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Problem set 5: Harmonic oscillator and second quantization & the use of a symbolic math program. The topic of this assignment concerns the vibrational problem, primarily in one dimension (diatomics), and the use of operator algebra to recast the problem and to find solutions. This provides a convenient introduction to the language of second quantization, that we will later on use also for the electronic structure problem. You will receive a set of hand written lecture notes. The problem set below focuses on operator algebra for vibrational problems. The last two questions require you to use Maple, Mathcad or python. We will go through a tutorial in class for those who need the introduction. Questions: 1. Show the following. You can use the commutation relations ˆ ˆ† ⎡ ⎤ b b, 1 = ⎣ ⎦ , the definitions † 1 1 ˆ ˆ ( ), ( ), ˆˆ ˆˆ 2 2 b q ip b q ip = − = + where p i ˆ q ∂ = − ∂ , as well as the Hermiticty of pˆ and qˆ in your proofs. a. The operators ˆ† b and ˆ b are Hermitian conjugates, i.e. * ˆ ˆ † φ ψ ψ φ φψ b b = ∀ , b. ˆˆ ˆ ˆ ˆ ˆ † †1 † 1 ,( ) ( ) , ( ) , ( ) n nn n b b nb b b nb − − ⎡⎤ ⎡⎤ = = ⎣⎦ ⎣⎦ 2. The orthonormal eigenfunctions of the harmonic oscillator are defined as 1 ˆ† ( )0 ! n n b n = , where 0 is normalized to unity. Using the relations discussed in the previous question show: a. n is indeed normalized to 1. b. ˆ† bn n n =++ 1 1 c. ˆ bn nn = −1 d. nm n m = 0, ≠ . From a. and d. it follows that the functions n are orthonormal. 3. Write the following operators in terms of the ˆ† b and ˆ b , remembering that these operators are defined in terms of the dimensionless coordinates m q x ω = h and the conjugate momentum operator i q ∂ ∂ . a. Position, x b. Momentum i x ∂ − ∂ h c. Kinetic energy operator 2 2 2 2m x ∂ − ∂ h d. Potential energy 1 2 2 2 m x ω e. The Hamiltonian 2 2 2 2m x ∂ − ∂ h + 1 2 2 2 m x ω 4. Consider the Hamiltonian of the following form 2 2 4 2 1 1 2 2 H qq q λ ∂ = − + + ∂ . We will use the linear variational principle using a finite set of harmonic oscillator basis states 1 ˆ† ( )0 ! n n b n = , to find approximate energy eigenvalues of Hˆ . We want to evaluate the general form for the matrix elements ˆ nHm . To this end we wish to express the Hamiltonian in terms of the operators ˆ b and ˆ† b . You can make use of the relations expressed under question 4, in particular: ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ † † †1 † † 1 ( ) ( ) ( ) , () () () nn nn nn bb b b nb b b b b nb − − =+ =+ . We want to write the operator 4 λq in a form that all of the ˆ† b operators in a product of operators are to the left of ˆ b . So the operators ˆ b act first. This is called “normal order”. Examples are given below. Let us do it in steps. 1 ˆ ˆ † ˆ ( ) 2 q bb = + 2 † † †† † † †† † 11 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ( )( ) ( ) ( 2 1) 22 2 q b b b b b b b b bb bb b b b b bb = + += + + + = + ++ 3 2 † †† † † ††† †† † † 1 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ( ) ( 2 1)( ) 2 22 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆˆˆ ... ( 3 3 3 3 ) 2 2 q q b b b b b b bb b b b b b b b b b b b bb bbb = += + ++ + == + + + + + Derive the formula for 3 qˆ for yourself, filling in the appropriate steps. Finally 4 3 † †3 †2 † † 2 3 † 1 1 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ( ) (( ) 3( ) 3 3 3 ( ) ( ) )( ) ... 2 4 q qb b b b b b b bb b b b = += + + ++ + += Now derive the complete formula for the Hamiltonian 2 2 4 2 1 1 2 2 H qq q λ ∂ = − + + ∂ in second quantization. As a check on your results confirm that your Hamiltonian is Hermitian. Next evaluate the matrix elements ˆ nHm , where you can make repeated use of the relations ˆ† bn n n =++ 1 1 , ˆ bn nn = −1 , nm n m = δ . For example: †† †† † , 1 ˆˆˆ ˆˆ ˆ 1 11 1 n m nbbbm m nbb m m nb m m m nm m m δ = − = = + += + + Again you can check your results by confirming that ˆ ˆ nHm mHn = Let us now consider the linear variational principle using the following expansion for the trial function 0 M n n n c = Ψ = ∑ . You can create the matrix elements for the Hamiltonian in Maple, in analogy to what is done in the sample Maple page on the web site. Choose λ = 0.1. Diagonalize the Hamiltonian for M=1,…, 10 and plot the eigenvalues as a function of the size of the basis set. You can see an example in the Maple sample file, of how to set things up. Confirm that your results obey the interleafing theorem. Also calculate the eigenvectors of the Hamiltonian in Maple. Do you observe something remarkable? In this context, why do about half the eigenvalues stay the same with every increase of the basis set? Or phrased differently why do all of the eigenvalues alternately lower , then stay the same as you increase the basis set 1 function at a time? You can generate a double well potential by using the potential 1 224 ( ) 2 Vq q q q = + − µ + λ with corresponding Hamiltonian 2 224 † 24 2 11 1 ˆ ˆ 22 2 H q q q bb q q q µ λ µ λ ∂ = − + − + = + − + ∂ . Taking µ =1 and λ = 0.05 defines a nice double well potential. Plot the total potential, including the pure harmonic part of the potential as a function of q . Next evaluate the matrix elements ˆ nHm as before and plot the new energy eigenvalues. Here you have obtained the numerical solutions of the double well potential. What happens to the eigenvalues if you increase the values for µ and/or decrease the value of λ . Always plot the potential to see if you are on the right track. Can you understand the “pairing of the eigenvalues” by sketching the eigenfunctions in a deep double well potential? This could be a model of the umbrella tunnelling vibration in the ammonia molecule. This full problem set is lengthy, but it shows you clearly how nice it can be to have a convenient Mathematics program! 5. Let us assume a ground state Hamiltonian 2 2 2 1 1 2 2 H q q ω ω ∂ = − + ∂ h h and the associated operators ˆ ˆ † b b, . An electronically excited state has the same harmonic frequency, but its equilibrium geometry is displaced by a distance 2d , in the reduced dimensionless coordinates. The excited state Hamiltonian is given by 2 2 2 1 1 ( 2) 2 2 HE q d q ω ω ∂ = Δ − + − ∂ h h a. Show that the excited state Hamiltonian in terms of the displaced operators ! b† = ˆ b† − d, ! b = ˆ b − d takes the form H! = ΔE + "ω ! b† ! b + 1 2 "ω . b. Show that the displaced operators satisfy the same commutation relations as ˆ ˆ † b b, , and express the vibrational eigenstates n! of the electronic excited state in terms of ! b† , ! 0 c. What are the excitation energies of the various vibrational levels in the electronically excited state w.r.t. the ground state vibrational level? What is the 0-0 energy? What is the vertical excitation energy? d. The intensities in the absorption spectrum are proportional to the squares of the Franck-Condon factors, which are defined as 0 n! . Show that 0 n! = (−d) n n! 0 0! . e. Use the completeness relation n! ∑ 0 n! n! 0 = 0 0 = 1 to derive 0 0 ! = e−d2 /2 . f. In the emission spectrum the transitions are given by ! 0 → n . What are the transition energies, and the corresponding intensities? g. Using Maple, and the following parameters: !ω = 1, ΔE = 10, d = 1.0 . Plot the absorption spectrum and the emission spectrum on the same energy scale. Also investigate what happens if the displacement is smaller (d=0.5), or larger (d=3.0).

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