2
$\begingroup$

Say I have a tensor product of four two dimensional spaces $\Lambda = S\otimes S\otimes S\otimes S$. The basis of $S$ is CB={{1,0},{0,1}} . I generate the representation of $\Lambda$ by

Lambda=ArrayFlatten[
 ArrayFlatten[
  ArrayFlatten[
   Table[TensorProduct[CB[[i]], CB[[j]], CB[[k]], CB[[m]]], 
   {i, 1, 2}, {j, 1, 2}, {k, 1, 2}, {m, 1, 2}]
  ]
 ]
]

The result is

$\Lambda=\left( \begin{array}{cccccccccccccccc} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right)$

Now, I have a problem to generate the corresponding basis vectors, which I thought are given by

Flatten[ArrayFlatten[TensorProduct[CB[[i]], CB[[j]], CB[[k]], CB[[m]]]]]

for $i,j,k,m\in\{1,2\}$. But, e.g.

Lambda.Flatten[ArrayFlatten[TensorProduct[CB[[2]], CB[[2]], CB[[1]], CB[[1]]]]]

results in the null vector. How to generate the correct basis vectors using ArrayFlatten[] or Flatten[]?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The basis vectors are

CB = {{1, 0}, {0, 1}};
Flatten@*KroneckerProduct @@@ Tuples[CB, 4]

(*    {{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0},
       {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}}    *)

I'm not sure what your matrix $\Lambda$ is: you said it was a vector space, but matrices are rather elements of this vector space, not the space itself.

I've written a book covering this topic: see chapter 2 of Using Mathematica for Quantum Mechanics: A Student's Manual.

Improve this answer
$\endgroup$
5
  • $\begingroup$ My problem is: how do I write e.g. the vector a=TensorProduct[CB[[1]],CB[[2]],CB[[1]],C[[1]]] in the same basis as Lambda, i.e., which of the basis vectors you wrote is a? Thank you for the link, I’ll have a look! $\endgroup$
    – pawel_winzig
    Commented Nov 4, 2019 at 22:02
  • $\begingroup$ a = Flatten@KroneckerProduct[CB[[1]], CB[[2]], CB[[1]], CB[[1]]] gives the basis vector {0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}. $\endgroup$
    – Roman
    Commented Nov 5, 2019 at 21:29
  • $\begingroup$ Shoudn't $\Lambda$ be in this case not a diagonal matrix? $\endgroup$
    – pawel_winzig
    Commented Nov 6, 2019 at 7:45
  • $\begingroup$ Again, Λ is a vector space, not a matrix. $\endgroup$
    – Roman
    Commented Nov 6, 2019 at 13:55
  • $\begingroup$ It seems my description of the question is misleading: See $\Lambda$ as an operator acting on four spins. The matrix is the representation of this operator. $\endgroup$
    – pawel_winzig
    Commented Nov 6, 2019 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.