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 $$ [A,B]:=AB-BA $$

  • $\begingroup$ 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


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 $ab$.


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.


You must log in to answer this question.

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