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 mω2 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 −mω0x /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 = −ω0X + F/m (3)

The corresponding wave function is

2 ˙ Ψ(t) = e−mω0(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 mω2X2 ! 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 ω0t) with X0 = F0/mω0. The action is mω X2 A = − 0 0 sin(2ω t) (7) 4 0

This has to be augmented by ω0t/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 † † coupled to the harmonic oscillator by the term F0xcˆ cˆ. The Greens function hcˆ(t)ˆc (0)i is given by hcˆ(t)ˆc†(0)i = hΨ(0)|Ψ(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 ! ωo F −iωot hΨ(0)|Ψ(t)i = exp −i t + 3 (e − 1) (9) 2 2mω0 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 ∂ ! ˆb† = √ x − (10) 2 ∂x 1 ∂ ! ˆb = √ x + . 2 ∂x

Here we simply the formulas by using units where m = 1 and ω0 = 1. The Hamiltonian is given by 1 F Hˆ = ˆb†ˆb + + √ (ˆb† + ˆb). (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 † n † z (ˆb ) |zi = ezˆb |0i = X |0i. (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 ˆb0 = √ ˆb − X/ 2. This is applied to the power series expansion in Eq. (12) using the relation ˆb(ˆb†)n|0i = n(ˆb†)n−1|0i to obtain √ ˆb0|zi = (z − X/ 2)|zi (13) √ 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 ˆn0 ˆ† n out using the relation h0|b (b ) |0i = n!δn0,n. The result for the displaced ground state is

2 !n √ √ 1 X 2 hX/ 2|X/ 2i = X = eX /2. (14) n n! 2 We can now derive the time dependence of the wave function, applying the the formula Hˆ (ˆb†)n|0i = (n + 1/2)(ˆb†)n|0i to the power series expansion. The result is

√ e−int X !n e−iHtˆ |X/ 2i = e−it/2 X √ (ˆb†)n|0i. (15) n n! 2 The Green’s function is obtained by taking the overlap with the initial state. Remembering the normalization, the result is √ √ hX/ 2|e−iHt|X/ 2i √ √ = exp(−it/2 + X2/2(e−it − 1)). (16) hX/ 2|X/ 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).

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