<p>PHYSICS IN A NUT-SHELL: TWO-STATE SYSTEMS D. Dubbers, WS 2008/2009</p><p>1. Prologue: Classical coupled vibrations </p><p>Demonstration: Double Pendulum: normal modes, beats</p><p>1.1 Symmetric coupled oscillator a) Free oscillator</p><p>Force law m˙x˙  kx , with spring constant k, has solution x(t)  x0 cos(ω0t  φ) with frequency ω0  k/m .</p><p>From initial cond's x(0), x˙(0) follows x0 , φ The total energy in the system is 1 2 E  2 kx0</p><p>(For mathematical pendulum:</p><p> k  mg/l : ω0  g/l is independent of m) </p><p>1.1 b) normal coordinates and eigenfrequencies</p><p>Spring constants k1  k2  k</p><p> 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</p><p> shorthand mx˙˙  Kx</p><p> x1  k  k'  k'  with x   , K    .  x2    k' k  k'</p><p>These differential equations can be decoupled by taking their sum and their difference, i.e. with 1 normal coordinates q  (x1  x2 ) 2</p><p> this becomes mq˙˙  kq</p><p> mq˙˙  (k  2k') q or mq˙˙  Dq ,</p><p>k 0  q  with diagonal matrix D    and q    . 0 k  2k' q  The solutions are the normal vibrations</p><p> q (t)  q (0)cos(ωt  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.</p><p>1.2 c) Initial conditions</p><p>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</p><p>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(ωt) 2 1 1 q 4. x1,2 (t)  (q  q )  x10[cos(ωt)  cos(ωt)] + 2 2</p><p> q − that is x1 (t)  x10 cos(½(ω  ω )t)  cos(½(ω  ω )t)</p><p> or x1 (t)  x10 cosωt  cos ω t and x2 (t)  x10 sin ωt  sinω t , with mean frequency ω  ½(ω  ω ) and beat frequency ω  ½(ω  ω ) .</p><p>1 2 The total energy E  2 kx0 of the system is exchanged between the two pendula with frequency Δω.</p><p>1.3 1.2 Asymmetric coupled oscillators (k1  k2 )</p><p>With k1  k  k , k2  k  k i.e. k  ½(k1  k2 ) , k  ½(k1  k2 ) :</p><p>k  k'  k'  k  k  k'  k'  1   Coefficient matrix K         k' k2  k'   k' k  k  k'</p><p>2 The eigenvalues e  mω of K are derived from the secular equation, with unit matrix I :</p><p> 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 ,</p><p>2 2 2 to mω  e  k  k' k' k .</p><p>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 : </p><p>The eigen-values repel each other near Δk = 0.</p><p>1.4 a) General procedure:</p><p>Equation of motion mx˙˙  Kx </p><p>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.</p><p>Because of KR  RD : </p><p>The rows ri of the transformation matrix R then are the eigenvectors of K: Kri  ei ri .</p><p>This diagonalization is achieved by a transformation of coordinates x  Rq , and q  R T x .</p><p>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) .</p><p>1st Example: Symmetric coupled oscillator</p><p> x1  1 q  q  1  1 1q  x            Rq ,  x2  2 q  q  2 1 1q  and the rows of R contain the coefficients of x with respect to the basis q.</p><p>1.5 2nd Example: Asymmetric coupled oscillator</p><p>This corresponds to the general 2-dim case: RT KR  D or KR  RD , with</p><p>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 </p><p>On both sides of KR  RD we can shorten out the common term (k  k')I.</p><p> cosφ sin φ  The most general orthogonal 2-dim matrix is R     sin φ cosφ</p><p>To obtain the variable φ we need to calculate only one element of the above matrix equation:  k  k'  c s  c s1 0  1        2 e  k'  k  s c  s c0 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 </p><p>1 2 k cosφ  k'sin φ  2 ecosφ s o</p><p> c 0.6 1 2 2 2 2 or (k  2 e) cos φ  k' (1 cos φ) d n a</p><p> 0.4 2</p><p> which, with the asym. parameter   k / k', n i s</p><p>  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  </p><p>1.6 CHECK of RT KR  D : Clearx x x  dk k'; a  ;  2 1 x cos  1a2 ; sin  1 a2 ; cos sin cos sin x 1 r  ; rt  ; matrixk  ; sin cossin cos  1 x diag  rt.matrixk.r; ApartPowerExpandFullSimplifydiagMatrixForm 1 x2 0  OK 0 1 x2 </p><p>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±.</p><p>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 .</p><p>1.7 b) Motion of the asymmetric coupled oscillator:</p><p>Initial conditions, with c  cosφ, s  sin φ :</p><p> x10  1. x(0)     0 </p><p> c s x10  c q (0) 2. q(0)  Rx(0)      x10       s c 0   s q (0)</p><p>q (0)cos(ωt) c cos(ωt) q (t) 3. q(t)     x10      q (0)cos(ωt)  s cos(ωt) q (t) c  sc 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(ωt)  s c cos(ωt)  s c cos(ωt) 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 ,</p><p>1 1 and k'/ k  5% relative coupling strength: x s e d u t i l</p><p> p 0 m a</p><p>1</p><p>0 20 40 60 80 100 time t</p><p>1.8 c) Extension to many coupled oscillators: </p><p> mx˙˙  Kx</p><p>2k k 0 0 0 k 2k k 0 0  matrixk  0 k 2k k 0 ; 0 0 k 2k k 0 0 0 k 2k k  1; SortEigenvaluesmatrixk, GreaterMatrixForm N</p><p>0.2679491924311228  1.  2. 3. 3.732050807568877 </p><p>1.9 c) Various notations</p><p>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</p><p>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</p><p>| x1  | q  q | x1   | q  q | x1     (q  x1 ) q | x   | q  q | x  general case i  j j j i</p><p>| q  q |  1 symbolically  j j j .</p><p>1.10</p>