Heavy Atom Quantum Diffraction by Scattering

from SurfacesEli Pollak

Chemical Physics DepartmentWeizmann Institute of Science

Coworkers:Dr. Jeremy M. Moix

Grants:Israel Science Foundation Weizmann-UK Joint Research

Program

Bill CasselmanUniversity of British Columbia, Vancouver, Canada

cass at math.ubc.ca

http://www.ams.org/featurecolumn/archive/rainbows.html

 

f

b

n·sin(r)= sin(ib/R=x, n=4/3

f(x) = 4r-2i = 4sin-1(x/n)-2sin-1(x)

df/dx=0 => x=√(20/27), f=42o

R4r

The rainbow angular distribution:

1

1

11 sin2sin42

1)( x

n

xdxP

Water: n=4/3 Perspex: n=3/2

i

f i

i

l

h(x)

xtan-1[h’(x)]~-h’(x)

f (x)=i+2ih’(x)

The rainbow angles are found when they arise from

the deflection points of the corrugation function.

0)(

2

2

dx

xhd

Rainbow scattering from a hard wall

The angular distribution for scattering from a hard wall

l

i xhdxl

p0

)('21

If the corrugation is sinusoidal:

l

xhxh

2sin)(

then the angular distribution is readily seen to be:

22

0 4

12cos

41)(

i

l

i

lhl

x

l

hdx

lp

l=30 mmh=0.42 mmR=i±10o

Experimental observations

Note:At 100 meV the de Broglie wavelength of Ar is 0.25 a.u.. The lattice length is ~ 8 a.u.. It is then not surprising that the measured distribution is classical.

I. In plane scattering of Ar atoms on a LiF(001) surface:

II. In plane scattering of Ar and Kr on a Ag(100) surface

Ar

Kr

III. In plane scattering of Ar on a 2H-W(100) surface

At 100 meV the de Broglie wavelength is 0.25 a.u.. Why then does one see

diffraction at Ei=27 meV? (The lattice length is ~ 10 a.u..)

Beam energy uncertainty of 1/30 implies a parallel coherence wavelength of ~7.5 a.u., enough to see the diffraction peak.

Krypton mass - ~84, at 35 meV, dBwl~0.3 a.u..Beam energy uncertainty of 1/35 implies a parallel coherence wavelength of ~10 a.u., enough to see the diffraction peak.

Heavy molecule interference experiments:

Two slit diffraction experiment -Waves at the two slits must be coherent

Fullerene speed of 100 m/s → 5.6 pico m de Broglie wavelength

Collimating the beam with 7 m slits separated by 1 m implies that transverse speed is less than 1.4·10-3 m/sec → ~400 nm de Broglie wavelength or an angular width of ~8 10-4 degrees.

Note the r2 dependent loss of signal due to the long path between the collimating slits.

High energy grazing collisions

A. Schueller, S. Wethekam, and H. Winter*Institut fuer Physik, Humboldt Universita¨t zu Berlin, Brook-Taylor-Strasse 6, D-12489 Berlin-Adlershof, Germany

Diffraction of Fast Atomic Projectiles during Grazing Scattering from a LiF(001) SurfacePRL 98, 016103 (2007)

Light atoms and molecules with energies from 300 eV to 25 keV are scattered under a grazing angle of incidence from a LiF(001) surface…. Experimental results for scattering of H, D, 3He, and 4He atoms as well as H2 and D2 molecules can be unequivocally referred to atom diffraction with de Broglie wavelengths as low as about 0.001 A°

Analysis: At 10 keV the de Broglie wavelength of 4He is 0.0027 a.u. a 1o grazing angle implies p┴~p║/60 so a perpendicular wavelength of 0.16 a.u.. The angular resolution is ~0.01o so that the coherence wavelength is ~16 a.u., enough to observe diffraction.

Theory – A typical diffraction pattern

A square surface

A hexagonal surface

At low energy quantum mechanics prevails

Consider again the Ar-LiF scattering experiment:

Classical and quantum scattering on frozen (model) surface.

Classical Wigner

Exact quantum

Ar energy of 705 meV → 5.5 pm de Broglie wavelength while LiFlattice length =400 pm. The angular width of the beam is 2 degrees so the perpendicular coherence wavelength is ~ 180 pm.

Reducing the uncertainty in the transverse momentum so that the angular width of the incident beam is 0.2o leads to the quantum mechanical diffraction pattern:

The typical lattice length is 1 nm, a factor of 100 smaller than the size of the grating used in the two slit experiment, implying a reduction of the collimation length by and thus a gain of 104 in the signal to noise ratio.

The final momentum distribution with an angular incident width of 0.2o

Decoherence via increase of surface temperature

Phonons modeled as single oscillator coupled to the vertical motion

Preliminary experimental results – scattering of Ar on Ru(0001) M. Minniti and D. Farias, Madrid.

Energy – 64 meV, de Broglie wavelength – 18 pm.Lattice length – 300 pmCoherence length – 180 pm (2 mm aperture)

400 pm (.75 mm aperture) 720 pm (0.4 mm aperture)

Summary:

• Quantum diffraction of heavy atoms is observable in surface scattering at low temperatures.

•Collimation of the incident beam leads to partial diffraction patterns.

•At sufficiently low surface temperature, surface phonons do not wipe out the coherence

•Decoherence may be studied quantitatively by varying the surface temperature

Future directions

• Use better theory (MCTDH) to include a realistic representation for the continuum of phonon modes.

• Study diffraction of molecular systems – do internal modes destroy the quantum diffraction through energy exchange?

• Can heavy atom diffraction be used to study surface interactions – phonons or electronic?

• Is it possible to use the quantum coherence to manipulate energy transfer with the surface?

• Can one conehrently control reactive processes on the surface?

Consider then the energy loss as a function of impact parameter for Ar-LiF scattering:

Energy loss rainbowsJ. Moix, E. Pollak and S. Miret-Artès, Phys. Rev. Lett. 104, 116103 (2010).

From the perturbation theory we found the Gaussian energy transfer kernel and the energy distribution of the scattered particle:

Joint energy and angular distribution for a 15O angle of incidence, 3eV incidence energy and surface temperature T=30K.

Have they been seen experimentally?

Rotational Rainbows

C2=0.01, C1=0

M

E

I

J Z2

20 2,1

For simplicity consider a diatomic molecule with angular momentum parallel to the surface. Let i denote the initial angle of the diatomic axis relative to the vertical to the surface:

For a smooth surface, the interaction potential will be a function of the angle The final angular momentum will then depend on the initial rotational angle.

Perturbation theory

',','sin2

1

2 12

2'2

'

22

ZV

pp

IZV

mm

ppH

BA

ZX

The Hamiltonian:

The unperturbed motion:

00

00'' '',2

',,0 ttI

JttJpp

1st order perturbation theory:

1''

0'201''

Vpp

JpJVpp

Energy conservation:

rotBA

ZZ EVpVpVpJImm

PP

2

1'2

1'1'00 2

2

1

222101

20

sin2

3cos12,

2

ZZ

ZZ

eCeCVZV

eeVZV → Morse potential

(C1=0 for a homonuclear diatomic)

The angular distribution:

Z

roti

Z

Z

Z

Xf

Z

Xi

E

E

P

P

P

P

P

P

4

2sin

1tan,tan00

0

0

0

When the molecule is not too slowly rotating one readily finds:

242 sin

1

Jr

P

→ The rotational rainbows cause maxima in the angular distribution!

Examples: