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In order to determine wavefunction, need to know ω (k)

In order to determine wavefunction, need to know ω (k). ω =vk v=c/n, n=1+ ω / ω o , with ω o =1.88E15 rad/s ω =ck/(1+ ω / ω o ) ω 2 + ω o ω -c ω o k=0 Solve quadratic eq. to get ω (k), discard negative solutions. Can now write Aisin(k i x- ω i t). For t=0, A i =1. Graphically measure

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In order to determine wavefunction, need to know ω (k)

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  1. In order to determine wavefunction, need to know ω(k) • ω=vk • v=c/n, n=1+ω/ωo, with ωo=1.88E15 rad/s • ω=ck/(1+ω/ωo) • ω2+ωoω-c ωok=0 • Solve quadratic eq. to get ω(k), discard negative solutions. • Can now write Aisin(kix-ωit)

  2. For t=0, Ai=1 Graphically measure Δx to be 6um, measuring at position where envelope goes to ½ its max For Δk=ko/5=1,260,000 ΔkΔx=7.56

  3. For t=1ps, Ai=1 Graphically measure Δx to be 11um, measuring at position where envelope goes to ½ its max For Δk=ko/5=1,260,000 ΔkΔx=13.8

  4. c) Determine velocity of wavepacket • Center of wavepacket in b) is about 134µm • Center of wavepacket in a) is 0µm • Δt=1ps=1E-12s • v=Δx/Δt=134E-6m/1E-12s=1.34E8=0.45c

  5. D) Analytical solution for vg • vg=dω/dk assume k=ko

  6. Δk=ko/25, t=0

  7. Δk=ko/25, t=0 Graphically measure Δx to be 3.6um, measuring at position where envelope goes to ½ its max For Δk=0.8ko=5,040,000 ΔkΔx=18.144

  8. Δk=ko/25, t=0

  9. Δk=ko/25, t=0.2ps

  10. Δk=ko/25, t=0.4ps

  11. Δk=ko/25, t=0.6ps

  12. Δk=ko/25, t=0.8ps

  13. Δk=ko/25, t=1ps

  14. Measuring ΔkΔx • can’t graphically measure Δx • Why? • Width in k values means that each wave travels with a much different velocity than it neighboring k-value. This mixes up the wavepackets high dispersion • ΔkΔx not really measurable • Therefore can’t measure shift in peak of wavepacket. • Analytical solution for group velocity shouldn’t change, though…

  15. Δk=ko/100, Ai Gaussian t=0 Graphically measure Δx to be 9um, measuring at position where envelope goes to ~½ its max Now, we should assume Δk=2σ=ko/10=630,000 ΔkΔx~6

  16. Δk=ko/100, Ai Gaussian t=0

  17. Δk=ko/100, Ai Gaussian t=0.2ps

  18. Δk=ko/100, Ai Gaussian t=0.4ps

  19. Δk=ko/100, Ai Gaussian t=0.6ps

  20. Δk=ko/100, Ai Gaussian t=0.8ps

  21. Δk=ko/100, Ai Gaussian t=1ps

  22. Δk=ko/100, Ai Gaussian t=1ps Graphically measure Δx to be 12um, measuring at position where envelope goes to ~½ its max Now, we should assume Δk=2σ=ko/10=630,000 ΔkΔx~8

  23. Why does Gaussian hold together better? • Most of the weight lies near ko, so the waves far from ko do not contribute as much, even though they travel with the most different velocity • The group velocity of the wavepacket will be the same as for the other wavepackets, as will the analytical solution

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