Keep Sz non-fixed in Canonical Hubbard

[sankrad]

Dear Developers,

Thank you very much for your great code!

Could you please help me with an issue? The manual reads that in the canonincal regime (expert mode) I migth chose not to specify total Sz. However when I am trying to do so, just skipping the line in modpara.def, I am getting an error.

My modpara.def:

--------------------
Model_Parameters
--------------------
Canonical Hubbard
--------------------
CDataFileHead   lgm
--------------------
--------------------
Nsite   4
Ncond 4
Lanczos_max   999
initial_iv -2
exct   1
LanczosEps   12
LanczosTarget   2

The error:

"Error: in sz.
imax = 2, Check.idim_max=1"

Kind regards, Anton.

[k-yoshimi]

@sankrad I'd like to check your input files. After compressing all the input files into a zip file, could you upload it here ?

[sankrad]

Dear k-yoshimi,

Thank you for the fast response!

hphi_configs.tar.gz

[k-yoshimi]

@sankrad Did you use 16 processes for your calculation ?

[sankrad]

Yes, exactly...

[k-yoshimi]

Please use 1 process or 4 processes for this calculation, since this system is too small for 16 processes parallelization.

In the following, the process element info (written in standard output) for16 processes parallelization is shown.

INTRA process site
    Site    Bit
       0       4
       1       4

  INTER process site
    Site    Bit
       2       4
       3       4

  Process element info
    Process       Dimension   Nup  Ndown  Nelec  Total2Sz   State
          0               1     0      0      4         0   0000
          1               4     0      0      3         0   0100
          2               4     0      0      3         0   1000
          3               6     0      0      2         0   1100
          4               4     0      0      3         0   0001
          5               6     0      0      2         0   0101
          6               6     0      0      2         0   1001
          7               4     0      0      1         0   1101
          8               4     0      0      3         0   0010
          9               6     0      0      2         0   0110
         10               6     0      0      2         0   1010
         11               4     0      0      1         0   1110
         12               6     0      0      2         0   0011
         13               4     0      0      1         0   0111
         14               4     0      0      1         0   1011
         15               1     0      0      0         0   1111

You can see 0 electron state at 15 process. In HPhi, the process number is used to represent the electron states. At 15 process, 4 electrons are assigned(15=3*4^1+3*4^0, here 3*4^n means doubloon state at n-th site). So the inner process cannot have any electron and HPhi fails the calculation.

[sankrad]

Oh, that was a stupid one! Thank you for the very fast assistance.