Mixing
Schur-Weyl duality for qubits
I am interested in applying Schur-Weyl duality to quantum information theory, specifically "qubits". But I have been stuck for some time on understanding how the Young symmetrizers work in this situation.
Let $V$ be a 2 dimensional complex vector space, a qubit. Let $S_3$ be the permutation group on three things, acting on $Votimes Votimes V$ by permuting the factors. Write a basis for $V$ as $v_0$ and $v_1$, and the corresponding basis of $Votimes Votimes V$ as $v_{000}, v_{001},...,v_{111}$. It's not too hard to see that this representation of $S_3$ is 8 dimensional, and breaks into 4 copies of the trivial representation:
begin{align*}
&v_{000},\
&v_{100} + v_{010} + v_{001},\
&v_{011} + v_{101} + v_{110},\
&v_{111},
end{align*}
and two copies of the 2 dimensional representation:
begin{align*}
&{v_{100}-v_{010}, 2v_{001}-v_{010}-v_{100}},\
&{v_{011}-v_{101}, 2v_{110}-v_{101}-v_{011}}.
end{align*}
Here I have written a basis for each of the 2 dimensional representations. So far so good. But now things start to get murky for me.
I would like to produce these 2 dimensional representations via
the action of the Young symmetrizer. This operator, $P$
is given by
$$
P v_{abc} = v_{abc} + v_{bac} - v_{cba} - v_{cab}.
$$
But feeding any of $v_{100}, v_{010}, v_{001}$ into $P$ results in a multiple of $v_{100}-v_{001}.$ For example
$$
P v_{100} = v_{100} + v_{010} - v_{001} - v_{010} = v_{100} - v_{001}.
$$
I have tried many variations on the definition of $P$, and also letting it act on "the left or the right" of the $v_{abc}.$ Nothing I do seems to produce more than a one dimensional subspace of the sought-after two dimensional space.
What I can do is produce the Specht modules from the group algebra of $S_3$ using the Young symmetrizers, but somewhere between the group algebra and $Votimes Votimes V$ I appear to be getting lost.
representation-theory symmetric-groups quantum-computation young-tableaux
asked Nov 18 '18 at 17:43
Simon Burton
1235
|
show 5 more comments
I am interested in applying Schur-Weyl duality to quantum information theory, specifically "qubits". But I have been stuck for some time on understanding how the Young symmetrizers work in this situation.
Let $V$ be a 2 dimensional complex vector space, a qubit. Let $S_3$ be the permutation group on three things, acting on $Votimes Votimes V$ by permuting the factors. Write a basis for $V$ as $v_0$ and $v_1$, and the corresponding basis of $Votimes Votimes V$ as $v_{000}, v_{001},...,v_{111}$. It's not too hard to see that this representation of $S_3$ is 8 dimensional, and breaks into 4 copies of the trivial representation:
begin{align*}
&v_{000},\
&v_{100} + v_{010} + v_{001},\
&v_{011} + v_{101} + v_{110},\
&v_{111},
end{align*}
and two copies of the 2 dimensional representation:
begin{align*}
&{v_{100}-v_{010}, 2v_{001}-v_{010}-v_{100}},\
&{v_{011}-v_{101}, 2v_{110}-v_{101}-v_{011}}.
end{align*}
Here I have written a basis for each of the 2 dimensional representations. So far so good. But now things start to get murky for me.
I would like to produce these 2 dimensional representations via
the action of the Young symmetrizer. This operator, $P$
is given by
$$
P v_{abc} = v_{abc} + v_{bac} - v_{cba} - v_{cab}.
$$
But feeding any of $v_{100}, v_{010}, v_{001}$ into $P$ results in a multiple of $v_{100}-v_{001}.$ For example
$$
P v_{100} = v_{100} + v_{010} - v_{001} - v_{010} = v_{100} - v_{001}.
$$
I have tried many variations on the definition of $P$, and also letting it act on "the left or the right" of the $v_{abc}.$ Nothing I do seems to produce more than a one dimensional subspace of the sought-after two dimensional space.
What I can do is produce the Specht modules from the group algebra of $S_3$ using the Young symmetrizers, but somewhere between the group algebra and $Votimes Votimes V$ I appear to be getting lost.
representation-theory symmetric-groups quantum-computation young-tableaux
asked Nov 18 '18 at 17:43
Simon Burton
1235
2
The problem is that the symmetric group cannot tell apart those two isomorphic representations you wrote down. Put another way, there are infinitely many ways of decomposing the sum of those two two-dimensional representations. (Similarly for the trivial representations). There's just no way that purely using the symmetric group you can get the decomposition you are looking for.
– Joppy
Nov 21 '18 at 12:57
1
What you need to do is add in the Schur-Weyl duality part, and use the action of the $operatorname{GL}_2$ to first project the problem down so that the symmetric group representations no longer occur with multiplicity.
– Joppy
Nov 21 '18 at 12:58
Here is a direct quote from Goodman and Wallach: "... Young symmetrizers. These elements of the group algebra of $C[S_k]$ act as projection operators onto $GL(n, C)$-irreducible invariant subspaces." So this is precisely what I have failed to do. Perhaps I am not explaining myself very well.
– Simon Burton
Nov 22 '18 at 14:36
1
So if you feed in all possible vectors into that projection operator, you will get a two-dimensional subspace out the other side, and it will be isomorphic to the two-dimensional irrep of $S_3$. It’s just not the irrep you are expecting. Morally, this is the same as decomposing $mathbb{R}^2$ Into a direct sum of the x and y axes, and then being given a projection-onto-a-line operator. There is almost no chance that projection will select one of the axes.
– Joppy
Nov 22 '18 at 15:00
1
In fact, the image of any element of $mathbb{C}[S_n]$ acting on $V^{otimes n}$ will be a GL-invariant subspace, just because the actions of GL and the symmetric group commute. Young symmetrisers are a bit more special since they yield irreducible GL representations. You almost already have the Schur-Weyl decomposition in terms of those two groups of 4-dimensional representations you wrote down (splitting into non-isomorphic representations is a canonical direct sum). The next thing to check is that each of these pieces is a GL representation of the same type.
– Joppy
Nov 22 '18 at 15:27
|
show 5 more comments
I am interested in applying Schur-Weyl duality to quantum information theory, specifically "qubits". But I have been stuck for some time on understanding how the Young symmetrizers work in this situation.
Let $V$ be a 2 dimensional complex vector space, a qubit. Let $S_3$ be the permutation group on three things, acting on $Votimes Votimes V$ by permuting the factors. Write a basis for $V$ as $v_0$ and $v_1$, and the corresponding basis of $Votimes Votimes V$ as $v_{000}, v_{001},...,v_{111}$. It's not too hard to see that this representation of $S_3$ is 8 dimensional, and breaks into 4 copies of the trivial representation:
begin{align*}
&v_{000},\
&v_{100} + v_{010} + v_{001},\
&v_{011} + v_{101} + v_{110},\
&v_{111},
end{align*}
and two copies of the 2 dimensional representation:
begin{align*}
&{v_{100}-v_{010}, 2v_{001}-v_{010}-v_{100}},\
&{v_{011}-v_{101}, 2v_{110}-v_{101}-v_{011}}.
end{align*}
Here I have written a basis for each of the 2 dimensional representations. So far so good. But now things start to get murky for me.
I would like to produce these 2 dimensional representations via
the action of the Young symmetrizer. This operator, $P$
is given by
$$
P v_{abc} = v_{abc} + v_{bac} - v_{cba} - v_{cab}.
$$
But feeding any of $v_{100}, v_{010}, v_{001}$ into $P$ results in a multiple of $v_{100}-v_{001}.$ For example
$$
P v_{100} = v_{100} + v_{010} - v_{001} - v_{010} = v_{100} - v_{001}.
$$
I have tried many variations on the definition of $P$, and also letting it act on "the left or the right" of the $v_{abc}.$ Nothing I do seems to produce more than a one dimensional subspace of the sought-after two dimensional space.
What I can do is produce the Specht modules from the group algebra of $S_3$ using the Young symmetrizers, but somewhere between the group algebra and $Votimes Votimes V$ I appear to be getting lost.
representation-theory symmetric-groups quantum-computation young-tableaux
asked Nov 18 '18 at 17:43
Simon Burton
1235
I am interested in applying Schur-Weyl duality to quantum information theory, specifically "qubits". But I have been stuck for some time on understanding how the Young symmetrizers work in this situation.
Let $V$ be a 2 dimensional complex vector space, a qubit. Let $S_3$ be the permutation group on three things, acting on $Votimes Votimes V$ by permuting the factors. Write a basis for $V$ as $v_0$ and $v_1$, and the corresponding basis of $Votimes Votimes V$ as $v_{000}, v_{001},...,v_{111}$. It's not too hard to see that this representation of $S_3$ is 8 dimensional, and breaks into 4 copies of the trivial representation:
begin{align*}
&v_{000},\
&v_{100} + v_{010} + v_{001},\
&v_{011} + v_{101} + v_{110},\
&v_{111},
end{align*}
and two copies of the 2 dimensional representation:
begin{align*}
&{v_{100}-v_{010}, 2v_{001}-v_{010}-v_{100}},\
&{v_{011}-v_{101}, 2v_{110}-v_{101}-v_{011}}.
end{align*}
Here I have written a basis for each of the 2 dimensional representations. So far so good. But now things start to get murky for me.
I would like to produce these 2 dimensional representations via
the action of the Young symmetrizer. This operator, $P$
is given by
$$
P v_{abc} = v_{abc} + v_{bac} - v_{cba} - v_{cab}.
$$
But feeding any of $v_{100}, v_{010}, v_{001}$ into $P$ results in a multiple of $v_{100}-v_{001}.$ For example
$$
P v_{100} = v_{100} + v_{010} - v_{001} - v_{010} = v_{100} - v_{001}.
$$
I have tried many variations on the definition of $P$, and also letting it act on "the left or the right" of the $v_{abc}.$ Nothing I do seems to produce more than a one dimensional subspace of the sought-after two dimensional space.
What I can do is produce the Specht modules from the group algebra of $S_3$ using the Young symmetrizers, but somewhere between the group algebra and $Votimes Votimes V$ I appear to be getting lost.
representation-theory symmetric-groups quantum-computation young-tableaux
representation-theory symmetric-groups quantum-computation young-tableaux
asked Nov 18 '18 at 17:43
Simon Burton
1235
asked Nov 18 '18 at 17:43
Simon Burton
1235
edited Nov 21 '18 at 10:45
asked Nov 18 '18 at 17:43
Simon Burton
1235
asked Nov 18 '18 at 17:43
Simon Burton
1235
asked Nov 18 '18 at 17:43
Simon Burton
1235
1235
2
The problem is that the symmetric group cannot tell apart those two isomorphic representations you wrote down. Put another way, there are infinitely many ways of decomposing the sum of those two two-dimensional representations. (Similarly for the trivial representations). There's just no way that purely using the symmetric group you can get the decomposition you are looking for.
– Joppy
Nov 21 '18 at 12:57
1
What you need to do is add in the Schur-Weyl duality part, and use the action of the $operatorname{GL}_2$ to first project the problem down so that the symmetric group representations no longer occur with multiplicity.
– Joppy
Nov 21 '18 at 12:58
Here is a direct quote from Goodman and Wallach: "... Young symmetrizers. These elements of the group algebra of $C[S_k]$ act as projection operators onto $GL(n, C)$-irreducible invariant subspaces." So this is precisely what I have failed to do. Perhaps I am not explaining myself very well.
– Simon Burton
Nov 22 '18 at 14:36
1
So if you feed in all possible vectors into that projection operator, you will get a two-dimensional subspace out the other side, and it will be isomorphic to the two-dimensional irrep of $S_3$. It’s just not the irrep you are expecting. Morally, this is the same as decomposing $mathbb{R}^2$ Into a direct sum of the x and y axes, and then being given a projection-onto-a-line operator. There is almost no chance that projection will select one of the axes.
– Joppy
Nov 22 '18 at 15:00
1
In fact, the image of any element of $mathbb{C}[S_n]$ acting on $V^{otimes n}$ will be a GL-invariant subspace, just because the actions of GL and the symmetric group commute. Young symmetrisers are a bit more special since they yield irreducible GL representations. You almost already have the Schur-Weyl decomposition in terms of those two groups of 4-dimensional representations you wrote down (splitting into non-isomorphic representations is a canonical direct sum). The next thing to check is that each of these pieces is a GL representation of the same type.
– Joppy
Nov 22 '18 at 15:27
|
show 5 more comments
2
The problem is that the symmetric group cannot tell apart those two isomorphic representations you wrote down. Put another way, there are infinitely many ways of decomposing the sum of those two two-dimensional representations. (Similarly for the trivial representations). There's just no way that purely using the symmetric group you can get the decomposition you are looking for.
– Joppy
Nov 21 '18 at 12:57
1
What you need to do is add in the Schur-Weyl duality part, and use the action of the $operatorname{GL}_2$ to first project the problem down so that the symmetric group representations no longer occur with multiplicity.
– Joppy
Nov 21 '18 at 12:58
Here is a direct quote from Goodman and Wallach: "... Young symmetrizers. These elements of the group algebra of $C[S_k]$ act as projection operators onto $GL(n, C)$-irreducible invariant subspaces." So this is precisely what I have failed to do. Perhaps I am not explaining myself very well.
– Simon Burton
Nov 22 '18 at 14:36
1
So if you feed in all possible vectors into that projection operator, you will get a two-dimensional subspace out the other side, and it will be isomorphic to the two-dimensional irrep of $S_3$. It’s just not the irrep you are expecting. Morally, this is the same as decomposing $mathbb{R}^2$ Into a direct sum of the x and y axes, and then being given a projection-onto-a-line operator. There is almost no chance that projection will select one of the axes.
– Joppy
Nov 22 '18 at 15:00
1
In fact, the image of any element of $mathbb{C}[S_n]$ acting on $V^{otimes n}$ will be a GL-invariant subspace, just because the actions of GL and the symmetric group commute. Young symmetrisers are a bit more special since they yield irreducible GL representations. You almost already have the Schur-Weyl decomposition in terms of those two groups of 4-dimensional representations you wrote down (splitting into non-isomorphic representations is a canonical direct sum). The next thing to check is that each of these pieces is a GL representation of the same type.
– Joppy
Nov 22 '18 at 15:27
2
2
The problem is that the symmetric group cannot tell apart those two isomorphic representations you wrote down. Put another way, there are infinitely many ways of decomposing the sum of those two two-dimensional representations. (Similarly for the trivial representations). There's just no way that purely using the symmetric group you can get the decomposition you are looking for.
– Joppy
Nov 21 '18 at 12:57
The problem is that the symmetric group cannot tell apart those two isomorphic representations you wrote down. Put another way, there are infinitely many ways of decomposing the sum of those two two-dimensional representations. (Similarly for the trivial representations). There's just no way that purely using the symmetric group you can get the decomposition you are looking for.
– Joppy
Nov 21 '18 at 12:57
1
1
What you need to do is add in the Schur-Weyl duality part, and use the action of the $operatorname{GL}_2$ to first project the problem down so that the symmetric group representations no longer occur with multiplicity.
– Joppy
Nov 21 '18 at 12:58
What you need to do is add in the Schur-Weyl duality part, and use the action of the $operatorname{GL}_2$ to first project the problem down so that the symmetric group representations no longer occur with multiplicity.
– Joppy
Nov 21 '18 at 12:58
Here is a direct quote from Goodman and Wallach: "... Young symmetrizers. These elements of the group algebra of $C[S_k]$ act as projection operators onto $GL(n, C)$-irreducible invariant subspaces." So this is precisely what I have failed to do. Perhaps I am not explaining myself very well.
– Simon Burton
Nov 22 '18 at 14:36
Here is a direct quote from Goodman and Wallach: "... Young symmetrizers. These elements of the group algebra of $C[S_k]$ act as projection operators onto $GL(n, C)$-irreducible invariant subspaces." So this is precisely what I have failed to do. Perhaps I am not explaining myself very well.
– Simon Burton
Nov 22 '18 at 14:36
1
1
So if you feed in all possible vectors into that projection operator, you will get a two-dimensional subspace out the other side, and it will be isomorphic to the two-dimensional irrep of $S_3$. It’s just not the irrep you are expecting. Morally, this is the same as decomposing $mathbb{R}^2$ Into a direct sum of the x and y axes, and then being given a projection-onto-a-line operator. There is almost no chance that projection will select one of the axes.
– Joppy
Nov 22 '18 at 15:00
So if you feed in all possible vectors into that projection operator, you will get a two-dimensional subspace out the other side, and it will be isomorphic to the two-dimensional irrep of $S_3$. It’s just not the irrep you are expecting. Morally, this is the same as decomposing $mathbb{R}^2$ Into a direct sum of the x and y axes, and then being given a projection-onto-a-line operator. There is almost no chance that projection will select one of the axes.
– Joppy
Nov 22 '18 at 15:00
1
1
In fact, the image of any element of $mathbb{C}[S_n]$ acting on $V^{otimes n}$ will be a GL-invariant subspace, just because the actions of GL and the symmetric group commute. Young symmetrisers are a bit more special since they yield irreducible GL representations. You almost already have the Schur-Weyl decomposition in terms of those two groups of 4-dimensional representations you wrote down (splitting into non-isomorphic representations is a canonical direct sum). The next thing to check is that each of these pieces is a GL representation of the same type.
– Joppy
Nov 22 '18 at 15:27
In fact, the image of any element of $mathbb{C}[S_n]$ acting on $V^{otimes n}$ will be a GL-invariant subspace, just because the actions of GL and the symmetric group commute. Young symmetrisers are a bit more special since they yield irreducible GL representations. You almost already have the Schur-Weyl decomposition in terms of those two groups of 4-dimensional representations you wrote down (splitting into non-isomorphic representations is a canonical direct sum). The next thing to check is that each of these pieces is a GL representation of the same type.
– Joppy
Nov 22 '18 at 15:27
|
show 5 more comments
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1
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I'll try to expand upon my answers in the comments here. You have the $mathbb{C}$-vector space $V = {v_0, v_1}$, and the tensor product $V^{otimes 3}$ has both a left action by $operatorname{GL}(V) cong operatorname{GL}_2(mathbb{C})$, $g cdot (v otimes u otimes w) = gv otimes gu otimes gw$, and a right action by $S_3$, given by the formula $(u_1 otimes u_2 otimes u_3) cdot sigma = u_{sigma(1)} otimes u_{sigma(2)} otimes u_{sigma(3)}$.
Now, by only viewing $V^{otimes 3}$ as a right $S_3$-module, it decomposes canonically into isotypic components. You've already found this decomposition, the part consisting of trivial representations:
$$begin{aligned}
&v_{000}, \
&v_{100} + v_{010} + v_{001}, \
&v_{110} + v_{101} + v_{011}, \
&v_{111} end{aligned}$$
and the part consisting of representations isomorphic to the 2-dimensional representation of $S_3$:
$$begin{aligned}
&v_{100} - v_{010}, \
&v_{001} - v_{010}, \
&v_{011} - v_{101}, \
&v_{110} - v_{101} end{aligned}$$
I'll point out that you can get this decomposition not by using Young projectors, but instead the projectors to isotypic components, which can be expressed as a weighted sum over characters of $S_3$, and are central idempotents of $mathbb{C}[S_3]$ (they commute with everything).
One consequence of Schur-Weyl duality is that this decomposition above is also automatically a decomposition into $operatorname{GL}(V)$-isotypic components. We can check this by looking at the weights of highest-weight elements. In the $S_3$-trivial subspace, the only highest-weight element is $v_{000}$ with weight $(3, 0)$, and so this subspace is the irreducible $operatorname{GL}(V)$ representation isomorphic to $S^3(V)$. In the other $S_3$-isotypic component, the subspace spanned by ${v_{100} - v_{010}, v_{001} - v_{010}}$ is highest-weight of weight $(2, 1)$, and so this $operatorname{GL}(V)$ irrep occurs with multiplicity $2$, which reflects the fact that the $S_3$-decomposition had multiplicity $2$.
It is possible to further decompose these spaces by making more choices in either $operatorname{GL}(V)$ or $S_3$ (the obvious one which I have already done above is $operatorname{GL}(V)$-weight spaces, you could also make choices of Young symmetrisers I guess?). But the point is that this is already the Schur-Weyl decomposition.
answered Nov 24 '18 at 23:22
Joppy
5,563420
add a comment |
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I'll try to expand upon my answers in the comments here. You have the $mathbb{C}$-vector space $V = {v_0, v_1}$, and the tensor product $V^{otimes 3}$ has both a left action by $operatorname{GL}(V) cong operatorname{GL}_2(mathbb{C})$, $g cdot (v otimes u otimes w) = gv otimes gu otimes gw$, and a right action by $S_3$, given by the formula $(u_1 otimes u_2 otimes u_3) cdot sigma = u_{sigma(1)} otimes u_{sigma(2)} otimes u_{sigma(3)}$.
Now, by only viewing $V^{otimes 3}$ as a right $S_3$-module, it decomposes canonically into isotypic components. You've already found this decomposition, the part consisting of trivial representations:
$$begin{aligned}
&v_{000}, \
&v_{100} + v_{010} + v_{001}, \
&v_{110} + v_{101} + v_{011}, \
&v_{111} end{aligned}$$
and the part consisting of representations isomorphic to the 2-dimensional representation of $S_3$:
$$begin{aligned}
&v_{100} - v_{010}, \
&v_{001} - v_{010}, \
&v_{011} - v_{101}, \
&v_{110} - v_{101} end{aligned}$$
I'll point out that you can get this decomposition not by using Young projectors, but instead the projectors to isotypic components, which can be expressed as a weighted sum over characters of $S_3$, and are central idempotents of $mathbb{C}[S_3]$ (they commute with everything).
One consequence of Schur-Weyl duality is that this decomposition above is also automatically a decomposition into $operatorname{GL}(V)$-isotypic components. We can check this by looking at the weights of highest-weight elements. In the $S_3$-trivial subspace, the only highest-weight element is $v_{000}$ with weight $(3, 0)$, and so this subspace is the irreducible $operatorname{GL}(V)$ representation isomorphic to $S^3(V)$. In the other $S_3$-isotypic component, the subspace spanned by ${v_{100} - v_{010}, v_{001} - v_{010}}$ is highest-weight of weight $(2, 1)$, and so this $operatorname{GL}(V)$ irrep occurs with multiplicity $2$, which reflects the fact that the $S_3$-decomposition had multiplicity $2$.
It is possible to further decompose these spaces by making more choices in either $operatorname{GL}(V)$ or $S_3$ (the obvious one which I have already done above is $operatorname{GL}(V)$-weight spaces, you could also make choices of Young symmetrisers I guess?). But the point is that this is already the Schur-Weyl decomposition.
answered Nov 24 '18 at 23:22
Joppy
5,563420
add a comment |
I'll try to expand upon my answers in the comments here. You have the $mathbb{C}$-vector space $V = {v_0, v_1}$, and the tensor product $V^{otimes 3}$ has both a left action by $operatorname{GL}(V) cong operatorname{GL}_2(mathbb{C})$, $g cdot (v otimes u otimes w) = gv otimes gu otimes gw$, and a right action by $S_3$, given by the formula $(u_1 otimes u_2 otimes u_3) cdot sigma = u_{sigma(1)} otimes u_{sigma(2)} otimes u_{sigma(3)}$.
Now, by only viewing $V^{otimes 3}$ as a right $S_3$-module, it decomposes canonically into isotypic components. You've already found this decomposition, the part consisting of trivial representations:
$$begin{aligned}
&v_{000}, \
&v_{100} + v_{010} + v_{001}, \
&v_{110} + v_{101} + v_{011}, \
&v_{111} end{aligned}$$
and the part consisting of representations isomorphic to the 2-dimensional representation of $S_3$:
$$begin{aligned}
&v_{100} - v_{010}, \
&v_{001} - v_{010}, \
&v_{011} - v_{101}, \
&v_{110} - v_{101} end{aligned}$$
I'll point out that you can get this decomposition not by using Young projectors, but instead the projectors to isotypic components, which can be expressed as a weighted sum over characters of $S_3$, and are central idempotents of $mathbb{C}[S_3]$ (they commute with everything).
One consequence of Schur-Weyl duality is that this decomposition above is also automatically a decomposition into $operatorname{GL}(V)$-isotypic components. We can check this by looking at the weights of highest-weight elements. In the $S_3$-trivial subspace, the only highest-weight element is $v_{000}$ with weight $(3, 0)$, and so this subspace is the irreducible $operatorname{GL}(V)$ representation isomorphic to $S^3(V)$. In the other $S_3$-isotypic component, the subspace spanned by ${v_{100} - v_{010}, v_{001} - v_{010}}$ is highest-weight of weight $(2, 1)$, and so this $operatorname{GL}(V)$ irrep occurs with multiplicity $2$, which reflects the fact that the $S_3$-decomposition had multiplicity $2$.
It is possible to further decompose these spaces by making more choices in either $operatorname{GL}(V)$ or $S_3$ (the obvious one which I have already done above is $operatorname{GL}(V)$-weight spaces, you could also make choices of Young symmetrisers I guess?). But the point is that this is already the Schur-Weyl decomposition.
answered Nov 24 '18 at 23:22
Joppy
5,563420
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I'll try to expand upon my answers in the comments here. You have the $mathbb{C}$-vector space $V = {v_0, v_1}$, and the tensor product $V^{otimes 3}$ has both a left action by $operatorname{GL}(V) cong operatorname{GL}_2(mathbb{C})$, $g cdot (v otimes u otimes w) = gv otimes gu otimes gw$, and a right action by $S_3$, given by the formula $(u_1 otimes u_2 otimes u_3) cdot sigma = u_{sigma(1)} otimes u_{sigma(2)} otimes u_{sigma(3)}$.
Now, by only viewing $V^{otimes 3}$ as a right $S_3$-module, it decomposes canonically into isotypic components. You've already found this decomposition, the part consisting of trivial representations:
$$begin{aligned}
&v_{000}, \
&v_{100} + v_{010} + v_{001}, \
&v_{110} + v_{101} + v_{011}, \
&v_{111} end{aligned}$$
and the part consisting of representations isomorphic to the 2-dimensional representation of $S_3$:
$$begin{aligned}
&v_{100} - v_{010}, \
&v_{001} - v_{010}, \
&v_{011} - v_{101}, \
&v_{110} - v_{101} end{aligned}$$
I'll point out that you can get this decomposition not by using Young projectors, but instead the projectors to isotypic components, which can be expressed as a weighted sum over characters of $S_3$, and are central idempotents of $mathbb{C}[S_3]$ (they commute with everything).
One consequence of Schur-Weyl duality is that this decomposition above is also automatically a decomposition into $operatorname{GL}(V)$-isotypic components. We can check this by looking at the weights of highest-weight elements. In the $S_3$-trivial subspace, the only highest-weight element is $v_{000}$ with weight $(3, 0)$, and so this subspace is the irreducible $operatorname{GL}(V)$ representation isomorphic to $S^3(V)$. In the other $S_3$-isotypic component, the subspace spanned by ${v_{100} - v_{010}, v_{001} - v_{010}}$ is highest-weight of weight $(2, 1)$, and so this $operatorname{GL}(V)$ irrep occurs with multiplicity $2$, which reflects the fact that the $S_3$-decomposition had multiplicity $2$.
It is possible to further decompose these spaces by making more choices in either $operatorname{GL}(V)$ or $S_3$ (the obvious one which I have already done above is $operatorname{GL}(V)$-weight spaces, you could also make choices of Young symmetrisers I guess?). But the point is that this is already the Schur-Weyl decomposition.
answered Nov 24 '18 at 23:22
Joppy
5,563420
I'll try to expand upon my answers in the comments here. You have the $mathbb{C}$-vector space $V = {v_0, v_1}$, and the tensor product $V^{otimes 3}$ has both a left action by $operatorname{GL}(V) cong operatorname{GL}_2(mathbb{C})$, $g cdot (v otimes u otimes w) = gv otimes gu otimes gw$, and a right action by $S_3$, given by the formula $(u_1 otimes u_2 otimes u_3) cdot sigma = u_{sigma(1)} otimes u_{sigma(2)} otimes u_{sigma(3)}$.
Now, by only viewing $V^{otimes 3}$ as a right $S_3$-module, it decomposes canonically into isotypic components. You've already found this decomposition, the part consisting of trivial representations:
$$begin{aligned}
&v_{000}, \
&v_{100} + v_{010} + v_{001}, \
&v_{110} + v_{101} + v_{011}, \
&v_{111} end{aligned}$$
and the part consisting of representations isomorphic to the 2-dimensional representation of $S_3$:
$$begin{aligned}
&v_{100} - v_{010}, \
&v_{001} - v_{010}, \
&v_{011} - v_{101}, \
&v_{110} - v_{101} end{aligned}$$
I'll point out that you can get this decomposition not by using Young projectors, but instead the projectors to isotypic components, which can be expressed as a weighted sum over characters of $S_3$, and are central idempotents of $mathbb{C}[S_3]$ (they commute with everything).
One consequence of Schur-Weyl duality is that this decomposition above is also automatically a decomposition into $operatorname{GL}(V)$-isotypic components. We can check this by looking at the weights of highest-weight elements. In the $S_3$-trivial subspace, the only highest-weight element is $v_{000}$ with weight $(3, 0)$, and so this subspace is the irreducible $operatorname{GL}(V)$ representation isomorphic to $S^3(V)$. In the other $S_3$-isotypic component, the subspace spanned by ${v_{100} - v_{010}, v_{001} - v_{010}}$ is highest-weight of weight $(2, 1)$, and so this $operatorname{GL}(V)$ irrep occurs with multiplicity $2$, which reflects the fact that the $S_3$-decomposition had multiplicity $2$.
It is possible to further decompose these spaces by making more choices in either $operatorname{GL}(V)$ or $S_3$ (the obvious one which I have already done above is $operatorname{GL}(V)$-weight spaces, you could also make choices of Young symmetrisers I guess?). But the point is that this is already the Schur-Weyl decomposition.
answered Nov 24 '18 at 23:22
Joppy
5,563420
answered Nov 24 '18 at 23:22
Joppy
5,563420
answered Nov 24 '18 at 23:22
Joppy
5,563420
answered Nov 24 '18 at 23:22
Joppy
5,563420
5,563420
add a comment |
add a comment |
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The problem is that the symmetric group cannot tell apart those two isomorphic representations you wrote down. Put another way, there are infinitely many ways of decomposing the sum of those two two-dimensional representations. (Similarly for the trivial representations). There's just no way that purely using the symmetric group you can get the decomposition you are looking for.
– Joppy
Nov 21 '18 at 12:57
1
What you need to do is add in the Schur-Weyl duality part, and use the action of the $operatorname{GL}_2$ to first project the problem down so that the symmetric group representations no longer occur with multiplicity.
– Joppy
Nov 21 '18 at 12:58
Here is a direct quote from Goodman and Wallach: "... Young symmetrizers. These elements of the group algebra of $C[S_k]$ act as projection operators onto $GL(n, C)$-irreducible invariant subspaces." So this is precisely what I have failed to do. Perhaps I am not explaining myself very well.
– Simon Burton
Nov 22 '18 at 14:36
1
So if you feed in all possible vectors into that projection operator, you will get a two-dimensional subspace out the other side, and it will be isomorphic to the two-dimensional irrep of $S_3$. It’s just not the irrep you are expecting. Morally, this is the same as decomposing $mathbb{R}^2$ Into a direct sum of the x and y axes, and then being given a projection-onto-a-line operator. There is almost no chance that projection will select one of the axes.
– Joppy
Nov 22 '18 at 15:00
1
In fact, the image of any element of $mathbb{C}[S_n]$ acting on $V^{otimes n}$ will be a GL-invariant subspace, just because the actions of GL and the symmetric group commute. Young symmetrisers are a bit more special since they yield irreducible GL representations. You almost already have the Schur-Weyl decomposition in terms of those two groups of 4-dimensional representations you wrote down (splitting into non-isomorphic representations is a canonical direct sum). The next thing to check is that each of these pieces is a GL representation of the same type.
– Joppy
Nov 22 '18 at 15:27