Linear Imaging Approximations

Notes to accompany the lectures delivered by David A. Muller at the Summer School on Electron : Fundamental Limits and New Science held at Cornell University, July 13-15, 2006.

Reading and References: Chapter 1-3 of “Advanced Computing in Electron Microscopy” by E. J. Kirkland 1 David Muller 2006 Scanning Transmission Electron Microscopy

200 kV Incident 1 atom wide (0.2 nm) beam is scanned Electron Beam across the sample to form a 2-D image ADF Signal Er M Edge (∆E=1 eV) 4 y x

3 Å

Elastic Scattering 00.511. ~ "Z contrast" Distance (nm)

Annular Dark Field (ADF) detector

Electron Energy Loss Spectrometer Increasing energy loss

Single atom P. Voyles, D. Muller, J. Grazul, P. Citrin, H. Gossmann, Nature 416 826 (2002) 2 DavidSensitivit Muller 2006y: U. Kaiser, D. Muller, J. Grazul, M. Kawasaki, Nature Materials, 1 102 (2002) ReciprocityReciprocity (or(or STEMSTEM vs.vs. CTEM)CTEM)

CTEM STEM Gun Detectors β Illumination Collector angle angle

Specimen Specimen

Objective Objective Aperture α Aperture

Viewing Screen Gun Reciprocity (for zero-loss images): A hollow-cone image in CTEM an annular-dark field image in STEM. However: In STEM, energy losses in the sample do not contribute to chromatic aberrations (Strong advantage for STEM in thick specimens) 3 David Muller 2006 Reciprocity

Reciprocity: Electron intensities and ray paths in the remain the same if (i) the direction of rays is reversed, and (ii) the source and detector are interchanged.

Proof follows from time-reversal symmetry of the electron trajectories and elastic scattering (to all orders).

Reciprocity does not hold for inelastic scattering: Sample is after probe forming optics in STEM - energy losses in sample do not cause chromatic blurring in the image

Sample is before the imaging optics in TEM – energy losses in the sample do cause chromatic blurring in the image. Imaging thick samples in TEM can be improved by energy filtering (so on the zero-loss image is recorded). This is not needed for STEM.

4 FromDavid MullerL. Reimer, 2006 Transmission Electron Microscopy Reciprocity

TEM STEM Condensor Collector aperture aperture Controls Objective Condensor aperture aperture (after sample) (before sample) Controls resolution

Image recorded Image recorded serially by In parallel scanning the source

5 FromDavid MullerL. Reimer, 2006 Transmission Electron Microscopy GeometricGeometric OpticsOptics –– AA SimpleSimple LensLens

Focusing: angular deflection of ray α distance from optic axis

α

x x

Object front Back image plane focal at z=0 focal plane plane plane 6 David Muller 2006 GeometricGeometric OpticsOptics –– AA SimpleSimple LensLens

Wavefronts in focal plane are the of the Image/Object

α1

α1

α1 x x

Object front Lens Back image plane focal at z=0 focal plane plane plane 7 David Muller 2006 Fourier Transforms

All plane waves at angle α r pass through the same point r α In the focal plane. k ≈ α λ If the component of ther wavevector in the focal plane is k ,then 1 a function in the back focal plane, F(k) k0 = is the Fourier transform of a function λ in the image plane f(x)

Forward Transform: (image- > diffraction) r ∞ r r F(k ) =F{}f (xr) = f (xr) ei2π k .x d 2 xr ∫−∞

Inverse Transform: (diffraction ->image) r ∞ r r r f (xr) =F − 1{F(k )}= F(k ) e−i2π k .x dsr ∫−∞

Note: in optics, we define k=1/d, while in physics k=2π/d 8 David Muller 2006 FT of A Circular Aperture

An ideal lens would have an aperture A(k)=1 for all k. However, there is a maximum

angle that can be accepted by the lens, αmax, and so there is a cut-off spatial frequency kmax= k0 αmax, Image Plane Focal Plane

FT

⎧ r 1, k < kmax 2J (2π k r) r ⎪ A(r) = π 1 max A(k ) = ⎨ r 0, k > k 2π kmaxr ⎩⎪ max

9 David Muller 2006 The Aperture Function in the Image Plane

2J1(2π kmaxr) A(r) = π A(r) 2 2π kmaxr

3 8

2 2 A(x) 1 |A(x)| 4

0 0 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 x x

First zero at 2πkmaxr = 0.61 2 The image of single point becomes blurred toA(r) (why?) 10 David Muller 2006 Linear Imaging (Kirkland chapter 3)

f(x) h(x) g(x)

object Blurring function image

Image = object convolved (symbol ⊗ ) with the blurring function, h(x) ∞ g()x = f (x') h(x − x')dx' ≡ f (x) ⊗ h(x) ∫−∞ Point spread function In Fourier Space, becomes multiplication (and visa-versa) G(k)= F(k)H(k) 11 David Muller 2006 Contrast Transfer Function Contrast Transfer Function The contrast transfer function (CTF) is the Fourier Transform of the point spread function (PSF)

The CTF describes the response of the system to an input plane wave. By convention the CTF is normalized to the response at zero frequency (i.e. DC level) ~ H(k) H()k = H(0) A low-pass filter A high-pass filter ~ ~ H()k H(k)

k k (multiplication by k in Fourier Space (division by k in Fourier Space ->differentiation in real space) ->integration in real space)

smoothing Edge-enhancing

12 David Muller 2006 High-pass filter

Fourier Transform A high-pass filter x Original image

Inverse Transform

=

Final image

13 David Muller 2006 Edge-enhancing Low pass filter

Fourier Transform A low-pass filter x Original image

Inverse Transform

=

Final image

14 David Muller 2006 smoothing The Aperture Function in the Image Plane 2 The image of single point becomes blurred toA(r) (why?)

2J1(2π kmaxr) A(r) = π A(r) 2 2π kmaxr

3 8

2 2 A(x) 1 |A(x)| 4

0 0 -3 -2 -1 0 1 2 3 -3 -2 -1 0 1 2 3 x x

First zero at 2πkmaxr = 0.61, with kmax = (2π/λ)θmax

i.e. r = 0.61/(2πkmax)~ 0.61 λ/θmax

15 David Muller 2006 Is this the best we can do?

ε

Adding a central beam-stop to the aperture increases the tails on the probe but only slighty narrows the central peak.

From Wyant and Creath 16 David Muller 2006 Coherent vs. Incoherent Imaging (Kirkland, Chapter 3.3)

obj Lateral Coherence of the Electron Beam 0.16λ ∆x ≈ for an angular spread βmax (Born&Wolf): coh α βmax λ Image resolution d ≈ β αmax ψ Combining these 2 formula we get:ψ Coherent imaging: β << 0.16α ψ max max ψ ∆x Wave Interference inside coh allows us to measureψ phaseψ changes 2 2 2 * ψ * as wavefunctions add: a + b = a + b + a b + bψ a

Incoherent imaging: βmax >> 0.16αmax (usuallyβmax > 3α max ) ψ 2 2 No interference, phase shifts are not detected. Intensities add a + ψ b 17 David Muller 2006 Coherent Imaging (Kirkland 3.1)

β~0

ψ (x) Lens has a PSF A(x) α object

ψ image (x)=ψ object (x) ⊗ A(x) ψ image (k)

ψ image (k)=ψ object (k)⋅ A(k)

2 g()x = ψ image x ()

We measure the intensity, not the wavefunction

2 g()x = ψ object (x) ⊗ A(x)

Convolve wavefunctions, measure intensities 18 David Muller 2006 Incoherent Imaging (Kirkland 3.4) Lost phase information, only work with intensities

2 β Lens has a PSF A(x)

ψ object (x) α ψ 2 2 2 image ()x = ψ object (x) ⊗ A(x) ψ image (k) ψ * * ψ image (k)= [ object (k) ⊗ψ object (k)]⋅[A(k) ⊗ A (k)]

2 g()x = ψ image x () CTF

We measure the intensity, not the wavefunction

2 2 g()x = ψ object (x) ⊗ A(x)

19 David Muller 2006 Convolve intensities, measure intensities Coherent vs Incoherent Imaging

Coherent Incoherent

Point Spread 2 Function A(x) A(x)

Phase Object Contrast Transfer Im[A(k)] * 2 Function A(k) ⊗ A (k) Amplitude Object Re[A(k)]

2 2 2 g()x = ψ (x) ⊗ A(x) We measure g()x = ψ object (x) ⊗ A(x) object

20 David Muller 2006 (from Linear Imaging Notes, Braun)

21 David Muller 2006 Resolution Limits Imposed by the Diffraction Limit

(Less diffraction with a large aperture – must be balanced against Cs)

Lens

d0

α0

Gaussian image plane The image of a point transferred through a lens with a circular aperture of 0.61λ 0.61λ semiangle α0 is an of diameter d0 = ≈ nsinα0 α0 (0.61 for incoherent imaging e.g. ADF-STEM, 1.22 for coherent or phase contrast,. E.g TEM)

(for electrons, n~1, and the angles are small) 22 David Muller 2006 Electron Wavelength vs. Accelerating Voltage

0.05

Relativistic Non-relativistic Accelerating v/c λ (Ǻ) 0.04 Voltage 1 V 0.0019784 12.264

0.03 100 V 0.0062560 1.2263

1 keV 0.062469 0.38763 0.02 (Angstroms) 10 keV 0.019194 0.12204 λ 100 keV 0.54822 0.037013

0.01 200 keV 0.69531 0.025078 300 keV 0.77653 0.019687 1 MeV 0.81352 0.0087189 0 0 200 400 600 800 1000 Electron Kinetic Energy (keV)

23 David Muller 2006 Resolution Limits Imposed by , C3 (Or why we can’t do subatomic imaging with a 100 keV electron)

Lens Plane of Least Confusion C3>0

C =0 3 dmin Gaussian image plane

For Cs>0, rays far from the axis are bent too strongly and come to a crossover before the gaussian image plane.

1 3 For a lens with aperture angle α, the minimum blur is d = C α min 2 3

Typical TEM numbers: C3= 1 mm, α=10 mrad → dmin= 0.5 nm 24 David Muller 2006 Balancing Spherical Aberration against the Diffraction Limit

(Less diffraction with a large aperture – must be balanced against Cs)

For a rough estimate of the optimum aperture size, convolve blurring terms -If the point spreads were gaussian, weλ could add in quadrature: 2 ⎛ 0.61α ⎞ 1 2 2 2 2 ⎜ ⎟ ⎛ 3 ⎞ dtot ≈ d0 + ds = ⎜ ⎟ + ⎜ C3α0 ⎟ ⎝ 0 ⎠ ⎝ 2 ⎠ 100

10 Optimal aperture And minimum d 0 Spot size

Probe Size(Angstroms) Probe d s 1 110d = 0.66C1/ 4λ3/ 4 min 3 25 David Muller 2006 α (mrad) Balancing Spherical Aberration against the Diffraction Limit

(Less diffraction with a large aperture – must be balanced against C3)

A more accurate wave-optical treatment, allowing less than λ/4 of phase shift across the lens gives

Minimum Spot size: 1/ 4 3/ 4 dmin = 0.43C3 λ

α 1/ 4 ⎛ 4λ ⎞ Optimal aperture: ⎜ ⎟ opt = ⎜ ⎟ ⎝ C3 ⎠

At 200 kV, λ=0.0257 Ǻ, dmin = 1.53Ǻ and αopt = 10 mrad

At 1 kV, λ=0.38 Ǻ, dmin = 12 Ǻ and αopt = 20 mrad

We will now derive the wave-optical case 26 David Muller 2006 Spherical Aberration (C3) as a Phase Shift

Lens ∆s = path difference due to wave aberration

α

C C

3 3 = > 0 0 Gaussian image plane Wavefronts (lines of equal phase) 2π Phase shift from lens aberrations: χ()α = ∆s()α λ (remember wave exp[α i(2π/λ) x] has a 2π phase change every λ) 1 4 For spherical aberration∆s()= C3α but there are other terms as well 4 27 David Muller 2006 An Arbitrary Distortion to the Wavefront can be expanded in a power series (Either or Seidel aberration coefficients) χ(ρ,θ ') =

Zernike Polynomials

28 DavidJ.C. Muller Wyant, 2006 K. Creath, APPLIED OPTICS AND OPTICAL ENGINEERING, VOL. Xl An Arbitrary Distortion to the Wavefront can be expanded in a power series Seidel Aberration Coefficients (Seidel 1856) EM community notation is similar:

Or more generally

C5,0 and C7,0 are also important

29 David Muller 2006 An Arbitrary Distortion to the Wavefront can be expanded in a power series Here are some terms plotted out

30 David Muller 2006 Phase Shift in a Lens (Kirkland, Chapter 2.4)

Electron wavefunction in focal plane of the lens

iχ (α ) χ ϕ(α )= e α π Where the phase shift from the lens is λ α 2 ⎛ 1 4 1 2 ⎞ ( ) = ⎜ C3 − ∆fα ⎟ ⎝ 4 2 ⎠

Keeping only spherical aberration and defocus

r ⎧eiχ ()k , k < k r ⎪ max A(k ) = ⎨ r 0, k > k ⎩⎪ max

31 David Muller 2006 Optimizing defocus and aperture size for ADF Goal is to get the smallest phase shift over the largest range of angles 4π

1/4C α4 3π 3

2π

π χ(α)

0 π/2 -π

Phase Shift (rad) 2 1/2∆fα α 0 -2π 0 2 4 6 8 10 12 α (mrad) Step 1: Pick largest tolerable phase shift: in EM λ/4=π/2, in light optics λ/10 Step 2: Use defocus to oppose the spherical aberration shift within the widest π/2 band Step 3: Place aperture at upper end of the π/2 band 32 David Muller 2006 Optimizing defocus and aperture size for ADF Goal is to get the smallest phase shift over the largest range of angles

Step 1: We assume a phase shift <λ/4=π/2 is small enough to be ignored

Step 2: Use defocus to oppose the spherical aberration shift within the widest π/2 band

1 Optimal defocus: ∆f = ()C λ 2 opt S α 1/ 4 ⎛ 4λ ⎞ ⎜ ⎟ Optimal aperture: opt = ⎜ ⎟ ⎝ C3 ⎠ Step 3: Place aperture at upper end of the π/2 band & treat as diffraction limited 0.61λ Minimum Spot size: dmin ≈ αopt

1/ 4 3/ 4 dmin = 0.43C3 λ

(The full derivation of this is given in appendix A of Weyland&Muller 33 David Muller 2006 Optimizing defocus and aperture size for TEM (derivation is different, given in Kirkland 3.1)

Look for a uniform phase shift of ±π/2 across lens

1 2 Optimal defocus: ∆fopt = ()0.5CS λ

α 1/ 4 ⎛ 6λ ⎞ ⎜ ⎟ Optimal aperture: opt = ⎜ ⎟ ⎝ C3 ⎠

1.22λ Minimum Spot size: dmin ≈ αopt

1/ 4 3/ 4 dmin = 0.77C3 λ

(The full derivation of this is given in appendix A of Weyland&Muller 34 David Muller 2006 Phase Shift in a Lens with an Aberration Corrector

Electronχ wavefunction in focal plane of the lens α π χ ϕ(α )= ei (α ) λ Where the phase shift from theα lens is α 2 ⎛ 1 2 1 4 1 4 ⎞ ( ) = ⎜− ∆f + C3 + C5α +K⎟ ⎝ 2 4 6 ⎠

5th order spherical aberration

35 David Muller 2006 O.L. Krivanek et al. Ultramicroscopy 96 (2003) 229 Optimizing Aperture size with a C3 Corrector Goal is to get the smallest phase shift over the largest range of angles 4

1/6C α6 3 5

2

1 1/2∆fα2 χ(α)

0 π/2

-1 4

Phase Shift (rad) -1/4C α 3 α 0 -2 0 5 10 15 20 25 30 α (mrad) Step 1: Pick largest tolerable phase shift: in EM λ/4=π/2, in light optics λ/10

Step 2: Use defocus and C3 (now negative) to balance C5 within the widest π/2 band Step 3: Place aperture at upper end of the π/2 band 36 David Muller 2006 Optimizing defocus and aperture size Goal is to get the smallest phase shift over the largest range of angles

Step 1: We assume a phase shift <λ/4=π/2 is small enough to be ignored

Step 2: Use C3 to oppose the C5 shift within the widest π/2 band 1 2 3 Optimal C3: C3 α = −(3λC ) opt λ5 1/ 6 1/ 6 3 ⎛ 3 ⎞ ⎛ λ ⎞ ⎜ ⎟ ⎜ ⎟ Optimal aperture: opt = ⎜ ⎟ =1.47⎜ ⎟ 2 ⎝ C5 ⎠ ⎝ C5 ⎠

Step 3: Place aperture at upper end of the π/2 band & treat as diffraction limited 0.61λ Minimum Spot size: dmin ≈ αopt

1/ 6 5/ 6 dmin = 0.42C5 λ

37 David Muller 2006 Contrast Transfer Functions of a lens with Aberrations Generated with ctemtf r ⎧eiχ ()k , k < k r ⎪ max Aperture function A(k ) = ⎨ r of a real lens 0, k > k ⎩⎪ max

Coherent Imaging CTF: Im[A(k)]= Sin[]χ(k)

E= 200keV, Cs= 1.2mm, df= 600A, Beta= 0mrad, ddf= 0A 1 0.8 coherent 0.6

0.4

0.2

0 MTF

-0.2

-0.4

-0.6 partially coherent -0.8

-1 0 0.1 0.2 0.3 0.4 0.5 0.6 Spatial Frequency (in 1/A) 38 David Muller 2006 Contrast Transfer Functions of a lens with Aberrations Generated with stemtf r ⎧eiχ ()k , k < k r ⎪ max Aperture function A(k ) = ⎨ r of a real lens 0, k > k ⎩⎪ max * 2 Incoherent Imaging CTF: A(k) ⊗ A (k)

Cs= 1.2mm, df= 400A, E= 200keV, OA= 10mrad, d0=0A 1.2

1 CTF for different defocii

0.8 Theorem: 0.6 Aberrations will never 550A

MTF Increase the MTF 0.4 For incoherent imaging

0.2 0A

0 900A

-0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 39 David Muller 2006 k in inv. Angstroms Phase vs. ADF Contrast

(JEOL 2010F, Cs=1mm) 1 STEM HRTEM 0.5

0

-0.5 Contrast Transfer Function Transfer Contrast

-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 k (Inverse Angstroms) TEM: Bandpass filter:low frequencies removed = artificial sharpening ADF : Lowpass filter: 3 x less contrast at 0.3 nm than HRTEM

David Muller 2006 Phase vs. ADF Contrast Random white noise

BF CTF ADF CTF

David Muller 2006 a-C support films look like this Phase vs. ADF Contrast

(JEOL 2010F, Cs=1mm, Scherzer aperture and focus) 0.4 STEM 0.3 -CTEM 0.2

PSF 0.1

0

ADF : 40% narrower FWHM, smaller probe tails

David Muller Muller, 2006 2000 Effect of defocus and aperture size on an ADF-STEM image

(200 kV, C3=1.2 mm)

CTF

PSF

43 David Muller 2006 What happens with a too-large aperture? ADF of [110] Si at 13 mr, C3=1mm

Strong {111} fringes Strong {311} fringes

2 clicks overfocus

Best 111 and 311 fringes occur at different focus settings David Muller 2006 If the aperture is too large Aperture Size is Critical (200 kV C3=1mm)

1

0.8 13 mrad 10 mrad 0.6

CTF 0.4 {220} Si 0.2

0 0 0.2{111} 0.4 0.6 0.8 Si k (1/A) 30% increase in aperture size ~50% decrease in contrast for Si {111} fringes 45 David Muller 2006 Aperture Size is Critical

(200 kV C3=1mm)

30

25

20 r = 0.8 nm 80% 15

10 r = 0.25 nm 5 80% 10 mrad Beam Current Enclosed (pA) r = 0.1 nm 13 mrad 64% 0 00.20.40.60.81 Radius (nm) All the extra probe current falls into the tails of the probe – reduces SNR 46 David Muller 2006 Finding the Aperture with the smallest probe tails (Kirkland, Fig 3.11)

1 2 ∆fmin = 0.8()CS λ

α 1/ 4 ⎛ λ ⎞ ⎜ ⎟ opt =1.22⎜ ⎟ ⎝ C3 ⎠

47 David Muller 2006 Depth of Field, Depth of Focus

d 0.61λ D0 = ≈ 2 tanα 0 α0

For d=0.2 nm, α=10 mrad, D0= 20 nm For d= 2 nm, α=1 mrad, D0= 2000 nm!

48 David Muller 2006 For d= 0.05 nm, α= 50 mrad, D0= 1 nm! Depth of Field in ADF-STEM: 3D Microscopy? (300 kV, 25 mrad)

6 25 mrad 10 mrad 5

4

3

D=6 nm: θ0 = 25 mr

Peak Intensity 2

1

0 -300 -200 -100 0 100 200 300 Depth (A)

49 David Muller 2006 Depth of Field in ADF-STEM: 3D Microscopy? Today: D~ 6-8 nm SuperSTEM: ~1 nm

100 kV 200 kV 300 kV 100 Cs>0

Cs~0

10 D=1nm: θ0 > 50mr

Depth Resolution (A) Resolution Depth C5~0

1 0 20406080100 Probe Forming Semi-Angle (mrad) 50 David Muller 2006 Does Channeling Destroy the 3D Resolution? (Multislice Simulation of [110] Si @ 200 kV, 50 mrad)

0.04

0.035 ) 0 0.03

0.025

0.02

0.015 Signal (Fraction of I (Fraction Signal Fringe contrast ∆f=20 nm 0.01 ∆f=1 nm

0.005

0 0 5 10 15 20 25 30 Thickness (nm) NO! (at least in plane – relative intensities between different depths are still out) 51 David Muller 2006 Summary

1/ 4 Contrast Transfer Functions: α ⎛ 6λ ⎞ 1/ 4 3/ 4 Coherent: ⎜ ⎟ opt = ⎜ ⎟ dmin = 0.77C3 λ ⎝ C3 ⎠ Lower resolution, higher contrast Easy to get contrast reversals with defocus Aperture size only affects cutoff in CTF

α 1/ 4 ⎛ 4λ ⎞ Incoherent: ⎜ ⎟ 1/ 4 3/ 4 opt = ⎜ ⎟ dmin = 0.43C3 λ ⎝ C3 ⎠ Higher resolution, lower contrast Harder to get contrast reversals with defocus Aperture size is critical – affects CTF at all frequencies

52 David Muller 2006