Smell in real noses: how the environment changesvibrations

Benjamin Yiwen Färber

May 16, 2014

Contents

1 Introduction 3

2 Relationship between spring constant and Morse potential 72.1 Euler-Lagrange equations . . . . . . . . . . . . . . . . . . . . . . 72.2 Coulomb interaction . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Vibrational frequency in a dimer . . . . . . . . . . . . . . . . . . 112.4 Polarisation effects . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Polarizability in the tight binding model 12

4 External effects on spring constant 16

5 Binding to the receptor 19

6 Conclusion 22

2

Chapter 1

Introduction

The functioning of the nose and the sensation of smell is described by the follow-ing biological model. Air containing odorous molecules is inhaled into the nasalpassage. The beginning of sensory signal transduction occurs in the olfactorycilia which are hairlike extensions of the receptor neurons. This 2.5 square cen-timeter patch contains about 50 million sensory receptor cells. Smelling startswhen the molecules bind to the receptor cells located on the cilia which stimulatethem to send signals to the brain. But how does the odorous molecule stimu-

Figure 1.1: Above the nasal passage is located the olfactory cilia where receptornerve cells are responsible for sensory transduction.

late the receptor cells to send a signal? According to Turin it occurs throughthe quantum mechanical inelastic transport of electrons (IETS) in the odorousmolecules. In inelastic transport of electrons (IETS), the vibrational frequencyof the odorous molecule causes an energy transition of the electrons. Thus thevibrational frequency of the odorous molecule leaves an information imprint onthe electrons responsible for signal transduction. In other words the vibrationalfrequency changes the energy levels of the transmitted electrons, thus alteringthe signals transmitted to the brain.The vibrational frequency therefore clearly affects sensory signal transduction tothe brain. In Turin’s theory the discrimination between two odorous molecules isthus dependent on their different vibrational frequencies. The odorous moleculeis characterized by its vibrational frequency.The olfactory receptor contains an electron donor (D) and an electron acceptor(A). In theory, when the molecule binds to the receptor, an electron tunnelsfrom source (D) to sink (A). Once the electron reaches (A), the signal is initi-

3

Figure 1.2: Inelastic transport is responsible for the energy transition of anelectron.

ated through the G-protein release mechanism. There is an energy separationbetween source (D) and sink (A) where we treat donor (D) and acceptor (A)as single molecular orbitals with energies εD and εA, coupled to each other bya weak hopping integral t. The electron tunnels from donor (D) to acceptor(A) by coupling to the phonon of the odorous molecule. This is the inelastictransport of the electrons (IETS) from donor (D) to acceptor (A). Bro10

It is important to verify Turin’s mechanism with a diatomic molecule sinceit is the simplest type of molecule with a vibrational frequency. The vibrationalfrequencies of diatomic molecules can be modeled classically as vibrational fre-quencies of a harmonic oscillator. Thus the simplest case applies, when theodorous molecule is a diatomic molecule (in general a molecule can containmany more atoms). It is therefore important to test the predictions of Turin’stheory in connection with diatomic molecules as it is the simplest way to verifythat odor sensation is dependent on the vibrational frequency of the molecule.

The nose is a messy environment and the frequency of the harmonic oscilla-tor might be different inside the nose as compared to being in vacuum. Thisproject is about how the frequency of the harmonic oscillator changes as it isaffected by the environment. In order to test results it would be preferable todirectly verify Turin’s mechanism of an electron tunneling inelastically from adonor (D) to an acceptor (A) by coupling to the phonon. However certain as-pects can be verified in the test tube by investigating how the cell fluid of thenose affects the frequency of the diatomic molecule. Then by gauging the fre-quencies of diatomic molecules within a narrow range, Turin’s mechanism canbe tested with a real receptor cell.It is important to investigate how diatomic molecules with different vibrationalfrequencies facilitate inelastic transport of electrons between donor (D) andacceptor (A) in the receptor cell. Thereby the vibrational frequency of the di-atomic molecule is affected by the cell fluid which is its environment.

In chapter 2 the parameters of the classical harmonic oscillator are linkedto quantum mechanical models. As the Morse potential describes diatomicmolecules better than the harmonic potential, we can use the harmonic po-tential well as an approximation when the separation between the two atomsdoes not depart too much from their equilibrium separation. We then expressthe spring constant k of the harmonic oscillator in terms of the well depth D

4

Figure 1.3: Turin’s theory requires a source of electrons or holes to allow chargeflow to take place. The precise biological origin is not known, but may wellconsist of reducing (oxidizing) species (X) in the cell fluid. These moleculesdiffuse through the aqueous medium and arrive with an average interval of τX .The charge then travels to the donor (D) in one transmembrane helix of theprotein over an average interval τL, from where it then hops to the acceptor(A) (possibly in a different helix) with either an average time τT0 for non-discriminating ("elastic") tunneling or an average time τT1 for discriminating("inelastic") tunneling. Only the inelastic contribution is sensitive to the odorant(M) oscillator frequency ω0, and so needs to dominate the elastic contribution(τT0 � τT1). The electron then travels from (A) to trigger the release of theG-protein over a time τR. Hor06

and the well width a of the Morse potential. In the next step it is supposedthat there is a strong contribution to the coupling between the electron and thephonon from interaction with external charges. In section 2.2 we therefore as-sume that external charges in the environment affect the vibrational frequencyof the molecule. By introducing internal charges onto the diatomic molecule wethus treat it as a dipole. We then calculate the spring constant change and vi-brational frequency change as a result of the introduced charge. Due to externalcharges surrounding the dipole, there are external electric fields which polarizethe molecule. The induced charges resulting from polarization cause the changein spring constant.In chapter 3 we connect the polarizability of the dipole to the parameters of thequantum mechanical tight binding model, which is our model for the diatomicmolecule. In chapter 4 we calculate the vibrational frequency of the dipole in theaqueous medium, modeled as a dielectric medium, in the presence of an externalcharge, through the method of image charges. In chapter 5 we consider thatthe diatomic molecule binds to the receptor and model this binding through asecond spring constant. By treating the receptor as a hard wall of infinite masswe derive the vibrational frequency in terms of the two spring constants andthe masses of the diatomic molecules.

5

Figure 1.4: Morse potential can be compared to a harmonic oscillator

6

Chapter 2

Relationship betweenspring constant and Morsepotential

We approximate the potential of the diatomic molecule with the Morse potentialto model the vibrational frequency spectrum of the dimer molecule. From theEuler-Lagrange equations for a classical harmonic oscillator we illustrate theconnection between the spring constant k of a harmonic oscillator and the Morsepotential’s parameters D, well depth, and a, well width. The advantage ofthis approach is that the dimer modeled with spring constant k can then berelated to the quantum mechanical parameters of the Morse potential. We havethen looked at different effects on the molecule’s vibration. The internal springconstant k is related to the internal charges of the molecule via the Coulombinteraction and the change in spring constant ∆k results from the Coulombinteraction. Thus we model the vibrational frequency of the dimer moleculewith internal charges. Also external effects acting on the spring constant, areconsidered through externally induced changes due to polarisation from theenvironment.

2.1 Euler-Lagrange equationsChoosing general coordinates q1 and q2 as the positions of the atoms, the sep-aration distance for a diatomic molecule r and the small distortion u fromequilibrium separation are defined as

r = q2 − q1 (2.1)u = r − r0 (2.2)

where r0 is the equilibrium separation of the harmonic oscillator. The kineticenergy T and potential energy V for a harmonic potential system with springconstant k can be expressed by the Lagrangian L.

T = 12(m1q

21 +m2q

22)

(2.3)

7

V = 12k (r − r0)2 = 1

2ku2 (2.4)

L = T − V (2.5)Inserting the above expressions for the kinetic energy T and potential energy Vinto the Lagrangian L gives:

L = 12(m1q

21 +m2q

22)− 1

2k (r − r0)2 (2.6)

The Euler Lagrange equations

d

dt

(∂L

∂qi

)− ∂L

∂qi= 0 (2.7)

are applied to the diatomic molecule yielding two simultaneous equations.

m1q1 − ku = 0m2q2 + ku = 0 (2.8)

Consider the potential V in a general Lagrangian for small distortions fromthe equilibrium separation which can be expressed in a Taylor series to secondorder.

V (r) = V (r0 + u) ≈ V (r0) + uV ′ (r0) + 12u

2V ′′ (r0) (2.9)

Inserting into the Lagrangian gives:

L = 12m1q

21 + 1

2m1q22 − V (r0 + u)

= 12m1q

21 + 1

2m1q22 − V (r0)− V ′ (r0) (q2 − q1 − r0)

− 12V′′ (r0) (q2 − q1 − r0)2 (2.10)

At equilibrium V ′ (r0) = 0. It is then shown that spring constant k is thesecond derivative of the potential V at equilibrium separation r = r0.

∂2V

∂r2 |r=r0u2 = ∂

∂r2

(12k (r − r0)2

)|r=r0u

2

= ∂

∂r2

(12k(r2 − 2rr0 + r2

0))|r=r0u

2

= ku2 (2.11)

This gives:∂2V

∂r2 |r=r0 = k (2.12)

Thus the two simultaneous equations from the Euler-Lagrange equation canbe expressed as:

m1q1 = −V ′′ (r0)um2q2 = V ′′ (r0)u (2.13)

8

The Morse potential can be written in two forms

VMorse (r) = D(e−a(r−r0) − 1

)2−D (2.14)

orVMorse (r) = D

(e−2a(r−r0) − 2e−a(r−r0)

)(2.15)

where r is the distance between the atoms, r0 is the equilibrium bond dis-tance, D is the well depth (defined relative to the dissociated atoms), and acontrols the ’width’ of the potential (the smaller a is, the larger the well). Wenow show that the spring constant is

kMorse = 2Da2 (2.16)

by differentiating equation (2.18) twice gives

V ′Morse (r) = −2Da(e−2a(r−r0) − e−a(r−r0)

)(2.17)

V ′′Morse (r) = D(

4a2e−2a(r−r0) − 2a2e−a(r−r0))

(2.18)

ThenkMorse = V ′′Morse (r0) = 2Da2 (2.19)

2.2 Coulomb interactionWe now include the Coulomb interaction. Introducing charges q and −q ontothe atoms of the diatomic molecule (induced by external charge, possibly imagecharges in a dielectric) modifies the potential to

V (r) = D(

exp (−2a (r − r0))− 2 exp (−a (r − r0)))− q2e2

4πε0r(2.20)

where e is the electron unit charge. We want to find the change in springconstant ∆k produced by this charge. Let us define the constant

b = q2e2

4πε0(2.21)

Expanding the potential for a small deviation ∆ from equilibrium distance r0with

r = r0 + ∆ (2.22)

in the Taylor expansion gives:

V (r0 + ∆) = − b

r0

(1 + ∆

r0

)−1+D

(e−2a∆ − 2e−a∆) (2.23)

9

Expanding the series then gives

V (r0 + ∆) = − b

r0

(1− ∆

r0+ ∆2

r20

+ . . .

)+D

(1− 2a∆ + 2a2∆2 − 2 + 2a∆− a2∆2 + . . .

)= − b

r0

(1− ∆

r0+ ∆2

r20

+ . . .

)+D

(−1 + a2∆2 + . . .

)= −

(b

r0+D

)+ b

r20

∆ +(− b

r30

+ a2D

)∆2 + . . . (2.24)

Differentiating the expanded potential with respect to the displacement ∆with the condition for the attractive force between the atoms:

∂V (r0 + ∆)∂∆ = 0 (2.25)

givesb

r20

+(−2br30

+ 2a2D

)∆0 = 0 (2.26)

The deviation ∆0, valid for equation (2.28) at equilibrium can then be expressedas

∆0 = − b

r20

1(− 2br3

0+ 2a2D

) (2.27)

To second order of the potential we derive

∂2

∂∆2V (r0 + ∆) =(−2br30

+ 2a2D

)(2.28)

Thus as the first derivative of the potential is zero at equilibrium we derivethe expression for the change in spring constant ∆k as the difference betweenthe spring constant k valid at any r and the spring constant kMorse valid atr = r0

∆k = k − kMorse (2.29)

As

k = ∂2

∂∆2V (r0 + ∆) (2.30)

we get the expression for the spring constant

k = kMorse −2br30

(2.31)

from equation (2.31). We then arrive at the expression for the change in springconstant ∆k due to the Coulumb interaction from the internal charge in thediatomic molecule.

∆k = k − kMorse = − 2q2e2

4πε0r30

(2.32)

10

2.3 Vibrational frequency in a dimerIt can be shown that that the change in vibrational frequency ∆ω is negativeif ∆k is negative and positive otherwise, showing that for opposite charges thevibrational frequency decreases. To first order we have:

ω2 = k

m(2.33)

2ω∆ω = ∆km

(2.34)

∆ωω

= ∆k2mω2 = ∆k

2mm

k= 1

2∆kk

(2.35)

A more accurate expression can be obtained as follows. ∆ω can be calculatedby substituting the vibrational frequency ω by ωnew = ω + ∆ω associated withknew = k + ∆k. Inserting these expressions into equations (2.36) gives

(ω + ∆ω)2 = k + ∆km

(2.36)

The solution is

∆ω =−2ω ±

√4ω2 + 4∆k

m

2 (2.37)

This expression can be summarized in

∆ωω

=√

1 + ∆kk− 1 (2.38)

For a diatomic molecule we assume a simple mechanical model of two massesconnected by a spring with vibrational frequency ω

ω =

√k (m1 +m2)

m1m2(2.39)

2.4 Polarisation effectsIt is assumed that in the biological environment of the nose the molecule issurrounded by an external charge Qi responsible for an external electric field Eiwhich polarizes the molecule.

Ei = Qie

4πε0r2 (2.40)

P = qed = αEi (2.41)where P is the dipole moment. This allows to estimate the induced charge q onthe atoms. If we insert equations (2.43) and (2.44) into equation (2.35) we get

∆k = −2(αEi

d

)24πε0r3

0= −

2(αQie

4πε0r2

)2

4πε0r30

(2.42)

We can plot equation (2.41) for constant spring constant k with E = ~∆ωand ∆EThermal = kBT . Point of interest is the plot E

∆EThermal= ~∆ω

kBTversus

∆EThermal = kBT .

11

Chapter 3

Polarizability in the tightbinding model

To help understand the origin of the polarizability of the dimer better, we nowconsider a simple model.

Figure 3.1: Morse potential for two hydrogen atoms

By combining two hydrogen atoms with electrons in their 1s ground state itmust be considered that their wave functions overlap. In the tight binding modela linear combination of atomic orbitals is allowed, giving the two combinationsΨA+ΨB and ΨA−ΨB . In the symmetric state an extra binding energy occurs.We state Schrödinger’s equation

HΨ = EΨ (3.1)

whereΨ = αφ1 + βφ2 (3.2)

12

Inserting the above expression into Schrödinger’s equation gives

H (αφ1 + βφ2) = E (αφ1 + βφ2) (3.3)

Multiplying by orbitals and taking integrals of both sides then gives,∫φ∗1H (αφ1 + βφ2) dr = E

∫φ∗1 (αφ1 + βφ2) dr∫

φ∗2H (αφ1 + βφ2) dr = E

∫φ∗2 (αφ1 + βφ2) dr (3.4)

where states φi are orthogonal (∫φ∗iφjdr = δij). Defining the Hamiltonian

operator H asHij =

∫φ∗i Hφjdr (3.5)

allows us to write the simultaneous equations using the orthogonality of thestates

H11α+H12β = Eα

H21α+H22β = Eβ (3.6)

or in matrix notation (H11 H12H21 H22

)(αβ

)= E

(αβ

)(3.7)

This only has solutions when∣∣∣∣H11 − E H12H21 H22 − E

∣∣∣∣ = 0 (3.8)

We define

H11 = H22 = ε

H12 = H21 = −V (3.9)

V are the overlap integrals which are dependent on the overlap between orbitalscentered at two neighbouring atoms. We get∣∣∣∣ε− E −V

−V ε− E

∣∣∣∣ = 0 (3.10)

HenceE = ε± V (3.11)

These are the energy levels of the hydrogen dimer. Now, when an electricfield is applied the Hamiltonian changes by ∆ε = −eEd, where d is the dis-tance between the two charges of the dimer. Thus we now have the followingHamiltonian

H =(

ε −V−V ε+ ∆ε

)(3.12)

orH =

(ε− 1

2∆ε −V−V ε+ 1

2∆ε

)(3.13)

13

The eigenstates satisfy(ε− E −V−V ε+ ∆ε− E

)(αβ

)=(

00

)(3.14)

Hence we get the eigenvalues

ε− E = −∆ε±√

∆ε2 + 4V 2

2 (3.15)

We get for the eigenvectors:

(ε− E)α− V β = 0−V α+ (ε+ δε− E)β = 0 (3.16)

Thus

V αβ = (ε− E)α2

V αβ = (ε− E)β2 + ∆εβ2 (3.17)

and(ε− E)

(α2 − β2) = ∆εβ2 (3.18)

Using α2 + β2 = 1 (2α2 − 1

)(ε− E) = ∆ε

(1− α2) (3.19)

Substituting the eigenvalue given in equation (3.18) for the bonding state gives

∆ε =(2α2 − 1

)√∆ε2 + 4V 2 (3.20)

where only the positive sign in equation (3.15) is taken into account.We connect the polarizability of the diatomic molecule to the parameters of thetight binding model. We do this by estimating the charge Q on one atom.

There are two electrons in the two atom system with wave function Ψ =(αβ

)and probabilities Pα = |〈α|Ψ〉|2 and Pβ = |〈β|Ψ〉|2. Thus the probability offinding an electron on one atom is 2|α|2. Then the total charge on one atom is

Q = Qe +Qn (3.21)

with Qe = −2|α|2e and Qn = e. Thus

Q = −e(2|α|2 − 1

)= eq (3.22)

The dipole has opposite equal charges q1 = q and q2 = −q. We now insert theexpression found in equation (3.20) in equation (3.22):

q = −∆ε√∆ε2 + 4V 2

(3.23)

As ∆ε = −eEd we get for the charge

q = eEd√e2E2d2 + 4V 2

(3.24)

14

The dipole moment is then

P = e2d2√e2E2d2 + 4V 2

E (3.25)

In the limit ed � 2V P = e2d2

2V E. Thus we find an expression for the polariz-ability

α = e2d2

2V (3.26)

15

Chapter 4

External effects on springconstant

We want to know what influence the surrounding dielectric medium might haveon the charge distribution and vibrational frequencies of our diatomic molecule.To discuss this problem we review the method of image charges. A point chargeq is embedded in a semi infinite dielectric ε1 at a distance d away from a planeinterface that separates medium ε1 from another semi-infinite dielectric ε2. Thesurface is taken as the plane z = 0.

Figure 4.1: The potential Φ at a point P of a charge q in a dielectric mediumand its image charge q’

16

At point P the potential Φ is given by:

Φ = 14πε1

(q

R1+ q′

R2

), z > 0 (4.1)

Because the electric field is linear, the electric field in the case of more thanone charge can be expressed as a sum of independent fields with single charges.We apply the boundary conditions for the charges of the dipole with an externalcharge in a dielectric medium and its image charges.

Figure 4.2: The dipole with opposite charges q and -q in a dielectric mediumwith image charges and an external charge Qexternaland its image charge

The equations of motion for the atoms in the dimer are

F1 = m1z1

F2 = m2z2 (4.2)

From the calculation of the force the vibrational frequency can be derived.The forces F1 on charge q and F2 on charge −q of the dipole are given by theCoulomb force between the real and image charges and the force FMorse fromthe Morse potential of the diatomic molecule:

F1 = FMorse−q2e2

4πε0d2 + qQexternale2

4πε0 (d+ a)2 + qe2

4πε0

[q

(2z)2 −q

(2z + d)2 + Qexternal

(2z + a+ d)2

](4.3)

F2 = FMorse−q2e2

4πε0d2−qQexternale

2

4πε0a2 + qe2

4πε0

[−q

(2z + d)2 + q

(2z + 2d)2 −Qexternal

(2z + 2d+ a)2

](4.4)

The first partial derivatives of the forces F1 and F2 with respect to z and dare given by

F ′1 = F ′Morse+ 2q2e2

4πε0d3−2qQexternale

2

4πε0 (d+ a)3 + qe2

4πε0

[−4q(2z)3 + 6q

(2z + d)3 −6Qexternal

(2z + d+ a)3

](4.5)

17

F ′2 = F ′Morse+ 2q2e2

4πε0d3 + qe2

4πε0

[6q

(2z + d)3 −8q

(2z + 2d)3 + 8Qexternal

(2z + 2d+ a)3

](4.6)

The vibrational frequency ω of the dipole is calculated by using the firstderivatives of the forces F1 and F2.

ω2 = k

(1m1

+ 1m2

)= V ′′ (r)

(1m1

+ 1m2

)= (F ′1 + F ′2)

(1m1

+ 1m2

)(4.7)

The corresponding vibrational frequency is

ω2 =(

2F ′Morse + 2 2q2e2

4πε0d3

)(1m1

+ 1m2

)+ 2 6q2e2

4πε0 (2z + d)3

(1m1

+ 1m2

)−

(4q2e2

4πε0 (2z)3 −6qQexternale

2

4πε0 (2z + d+ a)3 + 8qQexternale2

4πε0 (2z + 2d+ a)3 −8q2e2

4πε0 (2z + 2d)3

)(1m1

+ 1m2

)(4.8)

We can represent this as contributions to effective spring constants as:

kinternal = 2F ′Morse + 2 2q2e2

4πε0d3 (4.9)

kQ = 2 6q2e2

4πε0 (2z + d)3 (4.10)

kimage = 4q2e2

4πε0 (2z)3 −6qQexternale

2

4πε0 (2z + d+ a)3 + 8qQexternale2

4πε0 (2z + 2d+ a)3 −8q2e2

4πε0 (2z + 2d)3(4.11)

where kinternal is the contribution from the Morse potential and Coulombforce between the charges. kimage is the contribution from the image charges.It is discernible that the internal spring constant kinternal dominates the otherspring constants. The dominating effect of the internal spring constant has theresult that the vibrational frequency of the molecule is dominated to a largeextent by the Morse potential and induced charges.

18

Chapter 5

Binding to the receptor

Figure 5.1: Binding to a receptor

The odorous molecules, when they bind to the receptor, may experience achange of frequency because of that binding. Here we consider a simple model inwhich the odorous molecule is attached to the receptor (represented by anotheratom) by a weak spring. The simplest case is when the receptor has infinitemass, and the dimer atoms have equal mass, m1 = m2 = m. The dimer atomsare coupled by a spring with constant k, and then one of the atoms is coupled tothe receptor with a spring constant q. Then we get the Euler-Lagrange equationfor the coupled system:

L = 12m

(x2

1 + x22)− 1

2k (x2 − x1 − a)2 − 12q (x1 − a)2 (5.1)

The equations of motion are

mx1 = k (x2 − x1 − a)− q (x1 − a) (5.2)

mx2 = −k (x2 − x1 − a) (5.3)

Using the substituting variables u1 and u2 with x1 = a + u1, x2 = 2a + u2we get the new equations of motion

u1 = k

m(u2 − u1)− q

mu1 (5.4)

u2 = − km

(u2 − u1) (5.5)

Assuming the solution to be

u1 = A1eiωt (5.6)

u2 = A2eiωt (5.7)

19

we get the set of linear equations

−ω2A1 = k

mA2 −

(k

m+ q

m

)A1 (5.8)

−ω2A2 = − km

(A2 −A1) (5.9)

The eigenequation for the vibrational frequencies gives(ω2 − k

m −qm

km

km ω2 − k

m

)(A1A2

)=(

00

)(5.10)

From the determinant we get

ω4 −(

2 km

+ q

m

)+(kq

m2

)= 0 (5.11)

The solution is

ω2 =(k

m+ q

2m

)±

√(k

m

)2+( q

2m

)2(5.12)

For spring constants q → 0 k → 0 ω2 converges to

q →0 ω2 → k

m± k

m= 2km

or 0 (5.13)

k →0 ω2 → q

2m ±q

2m = q

mor 0 (5.14)

For a small spring constant q we get

ω2 =(k

m+ q

2m

)± k

m

(1 +

( q2k

)2) 1

2

(5.15)

ω2 ≈(k

m+ q

2m

)± k

m

(1 + 1

2

( q2k

)2)

+ . . . (5.16)

= 2 km

+ q

2m

(1 + q

4k

)(5.17)

or q

2m

(1− q

4k

)(5.18)

If we now give the receptor a finite mass m3, we get the following:

L = 12m

(x2

1 + x22)

+ 12m3x

23 −

12k1 (x1 − x2)2 − 1

2k2 (x2 − x3)2 (5.19)

The general Euler-Lagrange equations hold:

d

dt

(∂L

∂xi

)= ∂L

∂xi(5.20)

from which follow the equations of motion

mx1 = k1 (x2 − x1)mx2 = k2 (x1 − x2) + k2 (x3 − x2)m3x3 = k2 (x2 − x3) (5.21)

20

We make a substitution of variables:

qi =√mixi (5.22)

i.e. q1 =√mx1, q2 =

√mx2, q3 = √m3x3.

We get the following set of equations:

√mq1 = − k1√

mq1 + k1√

mq2

√mq2 = k1√

mq1 −

k1√mq2 −

k2√mq2 + k2√

m3q3

√m3q3 = − k2√

m3q3 + k2√

mq2 (5.23)

q1q2q3

=

−k1m

k1m 0

k1m −

(k1+k2m

)k2√mm3

0 k2√mm3

− k2m3

q1q2q3

(5.24)

This matrix is symmetric and can therefore be diagonalised by the eigenvec-tors

M−→q = ω2−→q (5.25)

(M − Iω2)−→q = −→0 (5.26)

|M − ω2I| = 0 (5.27)

Letting the receptor mass m3 go to infinite the problem becomes the sameas for the two mass spring system coupled to a wall. The determinant reducesto the result which was shown before.∣∣∣∣k1

m − ω2 −k1

m

−k1m

k1+k2m − ω2

∣∣∣∣ = ω4 −(

2k1

m+ k2

m

)ω2 + k1k2

m2 = 0 (5.28)

The solution again is

ω2 =(k1

m+ k2

2m

)±

√(k1

m

)2+(k2

2m

)2(5.29)

21

Chapter 6

Conclusion

The Morse potential describes the potential of a diatomic molecule through theparameters r0, the equilibrium bond distance, D, the well depth, and a, the’width’ of the potential.

VMorse (r) = D(e−a(r−r0) − 1

)2−D (6.1)

We used the Lagrangian formalism to elucidate the harmonic oscillator andthe connection between potential energy and spring constant. Using the Morsepotential for the potential for the diatomic molecule which we modeled as twoatoms connected through a spring, we found a new expression for the springconstant. The spring constant of the dimer molecule with a Morse potential is:

kMorse = 2Da2

Our classical model of the dimer molecule was now extended to quantum me-chanical parameters of the Morse potential. We then modified the potential toinclude the Coulomb interaction between two internal charges introduced to thetwo atoms of the diatomic molecule. We found that the spring constant thenchanges. The change in spring constant ∆k is produced by the charge on thetwo atoms of the diatomic molecule.

∆k = − 2q2e2

4πε0r30

Instead of only introducing the charges to the two atoms ad hoc, the chargeson the two atoms are induced by external charge surrounding the diatomicmolecule. The diatomic molecule is polarized by the external charge and thisresults eventually in the change in spring constant.

∆k = −2(αQie

4πε0r30

)4πε0r3

0

We then turned to the tight binding model to understand better the origin ofthe polarizability α of the dimer. Using the tight binding model we have foundthe equation

∆ε =(

2α2 − 1√

∆ε2 + 4V 2)

22

from which we derive an expression for the charge on one atom in the diatomicmolecule

q = − ∆ε√∆ε2 + 4V 2

In the limit ed � 2V we found an expression for the polarizability of themolecule in terms of the tight binding model

α = e2d2

2Vwhere d is the distance between the atoms of the dimer and V is the overlapintegral. This expression can be inserted into the equation for the change inspring constant. The change in spring constant thus results from external chargein the aqueous medium, the environment of the diatomic molecule as it polarizesthe molecule. In addition the polarization occurs also as a result of image chargesin the dielectric medium. We therefore derived an effective spring constantkimage. We then calculated the forces on the individual charged atoms of thedimer in the dielectric medium through the method of image charges:

F1 = FMorse−q2e2

4πε0d2 + qQexternale2

4πε0 (d+ a)2 + qe2

4πε0

[q

(2z)2 −q

(2z + d)2 + Qexternal

(2z + a+ d)2

]

F2 = FMorse−q2e2

4πε0d2−qQexternale

2

4πε0a2 + qe2

4πε0

[−q

(2z + d)2 + q

(2z + 2d)2 −Qexternal

(2z + 2d+ a)2

]

From these expressions we calculated the vibrational frequency in the dielec-tric medium:

ω2 =(

2F ′Morse + 2 2q2e2

4πε0d3

)(1m1

+ 1m2

)+ 2 6q2e2

4πε0 (2z + d)3

(1m1

+ 1m2

)−

(4q2e2

4πε0 (2z)3 −6qQexternale

2

4πε0 (2z + d+ a)3 + 8qQexternale2

4πε0 (2z + 2d+ a)3 −8q2e2

4πε0 (2z + 2d)3

)(1m1

+ 1m2

)(6.2)

The effective spring constants of the dimer are:

kinternal = 2F ′Morse + 2 2q2e2

4πε0d3 (6.3)

kQ = 2 6q2e2

4πε0 (2z + d)3 (6.4)

kimage = 4q2e2

4πε0 (2z)3 −6qQexternale

2

4πε0 (2z + d+ a)3 + 8qQexternale2

4πε0 (2z + 2d+ a)3 −8q2e2

4πε0 (2z + 2d)3(6.5)

The vibrational frequency of the molecule is dominated to a large extent by theMorse potential and induced charges. Thus the Coulomb term in the potentialin equation (2.20) is to be considered necessary to describe polarization effects.When the molecule binds to the receptor, it is modeled as binding to it througha weak spring constant. We therefore modeled the binding of the dimer to the

23

receptor as connecting it to a hard wall. We then calculated the vibrationalfrequency for a weak bond between the molecule and the receptor:

ω2 ≈(k

m+ q

2m

)± k

m

(1 + 1

2

( q2k

)2)

+ . . . (6.6)

= 2 km

+ q

2m

(1 + q

4k

)(6.7)

or q

2m

(1− q

4k

)(6.8)

24

Bibliography

[Bro10] Brookes,J.C., "Science is perception: what can our sense of smell tellus about ourselves and the world around us?" Phil. Trans. R. Soc. A 2010368, doi: 10.1098/rsta.2010.0117, published 5 July 2010

[Hor06] Brookes,J.C., Hartoutsiou,F., Horsfield,A.P., Stoneham,A.M.,"Could humans recognize odor by phonon assisted tunneling?,"arXiv:physics/0611205v1 physics.bio-ph, 22 Nov 2006

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