Hi Ryo IGARASHI,
First of all I would like to thank ALPS team for your collaboration and support. Thanks for your comments and explanations. Please find my questions below.
Thanks, Ruben
________________________________ From: Ryo IGARASHI rigarash@hosi.phys.s.u-tokyo.ac.jp To: comp-phys-alps-users comp-phys-alps-users@lists.phys.ethz.ch Sent: Tue, December 21, 2010 2:12:40 AM Subject: Re: [ALPS-users] anisotropic Heisenberg model setup
Hi, Ruben,
On Mon, Dec 20, 2010 at 8:36 PM, Ruben Ghulghazaryan ghulr@yahoo.com wrote:
By anisotropic model on a ladder I mean that we have two different interaction terms in Hamiltonian as shown in the file attached. Thus we have two bond types "0" and "1" in horizontal and vertical directions and we want to have different interaction terms in horizontal and vertical directions.
This is usual setting, I believe. [Ruben]---> Yes.
As far as I understand, according to "spin" model definition J0 parameter should be for bond type="0" and J1 for bonds type="1".
This is correct.
[Ruben]---> OK.
If we set J=1 only then (even if J1=0) all bonds irrespective of their type should get 1 as bond term, don't you?
This is simply wrong. J=1 just means J0=1. As Matthias said, all the J1 term is 0 (J1=Jz1=Jxy1=0) when only J=1 is set.
[Ruben]--->OK. Does it mean that all bond interaction terms will be the same both in horizontal and vertical directions? Or, does it mean we have only horizontal bonds with interaction J0=1 and vertical bonds terms = 0 and we have two non-interacting chains instead of a ladder?
In the case we specify J0=1 and J1=1 again all bonds should get 1 as bond term and the model results should be the same as for setting J only. In this case both isotropic and anisotropic models should give similar results.
J0=1 and J1=1 give the symmetric ladder. And thus, you will get the same results if you specify a symmetric ladder (i.e. all coupling constant is same IN THE LATTICE DESCRIPTION) with J=1.
[Ruben]---> I agree and this is what I expect too. I did an experiment with the following settings which as far as I understand both should set the same bond interaction term 1 for all bonds:
(a) LATTICE="ladder" MODEL="spin" local_S = 1 J = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4}
++++++++++++++++++++++++ (b)
LATTICE="ladder" MODEL="spin" local_S = 1 J0 = 1 J1 = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4}
========================== I run fulldiag then fulldiag_evaluate
fulldiag parm1.in.xml
fulldiag_evaluate --T_MIN 0.1 --T_MAX 10 --DELTA_T 0.1 parm1.task1.out.xml
The results of these two runs are different. I can not understand why the results should be different.
Does (a) defines a ladder were both horizontal and vertical bond terms are equal 1? Does (b) defines a ladder were both horizontal and vertical bond terms are equal since J0=J1=1? Please see the file attached for illustration. If these two parameter files use different settings but define the same lattice and model, then why the results of fulldiag runs are different?
Please let me now if you need more data. Thanks!
Hi, Ruben,
On Tue, Dec 21, 2010 at 5:31 PM, Ruben Ghulghazaryan ghulr@yahoo.com wrote:
This is simply wrong. J=1 just means J0=1. As Matthias said, all the J1 term is 0 (J1=Jz1=Jxy1=0) when only J=1 is set.
[Ruben]--->OK. Does it mean that all bond interaction terms will be the same both in horizontal and vertical directions? Or, does it mean we have only horizontal bonds with interaction J0=1 and vertical bonds terms = 0 and we have two non-interacting chains instead of a ladder?
Simply the latter case. Just check the result. You can check the result with chain lattice.
The coupling on the vertical bonds are defined by J1. Check the description of "ladder" lattice in lattices.xml you use.
In the case we specify J0=1 and J1=1 again all bonds should get 1 as bond term and the model results should be the same as for setting J only. In this case both isotropic and anisotropic models should give similar results.
J0=1 and J1=1 give the symmetric ladder. And thus, you will get the same results if you specify a symmetric ladder (i.e. all coupling constant is same IN THE LATTICE DESCRIPTION) with J=1.
[Ruben]---> I agree and this is what I expect too. I did an experiment with the following settings which as far as I understand both should set the same bond interaction term 1 for all bonds:
(a) LATTICE="ladder" MODEL="spin" local_S = 1 J = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4} ++++++++++++++++++++++++ (b) LATTICE="ladder" MODEL="spin" local_S = 1 J0 = 1 J1 = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4} ========================== I run fulldiag then fulldiag_evaluate fulldiag parm1.in.xml fulldiag_evaluate --T_MIN 0.1 --T_MAX 10 --DELTA_T 0.1 parm1.task1.out.xml The results of these two runs are different. I can not understand why the results should be different.
Does (a) defines a ladder were both horizontal and vertical bond terms are equal 1? Does (b) defines a ladder were both horizontal and vertical bond terms are equal since J0=J1=1? Please see the file attached for illustration. If these two parameter files use different settings but define the same lattice and model, then why the results of fulldiag runs are different?
I am repeatedly telling you that you should check the lattice description of "ladder" in lattices.xml, which shows that bond types are different from each other for horizontal and vertical bond. So the short answer for your questions are: (a) No. (b) Yes. (c) You are just simulating different setting.
Best regards,
Dear Ryo,
Indeed the results of ladder simulation with J=1 setting only corresponds to chain lattice simulation. It is clear now. Thanks a lot for your explanation.
With best regards, Ruben
________________________________ From: Ryo IGARASHI rigarash@hosi.phys.s.u-tokyo.ac.jp To: comp-phys-alps-users@lists.phys.ethz.ch Sent: Wed, December 22, 2010 7:28:46 AM Subject: Re: [ALPS-users] Fw: anisotropic Heisenberg model setup
Hi, Ruben,
On Tue, Dec 21, 2010 at 5:31 PM, Ruben Ghulghazaryan ghulr@yahoo.com wrote:
This is simply wrong. J=1 just means J0=1. As Matthias said, all the J1 term is 0 (J1=Jz1=Jxy1=0) when only J=1 is set.
[Ruben]--->OK. Does it mean that all bond interaction terms will be the same both in horizontal and vertical directions? Or, does it mean we have only horizontal bonds with interaction J0=1 and vertical bonds terms = 0 and we have two non-interacting chains instead of a ladder?
Simply the latter case. Just check the result. You can check the result with chain lattice.
The coupling on the vertical bonds are defined by J1. Check the description of "ladder" lattice in lattices.xml you use.
In the case we specify J0=1 and J1=1 again all bonds should get 1 as bond term and the model results should be the same as for setting J only. In this case both isotropic and anisotropic models should give similar results.
J0=1 and J1=1 give the symmetric ladder. And thus, you will get the same results if you specify a symmetric ladder (i.e. all coupling constant is same IN THE LATTICE DESCRIPTION) with J=1.
[Ruben]---> I agree and this is what I expect too. I did an experiment with the following settings which as far as I understand both should set the same bond interaction term 1 for all bonds:
(a) LATTICE="ladder" MODEL="spin" local_S = 1 J = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4} ++++++++++++++++++++++++ (b) LATTICE="ladder" MODEL="spin" local_S = 1 J0 = 1 J1 = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4} ========================== I run fulldiag then fulldiag_evaluate fulldiag parm1.in.xml fulldiag_evaluate --T_MIN 0.1 --T_MAX 10 --DELTA_T 0.1 parm1.task1.out.xml The results of these two runs are different. I can not understand why the results should be different.
Does (a) defines a ladder were both horizontal and vertical bond terms are equal 1? Does (b) defines a ladder were both horizontal and vertical bond terms are equal since J0=J1=1? Please see the file attached for illustration. If these two parameter files use different settings but define the same lattice and model, then why the results of fulldiag runs are different?
I am repeatedly telling you that you should check the lattice description of "ladder" in lattices.xml, which shows that bond types are different from each other for horizontal and vertical bond. So the short answer for your questions are: (a) No. (b) Yes. (c) You are just simulating different setting.
Best regards,
comp-phys-alps-users@lists.phys.ethz.ch