Commutators of tensor product of Pauli matricesPermutations of elements in a tensor productStructure of Parity Check Matrix of Non-Systematic Tensor Product CodesTensor Multiplication - Why should we use permutations?SO(2) group generator Lie AlgebraTensor product of $C^*$- algebrasNeed definition of symmetric and antisymmetric tensor representations of a Lie algebradirect sum and tensor product of representation of lie algebraDuality and tensor product of the Lie algebraKoszul sign convention and symmetric group action on the graded n-th tensor productNon-standard Generalizations of the Pauli Matrices that retain closedness, identity mapping & tracelessness?
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Commutators of tensor product of Pauli matrices
Permutations of elements in a tensor productStructure of Parity Check Matrix of Non-Systematic Tensor Product CodesTensor Multiplication - Why should we use permutations?SO(2) group generator Lie AlgebraTensor product of $C^*$- algebrasNeed definition of symmetric and antisymmetric tensor representations of a Lie algebradirect sum and tensor product of representation of lie algebraDuality and tensor product of the Lie algebraKoszul sign convention and symmetric group action on the graded n-th tensor productNon-standard Generalizations of the Pauli Matrices that retain closedness, identity mapping & tracelessness?
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
linear-algebra abstract-algebra lie-algebras tensor-products
asked Feb 16 '18 at 4:01
annie heartannie heart
673721
673721
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
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
$begingroup$
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 Answer
1
active
oldest
votes
$begingroup$
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 $ab$.
$endgroup$
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $ab$.
$endgroup$
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 $ab$.
$endgroup$
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 $ab$.
$endgroup$
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 $ab$.
edited Jun 1 '18 at 11:02
answered Jun 1 '18 at 8:33
agartthaagarttha
112
112
add a comment |
add a comment |
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$begingroup$
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