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Commutators of tensor product of Pauli matrices

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1

2

$begingroup$

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

linear-algebra abstract-algebra lie-algebras tensor-products

share|cite|improve this question

asked Feb 16 '18 at 4:01

annie heartannie heart

673721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just to be precise, I think your tensor product here also means the en.wikipedia.org/wiki/Kronecker_product
  $endgroup$
  – wonderich
  Feb 16 '18 at 4:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Mariano Suárez-Álvarez
  Feb 16 '18 at 4:07
  

  
  
  
  

add a comment | 

1

2

$begingroup$

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

linear-algebra abstract-algebra lie-algebras tensor-products

share|cite|improve this question

asked Feb 16 '18 at 4:01

annie heartannie heart

673721

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just to be precise, I think your tensor product here also means the en.wikipedia.org/wiki/Kronecker_product
  $endgroup$
  – wonderich
  Feb 16 '18 at 4:07
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Mariano Suárez-Álvarez
  Feb 16 '18 at 4:07
  

  
  
  
  

add a comment | 

$begingroup$

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

linear-algebra abstract-algebra lie-algebras tensor-products

share|cite|improve this question

asked Feb 16 '18 at 4:01

annie heartannie heart

673721

$endgroup$

share|cite|improve this question

asked Feb 16 '18 at 4:01

annie heartannie heart

673721

share|cite|improve this question

asked Feb 16 '18 at 4:01

annie heartannie heart

673721

asked Feb 16 '18 at 4:01

annie heartannie heart

673721

asked Feb 16 '18 at 4:01

annie heartannie heart

673721

1 Answer
1

active

oldest

votes

0

$begingroup$

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
$$
sigma^asigma^b = delta_abI+iepsilon_abcsigma^c
$$
you can expand the product $(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)$ as

beginalign
(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)
&= (sigma^asigma^b)otimes(sigma^csigma^d) \ &= (delta_abI+iepsilon_abesigma^e)otimes(delta_cdI+iepsilon_cdfsigma^f) \
&=delta_abdelta_cdI+iepsilon_abedelta_cd(sigma^eotimes I)+iepsilon_cdfdelta_ab(I otimes sigma^f)-epsilon_abeepsilon_cdf(sigma^eotimessigma^f).
endalign

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

$$
[sigma^a otimes sigma^c, sigma^b otimes sigma^d] = 2iepsilon_abedelta_cd(sigma^eotimes I)+2iepsilon_cdfdelta_ab(I otimes sigma^f).
$$

Note that the two terms are mutually exclusive since if $delta_cd=1$, then $epsilon_cdf=0$, and likewise for the pair of indices $ab$.

share|cite|improve this answer

edited Jun 1 '18 at 11:02

answered Jun 1 '18 at 8:33

agartthaagarttha

112

$endgroup$

add a comment | 

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0

$begingroup$

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
$$
sigma^asigma^b = delta_abI+iepsilon_abcsigma^c
$$
you can expand the product $(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)$ as

beginalign
(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)
&= (sigma^asigma^b)otimes(sigma^csigma^d) \ &= (delta_abI+iepsilon_abesigma^e)otimes(delta_cdI+iepsilon_cdfsigma^f) \
&=delta_abdelta_cdI+iepsilon_abedelta_cd(sigma^eotimes I)+iepsilon_cdfdelta_ab(I otimes sigma^f)-epsilon_abeepsilon_cdf(sigma^eotimessigma^f).
endalign

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

$$
[sigma^a otimes sigma^c, sigma^b otimes sigma^d] = 2iepsilon_abedelta_cd(sigma^eotimes I)+2iepsilon_cdfdelta_ab(I otimes sigma^f).
$$

Note that the two terms are mutually exclusive since if $delta_cd=1$, then $epsilon_cdf=0$, and likewise for the pair of indices $ab$.

share|cite|improve this answer

edited Jun 1 '18 at 11:02

answered Jun 1 '18 at 8:33

agartthaagarttha

112

$endgroup$

add a comment | 

0

$begingroup$

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
$$
sigma^asigma^b = delta_abI+iepsilon_abcsigma^c
$$
you can expand the product $(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)$ as

beginalign
(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)
&= (sigma^asigma^b)otimes(sigma^csigma^d) \ &= (delta_abI+iepsilon_abesigma^e)otimes(delta_cdI+iepsilon_cdfsigma^f) \
&=delta_abdelta_cdI+iepsilon_abedelta_cd(sigma^eotimes I)+iepsilon_cdfdelta_ab(I otimes sigma^f)-epsilon_abeepsilon_cdf(sigma^eotimessigma^f).
endalign

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

$$
[sigma^a otimes sigma^c, sigma^b otimes sigma^d] = 2iepsilon_abedelta_cd(sigma^eotimes I)+2iepsilon_cdfdelta_ab(I otimes sigma^f).
$$

Note that the two terms are mutually exclusive since if $delta_cd=1$, then $epsilon_cdf=0$, and likewise for the pair of indices $ab$.

share|cite|improve this answer

edited Jun 1 '18 at 11:02

answered Jun 1 '18 at 8:33

agartthaagarttha

112

$endgroup$

add a comment | 

$begingroup$

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
$$
sigma^asigma^b = delta_abI+iepsilon_abcsigma^c
$$
you can expand the product $(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)$ as

beginalign
(sigma^a otimes sigma^c)(sigma^b otimes sigma^d)
&= (sigma^asigma^b)otimes(sigma^csigma^d) \ &= (delta_abI+iepsilon_abesigma^e)otimes(delta_cdI+iepsilon_cdfsigma^f) \
&=delta_abdelta_cdI+iepsilon_abedelta_cd(sigma^eotimes I)+iepsilon_cdfdelta_ab(I otimes sigma^f)-epsilon_abeepsilon_cdf(sigma^eotimessigma^f).
endalign

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

$$
[sigma^a otimes sigma^c, sigma^b otimes sigma^d] = 2iepsilon_abedelta_cd(sigma^eotimes I)+2iepsilon_cdfdelta_ab(I otimes sigma^f).
$$

Note that the two terms are mutually exclusive since if $delta_cd=1$, then $epsilon_cdf=0$, and likewise for the pair of indices $ab$.

share|cite|improve this answer

edited Jun 1 '18 at 11:02

answered Jun 1 '18 at 8:33

agartthaagarttha

112

$endgroup$

share|cite|improve this answer

edited Jun 1 '18 at 11:02

answered Jun 1 '18 at 8:33

agartthaagarttha

112

answered Jun 1 '18 at 8:33

agartthaagarttha

112

answered Jun 1 '18 at 8:33

agartthaagarttha

112