vibronic problem. Choose the number of basis functions to be fairly large (10 or so) and set up the Hamiltonian matrix (it would have dimension 200x200 in this case. Calculate the eigenvalues and calculate the intensity of the transitions for each eigenstate νΨ , which

is proportional to 21 0,0,1; 2 0,0,2;( )x c x cν ν+ . Let us assume assumed that both states have the

same electronic transition moment here: 1 2 1x x= = . Plot the spectrum: eigenvalue vs. intensity. You will see that you get quite different spectra for the various sets of parameters. c) You can get a feel for the importance of vibronic coupling by setting the parameter

0µ = . Then the spectra reduce to 1 dimensional FC spectra for each state.

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9* (optional). E e⊗ Jahn-Teller problem. Consider the Hamiltonian

1 2

2 1

. .ˆ. .

E ho q qH

q E ho qλ λ

λ λ+ +⎡ ⎤

= ⎢ ⎥+ −⎣ ⎦,

where we assume again that the frequency of the Harmonic oscillator is 0.1, and we have only one parameter λ that fully characterizes the problem. Importantly, the energy E is the same for both electronic states, hence at 1 2 0q q= = the curves cross in a cone. You can pretty much follow the procedure in the previous problem. This is the quintessential Jahn-Teller problem for degenerate states. For these states which are degenerate at 1 2 0q q= = there is no single minimum for the excited states. Rather it is a circle of

minimum energies, in the linear approximation we consider here. a) Consider the following values for λ : 0.05, 0.1, 0.2, 0.4. Solve for the eigenvalues and eigenstates of the Hamiltonian, using a large enough basis set, so that you observe more or less converged results. b) Plot the potential energy curves as a function of 1q and 2q . c) Calculate the energy levels and monitor the degeneracy of the levels. d) Calculate the spectrum if you start from a non-degenerate harmonic ground state. The intensities are given by a similar formula as in the previous problem. You should add the intensities of degenerate states (see also below). e) If we have a pair of degenerate states 1 2,Ψ Ψ , having overlap factors 0i ia = Φ Ψ ,

show that the intensity 2 21 2a a+ is independent of the particular choice of orthonormal

states in the degenerate manifold. This is it!