Difference between revisions of "Perturbation Theory"

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:<math>E_g = E_g^\circ - \underbrace{\langle | \Psi_g | W | \Psi _g  \rangle}  \overrightarrow{F} + \,\!</math>
:<math>E_g = E_g^\circ - \underbrace{\langle | \Psi_g | W | \Psi _g  \rangle}  \overrightarrow{F} + \,\!</math>


:<math>\: \:\:\:\:\overrightarrow{\mu^\circ}\,\!</math>
:<math>\; \overrightarrow{\mu^\circ}\,\!</math>


A one to one correspondence can be made between the terms in the stark energy expression and the perturbation theory expression when the perturbation is minus μF.  
A one to one correspondence can be made between the terms in the stark energy expression and the perturbation theory expression when the perturbation is minus μF.  


As a side note, as you go to higher orders, things will look a bit more complicated because there are more summations over excited states. For example in 3rd order, there will be a double summation over excited states. In 4th order, there will be a triple summation over excited states. But it will always be products of matrix elements of this kind at the numerator, and the differences in energies of the states for the unperturbed molecule will be in the denominator. The expressions look more complex but by looking at the terms individually, notice that the same kind of terms come up.
As a side note, as you go to higher orders, things will look a bit more complicated because there are more summations over excited states. For example in 3rd order, there will be a double summation over excited states. In 4th order, there will be a triple summation over excited states. But it will always be products of matrix elements of this kind at the numerator, and the differences in energies of the states for the unperturbed molecule will be in the denominator. The expressions look more complex but by looking at the terms individually, notice that the same kind of terms come up.

Revision as of 11:57, 26 February 2010

Hellman-Feynman Theorem

The Hellman-Feynman Theorem, which expresses the dipole moment as minus derivative of the energy of the system with respect to the field. This equation expresses the response of a molecule which is 2nd order in terms of the energy of the molecule and first order in terms of the dipole moment of the molecule (1st order or 2nd order with respect to the field).

<math>\overrightarrow {\mu} = - \frac {\delta E} {\delta \overrightarrow{F}}\,\!</math>
<math>\overrightarrow{\mu} = \overrightarrow{\mu^\circ} + \alpha \overrightarrow{F} + \frac {1} 2! \beta \overrightarrow{F}\overrightarrow{F} + \frac {1} {3!} \gamma \overrightarrow{F}\overrightarrow{F}\overrightarrow{F}\,\!</math>


<math>\frac {\delta \overrightarrow{\mu}}{\delta \overrightarrow{F}} = \alpha + \beta \overrightarrow{F} + \frac {1} {2!} \gamma \overrightarrow{F}\overrightarrow{F} + ...\,\!</math>


Stark Energy Expression

Alpha is the linear polarizability or the first order polarizability. It describes how the molecule responds in terms of the modification of its ground state energy or of its dipole moment, in the presence of the field at the limit where the field goes to 0. Thus, alpha can be cast either as the 1st order derivative of the dipole moment with respect to the field when the field tends to 0 or minus the 2nd order derivative of the ground state energy of the molecule with respect to the field when the field goes to 0.

<math>\alpha = \left( \frac {\delta\overrightarrow{\mu}}{\delta \overrightarrow{F}}\right) \overrightarrow{F} \rightarrow 0 \,\!</math>
<math>= - \left( \frac {\delta^2 \overrightarrow{\mu}}{\delta \overrightarrow{F}^2}\right) \overrightarrow{F} \rightarrow 0 \,\!</math>

Beta is the 2nd order derivative of the dipole moment with respect to the field when the field goes to 0. Beta will be referred to as the 2nd order polarizability of the molecule. This can also be derived from the stark energy expression which is the 3rd order derivative of the energy with respect to the field when the field tends to 0.

<math>\beta = \left( \frac {\delta^2 \overrightarrow{\mu}}{\delta \overrightarrow{F}^2}\right) \overrightarrow{F} \rightarrow 0 \,\!</math>
<math>= - \left( \frac {\delta^3 \overrightarrow{\mu}}{\delta \overrightarrow{F}^3}\right) \overrightarrow{F} \rightarrow 0 \,\!</math>

Lastly, the gamma term corresponds to the 3rd order derivative of the dipole moment or minus the 4th order derivative of the energy with respect to the field at the limit where the field goes to 0. Gamma will be referred to as the 3rd order polarizability of the molecule.

<math>\gamma = \left( \frac {\delta^3 \overrightarrow{\mu}}{\delta \overrightarrow{F}^3}\right) \overrightarrow{F} \rightarrow 0 \,\!</math>
<math>= - \left( \frac {\delta^4 \overrightarrow{\mu}}{\delta \overrightarrow{F}^4}\right) \overrightarrow{F} \rightarrow 0 \,\!</math>

Stark energy describes the evolution of the energy of a system of particles in the presence of an electric field F. In the Stark energy expression, gamma corresponds to a 4th order term. However the in common terminology alpha is referred to as the linear polarizability, beta the 2nd order polarizability, and gamma the 3rd order polarizability. The Hellman-Feynman Theorem is the origin of these terms. Here it is presented as a Taylor series expansion, sometimes one uses a power series expansion.

<math>

E_g = E^\circ_g = \overrightarrow{\mu}^\circ \overrightarrow{F} - \frac {1} {2!} \alpha \overrightarrow{F}\overrightarrow{F} - \frac {1} {3!} \beta \overrightarrow{F}\overrightarrow{F}\overrightarrow{F} - \frac {1} {4!} \gamma \overrightarrow{F}\overrightarrow{F}\overrightarrow{F}\overrightarrow{F}\,\!</math>

<math>= E^\circ_g - \overrightarrow{\mu}\overrightarrow{F}\,\!</math>


Electric dipole approximation

The stark energy expression states that the ground state energy of the molecule in the presence of the field is the ground state energy in the absence of the field minus μF. Thus, μF is considered the interaction between the field and the molecule. The electric dipole approximation is that only the electric field component of light influences dipole moment. The magnetic component is ignored. In most instances in the literature, this approximation is assumed but not stated. The expression for the dipole moment is expressed as :

<math>\mu^\circ + \alpha F + \beta F^2 ... \,\!</math>

and so on, without paying any attention to where that expression comes from.

<math>- \overrightarrow{\mu}\overrightarrow{F}\,\!</math>

in dimensional analysis:

<math>\mu \equiv\,\!</math> charge * distance
<math>F \equiv\,\!</math> volt/distance
<math>\mu F \equiv\,\!</math> charge * volt :<math>\equiv\,\!</math> energy

After the electric dipole, the next two terms are the electric quadrupole and the magnetic dipole. 99% of the time only the electric dipole is considered. This represents the difference in energy between the perturbed state and the unperturbed state of the molecules. Thus, this must have an energy dimension. μ is the dipole moment which is the charge times the distance, and the electric field is volt over distance. When you multiply these terms, the expression has the dimension charge times volt or ev, and ev is an energy unit. This will be used in perturbation theory.


The stark energy expression states that the ground state energy of the molecule in the presence of the field is the ground state energy in the absence of the field minus μF. Thus, μF is considered the interaction between the field and the molecule. The electric dipole approximation is that only the electric field component of light influences dipole moment. The magnetic component is ignored. In most instances in the literature, this approximation is assumed but not stated. The expression for the dipole moment is expressed as μnot plus α * f plus β * f 2 and so on, without paying any attention to where that expression comes from.

<math>\overrightarrow{\mu} = e \sum(i) \overrightarrow{\pi}_i 9i\,\!</math>

After the electric dipole, the next two terms are the electric quadrupole and the magnetic dipole. 99% of the time only the electric dipole is considered. This represents the difference in energy between the perturbed state and the unperturbed state of the molecules. Thus, this must have an energy dimension. μ is the dipole moment which is the charge times the distance, and the electric field is volt over distance. When you multiply these terms, the expression has the dimension charge times volt or ev, and ev is an energy unit. This will be used in perturbation theory.

Perturbation theory

The perturbation theory will not be discussed at the quantum- mechanics level. The energy of the ground state of the system is the energy for the unperturbed system. Perturbation can be observed at different orders. At first order, the perturbation is referred as w. If that perturbation is the impact of the electric field of the light in the electric dipole approximation, the perturbation can be expressed as minus μF.

At the first order, the perturbation is operating on the unperturbed wave function of the ground state. Perturbation theory involves modification of systems due to the perturbation of all the wave functions for the unperturbed system. At first order, the perturbation is simply acting on the ground state wave function. Then it is integrated over space and the complex conjugate is taken. At second order, the wave functions of the excited state are taken into account to describe the modification of the system. An unperturbed system has a well defined wave function for the ground state and well defined wave functions for the excited states. The perturbed system is described on the basis of the wave functions of the unperturbed system.

<math>=\int \Psi* e \overrightarrow{\pi} \Psi dr\,\!</math>
<math>E_g = E^\circ_g + \langle \Psi_g | w | \Psi_g \rangle + \sum_p \frac {\langle \Psi_g | W | \Psi_p \rangle \langle \Psi_p | W | \Psi_g \rangle + ...}{ E^\circ_p - E^\circ _g}\,\!</math>
<math>E_g = E^\circ_g -\overrightarrow{\mu ^\circ} \overrightarrow{F} - \frac {1} {2!} \alpha \overrightarrow{F}^2 - ....\,\!</math>
<math>W = - \overrightarrow{\mu} \overrightarrow{F}\,\!</math>

In terms of non-linear optics, the perturbation theory expressions will show what the excited states are in your isolated molecule that will contribute to the linear polarizability, 2nd order polarizability, or the 3rd order polarizability and allow you to pinpoint exactly what excited states do play a major role in your optical response.

The complete set of wave functions for the unperturbed state will form the basic set for the perturbation expressions. In principle this includes all excited states. The 2nd order term in terms of perturbation and will correspond to alpha, the linear polarizability. In most conjugated systems, only the first excited state needs to be examined. This will often be the case for alpha and pi conjugated systems as well as for beta. But not for gamma in which two or more excited states must be taken into account. However, the number of states that need close attention can be heavily restricted.

<math>E_g = E_g^\circ - \underbrace{\langle | \Psi_g | W | \Psi _g \rangle} \overrightarrow{F} + \,\!</math>
<math>\; \overrightarrow{\mu^\circ}\,\!</math>

A one to one correspondence can be made between the terms in the stark energy expression and the perturbation theory expression when the perturbation is minus μF.

As a side note, as you go to higher orders, things will look a bit more complicated because there are more summations over excited states. For example in 3rd order, there will be a double summation over excited states. In 4th order, there will be a triple summation over excited states. But it will always be products of matrix elements of this kind at the numerator, and the differences in energies of the states for the unperturbed molecule will be in the denominator. The expressions look more complex but by looking at the terms individually, notice that the same kind of terms come up.