Propagation of a Gaussian Wave Packet

Propagation of a Gaussian wave packet

February 15, 2016

We study the time evolution of an initially Gaussian pulse.

1 Gaussian integrals

Before starting our example, we show how to ﬁnd the integral of a Gaussian curve,

∞ 2 2 I = dx e−(x−x0) /2σ ˆ −∞ First, change the integration variable to x − x ξ = √ 0 2σ so that ∞ √ 2 I = 2σ2 dξ e−ξ ˆ −∞ Now square I,

∞ ∞ 2 2 −ξ2 −ξ2 I = 2σ dξ e 1 dξ e 2 ˆ 1 ˆ 2 −∞ −∞ ∞ ∞ − ξ2+ξ2 = 2σ2 dξ dξ e ( 1 2 ) ˆ 1 ˆ 2 −∞ −∞

2 2 2 and change to polar coordinates, where ρ = ξ1 + ξ2 and dξ1dξ2 = ρdρdϕ. Then

2π ∞ 2 I2 = 2σ2 dϕ ρdρ e−ρ ˆ ˆ 0 0 ∞ 2 = 4πσ2 ρdρ e−ρ ˆ 0

2 ∞ −λ and another change of variable to λ = ρ gives a simple exponential, with 0 dλ e = 1, ´ ∞ I2 = 2πσ2 dλ e−λ ˆ 0 = 2πσ2

1 Therefore, ∞ 2 2 √ dx e−(x−x0) /2σ = 2πσ2 ˆ −∞

2 Initial conditions and the mode amplitude

We begin with an initially Gaussian pulse

2 2 u (x, 0) = e−x /2L eik0x

with zero initial rate of change, ∂u (x, 0) = 0 ∂t We use these initial data to ﬁnd A (k), then use A (k) to ﬁnd u (x, t). The solution for A (k) is found by integrating over the initial conditions

∞ 1 i ∂u A (k) = √ dx u (x, 0) − (x, 0) e−ikx 2π ˆ ω (k) ∂t −∞

∂u Substituting for u (x, 0) and ∂t (x, 0) gives a Gaussian integral,

∞ 1 2 2 A (k) = √ dx e−x /2L eik0xe−ikx 2π ˆ −∞

For the integral of a Gaussian, see the ﬁnal section. integral, we add and subtract a constant to complete the square in the exponent,

1 1 h 2 2i − x2 − 2iL2 (k − k ) x = − x2 − 2iL2 (k − k ) x + iL2 (k − k ) − iL2 (k − k ) 2L2 0 2L2 0 0 0 1 2 1 = − x − iL2 (k − k ) − L2 (k − k )2 2L2 0 2 0 Then, performing the Gaussian integral,

∞ ∞ 2 2 1 2 2 1 2 2 −x /2L −ikx ik0x − (x−iL (k−k0)) − L (k−k0) dxe e e = dx e 2L2 2 ˆ ˆ −∞ −∞ ∞ 1 2 2 1 2 2 − L (k−k0) − (x−iL (k−k0)) = e 2 dx e 2L2 ˆ −∞ √ − 1 L2(k−k )2 = 2πL2e 2 0

Therefore, the mode amplitudes follow a Gaussian as well,

1 √ 1 2 2 [iL (k−k0)] A (k) = √ 2πLe 2L2 2π 2 − L (k−k )2 = Le 2 0

2 3 Dispersion relation

The time dependence of the wave packet is now given by

∞ 1 1 h i u (x, t) = √ dk A (k) ei(kx−ωt) + A∗ (k) e−i(kx−ωt) 2 2π ˆ −∞ ∞ 2 1 L − L (k−k )2 i(kx−ωt) −i(kx−ωt) = √ dke 2 0 e + e 2 2π ˆ −∞ ∞ 2 L − L (k−k )2 i(kx−ωt) = √ Re dke 2 0 e 2π ˆ −∞ where ω = ω (k). Therefor, to continue, we need the form of of the dispersion relation, ω (k). Consider the case of high- frequency waves in plasma, for which we found q 2 2 2 ω = k c + ωp

Now we let ωP kc so that we may expand the square root,

2 c 2 ω ≈ ωP 1 + 2 k 2ωP 1 ≈ ω 1 + l2 k2 P 2 P

c where we deﬁne the plasma wavelength, lP = . This gives our dispersion relation. The group velocity is ωP dω v = = ω l2 k g dk P P so the group velocity evaluated at the center of the packet is

2 v0 ≡ ωP lP k0 and we may write 1 ω ≈ ω + kv P 2 g Notice that the approximation means that v 1 g = l2 ω k c 2c P P 1 = kc 2ωP 1

so the group velocity is much less than the speed of light. Notice that this form also works for a Klein-Gordon matter wave when the momentum due to mass is much greater than the wave momentum, mc ~k. Expanding, r m2c4 ω = + k2c2 ~2

3 r mc2 2k2c2 = 1 + ~ ~ m2c4 mc2 1 2 ≈ 1 + ~ k2 ~ 2 m2c2 1 = ω 1 + λ2k2 C 2 1 = ω + kv C 2 g

c where λC = ~ is the Compton wavelength and ωC = the corresponding frequency. We continue below mc λC with a wave packet in plasma.

4 Time evolution of the waveform

1 With this dispersion relation, ω = ωP + 2 kvg, the full time evolution of the wave is given by

∞ 2 L − L (k−k )2 i kx−ω + 1 kv t u (x, t) = √ Re dke 2 0 e ( P 2 g ) 2π ˆ −∞ Substititing for ω we have another Gaussian integral,

∞ 2 L − L (k−k )2 i kx− ω 1+ 1 l2 k2 t u (x, t) = √ Re dk e 2 0 e ( ( P ( 2 P )) ) 2π ˆ −∞ ∞ 2 L L 2 i 2 2 = √ Re dk exp − (k − k0) + ikx − iωP t − ωP l k t 2π ˆ 2 2 P −∞ Expanding the exponent,

L2 i L2 2 2 i 2 − (k − k )2 + ikx − iω t − ω l2 k2t = − (k − k )2 − ikx + iω t + ω l2 k2t 2 0 P 2 P P 2 0 L2 L2 P 2 L2 P P L2 2ix 2 iω l2 t = − k2 − 2kk + k2 − k + iω t + P P k2 2 0 0 L2 L2 P L2 L2 iω l2 t ix L2 2 = − 1 + P P k2 − 2 k + k − k2 + iω t 2 L2 0 L2 2 0 L2 P we complete the square,  2  2 2 r 2 ix ix iωP lP t 2 ix iωP lP t k0 + L2 k0 + L2 1 + k − 2 k0 + k = k 1 + − − 2 2  2 q 2  q 2  L L L iωP lP t iωP lP t 1 + L2 1 + L2 2 2 ix ! ix 2 iωP lP t k0 + L2 k0 + L2 = 1 + 2 k − 2 − 2 L iωP lP t iωP lP t 1 + L2 1 + L2 Two constant exponentials,

" 2 2 ix 2 # " 2 2 # L 2 2 L k0 + L2 1 2 2 1 k0L + ix exp − k0 + iωP t + 2 = exp − k0L + 2iωP t + 2 iωP l t 2 2 2 L 2 P 2 2 L + iωP lP t 1 + L2

4 come out of the integral, while the remaining Gaussian integral gives

∞  2 ix ! √ 1 2 2 k0 + L2 2π dk exp − L + iωP lP t k − 2  = iωP l t p 2 2 ˆ 2 1 + P L + iωP l t −∞ L2 P

Combining these,

" 2 # L 1 1 k L2 + ix u (x, t) = Re√ exp − k2L2 + 2iω t + 0 2 0 P 2 L + ilP ct 2 2 L + ilP ct where we have used lP ωP = c. We easily check that at t = 0 we recover the initial state,

" 2 # 1 1 k L2 + ix u (x, 0) = Re exp − k2L2 + 0 2 0 2 L2 " 2 # 1 1 k L2 + ix = Re exp − k2L2 + 0 2 0 2 L2

2 2 = e−x /2L eik0x

5 Finding the real part

Finally, we ﬁnd the real part of the waveform. Separating the real and imaginary parts of the argument of the exponential,

2 2 2 2 4 2 2 1 2 2 1 k0L + ix 1 2 2 1 L − ilP ct k0L + 2ixk0L − x − k0L + 2iωP t + 2 = − k0L + 2iωP t + 4 2 2 2 2 2 L + ilP ct 2 2 L + lP c t 2 4 2 2 2 2 1 2 2 1 k0L L + 2xk0L lP ct − x L = − k0L + 4 2 2 2 2 2 L + lP c t 4 2 2 4 i −2xk0L + lP x ct + k0L lP ct −iωP t + 4 2 2 2 2 L + lP c t The real part of this may be written as

2 4 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 1 k0L L + 2xk0L lP ct − x L 1 −k0L lP c t + 2xk0L lP ct − x L − k0L + 4 2 2 2 = 4 2 2 2 2 2 L + lP c t 2 L + lP c t 2 1 L 2 = − 4 2 2 2 (x − k0lP ct) 2 L + lP c t

Recalling that the group velocity, evaluated at the central wave number, is v0 = k0lpc, the real part of this has the form of a Gaussian with a time-dependent mean, ! (x − v t)2 exp − 0 2σ2 (t)

where r l2 σ (t) = L2 + P c2t2 L2

5 2 For the imaginary part, consider the limit as t becomes large, lP ct L near the center of the Gaussian, x ≈ v0t,

4 2 2 4 4 2 i −2xk0L + lP x ct + k0L lP ct i k0L (−2x + v0t) + x lP ct −iωP t + 4 2 2 2 → −iωP t + 4 2 2 2 2 L + lP c t 2 L + lP c t 1 v2 ≈ −i 1 − 0 ω t 2 c2 P

≈ −iωP t since the group velocity is much less than c. Finally, the amplitude of the exponential is L 1 √ = 2 q L + ilP ct lP ct 1 + i L2 1 = q 2 l ct l c2t2 i tan−1 P P L2 1 + L4 e 1 i tan−1 lP ct = e L2 q 2 2 2 lP c t 1 + L4 so at large t,

2 L L iπ √ ≈ e 2 2 L + ilP ct lpct Combining the results for large t,

2 2 ! L iπ (x − v0t) 2 −iωP t u (x, t) = Re e e exp − 2 lpct 2σ (t)

2 2 ! L (x − v0t) = exp − 2 sin ωP t lpct 2σ (t)

The peak of the wave moves with velocity v0 to the right, while the amplitude decreases linearly with time. The whole oscillates with frequency ωP .