Quantum Mechanics (Quantum Dynamics) Part 2

2.10 Derive the following expression S ol = m omega 2 sin( omega T) [(x 0 ^ 2 +2x T ^ 2 )cos( omega T)-x 0 x T ] for the classical action for a harmonic oscillator moving from the point x_{0} at t = bullet to the point x_{T} at f = T The Lagrangian for the one dimensional harmonic oscillator is given by mathcal L (x, hat x )= 1 2 m hat x ^ 2 - 4/2 * m * omega ^ 2 * x ^ 2 , (2.81) From the Lagrange equation we have partial mathcal L partial x - d dt partial mathcal L partial dot x =0^ (2.8t) Rightarrow - m * omega ^ 2 * x - d dt (m hat x )=0 Rightarrow tilde x + omega ^ 2 * x = 0 . which is the equation of motion for the system. This can be solved to give x(t) = Accs*omega*t + R * sin omega*t (2.83) with boundary conditions (2.82) (284) x(t = 0) (17) x_{0} = A r T = x_{0} * cos omega*T +B sin omega T Rightarrow B sin omega T=x T -x 0 cos omega T Rightarrow R = (x_{T} - x_{0} * cos omega*T)/(sin omega*T) (2.85) 54 So x(t) = x_{0} * cos omega*t + (x_{T} - x_{0} * cos omega*T)/(sin omega*T) * sin omega*t = (x_{0} * cos omega*t * sin omega*T + x_{T} * sin omega*t - x_{0} * cos omega*T * sin omega*t)/(sin omega*T) = (x_{T} * sin omega*t + x_{0} * sin omega * (T - t))/(sin omega*T) = dot x(t) = (x_{T}*omega * cos omega*t - x_{g}*omega*ccs*omega(T - t))/(sin omega*T) . (2.86) (2.87) With these at hand we have = int 0 ^ T dt mathcal L ( dot x , dot x )= int 0 ^ T dt( 1 2 m dot x ^ 2 - 1/2 * m * omega ^ 2 * x ^ 2 ) = int 0 ^ T dt[ 1 7 m d dt (x dot x )- 1 7 mx overline x - 1/7 * m * omega ^ 2 * x ^ 2 ] = - 1/2 * m * integrate 1 dt from 0 to T * x [ hat x + omega^ 2 x]+ m 2 x overline x | 0 ^ T = m 2 m 2 [x(T)ir(T)-. x(0) * ir(0) ] = m/2 * [(x_{T}*omega)/(sin omega*T) * (x_{T} * cos omega*T - x_{0}) - (x_{0}*omega)/(sin omega*T) * (x_{T} - x_{0} * cos omega*T)] = m omega 2 sin omega 0 omega T [x_{T} ^ 2 * cos omega*T - x_{0}*x_{2} - x_{0}*x_{1} + x_{v} ^ 2 * cos omega*T] = (m*omega)/(2sin omega*T) * [(x_{T} ^ 2 + x_{0} ^ 2) * cos omega*T - 2x_{0}*x_{T}] (2.88) 2.11 The Lagrangian of the single harmonic oscillator is mathcal L = 1 2 m dot x ^ 2 - 1/2 * m * omega ^ 2 * x ^ 2 (a) Show that langle w b t b [x a t a rangle= exp [ iS ad hbar ]G(0,t b ;0,t a ) where \mathcal{S}_{d} is the action along the classical path x cl from (x_{a}, l_{a}) to (x_{b}, t_{b}) and Gis G(0, t_{k}; 0, t_{a}) = 2. QUANTUM DYNAMICS 63 2.13 (a) Verify the relation [Pi_{i}, Pi_{j}] =( i hbar e c ) epsilon ijk B k where vec Pi equiv m vec f dt = vec p - (eA)/c and the relation m d vec x d * t ^ 2 = d vec Pi dt =e[ vec E + 1 2c ( d vec x dt * vec B - vec B * d vec x dt )] (b) Verify the continuity equation partial rho partial t + vec nabla ^ prime * vec rho = 0 vec j =( hbar m ) exists( psi^ * vec nabla ^ prime psi)- (e/(mc)) vec A | psi|^ 2 . with vec j given by (a) We have [Pi_{i}, Pi_{j}] = [p_{i} - (e*A_{i})/e, p_{j} - (e*A_{j})/e] = - e/e * [mu_{i}, A_{j}] - e/c * [A_{i}, p_{j}] = i hbar epsilon c partial A j partial x i - i hbar epsilon c partial A i partial x j = i hbar epsilon c ( partial A i partial x i - partial A i partial x j ) =( i hbar epsilon c )z ijk B k , (2.114) We have also that (d*x_{i})/(dt) = 1 t hbar [x i ,H]= 1 i hbar [x i , overline Pi ^ 2 2m +e phi]= 1 i hbar [x i , overline Pi ^ 2 2m ] = 1 i hbar2m \ \ x i ,0 j ]u j + Pi j \ x i , Pi j ]\ = 1 i hbar2m \{[x_{i}, p_{j}] * Pi_{j} + Pi_{j}[x_{i}, p_{j}]\} = 2i hbar 2i hbar m Pi j delta ij = Pi i m Rightarrow 64 1 = 2m +00 [n]+[['u'ul 'n+'n ['u"]} H The C 2mih 2me 2m²c C (B+, B)+ E εδό +(×)-(×)]→ dz (2.115) (b) The time-dependent Schrödinger equation is a (10:1)=(20,10:1)=( 1 2m 1 2m +(デ)(1) 1 2m ドア+(-)(1)+() A(P) 2m +000,10;1) 0,1)+c)(at) (7.1) 7)(1) ()()()(7.1)+(2)(1) + 1 = Multiplying the last equation by we get a