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Interplay Between Electronic and Nuclear Motion in the Photodouble Ionization of H 2 T J Reddish, J Colgan, P Bolognesi, L Avaldi, M Gisselbrecht, M Lavollée,

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Presentation on theme: "Interplay Between Electronic and Nuclear Motion in the Photodouble Ionization of H 2 T J Reddish, J Colgan, P Bolognesi, L Avaldi, M Gisselbrecht, M Lavollée,"— Presentation transcript:

1 Interplay Between Electronic and Nuclear Motion in the Photodouble Ionization of H 2 T J Reddish, J Colgan, P Bolognesi, L Avaldi, M Gisselbrecht, M Lavollée, M. S. Pindzola, and A Huetz DAMOP 2008

2 Ions escape much faster than molecular rotation. Detecting ion’s momenta gives: ‘fixed-in-space’ molecule.  Fully differential Cross Sections (FDCS) where are the polar angles of electrons 1 and 2 and the molecular axis, N, with respect to, and where and (with e = 1 or 2). h (76 eV) + H 2  H + + H + + e e 2 -    Photodouble Ionisation of H 2

3 Double Ionisation ‘threshold’: ~51 eV (R-dependent). Total energy is conserved by electron and ion pairs, (and the dissociation limit). Final ion pair “kinetic energy release” (KER) reflects internuclear separation (R) at moment of double ionisation. Filter data set via KER to map FDCS as a function of R. h (76 eV) + H 2  H + + H + + e e 2 - Fast ‘Coulomb Explosion’ R 0 = 1.4 a 0

4 (x e,y e,t e ) (x H,y H,t H ) H+H+ H+H+ e1-e1- e2-e2- Electric field Magnetic field Photon pxpx pypy pzpz x y t E   Momentum Imaging Apparatus Gisselbrecht et al, Rev Sci Instrum 76 (2005)

5 Coplanar Geometry: All 4 particles and  lie in the same plane. This configuration probes both electron-ion and electron-electron interactions. Coulomb repulsion favours “back-to-back emission”, yet PDI Selection rules: Node for back-to-back emission for “equal energy- electrons k e1 k e2 k  Walter and Briggs, Phys Rev Lett 85 (2000) 1630 k e1  Gisselbrecht et al Phys Rev Lett 96 (2006) (TDCC) Colgan et al, Phys Rev Lett 98 (2007) What happens to FDCS when R changes?

6 Coplanar FDCS “KER Effect”:  1 = 90º E 1 = E 2 = 12.5  10 eV,  N =  10º ‘Pure’  component shows no KER effect. ‘Pure’  component shows small KER effect. KER averaged at  N = 30º. Dramatic R-dependence at  N  20º, especially at large R where most yield is in 4 th quadrant. R ~ 1.6a 0 R ~ 1.2a 0 TDCC bandwidth averaged FDCS TDCC unaveraged FDCS

7 Coplanar FDCS “KER Effect”  1 = 0º E 1 = E 2 = 12.5  10 eV  N =  10º KER averaged at  N = 60, 30º. Again a dramatic movement of FDCS yield to the 2 nd quadrant as  N = 40 º  20º, but only for large internuclear separation. TDCC bandwidth averaged FDCS TDCC unaveraged FDCS

8 R ~ 1.6a 0 R ~ 1.2a 0 Coplanar FDCS “KER Effect”  1 = 60º A significant change In FDCS yield as a Function of R when  N = 20 º or 160º. All these FDCS have E 1 = E 2 = 12.5  10 eV: Therefore KER effects are not overly sensitive to electron energies. TDCC bandwidth averaged FDCS TDCC unaveraged FDCS Reddish et al Phys Rev Lett 100 (2008)  N =  10º

9 Why is  N ~ 20º (or 160 º) so critical to observe these KER effects? FDCS is the coherent sum of  and  components. We extract the  and  components, and cross term contributions, in TDCC FDCS. At  N = 20º, both components make significant contributions to the FDCS. Only the  component displays an appreciable dependence on R. Changes in sign (and shape) of the cross term with R ‘amplifies’ the small changes with R of the pure  component. FDCS Contributions: , , Cross term, Total. (  1,  N ) values are (20º, 160º) left, (60º, 20º) right. R = 1.6 (upper) and 1.2 (lower).

10 Why does the only the  amplitude have an angular dependence sensitive to R? Magnitude of the  amplitude decreases monotonically with R. Whereas  amplitude has a shallow minimum near R 0. This same behaviour is also seen in the photoionisation of H 2 +. Hence it is a feature of the axially symmetric nuclear potential, rather than electron correlation. ECS Horner et al Phys Rev Lett 98, (2007). Reddish et al Phys Rev Lett (2008) Colgan et al, J Phys B 41, (2008).

11 Photoionisation of H 2 +. The photoelectron angular distribution, with respect to the molecular axis.  u Final State A strong cancellation exists for the p-wave component for the  g   u transition, at a given (E k, R). * Like an Cooper minimum * Then the f-wave dominates  different angular distributions. p, f ‘mix’ is sensitive to R value. No such cancellation occurs for the  g  π u transition.  u Final State Colgan et al, J Phys B 41, (2008).

12 Summary and Conclusions Excellent agreement between TDCC and experiment. Dramatic changes in coplanar FDCS for  N ~20, 160º with internuclear separation, R, due to interference between  and  components, whose contributions have similar magnitudes at these  N values. Only  component has R dependence: larger ( 1, 2 ) are necessary for convergence of the TDCC  amplitude than for  particularly for large R. By our analogy with H 2 +, main R-dependent trends of the  and  amplitudes observed in PDI of H 2 are due to electron-ion rather than electron-electron interactions.

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14 KER Effect in Perpendicular Plane Weber et al, Nature (2004) ECS Vanroose et al, Science (2005) E 1  to E 2, R, and  “Frozen correlation” (  12 = 90º) We do not see a clear KER effect in this geometry. TDCC J. Colgan et al, J. Phys. B 40, 4391 (2007).

15 Photodouble Ionisation of H 2 e-e- e-e- H+H+ H+H+ h + H 2  H + + H + + e e 2 - Double electron escape in an axial symmetric potential Motivation:  Fundamental theoretical interest: Correlation and Dynamics  Angular distributions are sensitive probe (amplitude and phase)  Development of sensitive imaging techniques (  ++ ~ cm 2 )  Accurate test for theory in a ‘simple’ system

16 3D Momentum Imaging Apparatus Time-of-Flight and (x,y) ion and electron multihit position-sensitive detection. 4  Detection Solid Angles: Absolute . 10 Gauss magnetic field confines electrons up to 20 eV. Synchrotron radiation with well defined polarisation properties and high photon flux. ‘Complete’ kinematical description of ionization process.

17 H 2 Coplanar FDCS E 1 = E 2 = 12.5  2.5 eV,  1 = 90  15°,  12 =  20°,  N =  20°,  1N =  45°  Electron - electron distribution does depend on molecular alignment!  Symmetric two ‘lobes’ for  N = (a) 90  (  ), (d) 0  (  ).  Absolute  reduces by ~4 from  →  orientations. Gisselbrecht et al Phys Rev Lett 96 (2006) Weber et al Phys Rev Lett 92 (2004) What happens to FDCS when R changes?


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