## Transcribed Text

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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