Notes 19: The adiabatic approximation Consider a mechanical system, e.g. a simple harmonic oscillator or a simple pendulum, that has a characteristic ‘internal’ time scale, τ i , that depends on a system parameter, e.g. the spring constant or the pendulum length. If the system parameter changes on an ‘external’ time scale, τ e , that is long compared to τ i then the system will slowly evolve, maintaining its fundamental characteristics, e.g. simple harmonic motion, but with a change of internal time scale. In this case, the change is an adiabatic process. In contrast, if the system parameter is changed rapidly, e.g. the spring is cut, then the characteristics of the system will change dramatically in a diabatic process. The figure below illustrates the difference between an adiabatic change and a diabatic change. The spring constant of a simple harmonic oscillator changes in such a way that the period of oscillation increases exponentially on fixed time scale, τ e. Initially, the period of oscillation τ i is small compared to τ e. The system evolves adiabatically by oscillating with longer period and larger amplitude until ττie , at which oscillation ceases and the evolution becomes diabatic. 20 10 x 0 -10 -20 0 500 1000 1500 t Thus, provided ττie , we can analyze the system by first treating the system parameters as constant, and then let the system parameters evolve slowly with time. The quantum analog can be cast as a theorem. 1 The Adiabatic Theorem Previously, we have considered situations in which a time-dependent part of a Hamiltonian is small compared to the time-independent part of the Hamiltonian. Here we consider the situation in which the time-dependent part of the Hamiltonian cannot be treated as a perturbation to a time-independent Hamiltonian, but instead varies slowly. Consider a system that has one or more parameters and these parameters change slowly with time. We will assume that the system Hamiltonian always has non- degenerate discrete states. At some instant, let the Hamiltonian describing the system be Hi and at a later th time, Hf. The adiabatic theorem states that if the system is initially in the n stationary state of Hi, then at th the later time, it will be in the n stationary state of Hf. The proof of this theorem gives insight into the conditions required for it to be valid. At time t = 0, let the th l stationary state of Hi (treated as time-independent) have wave function ψ l .At later time, the wave function will be −itωl Ψ=ll(te) ψ , (19.1) where ωll= E . If the Hamiltonian changes with time, the eigenfunctions and eigenvalues become time dependent HttEtt( )ψψl( ) = ll( ) ( ), (19.2) but still form an orthonormal set ψψk(tt) l ( ) = δkl . (19.3) We can then write the solution of the time-dependent Schrödinger equation ∂ i Ψ=( t) Ht( ) Ψ( t), (19.4) ∂t as a linear combination of the eigenfunctions itθl ( ) Ψ=(t) ∑ cll( t)ψ ( te) . (19.5) l −itθ ( ) It is convenient to explicitly include the dynamic phase factor e l , where t θω= − ′′ ll(t) ∫ ( t) dt , (19.6) 0 as it would be present even if the Hamiltonian were time independent. Substituting the expression in equation (19.5) into equation (19.4), we get 2 θθ iill i∑∑( cllψ++ c ll ψ ic θψ lll) e = cHl( ψ l) e. (19.7) ll Because of equations (19.2) and (19.6), two terms in equation (19.7) cancel, and we find iθl ∑(cllψψ+= ce ll ) 0. (19.8) l Taking the inner product with ψ n , we get ii(θθln−−) (θθln) ccecn=−=−−∑∑ lψψ nlnψψ nn ce l ψψ nl . (19.9) l ln≠ th Now suppose as stated in the theorem that the system starts out in the n state, i.e. cn(0) =1 and cl(0) = 0 for l ≠ n. If the second term on the right hand side could be neglected, then ccn= − nψψ nn . (19.10) Now ψψnn(tt) ( ) =1implies ψψnn+= ψψ nn 2Re ψψnn = 0. (19.11) Hence ψψnn is pure imaginary. Then itγ n ( ) ctnn( ) = c(0,) e (19.12) where the geometric phase is t ∂ψ γψ(t) = i( t′) n ( t ′′) dt . (19.13) nn∫ ′ 0 ∂t 22 Since ctnn( ) = c(0) = 1, we must have cl(t) = 0 for l ≠ n. Then from equation (19.5) itθn( ) ititγθnn( ) ( ) Ψ=(t) cnn( t)ψψ( te) = n( te) e . (19.14) Except for a couple of phase factors, the wave function is the same, and so the system remains in the nth state. Thus for the adiabatic theorem to be valid, we need to find the conditions for which i(θθln− ) ∑celψψ nl (19.15) ln≠ 3 can be neglected. If the states are always non-degenerate, they will keep their energy eigenvalue order. i(θθ− ) Then e lnwill be oscillatory and its average over a long time will be close to zero. The dominant contribution to the expression in equation (19.15) will come from the terms ψψnl . (19.16) Differentiating equation(19.2), we find HHEEψψψψl+=+ l ll ll, (19.17) and on taking the inner product with ψ n , ψψnlH+=+ ψψ nl H EE lnllnl δ ψψ. (19.18) But H is an Hermitian operator, so that ψψnHE= nn. Then for l ≠ n, ψψnlH ψψnl = . (19.19) EEln− −1 th The internal time scale for the system is τωin= , where ωn is the frequency associated with the n energy eigenvalue. The external time scale is EEln− ∆ω τ e = ≈ , (19.20) H H where ∆ω is the smaller of the intervals between the neighboring state frequencies. The required condition for validity of the adiabatic theorem is then (approximately) H <∆ωωn . (19.21) Problem 11.36. The one-dimensional harmonic oscillator is subject to a driving force of form mω 2 ft( ). The Hamiltonian is 22 ∂ 1 22 2 Ht( ) =−+−mxωω mxft( ). (19.22) 22mx∂ 2 Assume that the driving force is turned on at t = 0. This problem can be solved exactly in both classical and quantum mechanics. (a) Determine the classical position of the oscillator, assuming it started at rest from the origin. Solution: The classical equation of motion is 4 x+=ωω22 x ft( ). To find the general solution, we start by looking for the solution of the homogeneous equation xx+=ω 2 0, which is xth ( ) = Acosωω tB + sin t , where A and B are constants. We next consider the inhomogeneous equation with a delta-function source term x+=−ωδ2 x( tt′), For tt< ′ and tt> ′, the solutions are of the same form as for the homogeneous equation. Thus, we can write A<<cosωω tB+< sin t , tt′ , xt( ) = A>>cosωω tB+> sin t , tt′ . where t′ > 0. Since the system is at rest at the origin before the driving force is switched on, we see that AB<<= = 0.The other two coefficients are related by continuity and jump conditions at tt= ′. These are xt><( ′′) =⇒+= xt( ) A >cosωω t ′ B > sin t ′ 0, and xt><( ′′) − xt( ) =⇒−1ωω( A> sin t′′ + B > cos ω t) = 1. Therefore sinωωtt′′ cos AB=−=,. >>ωω We have shown that the Green’s function for this problem is 1 0,tt< ′ , Gtt( , ′) = ω sinω (tt−>′′) , tt . Thus the solution for the classical oscillator is ∞ t = ′ ′′= ωω ′− ′′ xc ( t) ∫∫ f( t) G( t, t) dt f( t)sin( t t) dt . −∞ 0 5 (b) Show that the solution to the time dependent Schrödinger equation for this oscillator assuming it started out in the nth state of the undriven oscillator is im1 ω 2 t Ψ =ψω − − −+ ′ ′′ n( x, t) nc( x x)exp mx c x x c n t∫ x c( t) f( t) dt . 220 Solution: To solve the time dependent Schrödinger equation, we first change the variables from (x, t) to ( xt′′, ) where x′′=−= xxtc ( ),. t t (We are now solving the TDSE in the co-moving frame.) The partial derivatives in the two frames are related by ∂Ψ ∂Ψ ∂xt′′ ∂Ψ ∂ ∂Ψdx ∂Ψ =+=−+c , ∂t ∂∂ x′′ t ∂∂ t t ∂ x ′ dt ∂ t ′ ∂Ψ ∂Ψ ∂xt′′ ∂Ψ ∂ ∂Ψ =+=, ∂x ∂∂ xx′′′ ∂∂ tx ∂ x so that 22 ∂Ψ ∂Ψ dx ∂ 1 2 −c =− +ωω22′ +− ′′ + Ψ ih ih 2 m( x xcc) m( x x) f( t ) . ∂∂t′′ x dt22 m ∂ x ′ xg( t′) We then look for a solution of form Ψ=Φ( xt′′,,) e which leads to ∂Φ dg∂Φ dx 22∂ Φ ∂Φ +′ + Φ −c =− + +Φ 2 i ixx( c ) i 2 2 g g ∂t′dt ′′∂ x dt2 m ∂∂ x ′ x ′ 1 2 +mωω22( xx′ +−) m( xxft ′′ +) ( ) Φ. 2 cc Now we choose g so that the terms involving ∂Φ ∂x′ cancel, im dx g = c . dt′ Then 2 ∂Φ22 ∂ Φ 11dx d2 x =− +cc ++′′′ωω22 +− + Φ i 22m( xxcc) ( xx) ft( ) . ∂∂t′′22 m x dt ′ 2 dt′ Using the differential equation for xc to eliminate the acceleration term, this becomes 6 2 ∂Φ22 ∂ Φ 1 dx =−+ c −+ωω22 22′ Φ i 2 m xxc . ∂∂t′′22 m x dt ′ Because 2 dx ′ c 22 qt( ) = −ω xc dt′ is a function of t′, the partial differential equation is separable. Let Φ=( xt′′,.) X( xTt ′) ( ′) Then 22 11dT 1d X 122 i −=−+ mq( t′′) mω x . T dt′′22m X dx 2 2 Both sides of this equation must equal the same constant, which we will denote by E. Then 22 dX 1 22 −+mω x′ X = EX , 22m dx′2 which is the time independent Schrödinger equation for the quantum oscillator, and so E can only take the discrete values Enn =ω ( +12) .