ZMAT and Symmetry

 Posts: 10
 Joined: 11 Jan 2014, 13:59
 First name(s): Alfred
 Last name(s): Güthler
 Affiliation: privat
 Country: Germany
ZMAT and Symmetry
Hello DALTONusers,
I have problems using symmetry AND molfiles, especially with the ZMAT format, which I prefer. Using a simple 3atomic molecule, with C2vsym. , the program recognizes it as one with 4 atoms ? ! How do I circumvent that, so that the correct molecule is used ?
If I use cartesian coord. I must have to use less lines in the mol file, because of the symmetry. That means not three, but 2 ? How do I know, which line I have to drop ?
Thank you in advance for your help and answer.
best greetings
Alfred
I have problems using symmetry AND molfiles, especially with the ZMAT format, which I prefer. Using a simple 3atomic molecule, with C2vsym. , the program recognizes it as one with 4 atoms ? ! How do I circumvent that, so that the correct molecule is used ?
If I use cartesian coord. I must have to use less lines in the mol file, because of the symmetry. That means not three, but 2 ? How do I know, which line I have to drop ?
Thank you in advance for your help and answer.
best greetings
Alfred
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 optvib_B3_631gdp_zmtC2v.out
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 magnus
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Re: ZMAT and Symmetry
It's because you're using symmetry generators which will apply the specified symmetry operations. Instead remove the generators and let Dalton detect the symmetry.

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Re: ZMAT and Symmetry
Dear Magnus,
thank you for your reply !
Using your suggestion, the calculation works very well, and the symmetry D3h > C2v is correctly detected by DALTON2018.
I know however, that I have to use symmetry adapted coordinates, in connection with symmetry, at least with cartesians, because of the symmetry generators, but I wanted to now, how this work with zmatrices ?!
Do you have any idea and is there any way to to that ?! It is sometimes useful to calculate a molecule at a certain symmetry. Using tables that shows, how the symmetry is reduced  D3h > C2v  is also a possibility.
In the case of B3, the correct symmetry is D3h.
Thank you in advance for your answer.
best greetings
Alfred Güthler
thank you for your reply !
Using your suggestion, the calculation works very well, and the symmetry D3h > C2v is correctly detected by DALTON2018.
I know however, that I have to use symmetry adapted coordinates, in connection with symmetry, at least with cartesians, because of the symmetry generators, but I wanted to now, how this work with zmatrices ?!
Do you have any idea and is there any way to to that ?! It is sometimes useful to calculate a molecule at a certain symmetry. Using tables that shows, how the symmetry is reduced  D3h > C2v  is also a possibility.
In the case of B3, the correct symmetry is D3h.
Thank you in advance for your answer.
best greetings
Alfred Güthler
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 optvib_B3_631gstst_zmat.out
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 Posts: 252
 Joined: 27 Aug 2013, 16:42
 First name(s): Kenneth
 Last name(s): Ruud
 Affiliation: UiT The Arctic University of Norway
 Country: Norway
Re: ZMAT and Symmetry
Hi!
I am a little bit uncertain what your actual question is, but in general Dalton provides little support for the use of Zmatrices, and they are mainly provided for those that really do not want to work with Cartesian coordinates.
The internal handling of symmetry in Dalton is all based on simple bit operations for handling symmetry, which basically means we handle D2h and subgroups and only symmetry operations that change the sign of one or more of the Cartesian coordinates of a given atom and nothing more (programs able to handle nonAbelian symmetry often use petite lists).
Thus, you can very easily define symmetry generators in your input (e.g. Z, indicating that we will generate a symmetryrelated atom for any atom for which the Z coordinate =/= 0, that is, a mirror plane); or XY, indicating we will change simultaneously change the sign of the X and Y coordinates of atoms where one or both of these coordinates =/=0, that is, a rotation about the Z axis).
However, all this requires that the Cartesian axis system also constitute the rotation axes and mirror planes and that the molecule is oriented symmetrically with respect to these symmetry operations. That is also why there is no straightforward way of coupling a Z matrix input to a specific symmetry input. Instead, what the program does, as you have seen, is to move the molecular centre of mass to the origin of a Cartesian axes system, and align the moments of inertia tensor components along the different Cartesian coordinate axis as a way of detecting when symmetry is after all present.
As I said, I am uncertain whether this answers your question, but nevertheless hopefully of some help.
Best regards,
Kenneth
I am a little bit uncertain what your actual question is, but in general Dalton provides little support for the use of Zmatrices, and they are mainly provided for those that really do not want to work with Cartesian coordinates.
The internal handling of symmetry in Dalton is all based on simple bit operations for handling symmetry, which basically means we handle D2h and subgroups and only symmetry operations that change the sign of one or more of the Cartesian coordinates of a given atom and nothing more (programs able to handle nonAbelian symmetry often use petite lists).
Thus, you can very easily define symmetry generators in your input (e.g. Z, indicating that we will generate a symmetryrelated atom for any atom for which the Z coordinate =/= 0, that is, a mirror plane); or XY, indicating we will change simultaneously change the sign of the X and Y coordinates of atoms where one or both of these coordinates =/=0, that is, a rotation about the Z axis).
However, all this requires that the Cartesian axis system also constitute the rotation axes and mirror planes and that the molecule is oriented symmetrically with respect to these symmetry operations. That is also why there is no straightforward way of coupling a Z matrix input to a specific symmetry input. Instead, what the program does, as you have seen, is to move the molecular centre of mass to the origin of a Cartesian axes system, and align the moments of inertia tensor components along the different Cartesian coordinate axis as a way of detecting when symmetry is after all present.
As I said, I am uncertain whether this answers your question, but nevertheless hopefully of some help.
Best regards,
Kenneth

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 First name(s): Alfred
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Re: ZMAT and Symmetry
Dear Kenneth,
I have read your answer and I can understand that you are a bit uncertain, about my question, because I was not very clear at some points. However, as you hoped, you partially gave me an answer.
My Question was/is: how can one use a ZMATRIXinput in a molfile at a certain symmetry, if one already knows or guess the correct symmetry. ?
For instance, using H20 with the zmatrix at C2vsymmetry, one can be certain that the calculation is correct, because H20 HAS C2vsym.
Actually I could do any calculation without symmetrygenerators, but what would happen, if I use these "Generators = 2 X Z" ?
I will try this with H2CO.
Best Greetings
Alfred
I have read your answer and I can understand that you are a bit uncertain, about my question, because I was not very clear at some points. However, as you hoped, you partially gave me an answer.
My Question was/is: how can one use a ZMATRIXinput in a molfile at a certain symmetry, if one already knows or guess the correct symmetry. ?
For instance, using H20 with the zmatrix at C2vsymmetry, one can be certain that the calculation is correct, because H20 HAS C2vsym.
Actually I could do any calculation without symmetrygenerators, but what would happen, if I use these "Generators = 2 X Z" ?
I will try this with H2CO.
Best Greetings
Alfred

 Posts: 545
 Joined: 15 Oct 2013, 05:37
 First name(s): Peter
 Middle name(s): Robert
 Last name(s): Taylor
 Affiliation: Tianjin University
 Country: China
Re: ZMAT and Symmetry
It is inherent in the definitions of the terms in a Zmatrix that dealing with symmetry is complicated. It usually requires introduction of dummy atoms to ensure things like proper ring closure for symmetric rings like benzene, or with linear sections of a molecule. Dalton will detect the (highest subgroup of D_{2h}) symmetry from a Zmatrix, as other posters have said. I do not understand why you want to use generators, because they require your input to comprise only symmetryunique atoms.
So for H_{2}CO it would be possible to use a Zmatrix (and in this case no dummy atoms would be required), and if the geometry specified in the Zmatrix has C_{2v} symmetry (and the input does not contain the
I admit to very considerable prejudice here. I loathe Zmatrices and regard them as the worst thing Jimmy Stewart has contributed to our field. They were targetted at systems that have no symmetry, and/or for use in codes that do not use symmetry. I recognize that the most commonly used ab initio program uses Zmatrices and has popularized their use, but just because a majority of people do it does not make it sensible (a majority of people who voted supported the UK leaving the EU: this provides another illustration...). I suggest that if you want to exploit symmetry  and it is almost always advantageous to do so  you use Cartesian coordinates and specify the symmetry yourself with generators. The posting that appears at the top of the "Running Dalton" topic provides some description of this (full disclosure: I wrote it...).
Best regards
Pete
So for H_{2}CO it would be possible to use a Zmatrix (and in this case no dummy atoms would be required), and if the geometry specified in the Zmatrix has C_{2v} symmetry (and the input does not contain the
Nosymmetry
keyword) the program will detect this symmetry and use it. However, if you tell the program to use two generators to define C_{2v}, then the code expects only three atoms in the input because it will generate the second H atom. I have no idea how to do this with a Zmatrix, and to be honest I can't understand why one would want to.I admit to very considerable prejudice here. I loathe Zmatrices and regard them as the worst thing Jimmy Stewart has contributed to our field. They were targetted at systems that have no symmetry, and/or for use in codes that do not use symmetry. I recognize that the most commonly used ab initio program uses Zmatrices and has popularized their use, but just because a majority of people do it does not make it sensible (a majority of people who voted supported the UK leaving the EU: this provides another illustration...). I suggest that if you want to exploit symmetry  and it is almost always advantageous to do so  you use Cartesian coordinates and specify the symmetry yourself with generators. The posting that appears at the top of the "Running Dalton" topic provides some description of this (full disclosure: I wrote it...).
Best regards
Pete

 Posts: 10
 Joined: 11 Jan 2014, 13:59
 First name(s): Alfred
 Last name(s): Güthler
 Affiliation: privat
 Country: Germany
Re: ZMAT and Symmetry
Dear Users,
The reason why I use zmatrices is the easy us e of it and its conection with the internal coordinates. For Instance in the case of H20 one has: 2A_1+B_1, two different r's and one angle. That is exactly how one (can) write the zmatrix for water.
There is another reason: I can imagine the molecule in zmatrixformat at all, and not in cartesians. One had to "translate" the cartesian to bondings and angles, to get an imagination, about what happens. This is not possible without the help of other programs.
Nevertheless, thank you all for you valuable help.
Best Greetings
Alfred
The reason why I use zmatrices is the easy us e of it and its conection with the internal coordinates. For Instance in the case of H20 one has: 2A_1+B_1, two different r's and one angle. That is exactly how one (can) write the zmatrix for water.
There is another reason: I can imagine the molecule in zmatrixformat at all, and not in cartesians. One had to "translate" the cartesian to bondings and angles, to get an imagination, about what happens. This is not possible without the help of other programs.
Nevertheless, thank you all for you valuable help.
Best Greetings
Alfred

 Posts: 10
 Joined: 11 Jan 2014, 13:59
 First name(s): Alfred
 Last name(s): Güthler
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 Country: Germany
Re: ZMAT and Symmetry
Hello users,
I have to correct, what I have said before, concerning the vibrations.
The case H2O:
2A1 + B1.
There are two A1, namly the symmetric bonding, angle and one antisymmetric bonding, which is of B1.sym.
In the case B3:
D3h: A'_1+E' which is reduced in DALTON to C2vsym.
greetings
Alfred
I have to correct, what I have said before, concerning the vibrations.
The case H2O:
2A1 + B1.
There are two A1, namly the symmetric bonding, angle and one antisymmetric bonding, which is of B1.sym.
In the case B3:
D3h: A'_1+E' which is reduced in DALTON to C2vsym.
greetings
Alfred

 Posts: 326
 Joined: 27 Jun 2013, 18:44
 First name(s): Hans Jørgen
 Middle name(s): Aagaard
 Last name(s): Jensen
 Affiliation: Universith of Southern Denmark
 Country: Denmark
Re: ZMAT and Symmetry
Dear Alfred
To summarize:
1) When using ZMAT you must specify all atoms in the Z matrix and you must let Dalton determine the symmetry, i.e. it is not allowed to specify any symmetry generators. It is allowed to specify
2) You can see the definition of the symmetryadapted coordinates in the output.
To summarize:
1) When using ZMAT you must specify all atoms in the Z matrix and you must let Dalton determine the symmetry, i.e. it is not allowed to specify any symmetry generators. It is allowed to specify
NOSYMMETRY
. Any dummy atoms are eliminated, so that is not a problem.2) You can see the definition of the symmetryadapted coordinates in the output.

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Re: ZMAT and Symmetry
Hello Mr. Jensen and those who have written to me,
your answer is a bit pitty, however it (seems) that it is simply not possible in Dalton to use generators in connection with a zmatrix, not even with dummy atoms ?!
I have already written, why I am "fond of" zmatrices, although cartesians are prefered.
However your answer is the final one, so I have to look for an alternative.
Thank you and all !
Greetings
Alfred
your answer is a bit pitty, however it (seems) that it is simply not possible in Dalton to use generators in connection with a zmatrix, not even with dummy atoms ?!
I have already written, why I am "fond of" zmatrices, although cartesians are prefered.
However your answer is the final one, so I have to look for an alternative.
Thank you and all !
Greetings
Alfred

 Posts: 326
 Joined: 27 Jun 2013, 18:44
 First name(s): Hans Jørgen
 Middle name(s): Aagaard
 Last name(s): Jensen
 Affiliation: Universith of Southern Denmark
 Country: Denmark
Re: ZMAT and Symmetry
Alfred, but you specify all atoms in a zmatrix, right? As in your example. And as you can see in the second output you uploaded, when you let Dalton detect symmetry, it reduces automatically the total number of atoms to the number of symmetryindependent atoms and generates the remaining atoms with the found symmetry operations. In your example the third Batom is eliminated
and later restored by the symmetry operations (B 2/ 1 and B 2/ 2):
Dalton also tells you the symmetryadapted nuclear coordinates it is using in the C_2v subgroup (the coordinate numbers refers to the numbers in the list of Cartesian coordinates above):
I do not understand what else you want, but if this is not what you want, good luck with finding an alternative.
Code: Select all
Symmetry Independent Centres

B 2 5 : 0.98193050 1.70075351 0.00000000 Isotope 1
B 1 5 : 1.96386100 0.00000000 0.00000000 Isotope 1
Code: Select all
Cartesian Coordinates (a.u.)

Total number of coordinates: 9
B 1 : 1 x 1.9638609965 2 y 0.0000000000 3 z 0.0000000000
B 2/ 1 : 4 x 0.9819304983 5 y 1.7007535125 6 z 0.0000000000
B 2/ 2 : 7 x 0.9819304983 8 y 1.7007535125 9 z 0.0000000000
Code: Select all
Symmetry Coordinates

Number of coordinates in each symmetry: 3 3 2 1
Symmetry A1 ( 1)
1 B 1 x 1
2 B 2 x [ 4 + 7 ]/2
3 B 2 y [ 5  8 ]/2
Symmetry B1 ( 2)
4 B 1 y 2
5 B 2 x [ 4  7 ]/2
6 B 2 y [ 5 + 8 ]/2
Symmetry B2 ( 3)
7 B 1 z 3
8 B 2 z [ 6 + 9 ]/2
Symmetry A2 ( 4)
9 B 2 z [ 6  9 ]/2

 Posts: 10
 Joined: 11 Jan 2014, 13:59
 First name(s): Alfred
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Re: ZMAT and Symmetry
Dear Mr. Hans Jorgen Aagaard Jensen and other USERS,
you are definitely right. I simply have overlooked this fact, that the symmetry coordinates al already present and the unique's also. What I really wanted was, to calculate a molecule  B3  at a certain symmetry and let not Dalton find out the symmetry. That's all. At least I see and see now, that the alternative I was looking for, lies before me.
It would be very kind, if you could tell me, the way I can use the symmetry coordinates to generate the symmetryadapted equations  normal coordinates..
What I know is, that there should be 3 nomalcoordinates of A1 and B1 sym. The irreducible reps are: 2A1+B1, at least for an AB2 molecule.
Moreover if Ia = Ib < Ic, means that it is an oblate rotator. However the calc. at symmetry C2v, hints on an asymetric rotator, because of the moment of inertia at line 3483 which is: Ia < Ib < Ic ?!
Is this an effect of the choosen symmetry  OR ?
Actually I wanted to use D3h, which should be an oblate rotator, but the calculation uses C2v and gives an asymetric rotator. So I 'm a little bit confused about the results, mainly about the different kind of rotators
I am wondering also about the energy of the MOs. There should be also some degenerate orbs (> lines 2908). There are some ALMOST degenerate orbs, which should give E'orbs  but only almost. I expected some MOs of different sym. but with the same energy value. For Instance, E' splits into A1+B2 and A1+B2 of the same energy, must be an E' orb ?
My questions is, how does the reduced symmetry influences the outcome of the calculation ?!
Please give me some hints and tips. Thank you !
Best greetings
Alfred
you are definitely right. I simply have overlooked this fact, that the symmetry coordinates al already present and the unique's also. What I really wanted was, to calculate a molecule  B3  at a certain symmetry and let not Dalton find out the symmetry. That's all. At least I see and see now, that the alternative I was looking for, lies before me.
It would be very kind, if you could tell me, the way I can use the symmetry coordinates to generate the symmetryadapted equations  normal coordinates..
What I know is, that there should be 3 nomalcoordinates of A1 and B1 sym. The irreducible reps are: 2A1+B1, at least for an AB2 molecule.
Moreover if Ia = Ib < Ic, means that it is an oblate rotator. However the calc. at symmetry C2v, hints on an asymetric rotator, because of the moment of inertia at line 3483 which is: Ia < Ib < Ic ?!
Is this an effect of the choosen symmetry  OR ?
Actually I wanted to use D3h, which should be an oblate rotator, but the calculation uses C2v and gives an asymetric rotator. So I 'm a little bit confused about the results, mainly about the different kind of rotators
I am wondering also about the energy of the MOs. There should be also some degenerate orbs (> lines 2908). There are some ALMOST degenerate orbs, which should give E'orbs  but only almost. I expected some MOs of different sym. but with the same energy value. For Instance, E' splits into A1+B2 and A1+B2 of the same energy, must be an E' orb ?
My questions is, how does the reduced symmetry influences the outcome of the calculation ?!
Please give me some hints and tips. Thank you !
Best greetings
Alfred

 Posts: 545
 Joined: 15 Oct 2013, 05:37
 First name(s): Peter
 Middle name(s): Robert
 Last name(s): Taylor
 Affiliation: Tianjin University
 Country: China
Re: ZMAT and Symmetry
Can I start by suggesting you use A1 and B2 for the "inplane" irreducible representations? This is consistent with Mulliken's attempt to standardize these things in 1955. It is essential not to blindly accept the labels the program prints out in the C2v and D2h cases. These are not uniquely definable by the program and are in effect best guesses that are sometimes not consistent with Mulliken's proposal. (See the manual and my forum posting that I referred to before.)
Now, you are absolutely correct that the geometry is wandering away from D3h to C2v, and that the major cause of this relates to your starting with a geometry that would have a partly occupied degenerate orbital. Doing an SCF calculation here is problematic because the program will make a decision about which component of the degenerate orbital it occupies. This is then no longer an irreducible symmetry state (in D3h) and the calculation can only preserve the symmetry for this state, which is C2v. As you see you converge to a state that has a dipole moment.
One way to ensure that you maintain threefold symmetry is to do an MCSCF calculation using the SUPSYM (supersymmetry) option. When the code recognizes that a higher symmetry is present it will attempt to retain that symmetry, although sometimes this can require tightening the (already demanding) default convergence thresholds.
If you are simply experimenting, a possibility might be to do, say, B3+, which I think would be a closedshell system. In this case an SCF geometry optimization should preserve the D3h symmetry. But again this may require tightening some thresholds.
Best regards
Pete
Now, you are absolutely correct that the geometry is wandering away from D3h to C2v, and that the major cause of this relates to your starting with a geometry that would have a partly occupied degenerate orbital. Doing an SCF calculation here is problematic because the program will make a decision about which component of the degenerate orbital it occupies. This is then no longer an irreducible symmetry state (in D3h) and the calculation can only preserve the symmetry for this state, which is C2v. As you see you converge to a state that has a dipole moment.
One way to ensure that you maintain threefold symmetry is to do an MCSCF calculation using the SUPSYM (supersymmetry) option. When the code recognizes that a higher symmetry is present it will attempt to retain that symmetry, although sometimes this can require tightening the (already demanding) default convergence thresholds.
If you are simply experimenting, a possibility might be to do, say, B3+, which I think would be a closedshell system. In this case an SCF geometry optimization should preserve the D3h symmetry. But again this may require tightening some thresholds.
Best regards
Pete

 Posts: 10
 Joined: 11 Jan 2014, 13:59
 First name(s): Alfred
 Last name(s): Güthler
 Affiliation: privat
 Country: Germany
Re: ZMAT and Symmetry
Dear Mr Taylor and other DALTONUsers,
thank you for your valuable suggestions, but I was already aware of that fact, that a JTEffect could appear. My problem in this case is to distinguish between an JTeffect and the reduction of the D3hgroup to C2v by the the program. However, you wrote on the use of symmetry in the DALTON manual and I have to read this more deeply to understand the use of Generators correctly.
To be more concrete, of course I accept the labeling of an 3atomic Group: 2A_1+B_2, (C2v).
The Keyword "SUBSYM" is new to me. Does it prevent an "automatic" reduction of the D3hgroup AND keeps the calculation to be C2v, if it is really an JTEffect, otherwise it would be D3h ? Understanding this fact correctly, is essential to me, because I need to know if the reduction of symmetry is due to the JTeffect, or due to using a subgroup of D3h.
This effect appears only if degenerate orbitals are not equal occupied. I looked at the HOMOLUMO gap and saw that there are NO degenarte orbitals. Of course C2v has no degenerate orbitals, however if the energy are almost equal, despite the fact, that the symmetry is different, is this then an hint for the possibility of an JTEffect ?
It would be very kind to get answer to this.
Thank you !
Kind and best greetings
Alfred
thank you for your valuable suggestions, but I was already aware of that fact, that a JTEffect could appear. My problem in this case is to distinguish between an JTeffect and the reduction of the D3hgroup to C2v by the the program. However, you wrote on the use of symmetry in the DALTON manual and I have to read this more deeply to understand the use of Generators correctly.
To be more concrete, of course I accept the labeling of an 3atomic Group: 2A_1+B_2, (C2v).
The Keyword "SUBSYM" is new to me. Does it prevent an "automatic" reduction of the D3hgroup AND keeps the calculation to be C2v, if it is really an JTEffect, otherwise it would be D3h ? Understanding this fact correctly, is essential to me, because I need to know if the reduction of symmetry is due to the JTeffect, or due to using a subgroup of D3h.
This effect appears only if degenerate orbitals are not equal occupied. I looked at the HOMOLUMO gap and saw that there are NO degenarte orbitals. Of course C2v has no degenerate orbitals, however if the energy are almost equal, despite the fact, that the symmetry is different, is this then an hint for the possibility of an JTEffect ?
It would be very kind to get answer to this.
Thank you !
Kind and best greetings
Alfred
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