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.
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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
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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
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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<
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
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(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
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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) . The other equation
dT 1 i = E + mq( t′) T, dt′ n 2 has solution
im ′= −−ω ′ ′′ T( t) exp in t q( t) dt . 2 ∫
The solution for the TDSE is
im dx im t Ψ =ψ − c −− ωω2′ − 22 ′′ n( xt, ) nc xxt( ) exp xitn∫ xt c( ) xtdt c( ) . dt 2 0
Noting that
tt t t i x 2 tdtxx′′=−= xxdtxtxt ′ −= xxdtxtxt ′ −ω 22 xt′′′′ ft − x t dt, ∫∫c( ) [ cc]0 cc c( ) c( ) ∫ cc c( ) c( ) ∫c( ) ( ) c( ) 00 0 0 the solution can also be written as
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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
(c) Show that the eigenfunctions and eigenvalues of H(t) are
1122 ψnn( xt,,) =− ψ( x f) E n =+− nωω m f . 22
Solution: The Hamiltonian is
22 22 ∂∂1 112 =−+−ωω22 2 =−+−−ω2 ω2 2 Ht( ) 22m x m xft( ) m x ft( ) m f( t). 22mx∂∂22mx 2
By comparing with the ‘unforced’ Hamiltonian, we see that the eigenfunctions and eigenvalues of H(t) are
1122 ψnn( xt,,) =− ψ( x f) E n =+− nωω m f . 22
(d) Show that in the adiabatic approximation, the classical position reduces to xtc ( ) ≈ ft( ). Find the condition for adiabaticity.
Solution: If the driving force changes slowly enough, it can be balanced by the spring force so that the system evolves through equilibrium states. In this case, the solution in the adiabatic approximation is
xtc ( ) = ft( ). (19.23)
Since xc(0) = 0, the driving force must have f(0) = 0. Let
ft( ′′) = f (0) t + so that the exact solution gives
t 1 =ωω′− ′′ += −ω + xc ( t) f(0) ∫ t sin ( t t) dt f(0) t sin t . 0 ω
The adiabatic approximation will be valid if the second term in parenthesis is small compared to the first. It will be after a time τω −1. Another approach is note that the adiabatic solution will be valid if the acceleration is small, i.e. if
ff ω 2 .
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If the driving force is harmonic, then this condition requires than the drive frequency must be much less than the natural frequency of the system, i.e. the external time scale is long compared to the internal time scale.
(e) Confirm the adiabatic theorem for this example.
Solution: The exact solution to the TDSE is as shown in part (b),
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
In the adiabatic approximation, xc is replaced by f. Then
2 t im1 ω θγ Ψ =ψ − − −+ ωψ2 ′′= −iinn nn( xt, ) ( xf)exp mx ff/ n t∫ ftdt( ) n( xfee) , 220 where the dynamic phase is [equation (19.6)]
tt11mω 2 θω=−′′ =−+ ω + 2 ′ nn(t) ∫∫( t) dt n t f dt . 0022 and the geometric phase is
m 1 γ n (t) = x − ff/ . 2
We see that if the system starts in the nth state and evolves adiabatically it remains in the nth state but the center of the probability distribution is at x = f, i.e. it is fixed in the co-moving frame.
Note that equation (19.13) gives that the geometric phase should be zero (it is proportional to the expectation value of the momentum in the co-moving frame). The result here is non-zero but is close to zero if f is small enough (provided x is not too large).
The adiabatic series
The adiabatic approximation can be considered as the first term in a series for the coefficients cl(t) in equation (19.5). Suppose the system is initially in the nth state. In the adiabatic approximation, it remains in the nth state. However, the adiabatic approximation is not exact, and there is a non-zero probability of transitions to other states. We can find the transition probability in the near-adiabatic approximation by using the adiabatic approximation result
itγ n ( ) ctl( ) = δ ln e , (19.24) in the right hand side of the exact equation
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i(θθlm− ) ccm= −∑ lψψ ml e, (19.25) l
to get
itγ n i(θθnm− ) cem= − ψψmn e , (19.26)
which on integration with respect to time gives
t itγ ( ′) itθθ( ′′)− ( t) = − n ψψ′′nm ′ cm( t) c m(0.) ∫ emn( t) ( t) e dt (19.27) 0
As an example of the near-adiabatic approximation, consider the driven harmonic oscillator for which
11mmω 2 t 1 ψψ= − θ =−+ ω + 2 ′ γ= − ≈ nn( x, t) ( x f), n( t) n t∫ f dt, n( t) x f f/ 0. 220 2
The matrix element is
ψψmn =−−− ψ m( xf) ψn′ ( xf) f.
Now
∂ψ n i ψψm( xf−) n′( xf −=) ψ m( x′) ( x ′) = ψψmn( xp ′′) ( x ′) , ∂x′ where p′ is the momentum operator associated with the comoving position x′. We can now use the ladder operators to evaluate the matrix elements. We have
i1 mω mm ωω ψψmp n =− ψm aa+− − ψn =−++n1.δmn,1+ n δmn,1− 2 22
We see that the only possible transitions are to states n-1 and n+1. The transition probabilities can then be obtained from
t mω ′ = − −itω ′ cn−1 ( t) n∫ fe/ dt , 2 0
t mω ′ = + itω ′ cn+1 ( t) n1.∫ fe dt 2 0
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Geometric phase and Berry phase The geometric phase is given by
t ∂ψ γψ(t) = i( t′) n ( t ′′) dt . (19.28) nn∫ ′ 0 ∂t
Suppose the system has only one parameter R that can be time-dependent. Then
Rt( ) t ∂∂ψψdR γψ(t) = i nndt′ = i ψdR. (19.29) nn∫∫′ n 00∂∂R dt R( ) R
We see that if the system returns to its initial state after time T, i.e. RT( ) = R(0,) then the geometric phase is zero. However, the system may have a set of parameters Ri , that can vary with time. In this case, equation (19.29) generalizes to
R(t) γ= ψψ ∇⋅ n(ti) ∫ n Rn dR. (19.30) R(0)
For a closed loop in parameter space, we can write
γ(Ti) = ψψ ∇⋅ dR, (19.31) n ∫ n Rn which is not necessarily zero. This expression gives Berry’s phase, which depends only on the path taken in parameter space and not the time taken to complete the closed loop. In contrast the dynamic phase,
t θω= − ′′ ll(t) ∫ ( t) dt , (19.32) 0 does depend on time. The total phase (or a difference in phase) can be measured experimentally e.g. by exploiting interference between two beams, one of which passes through an adiabatically changing potential but the other does not.
The classic example of Berry’s phase is an electron subject to a magnetic field that has constant magnitude but changing direction. Let the direction of the magnetic field be described by spherical polar angles (θφ,,) so that the unit vector associated with this direction is
B bˆ ==++sinθ cos φ i sin θφ sin jk cos θ . (19.33) B
Then the spin matrix for this direction is
=sinθ cos φ ++ sin θφ sin cos θ . (19.34) Sb S xy SSz
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For the electron, the spin matrices are related to the Pauli matrices and so
Sb=(sinθ cos φσ xy ++ sin θφ sin σσ cos θ) 2 z cosθsin θ cos φ− i sin θφ sin = (19.35) 2 sinθ cos φ+−i sin θφ sin cos θ cosθθ sin e−iφ = iφ . 2 sinθθe − cos
The eigenvectors of this matrix correspond to spin up and spin down in the direction of bˆ. They are
θθ−iφ cos e sin (bb) 22( ) χχ+−= ,.= (19.36) iφ θθ e sin −cos 22
Suppose the system starts out in the spin up direction. The system parameters are the angles (θφ,.) As these change the direction vector traces out a trajectory on the surface of a sphere. To find the Berry (b) phase, we need to evaluate ∇R χ+ . We can use the expression for the gradient operator in spherical polar coordinates
∂∂11 ∂ ∇=rˆ +θˆˆ +φ , (19.37) ∂∂rrθ rsin θφ ∂ to get
θ −sin 0 (b) ˆˆ112 ∇=χ+ θ +φ θ . (19.38) R iφ 2rriφ θ sinθ ie sin e cos 2 2
Then
θ ∗ ˆˆ−sin 0 (bb) ( ) θθiφ θ 2 φ χχ++∇=cose sin + θ R iφ 2 2 2rriφ θ sinθ ie sin e cos 2 (19.39) 2 i sin2 (θ 2) = φˆ . r sinθ
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Now assume that the magnetic field vector sweeps out a closed loop on a sphere of radius r. The Berry phase is
1 sin2 (θ 2) γ (Td) =−⋅φˆ l, (19.40) n ∫ C r sinθ where dl is a path element on the loop, C. This integral can be evaluated by applying Stokes’ Theorem, ∫∫Al⋅dd = ∇× AS ⋅ , (19.41) CS
where S is a surface with boundary C. Now if A = Aφ φ, then
11∂∂ˆ ∇×Ar = (sinθ Aφφ) ˆ − ( rA )θ. rsinθθ∂∂rr
We find the Berry’s phase is
1sin2 (θ 2) 11 1 γ T=−∇× φˆ ⋅=−ddˆ ⋅=−Ω, (19.42) n ( ) ∫∫S2 rS SSrrsinθ 2 2 where Ω is the solid angle subtended by the closed loop at the origin, i.e. the solid angle swept out by the magnetic field vector.
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