Eigenmodes for coupled harmonic vibrations. Algebraic Method for Harmonic Oscillator.

Transcription

1 PHYS208 spring 2008 Eigenmodes for coupled harmonic vibrations. Algebraic Method for Harmonic Oscillator Adapted from the text Light - Atom Interaction PHYS261 autumn 2007 Go to list of topics

2 (1) Oscillator Topics: Eigenmodes for coupled harmonic vibrations. Algebraic Method for Harmonic Oscillator. Additional Topics of interest from the original text: Quantum treatment of extended systems - fields Density of States based on k-space

3 1 Generally on quantum treatment of extended systems - fields. Go to list of topics The system to be considered consists of an atom and the electromagnetic field. The field has in principle infinitely many degrees of freedom, i.e. the values of field variables in each point of space. The electromagnetic field can be treated by wave equations, it is thus a sort of harmonic system. The energy of a finite harmonic system can always be transformed to a sum of independent harmonic oscillators, each of them is in fact the eigenmode. Each harmonic oscillator can be then described using the equations for harmonic oscillator, with the algebraic method and number of quanta states - The creation and annihilation operators - click 3. The normal modes or eigenmodes for a general harmonic system are discussed in the section 2.

4 The prescription 1. identify the eigenmodes 2. quantize each of the modes as independent harmonic oscillator 3. use quanta of eigenmodes 4. express the general coordinates of the system using the eigenmode coordinates (inverse transformations to those discussed in section 2. Each coordinate χ i will then be replaced by its combination of the creation and annihilation operators as discussed in section 3.

5 2 Normal Coordinates for coupled harmonic vibrations. Go to list of topics Transformation to normal coordinates can be described as follows: Take the total energy, x represents the vector of all coordinates, x T represents the transposed vctor, M is the mass matrix, V is the matrix which gives the potential energy, including the couplings. Note that the mass matrix is written in a very general form, most often this matrix would be diagonal. E = 1 2ẋT Mẋ xt Vx First we transform the kinetic energy to a spherical form

6 1 2ẋT Mẋ = 1 2 ηt S T MS η = 1 2 ηt η This transformation does not conserve the lengths. The N-dimensional ellipsoid is transformed to an N-dim. sphere. S T MS = 1 If M ij = m j δ ij, it is quite easy to find S ij = m j 1/2 δ ij, Transformation S simplifies the kinetic energy, but the potential remains complicated. Therefore we use one more, from η to χ, 1 2 xt Vx = 1 2 ηt S T VSη = 1 2 χt U 1 S T VSUχ Transformation η = Uχ is a simple rotation obtained by diagonalizing the matrix of potential energy: ( U 1 S T VSU ) ij = Ω2 j δ ij = W ij

7 Therefore we took U 1 = U T. We can see that the length is conserved. 1 2 ηt η = 1 2 χt U 1 U χ = 1 2 χt χ E = 1 2ẋT Mẋ xt Vx = 1 2 χt χ χt Wχ This expression however is a sum over the new mutually independent degrees of freedom, since W is diagonal. N i=1 1 2 χt χ χt Wχ = 1 2 m iẋ 2 i N i,j=1 N i=1 V ij x i x j ( 1 2 χ2 i Ω2 i χ 2 i N i=1 ) ( 1 2 χ2 i Ω2 i χ 2 i )

8 Example A string of balls of mass m connected by springs of equal spring constant is described by displacements u i. We can label the displacements u i = u(x i ) where x i is the equilibrium position of the i-th ball. In the limit of infinitisemal small balls and short springs this would then lead to a wave equations for a continuous elastic string, with continuously observed displacement u(x). The total energy is: 1 2 m ( u(x i )) 2 i k i (u(x i ) u(x i + 1)) ku(x 1) ku(x N) 2 The last two terms represent the fixed end springs. The matrix M is simply a diagonal matrix with m as all diagonal elements. The matrix V is a band matrix (below). The transformation from u(x i ) to χ i can also be reversed. The reversed transformation in this case simply consists of eigenvectors of the matrix V, so that each of the eigenmodes is simply given by a function of time χ k (t) = χ k0 e iω kt

9 where χ k0 is the amplitude, and the actual displacements for the k-th mode can be written as u k (x i ) = S k i χ k (t) = S k i χ k0 e iω kt where S k i is the i-th component of the k-th eigenmode with the frequency Ω k The matrix V has the form The eigenvectors - giving the eigenmodes - are eigenvectors of this band matrix. The modes are standing waves, in the string limit. For infinite system - traveling waves. leads to Fourier expansions for fields ) see A(r) expressed in eigenmodes

10 Figure 1: Components of eigenmodes at positions x i. For N = 20 first and second node, fo N = 100 nodes 1,2, and 6.

11 3 Algebraic Method for Harmonic Oscillator. Go to list of topics In this section we show how one can work with the harmonic oscillator introducing so called ladder operators which move from state to state, the states being separated by the same quantum of energy. The classical hamiltonian can be transformed: 1 2m p2 + mω2 2 q2 transforms to 1 P = h m ω p With this the commutator [ q, p ] = i h becomes [ Q, P ] = i hω 2 Q = m ω h ( P 2 + Q 2) This simplifies the equation and brings it into a form where the energy is expressed in hω. q

12 The main transformation, however, is to go over to linear combinations of P and Q a = 1 2 Q + i P and a + = 1 Q 2 2 By a very simple algebra we find that for their commutator [ a, a + ] = 1 i 2 P and the energy is transferred to hω 2 Using the commutator a a + ( P 2 + Q 2) hω 2 ( a a + + a + a ) a + a = 1 we finally get H = hω a + a + hω 2 = hω N + hω 2 where we already include that N = a + a

13 will be a number operator. Why number? Let us play with [ a, a + ] = 1 or alternatively a a + a + a = 1 and the operator N. We quickly derive that [ N, a + ] = a + and [ N, a ] = a This is done simply by writing out [ N, a + ] = Na + a + N = (a + a)a + a + (a + a) = a + (a a + a + a) = a + [ a, a + ] = a + Since H = hω N + hω 2 we also have [ H, a + ] = hω a + and [ H, a ] = hω a These are the last equations which bring the physical interpretation. If there is a Q.M. state ψ(q) ψ such that H ψ = E ψ then H (a ψ ) = a H ψ hω a ψ = a E ψ hω a ψ = (E hω ) (a ψ )

14 that means that (a ψ ) is also an eigenstate. It has energy lower by hω. This we can continue again and again, getting eigenstates for E hω, E 2 hω, E 3 hω, E 4 hω, E 5 hω... Since this cannot go on infinitely, we get finally a state such that We quickly verify that this state has a ψ 0 = 0 E 0 = hω 2 further that the same thing is possible with a + only with opposite sign, so that we get energies E 0, E 0 + hω, E hω, E hω, E hω, E hω... Each of the eigenstates is an eigenstate of both H and N with obvious number of quanta. Therefore we call a + and a creation and annihilation operators. They make states with one more ore one less quantum hω

15 We can complete this treatment by a complete solution for the wavefunction: a ψ 0 = 0 a = 1 2 (Q + i P ) P = i Q Thus, writing ψ 0 ψ 0 (Q) a ψ 0 = 0 ( Q + ) ψ 0 (Q) = 0 Q The solution of 1. order differential eq. to the left is easy ψ 0 (Q) = C 0 e Q2 /2 and using the expression for a + and the ψ n ( ψ 1 (Q) = C 1 Q ) Q e Q2 /2 ( ψ 2 (Q) = C 2 Q ) ( Q Q ) Q e Q2 /2 and n times for general ψ n. cf. recursion relations for Hermite polynomials.

16 4 Density of States Go to list of topics We consider the electromagnetic field enclosed in a box with volume V. For a finite volume there is a discrete number of modes satisfying the imposed boundary conditions. Therefore the sum over final states is an ordinary sum. As this volume approaches infinite size, the summation over k, will be approaching an integral. The density of states factor is found from performing such a limiting process, using following relations. The summation will be replaced by an integral ρ( k) d k (2) ρ( k) is the density of states in the K-space. k The allowed discrete values of k are obtained by combinations of components k (nx) x = 2π L n x k (ny) y = 2π L n y k (nz) z = 2π L n z (3)

17 where the numbers n x,n y,n z are positive and negative integers. It means that each of the allowed vectors k occupies a small volume of the K-space ( ) 2π 3 k x k y k z = = (2π)3 L V The density of states in the K-space is thus a constant (i.e. one per the above small k-space volume), and the above relation can be written as V (2π) 3 k (4) d k (5) Since the derivation of the golden rule assumes integration over frequencies, or energies, we shall transform this integral over momenta (i.e. wave numbers k) to integral over energy, V d (2π) k ρ(e)de 3 V (2π) 3 d k = V dω (2π) 3 k Ω k k 2 dk = Ω k [ ] V dk k2 (2π) 3 de dω k de (6)

18 so that the ρ(e) can be identified as V dk k2 dω (2π) 3 k (7) de Ω k If the processes depend on the direction of the wave vector, which is true in photon emission case, we must keep the angular information inside of the density of states, and perform the angular integration afterwards. We must now evaluate the above density of states in terms of energy only, using k = ω E = hkc c Setting these relations of k and E, we obtain the expression for ρ(e) ρ(e) = V (2π) E 2 3 ( hc) dω 3 k = V (2π) 1 3 h ω2 c dω 3 k (8)