Mixing
Vectorization identity: Weyl matrices
$begingroup$
Take a ring $Z_n = {0,1,...,n-1}$. All operations occur modulo $n$. Consider the $n^{th}$ roots of unity $omega_n = expleft(frac{2pi i}{n}right)$. Now define
$$X = sum_a vert{a+1}ranglelangle{a}vert$$
$$Z = sum_a omega_a vert{a}ranglelangle{a}vert,$$
where $vert arangle$ is a column vector of an orthonormal basis and $langle avert$ is the corresponding basis row vector.
These matrices follow the commutation $XZ = omega ZX$. The Weyl matrices are defined as $W_{a,b} = X^aZ^b$.
Vectorization of a matrix $A=sum A_{i,j}vert i ranglelangle jvert$ is defined by vec$(A) = sum A_{i,j}vert i ranglevert jrangle$
I am given an identity vec$(mathbb{I})$vec$(mathbb{I})^* = frac{1}{n}sumlimits_{c,d}overline{W_{c,d}}otimes W_{c,d}$. Could someone point out how I could prove this?
linear-algebra weyl-group vectorization
asked Feb 1 at 19:40
user1936752user1936752
5841515
$endgroup$
|
show 3 more comments
$begingroup$
Take a ring $Z_n = {0,1,...,n-1}$. All operations occur modulo $n$. Consider the $n^{th}$ roots of unity $omega_n = expleft(frac{2pi i}{n}right)$. Now define
$$X = sum_a vert{a+1}ranglelangle{a}vert$$
$$Z = sum_a omega_a vert{a}ranglelangle{a}vert,$$
where $vert arangle$ is a column vector of an orthonormal basis and $langle avert$ is the corresponding basis row vector.
These matrices follow the commutation $XZ = omega ZX$. The Weyl matrices are defined as $W_{a,b} = X^aZ^b$.
Vectorization of a matrix $A=sum A_{i,j}vert i ranglelangle jvert$ is defined by vec$(A) = sum A_{i,j}vert i ranglevert jrangle$
I am given an identity vec$(mathbb{I})$vec$(mathbb{I})^* = frac{1}{n}sumlimits_{c,d}overline{W_{c,d}}otimes W_{c,d}$. Could someone point out how I could prove this?
linear-algebra weyl-group vectorization
asked Feb 1 at 19:40
user1936752user1936752
5841515
$endgroup$
$begingroup$
How are the bras and kets defined?
$endgroup$
– md2perpe
Feb 1 at 20:00
$begingroup$
@md2perpe Sorry, will add it to the question. The bras and kets are row and column vectors respectively. Hence $vert aranglelangle a vert$ would be a projector matrix, for instance.
$endgroup$
– user1936752
Feb 1 at 21:53
$begingroup$
Is the basis arbitrary or should it be orthonormal?
$endgroup$
– md2perpe
Feb 2 at 7:01
$begingroup$
@md2perpe, I believe it is an orthonormal basis.
$endgroup$
– user1936752
Feb 2 at 10:17
$begingroup$
Could also be that ${ vert a rangle }$ is any basis and that ${ langle a vert }$ is the dual basis such that $langle a' vert a'' rangle = delta(a', a''),$ i.e. $=1$ if $a'=a'',$ and $=0$ otherwise.
$endgroup$
– md2perpe
Feb 2 at 11:16
|
show 3 more comments
$begingroup$
Take a ring $Z_n = {0,1,...,n-1}$. All operations occur modulo $n$. Consider the $n^{th}$ roots of unity $omega_n = expleft(frac{2pi i}{n}right)$. Now define
$$X = sum_a vert{a+1}ranglelangle{a}vert$$
$$Z = sum_a omega_a vert{a}ranglelangle{a}vert,$$
where $vert arangle$ is a column vector of an orthonormal basis and $langle avert$ is the corresponding basis row vector.
These matrices follow the commutation $XZ = omega ZX$. The Weyl matrices are defined as $W_{a,b} = X^aZ^b$.
Vectorization of a matrix $A=sum A_{i,j}vert i ranglelangle jvert$ is defined by vec$(A) = sum A_{i,j}vert i ranglevert jrangle$
I am given an identity vec$(mathbb{I})$vec$(mathbb{I})^* = frac{1}{n}sumlimits_{c,d}overline{W_{c,d}}otimes W_{c,d}$. Could someone point out how I could prove this?
linear-algebra weyl-group vectorization
asked Feb 1 at 19:40
user1936752user1936752
5841515
$endgroup$
Take a ring $Z_n = {0,1,...,n-1}$. All operations occur modulo $n$. Consider the $n^{th}$ roots of unity $omega_n = expleft(frac{2pi i}{n}right)$. Now define
$$X = sum_a vert{a+1}ranglelangle{a}vert$$
$$Z = sum_a omega_a vert{a}ranglelangle{a}vert,$$
where $vert arangle$ is a column vector of an orthonormal basis and $langle avert$ is the corresponding basis row vector.
These matrices follow the commutation $XZ = omega ZX$. The Weyl matrices are defined as $W_{a,b} = X^aZ^b$.
Vectorization of a matrix $A=sum A_{i,j}vert i ranglelangle jvert$ is defined by vec$(A) = sum A_{i,j}vert i ranglevert jrangle$
I am given an identity vec$(mathbb{I})$vec$(mathbb{I})^* = frac{1}{n}sumlimits_{c,d}overline{W_{c,d}}otimes W_{c,d}$. Could someone point out how I could prove this?
linear-algebra weyl-group vectorization
linear-algebra weyl-group vectorization
asked Feb 1 at 19:40
user1936752user1936752
5841515
asked Feb 1 at 19:40
user1936752user1936752
5841515
edited Feb 2 at 10:17
user1936752
asked Feb 1 at 19:40
user1936752user1936752
5841515
asked Feb 1 at 19:40
user1936752user1936752
5841515
asked Feb 1 at 19:40
user1936752user1936752
5841515
5841515
$begingroup$
How are the bras and kets defined?
$endgroup$
– md2perpe
Feb 1 at 20:00
$begingroup$
@md2perpe Sorry, will add it to the question. The bras and kets are row and column vectors respectively. Hence $vert aranglelangle a vert$ would be a projector matrix, for instance.
$endgroup$
– user1936752
Feb 1 at 21:53
$begingroup$
Is the basis arbitrary or should it be orthonormal?
$endgroup$
– md2perpe
Feb 2 at 7:01
$begingroup$
@md2perpe, I believe it is an orthonormal basis.
$endgroup$
– user1936752
Feb 2 at 10:17
$begingroup$
Could also be that ${ vert a rangle }$ is any basis and that ${ langle a vert }$ is the dual basis such that $langle a' vert a'' rangle = delta(a', a''),$ i.e. $=1$ if $a'=a'',$ and $=0$ otherwise.
$endgroup$
– md2perpe
Feb 2 at 11:16
|
show 3 more comments
$begingroup$
How are the bras and kets defined?
$endgroup$
– md2perpe
Feb 1 at 20:00
$begingroup$
@md2perpe Sorry, will add it to the question. The bras and kets are row and column vectors respectively. Hence $vert aranglelangle a vert$ would be a projector matrix, for instance.
$endgroup$
– user1936752
Feb 1 at 21:53
$begingroup$
Is the basis arbitrary or should it be orthonormal?
$endgroup$
– md2perpe
Feb 2 at 7:01
$begingroup$
@md2perpe, I believe it is an orthonormal basis.
$endgroup$
– user1936752
Feb 2 at 10:17
$begingroup$
Could also be that ${ vert a rangle }$ is any basis and that ${ langle a vert }$ is the dual basis such that $langle a' vert a'' rangle = delta(a', a''),$ i.e. $=1$ if $a'=a'',$ and $=0$ otherwise.
$endgroup$
– md2perpe
Feb 2 at 11:16
$begingroup$
How are the bras and kets defined?
$endgroup$
– md2perpe
Feb 1 at 20:00
$begingroup$
How are the bras and kets defined?
$endgroup$
– md2perpe
Feb 1 at 20:00
$begingroup$
@md2perpe Sorry, will add it to the question. The bras and kets are row and column vectors respectively. Hence $vert aranglelangle a vert$ would be a projector matrix, for instance.
$endgroup$
– user1936752
Feb 1 at 21:53
$begingroup$
@md2perpe Sorry, will add it to the question. The bras and kets are row and column vectors respectively. Hence $vert aranglelangle a vert$ would be a projector matrix, for instance.
$endgroup$
– user1936752
Feb 1 at 21:53
$begingroup$
Is the basis arbitrary or should it be orthonormal?
$endgroup$
– md2perpe
Feb 2 at 7:01
$begingroup$
Is the basis arbitrary or should it be orthonormal?
$endgroup$
– md2perpe
Feb 2 at 7:01
$begingroup$
@md2perpe, I believe it is an orthonormal basis.
$endgroup$
– user1936752
Feb 2 at 10:17
$begingroup$
@md2perpe, I believe it is an orthonormal basis.
$endgroup$
– user1936752
Feb 2 at 10:17
$begingroup$
Could also be that ${ vert a rangle }$ is any basis and that ${ langle a vert }$ is the dual basis such that $langle a' vert a'' rangle = delta(a', a''),$ i.e. $=1$ if $a'=a'',$ and $=0$ otherwise.
$endgroup$
– md2perpe
Feb 2 at 11:16
$begingroup$
Could also be that ${ vert a rangle }$ is any basis and that ${ langle a vert }$ is the dual basis such that $langle a' vert a'' rangle = delta(a', a''),$ i.e. $=1$ if $a'=a'',$ and $=0$ otherwise.
$endgroup$
– md2perpe
Feb 2 at 11:16
|
show 3 more comments
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$begingroup$
How are the bras and kets defined?
$endgroup$
– md2perpe
Feb 1 at 20:00
$begingroup$
@md2perpe Sorry, will add it to the question. The bras and kets are row and column vectors respectively. Hence $vert aranglelangle a vert$ would be a projector matrix, for instance.
$endgroup$
– user1936752
Feb 1 at 21:53
$begingroup$
Is the basis arbitrary or should it be orthonormal?
$endgroup$
– md2perpe
Feb 2 at 7:01
$begingroup$
@md2perpe, I believe it is an orthonormal basis.
$endgroup$
– user1936752
Feb 2 at 10:17
$begingroup$
Could also be that ${ vert a rangle }$ is any basis and that ${ langle a vert }$ is the dual basis such that $langle a' vert a'' rangle = delta(a', a''),$ i.e. $=1$ if $a'=a'',$ and $=0$ otherwise.
$endgroup$
– md2perpe
Feb 2 at 11:16