We want to use the Laplace transform in order to construct the prototype of the Feynman propagator. The idea is to describe mathematically the response of a physical system to an external force which is switched on at the initial timet= 0.
The causal initial-value problem for the harmonic oscillator.As a simple example, let us consider the following initial-value problem
m¨x(t) =−κ0x(t) +F(t), t >0,
x(+0) =x0, x(+0) =˙ v0 (2.18)
along with x(t) = 0 for all t < 0. Here, the function x = x(t) describes the motion of a classical particle of mass m >0 on the real line under the influence of
• the repulsive force−κ0xwith the coupling constantκ0>0,and
• the external forceF which is switched on at the initial timet= 0,that is, F(t) = 0 for all t <0. We assume that the given functionF : R→ Ris continuous on the time interval [0,∞[.
We are given the initial positionx0and the initial velocityv0 of the particle at timet= 0.We are looking for a solutionx:R→Rwhich is differentiable twice on the open time interval ]0,∞[. In addition, we postulate that the functions xand ˙xcan be continuously extended to the closed time interval [0,∞[.We setω0:=
κ0/m.
Theorem 2.3 The initial-value problem (2.18) has the unique solution x(t) =θ(t) x0cosω0t+ v0
ω0 ãsinω0t+ 1 mω0
t 0
sinω0(t−τ)ãF(τ)dτ
for all timest∈R.
Proof.To simplify notation, setm:= 1.
(I) Existence. Lett >0.Differentiation with respect to timetyields
˙
x(t) =−ω0x0sinω0t+v0cosω0t+ t
0
cosω0(t−τ)ãF(τ)dτ.
Note that the differentiation of the integral yields the additional term sinω0(t−t)ãF(t) which vanishes. Furthermore,
¨
x(t) =−ω20x0cosω0t−ω0v0sinω0t−ω0
t 0
sinω0(t−τ)ãF(τ)dτ +F(t).
Hence ¨x(t) =−ω20x(t) +F(t) ift >0.
(II) Uniqueness. Suppose that the two functions x1, x2 : R → R are solutions of (2.18). Define x(t) := x1(t)−x2(t). The function x : R → R satisfies then problem (2.18) in the special case where x0 = v0 = 0 and F(t)≡0.By a standard result, this impliesx(t)≡0. 2
The retarded Green’s function. Set G(t) :=
sinω0t
mω0 if t≥0, 0 if t <0.
If x0 = v0 = 0, that is, the particle rests at time t = 0, then the solution from Theorem 2.3 can be written as
x(t) = ∞
−∞G(t−τ)F(τ)dτ for all t∈R. (2.19) Synonymously, the functionG is called
• the response function,
• the Feynman propagator, and
• the retarded Green’s function
of the causal initial-value problem (2.18) for the harmonic oscillator on the real line. In terms of physics, the function G describes the response of the harmonic oscillator to the action of the external force F which is switched on at time t = 0. Here, the harmonic oscillator rests on the time interval ]− ∞,0].Observe that the construction of the functionG critically depends on causality, that is, this function refers to the behavior of the harmonic oscillator in the future.
Response functions play a fundamental role in quantum field theory.
We will study this in Chap. 14.
The Dirac delta function.Physicists like to use the Dirac delta func- tionδ =δ(t) in order to introduce the Green’s functionG in a formal, but elegant way. To this end, they choose the external force
F(t) :=δ(t) for all t∈R
which only acts at the initial time t = 0. Using the characteristic property (11.19) of the Dirac delta function on page 590, it follows from (2.19) that
x(t) = ∞
−∞G(t−τ)δ(τ)dτ =G(t) for all t∈R.
Intuitively, the retarded Green’s functiont→ G(t) describes the motion of a harmonic oscillator on the real line
• which rests on the time interval ]− ∞,0],and
• which starts to move under the influence of a large external force that acts precisely at the time pointt= 0.
Observe that the Dirac delta function is not a classical function, but a gener- alized function also called distribution. The rigorous theory of distributions will be studied in Chap. 11.
Application of the Laplace transform to the harmonic oscillator.
We want to show how the solution from Theorem 2.3 can be obtained by using the Laplace transform. To display the elegant basic idea as clearly as possible, we restrict ourselves to a formal argument by ignoring the range of validity of the following formulas. Moreover, in order to simplify notation we setm:= 1 and := 1. Then E = ω. We will use the basic rules (i), (ii), (iii) for the Laplace transform considered on page 93.
To begin with, suppose that the functionx = x(t) is a solution of the causal initial-value problem (2.18) above. By the derivative rule (ii), the Laplace transformLxsatisfies the equation
−ω2Lx+ iωx0−v0+ω02Lx=LF.
Hence
Lx=− iωx0
ω02−ω2 + v0
ω02−ω2+ LF ω20−ω2.
This means that the Laplace transform Lx of the motion possesses a very simple structure in the frequency space. This solution depends critically on the function
R(ω) := 1 ω20−ω2
which is called the response function of the harmonic oscillator in the fre- quency space. The singularities of the function R at the points ω = ±ω0
correspond to the eigenoscillationst→e±iω0tof the harmonic oscillator. We will see at the end of the proof that the functionRis the Laplace transform of the retarded Green’s function G. It remains to reconstruct the function x=x(t) from its Laplace transform. Let us discuss this.
Using partial fractions and the wave rule (i), we get iω
ω2−ω02 =1 2
i
ω−ω0 + i ω+ω0
=L e−iω0t+ eiω0t 2
=L(cosω0t).
Similarly, 1
ω20−ω2 = 1 2ω0
1
ω+ω0 − 1 ω−ω0
=L eiω0t−e−iω0t 2iω0
=L sinω0t ω0
.
Finally, the convolution rule (iii) tells us that 1
ω20−ω2 ã LF =L sinω0t ω0 ∗F
.
Summarizing, for all timest≥0, x(t) =x0cosω0t+ v0
ω0 ãsinω0t+ t
0
sinω0(t−τ)
ω0 ãF(τ)dτ.
This is precisely the solution from Theorem 2.3 on page 95.