Tensor power of the natural representation of Sn The symmetric group $S_n$ acts over $V=\mathbb{R}^n$ by permuting the canonical basis.
So it acts over $V^{\otimes p}$ with a diagonal action (acts the same over each element of the tensor product).
I'd like to find the decomposition into irreps of it. As I'm a physicist, I'm first interested in the simple cases $p=2$ (and if this is still difficult, $n=3,4$), even if something general would be nice!
I also need something constructive: I guess the proof will be constructive, but I'd prefer a reference where the basis vectors are easily tractable.
Thanks!
 A: This question was recently completely solved in this paper.
As explained on page 15 of the paper, letting $v_1,...,v_n$ be a basis for $V$, the standard basis vectors $v_{i_1} \otimes \cdots \otimes v_{i_p}$ for $V^{\otimes p}$ naturally correspond to partitions of the set $\{1,...,p\}$ into at most $n$ blocks.  The submodule corresponding to a particular partition into $t$ blocks is isomorphic to the permutation module $H^{(n-t,1^t)}=Ind_{S_{n-t}}^{S_n} 1$.  Multiplicities of irreducibles in $H^{\lambda}$ are well-known to be Kostka numbers $K_{\lambda, \mu}$, so we get that the multiplicity of the irreducible $S_{\lambda}$ in $V^{\otimes p}$ is $$\sum_{t=0}^n S(p,t) K_{\lambda, (n-t,1^t)}$$
Where $S(p,t)$ denotes the Stirling number.  The paper also gives a bijective proof of this equality using paths in the relevant Bratteli diagram.
The fact that you are working over $\mathbb{R}$ rather than $\mathbb{C}$ does not matter since all complex irreducible representations of $S_n$ can in fact be realized over $\mathbb{Z}$.
