Comp-phys-alps-users December 2010

comp-phys-alps-users@lists.phys.ethz.ch

14 participants
13 discussions

stiffness?
by Xiao Quan Yu 27 Dec '10

27 Dec '10

Dear all, I am now studying the 3D bose glass model. I got some confusions about the stiffness. By using SSE, I got the stiffness is 2.2427 from the following setting: LATTICE_LIBRARY="lattices.xml"; LATTICE="inhomogeneous simple cubic lattice"; L=5; MODEL_LIBRARY="models.xml"; MODEL="hardcore boson"; t = 1.0; T = 0.1; delta=10.0; SWEEPS=500000; THERMALIZATION=10000; EPSILON=0.25; { DISORDERSEED = 34275; mu=delta*2*(random()-0.5); } My question is: if this stiffness is rescaled one, namely ρs*L^4=2.2427 ? The power 4th is from Fisher argument. Thank you in advance! Best, Xiaoquan ---------------------------------------------------------------- SISSA Webmail https://webmail.sissa.it/ Powered by SquirrelMail http://www.squirrelmail.org/

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Re: [ALPS-users] what is the prefactor of stiffness?
by Fan Xiao 23 Dec '10

23 Dec '10

As my understanding, the "stiffness" extracted is actually (rho*L^{d-1}) for square lattice or 3d simple cubes. The real one should be 2.2427/5 in your case. But I have a question about the stiffness as well. I used a 8x8x2 lattice, how is the simulated "stiffness" related to the actual one? (if it is 8x8x8, I know it is rho*8, but this is not a simple cubic, is there an expression to calculated it?)

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Fw: anisotropic Heisenberg model setup
by Ruben Ghulghazaryan 22 Dec '10

22 Dec '10

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(a)hosi.phys.s.u-tokyo.ac.jp> To: comp-phys-alps-users <comp-phys-alps-users(a)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(a)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! -- Ryo IGARASHI, Ph.D. rigarash(a)hosi.phys.s.u-tokyo.ac.jp OpenPGP fingerprint: BAD9 71E3 28F3 8952 5640 6A53 EC79 A280 6A19 2319

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anisotropic Heisenberg model setup
by Ruben Ghulghazaryan 21 Dec '10

21 Dec '10

Hi ALPS team, I am trying to make simple simulations with anisotropic Heisenberg model on a ladder with ALPS2 rc2. According to models.xml and full diagonalization web page if I specify J0 and J1 parameters then for different bond types "0" and "1" I will have J0 and J1 coefficients in Heisenberg ("spin") model. I did a simple experiment by performing fulldiag two simulations: one by using J=1 and the other J0=1,J1=1 and compared the results. I expected that the results should be the same according to "spin" model definition: defaults for J0and J1 is J if they are not specified. But I found that the results of energy and other quantities are different. below are my two parameter files: LATTICE="ladder" MODEL="spin" local_S = 1 J = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4} ++++++++++++++++++++++++ LATTICE="ladder" MODEL="spin" local_S = 1 J0 = 1 J1 = 1 CONSERVED_QUANTUMNUMBERS="Sz" {L = 4} ========================== I run fulldiag then fulldiag_evaluate fulldiag_evaluate --T_MIN 0.1 --T_MAX 10 --DELTA_T 0.1 parm6a.task1.out.xml as given in the web page http://alps.comp-phys.org/mediawiki/index.php/ALPS_2_Tutorials:ED-06_FullDi… Would you please help me to understand why the results are different? Thanks, Ruben

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what is the prefactor of stiffness?
by Xiao Quan Yu 14 Dec '10

14 Dec '10

Dear all, I am now studying the 3D bose glass model. By using SSE, I got the stiffness is 2.2427 from the following setting: LATTICE_LIBRARY="lattices.xml"; LATTICE="inhomogeneous simple cubic lattice"; L=5; MODEL_LIBRARY="models.xml"; MODEL="hardcore boson"; t = 1.0; T = 0.1; delta=10.0; SWEEPS=500000; THERMALIZATION=10000; EPSILON=0.25; { DISORDERSEED = 34275; mu=delta*2*(random()-0.5); } My question is: why this stiffness is larger than one? Is there some scaling prefactor in front? Thank you in advance! Best, Xiaoquan ---------------------------------------------------------------- SISSA Webmail https://webmail.sissa.it/ Powered by SquirrelMail http://www.squirrelmail.org/

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Can the DMRG packet of ALPS calculate the finite temperature problem?
by Chen Ahai 13 Dec '10

13 Dec '10

dear all my question just as title. thank you -- -------------------------------------------------------------------------- Chen Ahai Department of Physics, Zhejiang Normal University Jinhua, Zhejiang Province, 321004, China e-mail: chenahaiphysics(a)gmail.com; chenahai(a)yahoo.cn

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ALPS 2.0.0rc4 packages for Debian (amd64)
by Ryo IGARASHI 13 Dec '10

13 Dec '10

Hi, all, As always, I have created binary packages of ALPS 2.0.0rc4 for Debian GNU/Linux (sid/unstable) for amd64. The apt line is as follows: deb http://alps.comp-phys.org/static/software/debian/sid ./ deb-src http://alps.comp-phys.org/static/software/debian/sid ./ The largest change from previous versions is 'dirloop_sse'. As dirloop_sse do not depend upon lp_solve any more, I can easily include it in the package. If you find any problems regarding this package, please let me know. Best regards, -- Ryo IGARASHI, Ph.D. rigarash(a)hosi.phys.s.u-tokyo.ac.jp OpenPGP fingerprint: BAD9 71E3 28F3 8952 5640 6A53 EC79 A280 6A19 2319

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Problem with ALPS installation
by MSzynisz 12 Dec '10

12 Dec '10

Hello, I have a problem installing the ALPS. Ubuntu version: 10.10 ALPS version: 1.3.5 Boost version: 1.45.0 gcc version 4.4.5 (Ubuntu/Linaro 4.4.4-14ubuntu5) Streamed what I can see in terminal in attachment. Used command sudo make Please help. Best, Marcin

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Fwd: Help about spin Hamiltonian set up
by Wei Wu 12 Dec '10

12 Dec '10

Dear Sir/Madam, Could you help me with this? I need to define a two-dimensional lattice with canted spin structure. In each chain, we have stronger FM interaction, and weak AFM interaction between chains, and each column has different anisotropic directions one is X and the other is Z, i.e., they are vertical to each. How could I define the lattice and model for this system. I hope I have made clear my question. Looking forward to your reply! Best regards, Wei ---------- Forwarded message ---------- From: Fabien Alet <alet(a)irsamc.ups-tlse.fr> Date: 6 December 2010 14:13 Subject: Re: Help about spin Hamiltonian set up To: Wei Wu <wuwei515(a)googlemail.com> Hi, I suspect that you just have to define a new lattice with two sites per unit cell, and use two different sets of Jx,Jy,Jz for each unit cell. But since I'm not sure exactly what you wish to do, and since this appears to be a generic ALPS question, I suggest you send a (maybe more precise) question to the ALPS users mailing list. comp-phys-alps-users(a)lists.phys.ethz.ch Best Fabien Le 6 déc. 10 à 15:06, Wei Wu a écrit : > Hi, > > Yeah, I mean ALPS. > > Thanks, > Wei > > On 6 December 2010 08:43, ALET Fabien <alet(a)irsamc.ups-tlse.fr> wrote: >> >> Hi, >> >> are you talking about doing this with ALPS, or is it about something else ? >> >> Best >> Fabien >> >> >> On 12/03/2010 08:14 PM, Wei Wu wrote: >>> >>> Dear Dr. Alet Fabien, >>> >>> My name is Wei Wu. I am a postdoc researcher at Imperial College >>> London, UK. I am trying to set up a spin Hamiltonian of spin ladder or >>> coupled spin ladder with canted spins, i.e., the spin easy axes on two >>> legs are different and they are perpendicular to each other. Do you >>> think I can set up a Hamiltonian for this? and if possible, how can I >>> set that up? >>> >>> Many thanks! >>> >>> Best regards, >>> Wei >>> >>> >> >>

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Defining pair correlation for fermions?
by Meng Cheng 04 Dec '10

04 Dec '10

Hi, I'm a new user of ALPS and I have some trouble with defining pair correlations. The problem I'm studying is spin-1/2 fermions in one-dimension. I want to measure the pair correlation, namely the correlation function of c_up(i) c_down(i). I learnt from the ALPS tutorial that to measure the correlation of certain operator I have to define it in "sitebasis" first. Naively I can define the pair operator as <OPERATOR name="pair" matrixelement="1"> <CHANGE quantumnumber="Nup" change="-1"/> <CHANGE quantumnumber="Ndown" change="-1"/> </OPERATOR> and I add this to the definition of sitebasis of "fermion". It seems the only way to implement this operator, but what confuses me is the sign: how do this definition know that c_up and c_down are anti-commuting? It seems to me that the matrixelement can be 1 or -1, depending on how the basis state is defined. So anyone can tell me whether this is the right way to go? Thanks a lot. with all best, Meng Cheng Graduate Research Assistant Department of Physics University of Maryland College Park http://terpconnect.umd.edu/~chmeng

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Comp-phys-alps-users December 2010