Forced harmonic oscillator Notes by G.F. Bertsch, (2014) 1. The time-dependent wave function The evolution of the ground state of the harmonic oscillator in the presence of a time- dependent driving force has an exact solution. It is useful to exhibit the solution as an aid in constructing approximations for more complicated systems. The Hamiltonian to be considered is 1 @2 m!2 H = − + 0 x2 + F x: (1) 2m @x2 2 Here the force F ≡ F (t) is an arbitrary function of time. The (unnormalized) ground state wave function is 2 −m!0x =2 Ψ0 = e : (2) To derive the solution of the time-dependent equation, we are guided by the close connection between the classical and quantum oscillators. We shall try a wave function that has a displacement X(t) and a momentum mX_ (t) determined by the classical equation of motion, ¨ 2 X = −!0X + F=m (3) The corresponding wave function is 2 _ Ψ(t) = e−m!0(x−X) =2+imXx−iΦ (4) Two comments: • The functions X and X_ may or may not be expressible analytically, depending on the form of the force function F (t). However, they can always be expressed as an integral over F using the Green's function for the harmonic oscillator. • We have added a phase angle Φ(t) which is in principle beyond the scope of classical physics. However, it turns out it is almost equal to the classical action A associated with the motion, Z t mX_ 2 m!2X2 ! A(t) = dt − 0 (5) 2 2 We now verify the solution by substituting in the time-dependent Hamiltonian equation, @Ψ HΨ = i : (6) @t 1 The simplest nontrivial case is to take a constant force, F (t) = F0. The classical solution 2 is X(t) = X0(1 − cos !0t) with X0 = F0=m!0. The action is m! X2 A = − 0 0 sin(2! t) (7) 4 0 This has to be augmented by !0t=2 to get the phase angle Φ. The algebra is a bit tedious, so let's speed up the derivation using Mathematica. The following script evaluates the left-hand and right-hand sides of Eq. (6) and then compare them: Psi = E^( -om (x - x0 Cos[om t])^2/2 - I om x0 x Sin[om t]- I om/2 t +I om x0^2/4 Sin[2 om t]) Psi0 = E^(-om (x-x0)^2/2) LHS = Simplify[-1/2 D[D[Psi,x],x] + om ^2 x^2/2 Psi] RHS = I D[Psi,t] Simplify[LHS-RHS] When you run this script, the final answer is zero, since the two sides are equal. A useful application of the result Eqs. (4,7) is to the Green's function for a particle y y coupled to the harmonic oscillator by the term F0xc^ c^. The Greens function hc^(t)^c (0)i is given by hc^(t)^cy(0)i = hΨ(0)jΨ(t)i: (8) The following Mathematica script evaluates the right-hand side of this equation: Psi = E^( -om (x - x0 Cos[om t])^2/2 - I om x0 x Sin[om t]- I om/2 t +I om x0^2/4 Sin[2 om t]) Psi0 = E^(-om (x-x0)^2/2) olp = Integrate[ Sqrt[om/Pi] Psi Psi0,{x,-Infinity,Infinity}, Assumptions -> om > 0] The final answer for this script is 2 ! !o F −i!ot hΨ(0)jΨ(t)i = exp −i t + 3 (e − 1) (9) 2 2m!0 The second term in the exponent is the cumulant term introduced into X-ray absorption theory by Langreth [1]. 2 2. Operator method We can derive Eq. (9) more easily using the harmonic oscillator creation and anhiliation operators. Recall the definition 1 @ ! ^by = p x − (10) 2 @x 1 @ ! ^b = p x + : 2 @x Here we simply the formulas by using units where m = 1 and !0 = 1. The Hamiltonian is given by 1 F H^ = ^by^b + + p (^by + ^b): (11) 2 2 For a constant F , the Hamiltonian is transformed to one with F = 0 by changing variable x ! x0 + X with X = F . We now examine the properties of the coherent state defined by n y n y z (^b ) jzi = ez^b j0i = X j0i: (12) n n! We first verify that a displaced harmonic oscillator ground state can be expressed as a coherent state by applying the annihilation operator to it. The displaced operator is ^b0 = p ^b − X= 2. This is applied to the power series expansion in Eq. (12) using the relation ^b(^by)nj0i = n(^by)n−1j0i to obtain p ^b0jzi = (z − X= 2)jzi (13) p Thus the state is annihilated if we take z = X= 2. The next task is to find the normalization of the coherent state. This is easily carried ^n0 ^y n out using the relation h0jb (b ) j0i = n!δn0;n. The result for the displaced ground state is 2 !n p p 1 X 2 hX= 2jX= 2i = X = eX =2: (14) n n! 2 We can now derive the time dependence of the wave function, applying the the formula H^ (^by)nj0i = (n + 1=2)(^by)nj0i to the power series expansion. The result is p e−int X !n e−iHt^ jX= 2i = e−it=2 X p (^by)nj0i: (15) n n! 2 The Green's function is obtained by taking the overlap with the initial state. Remembering the normalization, the result is p p hX= 2je−iHtjX= 2i p p = exp(−it=2 + X2=2(e−it − 1)): (16) hX= 2jX= 2i 3 This is the same as result as we found in Eq. (9), apart from dimensional factors. [1] D.C. Langreth, Phys. Rev. B 1 471 (1970). 4.