# User:Guy vandegrift/btvs/Effort 5

## Calculation.

Longitudinal modes in a lattice of springs and masses near the ground state form a quantized phonon field that interacts with the atom via a weak linear coupling spring at the center.

Place springs at ℓ = 0 ±Δℓ, ±2Δℓ, ... ±L, so that the length of the string is 2L. Often we shall count from the left, with the center at (N+1)Δℓ, so that there are 2N-1 little mass that make up the string, plus one "atomic" mass. The classical calculation can be done using matrices and either the energy or the force law. We begin with the force law.

It is convenient to define a spring constant for the string, KS, based on the tension, T, and string length, 2Δℓ:

${\displaystyle K_{S}=T/\Delta \ell }$

### Force law

Counting from the left:

{\displaystyle {\begin{aligned}M_{S}{\ddot {y}}_{1}=-\omega ^{2}M_{S}\cdot y_{1}&=-K_{S}\cdot y_{1}-K_{S}\cdot (y_{1}-y_{2})\\&=-2K_{S}\cdot y_{1}+K_{S}\cdot y_{2}\end{aligned}}}

where ${\displaystyle K_{S}=T/\Delta \ell }$. For atoms not touching the wall or the coupling spring connecting to the "atom":

${\displaystyle M_{S}{\ddot {y}}_{n}=-\omega ^{2}M_{S}\cdot y_{n}=K_{S}\cdot y_{n-1}-2K_{S}\cdot y_{n}+K_{S}\cdot y_{n+1}}$

The N-th mass is attached to the atom, and obeys:

${\displaystyle M_{S}{\ddot {y}}_{N}=-\omega ^{2}M_{S}\cdot y_{N}=K_{S}\cdot y_{N-1}-2K_{S}\cdot y_{N}+K_{S}\cdot y_{N+1}+K_{C}\cdot (y_{A}-y_{N})}$

and the atom obeys:

${\displaystyle M_{A}{\ddot {y}}_{A}=-\omega ^{2}M_{A}\cdot y_{N}=-K_{A}\cdot y_{A}+K_{C}\cdot (y_{N}-y_{A})}$

### System with 3 masses

One atom. Two masses plus one massless object on the string. N=2 and 2N-1=3 masses on the string. There are 4 masses and we need 4 equations:

Constraint for symmetric modes: ${\displaystyle 0=(y_{1}+y_{3}-2y_{2})K_{S}+(y_{A}-y_{2})K_{S}=0=(2y_{1}-2y_{2})K_{S}+(y_{A}-y_{2})K_{C}}$

${\displaystyle {\begin{matrix}-M_{S}\omega ^{2}\cdot y_{1}&=&-2K_{S}\cdot y_{1}&+K_{S}\cdot y_{2}&\\\;\qquad 0\cdot y_{2}&=&+2K_{S}\cdot y_{1}&(-2K_{S}-K_{C})\cdot y_{2}&+K_{C}\cdot y_{A}\\-M_{A}\omega ^{2}\cdot y_{A}&=&&+K_{C}\cdot y_{2}&-(K_{A}+K_{C})\cdot y_{A}&\end{matrix}}}$

Define: ${\displaystyle x=\omega ^{2},\quad \sigma =K_{S}/M_{S},\quad \alpha =K_{A}/M_{A}}$    ${\displaystyle \gamma =K_{C}/M_{S},\quad \gamma \beta =K_{C}/M_{A}}$, where ${\displaystyle \quad \beta =M_{S}/M_{A}}$

#### Algebra

${\displaystyle {\begin{matrix}-x\cdot y_{1}&=&-2\sigma \cdot y_{1}&+\sigma \cdot y_{2}&\\\qquad 0\qquad &=&+2\sigma \cdot y_{1}&(-2\sigma -\gamma )\cdot y_{2}&+\gamma \cdot y_{A}\\-x\cdot y_{A}&=&&+\gamma \beta \cdot y_{2}&-(\alpha +\gamma \beta )\cdot y_{A}&\end{matrix}}}$

Define: ${\displaystyle x=\omega ^{2},\quad \sigma =K_{S}/M_{S},\quad \alpha =K_{A}/M_{A}}$    ${\displaystyle \gamma =K_{C}/M_{S},\quad \gamma \beta =K_{C}/M_{A}}$, where ${\displaystyle \quad \beta =M_{S}/M_{A}}$

${\displaystyle {\begin{matrix}0=&\left(x-2\sigma \right)y_{1}&\quad +\sigma \cdot y_{2}&\\0=&\quad +2\sigma \cdot y_{1}&(-2\sigma -\gamma )\cdot y_{2}&\;\quad +\gamma \cdot y_{A}\\0=&&+\gamma \beta \cdot y_{2}&+\left(x-\alpha -\gamma \beta \right)\cdot y_{A}\end{matrix}}}$

#### Match frequencies at zero coupling

${\displaystyle {\begin{vmatrix}x-2\sigma &+\sigma &0\\+2\sigma &-2\sigma -\gamma &+\gamma \\0&+\gamma \beta &x-\alpha \end{vmatrix}}=0}$

If ${\displaystyle \gamma =0}$ the solutions are ${\displaystyle x=\alpha }$ and ${\displaystyle x=\sigma }$

matlab output
gamma =  0
M =
[ x - 2*sigma,    sigma,         0]
[     2*sigma, -2*sigma,         0]
[           0,        0, x - alpha]

deter=collect(-det(M),x)
deter =
(2*sigma)*x^2 + (- 2*sigma^2 - 2*alpha*sigma)*x + 2*alpha*sigma^2

solutions = solve(deter)
alpha
sigma

Henceforth ${\displaystyle \sigma =\alpha }$

#### Eigenfrequencies

matlab code
clc;close all;clear all;
syms sigma beta gamma alpha y1 y2 y3 y4 x
sigma = alpha
M=[x-2* sigma   ,  + sigma           ,    0      ;
+2*sigma    ,   -2*sigma - gamma          ,  + gamma   ;
0           ,  + gamma* beta    ,  x- alpha-gamma*beta;]
deter=collect(-det(M),x)
solutions = solve(deter)

Results:
M =

[ x - 2*alpha,             alpha,                      0]
[     2*alpha, - 2*alpha - gamma,                  gamma]
[           0,        beta*gamma, x - alpha - beta*gamma]

deter =
(2*alpha + gamma)*x^2 + (- 3*alpha*gamma - 4*alpha^2 - 2*alpha*beta*gamma)*x + 2*alpha^2*gamma + 2*alpha^3 + 2*alpha^2*beta*gamma
solutions =

alpha
(2*alpha*gamma + 2*alpha^2 + 2*alpha*beta*gamma)/(2*alpha + gamma)

${\displaystyle 0=(2\alpha +\gamma )x^{2}-(4\alpha ^{2}+3\alpha \gamma +2\alpha \beta \gamma )x+2\alpha ^{2}\gamma +2\alpha ^{3}+2\alpha ^{2}\beta \gamma }$

{\displaystyle {\begin{aligned}x&=\alpha \\&={\frac {2\alpha ^{2}+2\alpha \gamma +2\alpha \beta \gamma }{2\alpha +\gamma }}\end{aligned}}}

#### Eigenvectors

${\displaystyle {\begin{matrix}0=&\left(x-2\alpha \right)y_{1}&\quad +\alpha \cdot y_{2}&\\0=&\quad +2\alpha \cdot y_{1}&(-2\alpha -\gamma )\cdot y_{2}&\;\quad +\gamma \cdot y_{A}\\0=&&+\gamma \beta \cdot y_{2}&+\left(x-\alpha -\gamma \beta \right)\cdot y_{A}\end{matrix}}}$

##### Mode 1
${\displaystyle x=\alpha }$
${\displaystyle y_{1}=1\quad y_{A}=1\quad y_{3}=1}$
##### Mode 2
${\displaystyle x={\frac {2\alpha ^{2}+2\alpha \gamma +2\alpha \beta \gamma }{2\alpha +\gamma }}}$
algebra

${\displaystyle x-2\alpha ={\frac {2\alpha ^{2}+2\alpha \gamma +2\alpha \beta \gamma }{2\alpha +\gamma }}-{\frac {(2\alpha )(2\alpha +\gamma )}{2\alpha +\gamma }}={\frac {2\alpha \beta \gamma }{2\alpha +\gamma }}}$

${\displaystyle {\frac {1}{x-2\alpha }}={\frac {2\alpha +\gamma }{2\alpha \beta \gamma }}}$

• ${\displaystyle y_{2}=1}$
${\displaystyle (x-2\alpha )y_{1}=-\alpha \cdot y_{2}=-\alpha }$
${\displaystyle (x-2\alpha )y_{1}\left({\frac {1}{x-2\alpha }}\right)=-\alpha \cdot {\frac {2\alpha +\gamma }{2\alpha \beta \gamma }}={\frac {2\alpha +\gamma }{2\beta \gamma }}}$
• ${\displaystyle y_{1}={\frac {2\alpha +\gamma }{2\beta \gamma }}=y_{3}}$

{\displaystyle {\begin{aligned}x-\alpha -\gamma \beta &={\frac {2\alpha ^{2}+2\alpha \gamma +2\alpha \beta \gamma }{2\alpha +\gamma }}-\alpha {\frac {2\alpha +\gamma }{2\alpha +\gamma }}-\gamma \beta {\frac {2\alpha +\gamma }{2\alpha +\gamma }}\\&={\frac {2\alpha ^{2}+2\alpha \gamma +2\alpha \beta \gamma }{2\alpha +\gamma }}+{\frac {-2\alpha ^{2}-\alpha \gamma }{2\alpha +\gamma }}+{\frac {-2\gamma \beta \alpha -\gamma ^{2}\beta }{2\alpha +\gamma }}\\&={\frac {\alpha \gamma -\gamma ^{2}\beta }{2\alpha +\gamma }}\end{aligned}}}

${\displaystyle {\frac {1}{x-\alpha -\gamma \beta }}={\frac {2\alpha +\gamma }{\alpha \gamma -\gamma ^{2}\beta }}}$

From the eigenvalue equation above:

• ${\displaystyle y_{A}={\frac {-\gamma \beta }{x-\alpha -\gamma \beta }}\cdot y_{2}={\frac {-\gamma \beta }{x-\alpha -\gamma \beta }}={\frac {-\gamma \beta }{1}}{\frac {2\alpha +\gamma }{\alpha \gamma -\gamma ^{2}\beta }}==-{\frac {2\beta \alpha +\gamma \beta }{\alpha -\gamma \beta }}={\frac {2\beta \alpha +\gamma \beta }{\gamma \beta -\alpha }}}$
${\displaystyle y_{1}={\frac {2\alpha +\gamma }{2\beta \gamma }}\qquad y_{A}={\frac {2\beta \alpha +\gamma \beta }{\gamma \beta -\alpha }}\qquad y_{3}={\frac {2\alpha +\gamma }{2\beta \gamma }}}$
##### Mode 3
${\displaystyle x=2\alpha }$
${\displaystyle y_{1}=1\qquad y_{A}=0\qquad y_{3}=-1}$