Multiply out product of sums

Question

For a specific quantum mechanical problem I need to multiply out operators in order to calculate a trace by hand. For example I need a Hamiltonian squared with $H^2$. The Hamiltonian contains of a few single terms and flip-flop terms embedded in sums like $\frac{1}{2} \sum_{m=1}^N S_0^+ S_m^- + S_0^- S_m^+$.

My first idea was simply to write down the sums and ask Mathematica to multiply them out by expanding them. Unfortunately, this yields a result I could have imagined myself with $\sum_m (...) \times \sum_n (...)$. What I need is a sum of those product terms for example of the form $\sum_{m=1,n=1}^N S_0^+ S_m^-S_0^+ S_n^- + \sum_{m=1,n=1}^N (...)$ as I want to evaluate each operator separately by hand.

Changing the sum parenthesis doesn't seem to have any effect. Expand refuses to work.

Is there a possibility to let Mathematica multiply out all those terms individually?

Remark: As specific operators don't commute, e.g. $[S^+,S^-]$ for the same index I need Mathematica to give out each term. Simplifying by changing operator positions could lead to false results, so something like $(a+b)*(a+b)=a^2+ab+ba+b^2$ is needed here.

Edit: Here is a minimum example of what doesn't work as expected:

Expand[(-h Subscript[S^z, 0] - 
Sum[Subscript[J, 
  m] (Subscript[S, 0]^z Subscript[S, m]^z + 
    1/2 SuperPlus[Subscript[S, 0]] SuperMinus[Subscript[S, m]] + 
    1/2 SuperMinus[Subscript[S, 0]] SuperPlus[Subscript[S, 
     m]]), {m, 1, NN}]) (-h Subscript[S^z, 0] - 
Sum[Subscript[J, 
  n] (Subscript[S, 0]^z Subscript[S, n]^z + 
    1/2 SuperPlus[Subscript[S, 0]] SuperMinus[Subscript[S, n]] + 
    1/2 SuperMinus[Subscript[S, 0]] SuperPlus[Subscript[S, 
     n]]), {n, 1, NN}])]

If you can provide a minimum (non-)working example it would be great. — Sektor, Apr 22 '15 at 15:56
Welcome to Mathematica.SE! I suggest the following: 0) Browse the common pitfalls question 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the [faq]! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! — Dr. belisarius, Apr 22 '15 at 16:15
Related: Pull out scalars from NonCommutativeMultiply in commutator of SU2 spin algebra, and Traces of products of Pauli matrices — Jens, Apr 22 '15 at 19:40
@chuy It has to remain that way. The variable is not known beforehand. — pbx, Apr 23 '15 at 11:34
Note that even if the operators commuted, products of "simple" sums without definite bounds aren't automatically expanded by Expand either. (I think this is what @chuy was driving at.) For example, Expand[Sum[Subscript[a, i], {i, 1, NN}] Sum[Subscript[b, i], {i, 1, NN}]] just returns the product of the two sums. So there are two separate issues here: expanding an indefinite sum, and making sure operator order is preserved in the process. — Michael Seifert, Aug 20 '15 at 20:53

score 1 · Answer 1 · answered Apr 22 '15 at 16:44

1

Try to read this. I used the package some time ago and it works quite nicely.

answered Apr 22 '15 at 16:44

atapaka

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Multiply out product of sums

1 Answers1