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additional update
by tesfaye mamuye 10 Sep '14

10 Sep '14

Dear ALPS Developers/users I am using alps CT-QMC(Hybridization expansion) as impurity solver. But the total acceptance rate is getting low and low, and I'm wondering if it may affect the final result. In order to overcome this, I'm thinking to use additional update: 'shift of the segment end point' which is used to enhance the sampling efficiency (to increase the acceptance ratio and reduce auto-correlation times) which do not change the expansion order. However, I didn't get in the documentation on how to implement this additional update as control parameter 'SPINFLIP'={0.1} : specify to whether to perform spin flip update. Is it possible to use for shift of the segment end point update 'N_SHIFT'=1 ? I need help on how to implement this additional update. here is the sample |------ Simulation details (master only) after 2665880 sweeps ------| Total acceptance rate = 22.92% Individual acceptance rate for update change zero state = 0.00% (proposal rate = 10.00%) insert segment = 6.37% (proposal rate = 25.00%) remove segment = 6.37% (proposal rate = 25.00%) insert anti-segment = 5.09% (proposal rate = 20.00%) remove anti-segment = 5.09% (proposal rate = 20.00%) swap segment = 0.00% (proposal rate = 0.00%) |-----------------------------------------------------------------| Thank you for your time

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Help with LiveALPS
by Niraj Ghimire 10 Sep '14

10 Sep '14

Dear ALPS users, I have been running LiveALPS from my USB flash drive and I can't get access to the internet. I could not find any information anywhere on how to resolve this issue. Would you be able to help me out ? ~Niraj

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(no subject)
by hamid mosadegh 07 Sep '14

07 Sep '14

Dear All What is the momentum of , E_1(N_{\uparrow},(N_{\downarrow} ,L), the lowest-excitation energy in the Hubbard model in the S=0 sector. I means, I want to evaluate the velocity of charge excitation from E_1(N_{\uparrow},(N_{\downarrow} ,L)- E_0(N_{\uparrow},(N_{\downarrow} ,L) where E_1(N_{\uparrow},(N_{\downarrow} ,L) is the lowest-excitation energy with momentum k=2\pi/L and total spin S=0. If i run MODEL="fermion Hubbard" CONSERVED_QUANTUMNUMBERS="Nup,Ndown" Ndown_total=10;Nup_total=10 NUMBER_EIGENVALUES=2 what is the momentum of first excitation. is it possible to determine or change it? thanks H. Mosadeq Shahr-e-Kord University (SKU), Shahr-e-Kord , Iran

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the parity sector in DMRG
by hamid mosadegh 03 Sep '14

03 Sep '14

Dear all Is it possible to do the DMRG or ED calculations of fermion Hubbard model in the odd(even) parity sector using ALPS? I need to determine the minimum energies for eigenstates in the even and odd parity sectors. Thanks -- H. Mosadeq Shahr-e-Kord University (SKU), Shahr-e-Kord , Iran

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parallel calculation on Windows
by orchidwood 01 Sep '14

01 Sep '14

Dear all, Can the ALPS on 64-bit Windows 7 do parallel calculation with Vistrail system? If it can do this, how?Thanks,Mu Lan

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Re: [ALPS-users] Comp-phys-alps-users Digest, Vol 101, Issue 14
by bonanjiang＠gmail.com 26 Aug '14

26 Aug '14

Hi, Matthias, Since you said sparsediag was the code to use, then how to set the number of eignstates to calculate in the parameter file or command line? I can not find the answer in the turtorials. Bo-Nan bonanjiang(a)gmail.com From: comp-phys-alps-users-request Date: 2014-08-26 18:00 To: comp-phys-alps-users Subject: Comp-phys-alps-users Digest, Vol 101, Issue 14 Send Comp-phys-alps-users mailing list submissions to comp-phys-alps-users(a)lists.phys.ethz.ch To subscribe or unsubscribe via the World Wide Web, visit https://lists.phys.ethz.ch/listinfo/comp-phys-alps-users or, via email, send a message with subject or body 'help' to comp-phys-alps-users-request(a)lists.phys.ethz.ch You can reach the person managing the list at comp-phys-alps-users-owner(a)lists.phys.ethz.ch When replying, please edit your Subject line so it is more specific than "Re: Contents of Comp-phys-alps-users digest..." Today's Topics: 1. Help with LiveALPS (Niraj Ghimire) 2. a question about fulldiag (bonanjiang(a)gmail.com) 3. Re: a question about fulldiag (Matthias Troyer) ---------------------------------------------------------------------- Message: 1 Date: Mon, 25 Aug 2014 12:08:57 -0400 From: Niraj Ghimire <niraj.ghimire(a)gmail.com> To: "comp-phys-alps-users(a)lists.comp-phys.org" <comp-phys-alps-users(a)lists.comp-phys.org> Subject: [ALPS-users] Help with LiveALPS Message-ID: <CACMdJOXGtHsttzbF-+1LLxL3OmvdnntrTvcQjiJ9fsCXUGk1jQ(a)mail.gmail.com> Content-Type: text/plain; charset="utf-8" Dear ALPS users, I have been running LiveALPS from my USB flash drive and I can't get access to the internet. I could not find any information anywhere on how to resolve this issue. Would you be able to help me out ? ~Niraj -------------- next part -------------- An HTML attachment was scrubbed... URL: <https://lists.phys.ethz.ch/pipermail/comp-phys-alps-users/attachments/20140…> ------------------------------ Message: 2 Date: Tue, 26 Aug 2014 14:41:48 +0800 From: "bonanjiang(a)gmail.com" <bonanjiang(a)gmail.com> To: comp-phys-alps-users <comp-phys-alps-users(a)lists.phys.ethz.ch> Subject: [ALPS-users] a question about fulldiag Message-ID: <201408261441449660406(a)gmail.com> Content-Type: text/plain; charset="us-ascii" Hi, all I currently tend to calculate the energy of the lowest 4 states of a fermion system with fulldiag of ALPS. However, since the system has 12 eigenstates in total and fulldiag gives all of them, the time it takes is a little bit long. So, my question is: is there any way to limit the number of eigenstates that fulldiag calculates? for example, the lowest 4 states. Bo-Nan bonanjiang(a)gmail.com -------------- next part -------------- An HTML attachment was scrubbed... URL: <https://lists.phys.ethz.ch/pipermail/comp-phys-alps-users/attachments/20140…> ------------------------------ Message: 3 Date: Tue, 26 Aug 2014 08:56:54 +0200 From: Matthias Troyer <troyer(a)phys.ethz.ch> To: "comp-phys-alps-users(a)lists.phys.ethz.ch" <comp-phys-alps-users(a)lists.phys.ethz.ch> Subject: Re: [ALPS-users] a question about fulldiag Message-ID: <362E955B-595C-4AE6-A412-27EA6F8EEA63(a)phys.ethz.ch> Content-Type: text/plain; charset="us-ascii" sparsediag is the code to use in this case > On Aug 26, 2014, at 8:41, "bonanjiang(a)gmail.com" <bonanjiang(a)gmail.com> wrote: > > Hi, all > I currently tend to calculate the energy of the lowest 4 states of a fermion system with fulldiag of ALPS. However, since the system has 12 eigenstates in total and fulldiag gives all of them, the time it takes is a little bit long. So, my question is: is there any way to limit the number of eigenstates that fulldiag calculates? for example, the lowest 4 states. > > Bo-Nan > bonanjiang(a)gmail.com -------------- next part -------------- An HTML attachment was scrubbed... URL: <https://lists.phys.ethz.ch/pipermail/comp-phys-alps-users/attachments/20140…> End of Comp-phys-alps-users Digest, Vol 101, Issue 14 *****************************************************

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a question about fulldiag
by bonanjiang＠gmail.com 26 Aug '14

26 Aug '14

Hi, all I currently tend to calculate the energy of the lowest 4 states of a fermion system with fulldiag of ALPS. However, since the system has 12 eigenstates in total and fulldiag gives all of them, the time it takes is a little bit long. So, my question is: is there any way to limit the number of eigenstates that fulldiag calculates? for example, the lowest 4 states. Bo-Nan bonanjiang(a)gmail.com

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Help with LiveALPS
by Niraj Ghimire 20 Aug '14

20 Aug '14

Dear alps users, I am running LiveALPS on my computer and I cannot get access to the internet. I could not find any information anywhere on it. Could you please let me know how I can fix this issue ? ~Niraj

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Computing correlations with sparsediag
by Eoin Quinn 18 Aug '14

18 Aug '14

Dear ALPS users, I have a question regarding the calculation of correlations with sparsediag. Let me take for an example a one-dimensional ferromagnetic spin chain of length 4. I will not impose ‘Sz’ conservation as the problem I am having arises when ‘Sz’ is not conserved, once I impose it all works well. The following are my input parameters for the calculation: <PARAMETERS> <PARAMETER name="MEASURE_CORRELATIONS[Offdiagonal spin correlations]">Splus:Sminus</PARAMETER> <PARAMETER name="NUMBER_EIGENVALUES">1</PARAMETER> <PARAMETER name="BC">periodic</PARAMETER> <PARAMETER name="local_S">0.5</PARAMETER> <PARAMETER name="J">-1</PARAMETER> <PARAMETER name="L">4</PARAMETER> <PARAMETER name="LATTICE">chain lattice</PARAMETER> <PARAMETER name="SEED">911696590</PARAMETER> <PARAMETER name="W">1</PARAMETER> <PARAMETER name="MODEL">spin</PARAMETER> <PARAMETER name="MODEL_LIBRARY">aux/models.xml</PARAMETER> <PARAMETER name="LATTICE_LIBRARY">aux/lattices.xml</PARAMETER> <PARAMETER name="MEASURE_CORRELATIONS[Diagonal spin correlations]">Sz</PARAMETER> <PARAMETER name="CONSERVED_QUANTUMNUMBERS"></PARAMETER> <PARAMETER name="TOTAL_MOMENTUM">4.712388980384689674</PARAMETER> </PARAMETERS> The output for the ground state is then <EIGENSTATES number="1"> <QUANTUMNUMBER name="TOTAL_MOMENTUM" value="0"/> <EIGENSTATE number="0"> <SCALAR_AVERAGE name="Energy"><MEAN>-1</MEAN></SCALAR_AVERAGE> <VECTOR_AVERAGE name="Diagonal spin correlations"> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 0 )"><MEAN>0.24999999999999997</MEAN></SCALAR_AVERAGE> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 1 )"><MEAN>0.0078869096162297037</MEAN></SCALAR_AVERAGE> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 2 )"><MEAN>0.0078869096162296551</MEAN></SCALAR_AVERAGE> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 3 )"><MEAN>0.0078869096162297071</MEAN></SCALAR_AVERAGE> </VECTOR_AVERAGE> <VECTOR_AVERAGE name="Offdiagonal spin correlations"> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 0 )"><MEAN>0.55359228338317501</MEAN></SCALAR_AVERAGE> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 1 )"><MEAN>0.24211309038377027</MEAN></SCALAR_AVERAGE> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 2 )"><MEAN>0.2421130903837703</MEAN></SCALAR_AVERAGE> <SCALAR_AVERAGE indexvalue="( 0 ) -- ( 3 )"><MEAN>0.24211309038377027</MEAN></SCALAR_AVERAGE> </VECTOR_AVERAGE> </EIGENSTATE> </EIGENSTATES> The problem I am having is that the diagonal and off diagonal components of the ( 0 ) -- ( 0 ) spin correlations do not sum to 3/4 = S(S+1) = \vec{S}.\vec{S}, whereas I would expect them to. Do I understand correctly that the "Diagonal spin correlations" give <S^z(i) S^z(j)> and the "Offdiagonal spin correlations" give <S^x(i) S^x(j) + S^y(i) S^y(j)>? With best regards, Eoin

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query regarding building of structure and its consideration in the simulation
by D´Souza, Sunil Wilfred 17 Aug '14

17 Aug '14

Dear ALPS users, I am just a beginner in the ALPS simulations.The description of my question is as follows: As a practice exercise, I want to run the quantum montecarlo simulations using the spin ½ model defined in the ALPS library to calculate the Curie temperature (Tc) of bulk Iron metal by considering the BCC structure with a = 286.65 pm. Therefore, I have generated a BCC structure with first and second nearest neighbors. I have attached the structure related xml file “Fe.xml” along with this mail for your consideration. Firstly, It is well known that the BCC structure has 8 first nearest neighbors placed at ( ±½, ±½, ±½ ) positions with respect to the vertex at (0, 0, 0). In the xml file I have defined only two nearest neighbor edges connecting the (0, 0, 0) to (½, ½, ½ ) and (-½, ½, ½ )positions in the unit cell. Thereafter, when i viewed the structure by using lattice-preview program in the periodic boundary condition with L=2,it is showing all 8 nearest neighbors. My question is that due to the definition of periodicity does the simulation takes into account all 8 nearest neighbors even though if I don't define the 6 other nearest neighbors explicitly in the unit cell. Secondly, along with the above mentioned issue on the definition of the structure, I am kindly requesting you to provide me some insights regarding how to calculate the Curie temperature of Iron metal by using the ALPS simulation in terms of which programs I have to use in order to accomplish this task. A simple example describing this task would be of very great help. If you are finding my questions being trivial ones, I apologize to you for that. But, these confusions and problems are stopping me to proceed further, I read through the web documentation of ALPS, but could not solve them. Hence, I am kindly looking forward for your suggestions and help. I thank you in advance for your time and consideration. With best regards Sunil Max Planck Institute, Dresden

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