Download The Transition Dipole Moment

The Transition Dipole Moment
Interaction of Light with Matter
• The probability that a molecule absorbs or emits light and undergoes a
transition from an initial to a final state is given by the Einstein
coefficient, Bfi
Bif = B fi =
µif
2
6ε 0 ! 2
Equation 1
o More details connecting an absorbance measurement to the Einstein
coefficient are given in the OPTIONAL additional notes.
• the only component not a constant in Equation 1 is the transition dipole
moment µfi
o the transition dipole moment is a quantum mechanical quantity
o it couples the total molecular wavefunction of the initial Ψi and final
Ψf states of the molecule via the molecular dipole moment µ.
IMPORTANT
!
µ fi = ∫ Ψ f µΨ i dτ = f µ i ≠ 0
Equation 2
o the transition dipole moment must be non-zero for a transition to occur (or
for a peak to be present in an experimental spectrum)
o the transition dipole moment can be regarded as a measure of the size of
the electromagnetic jolt given to a system
IR Spectroscopy and the Dipole Moment
• What does the dipole moment have to do with light interacting with a
molecule?
o the static dipole moment (µ0) is a vector with components in the x, y and
z directions. We can think of the dipole moment as "summarising" the
charge distribution within a molecule, Equation 3,
o where rα is the distance from the CoM of the molecule to the charge qα
and has two components, an electronic (e) and nuclear (Zn) component.
⎛ µx ⎞
⎜
⎟
µ = ⎜ µ y ⎟ = e∑ Z n Rn + ∑ ei ri = ∑ qα rα
n
i
α
⎜ µ ⎟
⎝ z ⎠
Equation 3
o light is electromagnetic radiation and is represented by an oscillating
electric field ( ε̂ )
o this electric field interacts with the charges distributed throughout a
molecule, the oscillating electric field of the photon induces oscillations in
the electron density (negative) and nuclei (positive charges).
o this oscillation of charges within the molecule can couple final and initial
states, that is cause the system to change from an initial to a final state
o electromagnetic radiation also has an oscillating magnetic field, however
the magnitude of the interaction is much smaller and a transition is 104
times less likely to occur.
o More details on the form of the CoM motion and the transition dipole
moment are given in the OPTIONAL additional notes.
Hunt / Lecture 4
1
Raman Spectroscopy and the Polarizability
• Raman spectroscopy is an indirect technique. The molecule is irradiated with
very high intensity laser radiation, most of which is scattered elastically (ie
with no energy transfer between photons and molecule) back towards the
source, producing the Rayleigh line in spectra, Figure 1.
Figure 1 Raman spectrum of CCl4 (488.0 nm excitation)1
• a very small number of inelastic interactions occur where energy is
transferred between a photon and the molecule. These very rare events,
approximately 1 in every 107 photons, and are very difficult to observe in the
strong background of the elastically scattered photons.
• when the light interacts with the molecule it enters a “virtual energy state”
(light + molecule) for a short period after which the light is re-emitted.
During this process the photon can transfer a (vibrational) quanta of energy
into the molecule, loosing this quanta itself, and producing the Stokes lines
in a Raman spectrum. The anti-Stokes lines are due to a quanta of energy
being removed from the molecule and being gained by the photon.
• because there are more molecules in the ground vibrational state
(Boltzmann’s distribution) the Stokes lines are much stronger than the antiStokes lines and it is customary to measure and report the Stokes lines.
Because of the dependence on the population of vibrational states, Raman
spectroscopy is sensitive to temperature.
hν0"
hν0'
hν0
scattered light
h(ν0"- ν1 )
h(ν0'-ν1 )
h(ν0-ν1 )
scattered light
minus 1quanta of
vibrational energy
v=1
v=0
Figure 2 Stokes transitions for Raman spectroscopy
1
from Nakamoto Infrared and Raman Spectra of Inorganic and Coordination Compounds, 5th
Edition (1997), John Wiley & Sons, New York, PartA, p10, Fig I-6.
Hunt / Lecture 4
2
• in addition totally symmetric Raman active modes leave plane polarised light
polarised, otherwise the light will be depolarised.
o polarised light is light that oscillates in only one plane (normal light
oscillates in all planes)
o when plane polarised light initially interacts with the molecule it is with
components of the vibration that are aligned to plane of oscillation.
o however the molecules in a liquid sample are moving and randomly reorientating, the emitted photon will be in a different orientation. If the
active vibration is not-totally symmetric light will be depolarised.
o if the vibration totally symmetric, the orientation of the molecule does not
matter and the scattered light remains polarised.
o this is a very simplified justification of a much more complex phenomena
beyond the scope of this course to discuss further.
light wave
direction of
propagation
arrows represent the electric field oscillating
up and down
looking down
the center line
vertical polarizer
unpolarized light
unpolarized light
ONLY vertically
polarized light
the electric field oscillates
up and down
horizontal polarizer
no light!
Figure 3 Stokes transitions for Raman spectroscopy
• as the electric field of light is oscillating the nuclei and electrons get pulled
first one way and then the other.
o a transient dipole moment depends on the electromagnetic wave can
be induced in the sample, Figure 4. µ = αε where ε is the magnitude of
the electric field and α is the polarizability.
o α measures how easy it is to deform the electronic structure, typically
"soft" elements and ligands are said to be highly polarizable. Examples
include iodine atoms or sulphur containing ligands
o Raman spectroscopy depends on the induced transient dipole moment and
a change in the electric polarizability with respect to the vibrational mode
is required for a mode to be Raman active.
δ-
δ+
hν
δ+
δ-
oscillating electric field
Figure 4 Cartoon representing the formation of transient dipoles under the influence of an
oscillating electric field, the dipole moment vector and polarizability tensor
Hunt / Lecture 4
3
• This can be expressed more formally such that if we take the molecular
dipole and expand it in a Taylor series with respect to the electric field
( ε̂ ) we obtain the following:
# ∂µ &
1 # ∂2 µ &
µ = µ 0 + % ( ε + % 2 ( ε 2 −!
$ ∂ε '0
2 $ ∂ε '0
µ = αε
⎡ µ x ⎤ ⎡ α xx α xy α xz
⎢
⎥ ⎢
⎢ µ y ⎥ = ⎢ α yx α yy α yz
⎢ µ ⎥ ⎢ α
⎢⎣ z ⎥⎦ ⎢⎣ zx α zy α zz
# ∂2 µ &
β =% 2 (
$ ∂ε '0
1
µ = µ 0 + αε + βε 2
2
# ∂µ &
α =% (
$ ∂ε '0
⎤⎡ ε
⎥⎢ x
⎥ ⎢ εy
⎥⎢
⎥⎦ ⎢⎣ ε z
⎤
⎥
⎥
⎥
⎥⎦
Equation 4 the polarizability
o µ0 is the permanent electric dipole independent of the electric field, α is
the polarizability and β is the first hyperpolarizability. Both α and β are
dependent on the electric field.
• the dipole moment is a vector, as is the electric field, this means that the
polarizability must be a matrix, in this case a special type of matrix called a
tensor.
o notice the subscripts (they are similar to the binary functions!)
o it is assumed that there is no permanent dipole moment, and that we are
taking the expansion in the electric field only to first order.
Where does the transition dipole moment come from?
• to find the origins of the transition dipole moment we need to go right back
to quantum mechanics and probabilities
• we know that the probability of being in a particular state:
ρ = ∫ ψ 2 dτ ψ = ∑ akϕ k
k
= ∫ ∑ akϕ k ∑ amϕ m dτ = ∑ ak am ∫ ϕ kϕ m dτ = ∑ ak2
k
m
k,m
Equation 5
k,m
o here k=0 is the ground state and k>0 are excited states, thus to find the
probability of the molecule being in an excited state we need to determine
the coefficients ak
o we need to determine how has the light has affected the initial state to
""
→Ψ f
produce the final state, Ψ i "light
• setting up the problem
o in the basic Schrödinger equation we use the molecular Hamiltonian
(Hmol=Te+Tn+Vee+Vne+Vnn) which is independent of time, H Ψ = EΨ
o now we need to introduce a perturbation due to the lightwave, "pert"
below stands for perturbation (the molecule is perturbed by the
electromagnetic wave of the light)
H = H mol + H pert
Equation 6
o the light interacting with the molecule is mathematically represented by
the dipole moment of the molecule interacting with the electric field (ε) of
the lightwave
H pert = − µε
Equation 7
Hunt / Lecture 4
4
o lightwaves are oscillating in time so we should use a perturbation
expression that represents this oscillating nature (time dependent
perturbation theory), it is traditional to use a cos function for this (see
below, the extra 2 is to make some of the mathematics easier) and we
should use the time dependent Schrödinger equation!
V = − µε (t)
(
H (t) pert = 2V cos(wt) = 2V eiwt + e−iwt
)
Equation 8
∂Ψ
H Ψ = i!
∂t
o teaching you time dependent perturbation theory is beyond the scope of
this course. What I will do is introduce you to some time-independent
(non-degenerate) perturbation theory so you can obtain an understanding
of where the transition dipole moment comes from (and so that you can
move onto time-dependent perturbation theory more easily if you want to
take this further).
o time dependent perturbation is a core part of quantum mechanics and if
you are interested in quantum mechanics I suggest you take a look at this
theory in Atkins and Friedman "Molecular Quantum Mechanics".
(Non-degenerate) Time Independent Perturbation Theory
• Start with a general expansion of the Hamiltonian, H(0) is our unperturbed
Hamiltonian, H(n) are perturbations of decreasing magnitude and of differing
types, the superscript gives the "order" of the term, and λ is just a
mathematical tool for a neat trick (and represents a coefficient for controlling
the strength of a perturbation) …
H = H (0) + λ H (1) + λ 2 H (2) + !
Equation 9
• now if the Hamiltonian is changing, so too must the wavefunctions
(solutions) and the energy, but they will both be perturbed in a similar
manner to the Hamiltonian, so we use a similar expansion
ψ = ψ (0) + λψ (1) + λ 2ψ (2) + !
E = E (0) + λ E (1) + λ 2 E (2) + !
Equation 10
• if we pack it all back into the Schrödinger equation
Hψ = Eψ
(H
(0)
)(
)
+ λ H (1) + λ 2 H (2) ψ (0) + λψ (1) + λ 2ψ (2) + !
(
)(
)
= E = E (0) + λ E (1) + λ 2 E (2) + ! ψ (0) + λψ (1) + λ 2ψ (2) + !
Equation 11
• collect all the terms with the same λ coefficients
(H
ψ (0) )
(0)
+ λ ( H (0)ψ (1) + H (1)ψ (0) +!)
= ( E (0)ψ (0) )
+ λ ( E (0)ψ (1) + E (1)ψ (0) +!)
+ λ 2 ( H (0)ψ (2) + H (1)ψ (1) + H (2)ψ (0) +!) + λ 2 ( E (0)ψ (2) + E (1)ψ (1) + E (2)ψ (0) +!)
Equation 12
Hunt / Lecture 4
5
• these must be equal and so we obtain n separate equations and we throw away
the λ's as they have done their job
• the first equation is the equation for the unperturbed system, the second
equation is called the first order correction … and so on
H (0)ψ (0) = E (0)ψ (0)
H (0)ψ (1) + H (1)ψ (0) = E (0)ψ (1) + E (1)ψ (0)
H (0)ψ (2) + H (1)ψ (1) + H (2)ψ (0) = E (0)ψ (2) + E (1)ψ (1) + E (2)ψ (0)
Equation 13
• we now focus on the first order correction, we make the assumption that the
slightly perturbed wavefunctions ψ n(1) can be expanded in terms of the
unperturbed wavefunctions ψ n(0)
{H
(0)
}
{
}
− E (0) ψ (1) = E (1) − H (1) ψ (0)
(0)
ψ (1) = ∑ anψ n
Equation 14
n
• in the unperturbed system there will be a number of solutions, each with a
different energy, the lowest energy state E0(0) will have the wavefunction ψ0(0) ,
the next highest state will have E1(0) and ψ1(0) and so on
Hψ 0(0) = E0(0)ψ 0(0)
Hψ 1(0) = E1(0)ψ 1(0)
Hψ 2(0) = E2(0)ψ 2(0)
Equation 15
o The unperturbed ground state is described by E0(0) and ψ0(0) . This state is
perturbed slightly to become ψ0(1) . One way to think of this is as a mixing
of small amounts of the original excited states into the ground state
wavefunction, ie adding a small amount ψ1(0) and ψ2(0) etc into the original
unperturbed ground state ψ 0(0) .
• Now we substitute for ψ0(1) into our perturbation expression, I have now
added subscripts related to the different n explicitly
{ H (0) − E0(0) }ψ0(1) = {E0(1) − H (1) }ψ0(0)
Equation 16
o starting with the left hand side (LHS)
LHS = { H (0) − E0(0) } ∑ anψ n(0)
= ∑ an { E
n
(0)
n
n
− E0(0) }ψ n(0)
since H (0)ψ n(0) = En(0)ψ n(0)
Equation 17
o putting the equation back together again:
∑ a {E
n
(0)
n
− E0(0) }ψ0(0) = { E0(1) − H (1) }ψ0(0)
n
Equation 18
• now we pre-multiply and integrate with a single unperturbed wavefunction
%
∫ ψ '&∑ a {E
(0)
k
n
n
(0)
n
(
− E0(0) }ψn(0) * dτ = ∫ ψk(0) %&{ E0(1) − H (1) }ψ0(0) () dτ
)
Equation 19
Hunt / Lecture 4
6
• we pull constants out of the integration, and look at the integrals, we know
that the unperturbed solutions are orthogonal and normalised, so we can
simplify the equations,
∑ a {E
n
(0)
n
n
− E0(0) } ∫ ψk(0)ψn(0)dτ = E0(1) ∫ ψk(0)ψ0(0)dτ − ∫ ψk(0) H (1)ψ0(0)dτ
!#"#$
!#"#$
=0 if k≠n
=1 if k=n
=0 if k≠0
=1 if k=0
Equation 20
• however on the RHS we find that we must have k=0 if we want to find the
first order correction to the energy E(1), so setting k=0 for the whole equation
∑ a {E
n
(0)
n
n
− E0(0) } ∫ ψ0(0)ψn(0)dτ = E0(1) ∫ ψ0(0)ψ0(0)dτ − ∫ ψ0(0) H (1)ψ0(0)dτ
!#"#$
=0 if k≠n
=1 if k=n
Equation 21
o we can see that the integral on the LHS will only be non-zero if k=n so
setting k=n=0, the LHS energy term becomes zero!
a0 { E0(0) − E0(0) } ∫ ψ0(0)ψ0(0)dτ = E0(1) ∫ ψ0(0)ψ0(0)dτ − ∫ ψ0(0) H (1)ψ0(0)dτ
!#"#$
!#"#$ !#"#$
=1
=0
(1)
0
(0)
0
(1)
=1
(0)
0
0 = E − ∫ ψ H ψ dτ
Equation 22
o we obtain the result that the first order correction to the energy depends
only on the unperturbed wavefunctions and on the perturbation in the
Hamiltonian.
E0(1) = ∫ ψ0(0) H (1)ψ0(0)dτ
Equation 23
o however we are also after the correction to the wavefunction, so if we
focus on the LHS and set n=k≠0
∑ a {E
n
n
(0)
n
− E0(0) } ∫ ψ k(0)ψ n(0)dτ = E0(1) ∫ ψ k(0)ψ 0(0)dτ − ∫ ψ k(0) H (1)ψ 0(0)dτ
!#"#$
!#"#$
=0 if k≠n
=1 if k=n
=0 if k≠0
=1 if k=0
ak { Ek(0) − E0(0) } = 0 − ∫ ψ k(0) H (1)ψ 0(0)dτ
ak =
− ∫ ψ k(0) H (1)ψ 0(0)dτ
(E
∫ψ
=
(0)
k
(E
=
(0)
k
− E0(0) )
H (1)ψ 0(0)dτ
(0)
0
− Ek(0) )
H k(1)0
ω k0
Equation 24
o we obtain the first order correction to the wavefunction
Hunt / Lecture 4
7
ψ0(1) = ∑ anψ0(0)
n
ψ0(1) = a0ψ0(0) +
H k(1)0 (0)
∑ ω ψk
k0
k,k≠0
Equation 25
• what does this mean?
o after the system has been perturbed there is some probability of finding
the system in a higher energy state, this probability is related to the size of
the electric field, the transition dipole moment, and to the energy gap
between the states (the larger the gap the smaller the chance of a
transition)
2
H k(1)0
εµ
a =
= k0
ω k0
ω k0
2
2
k
Equation 26
o hence it is possible to see why Βfi is dependent on the square modulus of
the transition dipole moment.
B fi =
8π 3 µ fi
2
3h 2
Equation 27
• the above derivation is sound for a time independent perturbation between
non-degenerate states, and provides linking justification for the form of the
transition dipole moment
• a full treatment of time dependent perturbation theory and subsequent
derivation of the Einstein coefficient and Fermi's Golden Rule can be found
in Atkins and Friedman's Molecular Quantum Mechanics.
Key Points
• be able to write the equation for the Einstein coefficient Bfi
• be able to define and discuss the importance of the transition dipole moment
• be able to briefly describe key features of Raman spectroscopy
• be able to define the dipole moment and polarizability and describe how they
relate to light interacting with matter and IR and Raman spectroscopy
• be able to present the derivations for time independent non-degenerate
perturbation theory
• be able to derive the first order correction to the energy and the first order
correction to the wavefunction
Hunt / Lecture 4
8