Cfrgtkky

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.

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.