Physics in a Nut-Shell: Two-State Systems
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PHYSICS IN A NUT-SHELL: TWO-STATE SYSTEMS D. Dubbers, WS 2008/2009
1. Prologue: Classical coupled vibrations
Demonstration: Double Pendulum: normal modes, beats
1.1 Symmetric coupled oscillator a) Free oscillator
Force law m˙x˙ kx , with spring constant k, has solution x(t) x0 cos(ω0t φ) with frequency ω0 k/m .
From initial cond's x(0), x˙(0) follows x0 , φ The total energy in the system is 1 2 E 2 kx0
(For mathematical pendulum:
k mg/l : ω0 g/l is independent of m)
1.1 b) normal coordinates and eigenfrequencies
Spring constants k1 k2 k
m˙x˙ kx k'(x x ) Force law 1 1 2 1 m˙x˙ kx k'(x x ) 2 2 1 2 k k' k
shorthand mx˙˙ Kx
x1 k k' k' with x , K . x2 k' k k'
These differential equations can be decoupled by taking their sum and their difference, i.e. with 1 normal coordinates q (x1 x2 ) 2
this becomes mq˙˙ kq
mq˙˙ (k 2k') q or mq˙˙ Dq ,
k 0 q with diagonal matrix D and q . 0 k 2k' q The solutions are the normal vibrations
q (t) q (0)cos(ωt φ ) with the eigen-frequencies of the normal modes: a low frequency ω k / m ω0 = in-phase frequency = frequency of free pendulum a high frequency ω (k 2k') / m = opposite-phase frequency, under double stretch of spring.
1.2 c) Initial conditions
Recipe: 1. define initial conditions for x and x˙ 2. translate to initial conditions for normal modes q and q˙ 3. write solution for q(t) 4. transform solution back to x(t) x Example "beats": 1
1. x1 (0) x10 , x2 (0) 0, x˙1 (0) x˙ 2 (0) 0 x 2 2. q (0) q (0) x10 / 2, q˙ (0) q˙ (0) 0 φ 0 1 3. q (t) x10 cos(ωt) 2 1 1 q 4. x1,2 (t) (q q ) x10[cos(ωt) cos(ωt)] + 2 2
q − that is x1 (t) x10 cos(½(ω ω )t) cos(½(ω ω )t)
or x1 (t) x10 cosωt cos ω t and x2 (t) x10 sin ωt sinω t , with mean frequency ω ½(ω ω ) and beat frequency ω ½(ω ω ) .
1 2 The total energy E 2 kx0 of the system is exchanged between the two pendula with frequency Δω.
1.3 1.2 Asymmetric coupled oscillators (k1 k2 )
With k1 k k , k2 k k i.e. k ½(k1 k2 ) , k ½(k1 k2 ) :
k k' k' k k k' k' 1 Coefficient matrix K k' k2 k' k' k k k'
2 The eigenvalues e mω of K are derived from the secular equation, with unit matrix I :
k k'e k k' det(K eI) 0 k' k k'e k or (k k'e) 2 k 2 k'2 e 2 2(k k')e k 2 2kk'k 2 0 ,
2 2 2 to mω e k k' k' k .
The splitting of the eigenvalues e+ − e− increases with increasing asymmetry Δk of the configuration, shown for the example k 0.8, k' 0.2 :
The eigen-values repel each other near Δk = 0.
1.4 a) General procedure:
Equation of motion mx˙˙ Kx
K is symmetric (actio = reactio), can be diagonalized such that RT KR D with diagonal matrix D, with the eigenvalues ei on its diagonal, and with orthogonal matrix R: RR T I , with transposed matrix RT.
Because of KR RD :
The rows ri of the transformation matrix R then are the eigenvectors of K: Kri ei ri .
This diagonalization is achieved by a transformation of coordinates x Rq , and q R T x .
New equation of motion mq˙˙ mR T ˙x˙ RT K x expanded with R R T I to mq˙˙ RT K(RRT )x (R T KR)(RT x) gives mq˙˙ Dq with oscillatory solutions q(t), which can be back-transformed to give x(t) Rq(t) .
1st Example: Symmetric coupled oscillator
x1 1 q q 1 1 1q x Rq , x2 2 q q 2 1 1q and the rows of R contain the coefficients of x with respect to the basis q.
1.5 2nd Example: Asymmetric coupled oscillator
This corresponds to the general 2-dim case: RT KR D or KR RD , with
k k k' k' k k' k'2 k 2 0 K and D 2 2 k' k k k' 0 k k' k' k
On both sides of KR RD we can shorten out the common term (k k')I.
cosφ sin φ The most general orthogonal 2-dim matrix is R sin φ cosφ
To obtain the variable φ we need to calculate only one element of the above matrix equation: k k' c s c s1 0 1 2 e k' k s c s c0 1 STATE MIXING 1 2 2 with abbrev. c cosφ, s sin φ, e e e 2 k' k . 0.8 The (1,1)-element for instance, gives
1 2 k cosφ k'sin φ 2 ecosφ s o
c 0.6 1 2 2 2 2 or (k 2 e) cos φ k' (1 cos φ) d n a
0.4 2
which, with the asym. parameter k / k', n i s
0.2 2 1 leads to cos φ 2 1 2 1 4 2 0 2 4 2 1 ξ = k k' and sin φ 2 1 2 1
1.6 CHECK of RT KR D : Clearx x x dk k'; a ; 2 1 x cos 1a2 ; sin 1 a2 ; cos sin cos sin x 1 r ; rt ; matrixk ; sin cossin cos 1 x diag rt.matrixk.r; ApartPowerExpandFullSimplifydiagMatrixForm 1 x2 0 OK 0 1 x2
For a given pendulum, for instance pendulum no. 1 with trajectory x1 (t) q (t)cosφ q (t)sin φ , the mixing coefficients cosφ and sin φ give the relative amplitudes of the normal modes q±(t) with frequencies ω±.
2 2 cos φ and sin φ then are the fraction of energy in the normal modes q+(t), and q−(t). 2 2 1 For the symmetric double pendulum, with Δk = 0, cos φ cos φ 2 .
1.7 b) Motion of the asymmetric coupled oscillator:
Initial conditions, with c cosφ, s sin φ :
x10 1. x(0) 0
c s x10 c q (0) 2. q(0) Rx(0) x10 s c 0 s q (0)
q (0)cos(ωt) c cos(ωt) q (t) 3. q(t) x10 q (0)cos(ωt) s cos(ωt) q (t) c sc cos(ω t) c 2 cos(ω t) s 2 cos(ω t) x(t) RT q(t) x x 4. 10 10 s c s cos(ωt) s c cos(ωt) s c cos(ωt) for: 10% ASYMM. COUPLED OSCILL., k k' 2 3 k 1 k 0.1 k' 0.05 2 2 i.e. k / k = 10% asymmetry between spring constants k1, k2, x ,
1 1 and k'/ k 5% relative coupling strength: x s e d u t i l
p 0 m a
1
0 20 40 60 80 100 time t
1.8 c) Extension to many coupled oscillators:
mx˙˙ Kx
2k k 0 0 0 k 2k k 0 0 matrixk 0 k 2k k 0 ; 0 0 k 2k k 0 0 0 k 2k k 1; SortEigenvaluesmatrixk, GreaterMatrixForm N
0.2679491924311228 1. 2. 3. 3.732050807568877
1.9 c) Various notations
N.B.: The normal modes q± form an orthonormal system: T 2 2 (q q ) cos (t ) dt 1 , and T 0 2 T (q q∓) cos(t ) cos(∓t ∓) dt 0 , for large T. T 0
As the q and q form an orthogonal system, so do the x and x . + − 1 2 x1 q cosφ q sin φ q alternative expression x1 q (q x1 ) q (q x1 ) x2 x q (q x ) x1 general case i j j j i (q x1 ) the same in quantum mechanics: φ
| x1 | q q | x1 | q q | x1 (q x1 ) q | x | q q | x general case i j j j i
| q q | 1 symbolically j j j .
1.10