Recasting a problem in Fourier space Amplitude as a function of time - - PowerPoint PPT Presentation

recasting a problem in fourier space
SMART_READER_LITE
LIVE PREVIEW

Recasting a problem in Fourier space Amplitude as a function of time - - PowerPoint PPT Presentation

Recasting a problem in Fourier space Amplitude as a function of time for plucked guitar: Same information is contained in amplitude of frequencies Energy by adding up frequencies/wavelengths Describe energy by where particle is and how fast it is


slide-1
SLIDE 1

Recasting a problem in Fourier space

Amplitude as a function of time for plucked guitar: Same information is contained in amplitude of frequencies

slide-2
SLIDE 2
slide-3
SLIDE 3

Energy by adding up frequencies/wavelengths

Describe energy by where particle is and how fast it is moving

  • r by adding up energy in each wavelength given the

distribution of wavelengths

slide-4
SLIDE 4

2D Fourier transforms of images

slide-5
SLIDE 5

Expanding wave functions in an HO basis

Single-particle radial wf ψ(r) Expand in harmonic oscillator wfs: ψNmax(r) =

Nmax

  • α=0

cαφα(r) Find cαs by diagonalizing HΨ = EΨ

]=

0.5 1.0 1.5 2.0 2.5 r

  • 5

5 10 15 V@rD

Eexact'=')1.51'

ψexact(r), ψ0(r), 0.5 ∗ φ0 ψexact(r), ψ0(r), 0.5 ∗ φ0

]=

0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf 0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf

Nmax = 0, E0 = −1.30 Nmax = 0, E0 = +5.23

slide-6
SLIDE 6

Expanding wave functions in an HO basis

Single-particle radial wf ψ(r) Expand in harmonic oscillator wfs: ψNmax(r) =

Nmax

  • α=0

cαφα(r) Find cαs by diagonalizing HΨ = EΨ

]=

0.5 1.0 1.5 2.0 2.5 r

  • 5

5 10 15 V@rD

Eexact'=')1.51'

ψexact(r), ψ2(r), 0.5 ∗ φ2 ψexact(r), ψ2(r), 0.5 ∗ φ2

]=

0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf 0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf

Nmax = 2, E2 = −1.46 Nmax = 2, E2 = −0.87

slide-7
SLIDE 7

Expanding wave functions in an HO basis

Single-particle radial wf ψ(r) Expand in harmonic oscillator wfs: ψNmax(r) =

Nmax

  • α=0

cαφα(r) Find cαs by diagonalizing HΨ = EΨ

]=

0.5 1.0 1.5 2.0 2.5 r

  • 5

5 10 15 V@rD

Eexact'=')1.51'

ψexact(r), ψ4(r), 0.2 ∗ φ4 ψexact(r), ψ4(r), 0.2 ∗ φ4

]=

0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf 0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf

Nmax = 4, E4 = −1.46 Nmax = 4, E4 = −1.04

slide-8
SLIDE 8

Expanding wave functions in an HO basis

Single-particle radial wf ψ(r) Expand in harmonic oscillator wfs: ψNmax(r) =

Nmax

  • α=0

cαφα(r) Find cαs by diagonalizing HΨ = EΨ

]=

0.5 1.0 1.5 2.0 2.5 r

  • 5

5 10 15 V@rD

Eexact'=')1.51'

ψexact(r), ψ6(r), 0.2 ∗ φ6 ψexact(r), ψ6(r), 0.2 ∗ φ6

]=

0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf 0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf

Nmax = 6, E6 = −1.50 Nmax = 6, E6 = −1.40

slide-9
SLIDE 9

Expanding wave functions in an HO basis

Single-particle radial wf ψ(r) Expand in harmonic oscillator wfs: ψNmax(r) =

Nmax

  • α=0

cαφα(r) Find cαs by diagonalizing HΨ = EΨ Extend to many-body system

]=

0.5 1.0 1.5 2.0 2.5 r

  • 5

5 10 15 V@rD

Eexact'=')1.51'

ψexact(r), ψ8(r), 0.2 ∗ φ8 ψexact(r), ψ8(r), 0.2 ∗ φ8

]=

0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf 0.5 1.0 1.5 2.0 2.5 r

  • 0.5

0.5 1.0 1.5 2.0 2.5 wf

Nmax = 8, E8 = −1.50 Nmax = 8, E8 = −1.43