12
$\begingroup$

Given tensor product of rank-2 Pauli matrices $\sigma^a$. Each $\sigma^a$ is related to the generator of SU(2) Lie algebra.

We know they satisfy

$$[\sigma^a, \sigma^b ] = 2 i \epsilon^{abc} \sigma^c$$

Do you know any equality/identity to simplify: $$ [\sigma^a \otimes \sigma^c, \sigma^b \otimes \sigma^d] = ? $$ also $$ [\sigma^a \otimes \sigma^c \otimes \sigma^e, \sigma^b \otimes \sigma^d \otimes \sigma^f] = ? $$ $$ [\sigma^a \otimes \sigma^c \otimes \sigma^e \otimes \sigma^g, \sigma^b \otimes \sigma^d \otimes \sigma^f \otimes \sigma^h] = ? $$ so that the final answers have no commutators?

Commutator is defined by default as $$ [A,B]:=AB-BA $$

$\endgroup$
3
  • $\begingroup$ just to be precise, I think your tensor product here also means the en.wikipedia.org/wiki/Kronecker_product $\endgroup$
    – wonderich
    Feb 16, 2018 at 4:07
  • 4
    $\begingroup$ There is no reason to expect anything nice for those formulas. The reason is that the tensor product of lie algebras is not a lie algebra in any sensible way. $\endgroup$ Feb 16, 2018 at 4:07
  • $\begingroup$ Supposedly always either the commutator or the anticommutator is zero (unfortunately I only read the result without proof) $\endgroup$
    – lalala
    Aug 12, 2021 at 16:46

2 Answers 2

12
$\begingroup$

If you check out Kronecker Product you will see that it has the mixed-product property:

$$ (\mathbf {A} \otimes \mathbf {B} )(\mathbf {C} \otimes \mathbf {D} )=(\mathbf {AC} )\otimes (\mathbf {BD} ). $$

Using this property and the fact that $$ \sigma^a\sigma^b = \delta_{ab}I+i\epsilon_{abc}\sigma^c $$ you can expand the product $(\sigma^a \otimes \sigma^c)(\sigma^b \otimes \sigma^d)$ as

\begin{align} (\sigma^a \otimes \sigma^c)(\sigma^b \otimes \sigma^d) &= (\sigma^a\sigma^b)\otimes(\sigma^c\sigma^d) \\ &= (\delta_{ab}I+i\epsilon_{abe}\sigma^e)\otimes(\delta_{cd}I+i\epsilon_{cdf}\sigma^f) \\ &=\delta_{ab}\delta_{cd}I+i\epsilon_{abe}\delta_{cd}(\sigma^e\otimes I)+i\epsilon_{cdf}\delta_{ab}(I \otimes \sigma^f)-\epsilon_{abe}\epsilon_{cdf}(\sigma^e\otimes\sigma^f). \end{align}

Since the first and last terms in this expression are symmetric when the indices $ab$ and $cd$ are permuted, the first commutator you ask for simplifies to

$$ [\sigma^a \otimes \sigma^c, \sigma^b \otimes \sigma^d] = 2i\epsilon_{abe}\delta_{cd}(\sigma^e\otimes I)+2i\epsilon_{cdf}\delta_{ab}(I \otimes \sigma^f). $$

Note that the two terms are mutually exclusive since if $\delta_{cd}=1$, then $\epsilon_{cdf}=0$, and likewise for the pair of indices $ab$.

$\endgroup$
1
$\begingroup$

One can also derive the really neat expression for this sort of thing:

$$ [a_1\otimes a_2,b_1\otimes b_2]= [a_1,b_1]\otimes\{a_2,b_2\}+\{a_1,b_1\}\otimes[a_2,b_2]\ $$ and $$ \{a_1\otimes a_2,b_1\otimes b_2\}= \{a_1,b_1\}\otimes\{a_2,b_2\}+[a_1,b_1]\otimes[a_2,b_2]\ $$

Which can be recursively applied to the right hand side if you have more tensor products, for example if $a_1$ and $b_1$ are also tensor products of something else:

$$ [a_1,b_1] = [a_{11}\otimes a_{12},b_{11}\otimes b_{12}] $$ and $$ \{a_1,b_1\} = \{a_{11}\otimes a_{12},b_{11}\otimes b_{12}\} $$

If you prefer to write commutators of tensor products as tensor products of commutators.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .