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 , USA 1

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

x CaCa 11 22 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) LAM torsion  not shown 3 stretch SAVs not shown 3

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

CaCa 11 22 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

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 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 (a) (b) (c) (d)

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 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 (a) (b) (c) (d)

Baraban, Beck, Steeves, Stanton, Field, JCP 134 (2011) % 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

9 Baraban, Beck, Steeves, Stanton, Field, JCP 134 (2011) % 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.

10 Baraban, Beck, Steeves, Stanton, Field, JCP 134 (2011) % 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.

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 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 +2  /3 0  2  /3 22 +2  /3 0  2  /3 1 1 (a) (b) ??? (c) (d)

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

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

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

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

A trans-well function (  = B us in G 4 (2) = G 8 at E = 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) 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

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

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 11 22 H2H2 H1H1 x z CbCb +2  /3 -2  /3 +2  /3 -2  /3 1=01=0 2=02=0