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Multi-valued versus single-valued large-amplitude bending-torsional-rotational coordinate systems for simultaneously treating trans-bent and cis-bent acetylene in its S 1 state Jon T. Hougen Sensor Science Division, National Institute of Standards and Technology, Gaithersburg, MD 20899-8441, USA 1
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Purpose of the present talk is to show how LAM treatments of trans-bent and cis-bent acetylene using multi-valued coordinates and multi-fold extended PI groups are related mathematically to LAM treatments using single-valued coordinates and ordinary PI groups. Short answer: Symmetry properties in the multi-valued coordinate systems become boundary conditions in the single-valued coordinate system. 2
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x CaCa 11 22 H2H2 H1H1 z CbCb Trans and cis acetylene No C-H bond breaking LAM CCH bends 1, 2 Motion on two circles centered on the C atoms -2 /3 < 1, 2 < + 2 /3 H cannot enter shaded areas | | < 120 is arbitrary. | | < 100 or 170 are OK. | | < 181 is no good, because H can pass through C-C bond. Review the global LAM coordinates from Hougen & Merer: 2011 Columbus + JMS 267 (2011) 200-221. LAM torsion not shown 3 stretch SAVs not shown 3
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These LAM bending and torsional angles lead to multiple-valued coordinate systems. (Connect to single-valued coordinate system later.) The coordinates { , , 1, 2 } = {K-rotation, HCCH torsion, HCC bend, CCH bend} for a given configuration in space are not unique. (See this in next slide.) A multiple valued coordinate system leads to “extra” minima on the potential surface. (This is what some people find objectionable, and is why we are giving this talk today.) 4
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CaCa 11 22 H2H2 H1H1 x z CbCb CaCa -1-1 -2-2 H2H2 H1H1 x z CbCb 1, 2 - 1, - 2 The lab configuration above is described by 8 different sets of coordinates. 1. 1, 2, , 2.- 1, - 2, + , 3.- 1, - 2, , + 4.… There are 8 identical trans minima in the 1, 2, , coordinate system. There are 2 identical trans minima in the 1, 2, coordinate system (no torsion ). 5
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6 Allowed domains for bending 1 and 2 double- valued & single-valued coordinates (no torsion), can be shown as a square or parts of a square. Domain of double-valued coordinate system (a) contains two trans and two cis wells. Domains of single-valued coordinate systems (b),(c),(d) contain one trans and one cis well. +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 (a) (b) (c) (d)
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7 Boundary conditions for domains of 1 and 2 In double-valued & single-valued coordinates Boundary conditions for are simple on the thick black borders: = 0 because V = . Boundary conditions are complicated = ??? on the dotted - - - borders because V . +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 (a) (b) (c) (d)
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Baraban, Beck, Steeves, Stanton, Field, JCP 134 (2011) 244311. 69% trans + 31% cis wavefunction. n=77, =A gs, E~4397 cm -1. Note the visually perfect symmetry about both diagonals. Taylor’s expansion: = 0 + (d /dx)x + ½(d 2 /dx 2 )x 2 + … then gives: Even reflection symmetry: 0 on the diagonal, but d /d = 0 across diagonal. Odd reflection symmetry: = 0 on the diagonal, but d /d 0 across diagonal. These symmetry properties have the form of boundary conditions. 8
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9 Baraban, Beck, Steeves, Stanton, Field, JCP 134 (2011) 244311. 69% trans + 31% cis wavefunction. n=77, =A gs, E~4397 cm -1. Note the visually perfect symmetry about both diagonals. Taylor’s expansion: = 0 + (d /dx)x + ½(d 2 /dx 2 )x 2 + … then gives: Even reflection symmetry: 0 on the diagonal, but d /d = 0 across diagonal. Odd reflection symmetry: = 0 on the diagonal, but d /d 0 across diagonal. These symmetry properties have the form of boundary conditions.
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10 Baraban, Beck, Steeves, Stanton, Field, JCP 134 (2011) 244311. 69% trans + 31% cis wavefunction. n=77, =A gs, E~4397 cm -1. Note the visually perfect symmetry about both diagonals. Taylor’s expansion: = 0 + (d /dx)x + ½(d 2 /dx 2 )x 2 + … then gives: Even reflection symmetry: 0 on the diagonal, but d /d = 0 across diagonal. Odd reflection symmetry: = 0 on the diagonal, but d /d 0 across diagonal. These symmetry properties have the form of boundary conditions.
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11 In the two triangular domains (c) and (d), the roles of the diagonals are interchanged: domain boundary symmetry axis. This helps to understand intuitively the boundary conditions that must be applied: = 0, d /d 0 or 0, d /d = 0, but a rigorous mathematical proof is also possible. +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 +2 /3 0 2 /3 22 +2 /3 0 2 /3 1 1 (a) (b) ??? (c) (d)
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Now repeat all this for 3 LAMs = 2 bends 1, 2 and 1 torsion . 3 LAMs mean square cube for domain display Look at 1 domain slide (i)Big cube = octuple-valued coordinate domain with 8 trans and 8 cis wells. (ii)Parts of big cube = single-valued coordinate domains with 1 trans and 1 cis well. Look at 1 boundary value slide (i) = 0 on faces where V = . (ii) = complicated on faces where V , but given by symmetry properties in big cube. 12
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13 (a) (b) (c) +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 (d) (e) (f) +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 octuple-valued domain single-valued domains 8 trans + 8 cis minima one trans + one cis minimum all single-valued domains, with one trans + one cis minimum
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14 (a) (b) (c) +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 +2 /3 0 2 /3 +2 /3 0 2 /3 2 0 1 1 2 2 = 0 on 4 faces (V = ) on 2 faces on 3 faces = periodic ( = 0 & 2 ) = ??? on inner faces (V ) Boundary conditions on all boundary faces of the octuple-valued coordinate domain are very simple. Boundary conditions on some boundary faces of the single-valued coordinate domain are complicated. Boundary conditions are different on different faces
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15 One remaining difficult question (not yet treated) is: In what mathematical way does this procedure fail for the bending vibration in triatomic molecules, where a LAM bend MUST pass through the linear configuration where A ? Summary: For HCCH, the multiple-valued-coordinate and extended-group treatment is just a convenient way of putting the correct boundary conditions on a single- valued treatment. There is no need to be afraid of the “extra, non-physical” minima that arise.
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1. Tunneling path is nearly circular in 1, 2 space 2. Note very high barrier to linear configuration. Consider only bent forms of HCCH This allows us to avoid quasi-linear molecule complications 17
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A trans-well function ( = B us in G 4 (2) = G 8 at E = 779.968 cm -1 ) with perfect symmetry across the principal diagonal \, but with imperfect symmetry across the other diagonal /. All three wavefunctions below from: “Reduced dimension discrete variable representation study of cis-trans isomerization in the S 1 state of C 2 H 2 Baraban, Beck, Steeves, Stanton, Field, J. Chem. Phys. 134 (2011) 244311. Vibrational state is trans v cis-bend =1 (one node across \) v trans-bend =0 (no node \) and has odd trans-trans tunneling parity (a node across /) 18
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Summary of Extended Group Results For cis and trans acetylene with only 2 LAM bends: G 4 (2) For cis and trans acetylene with only 1 LAM torsion: G 4 (2) For cis and trans acetylene with all 3 LAMs: G 4 (8) cis/trans acetylene + vinylidene with 2 LAM bends: G 8 (2) cis/trans acetylene + vinylidene with all 3 LAMs: G 8 (8) 19
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CaCa -1-1 -2-2 H2H2 H1H1 x z CbCb 1, 2 - 1, - 2 1. Apply also + “Limited Identity” 2. Apply also + “Limited Identity” There is 1 real identity and 7 limited identities = identity in PI group G 4, but not for wavefunction There are 8 identical trans minima. 20 CaCa 11 22 H2H2 H1H1 x z CbCb +2 /3 -2 /3 +2 /3 -2 /3 1=01=0 2=02=0
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