What is $\mathrm{dim}(\mathrm{Sym}(\mathrm{Herm}(H)^{\otimes N})$?

Question

The totally symmetric subspace of $(H^k)^{\otimes N}$, with $H^K$ a $k$-dimensional Hilbert space, has dimension $\binom{N+k-1}{k-1}$. But I now want to know the dimension of the totally symmetric subspace of the $N$-fold tensor product of Hermitian operators on $H^k$, i.e. $\mathrm{dim}(\mathrm{Sym}(\mathrm{Herm}(H^k)^{\otimes N}))$. Is this dimension equal to the dimension of the totally symmetric subspace of density operators on $(H^k)^{\otimes N}$, and if not, is there an explicit expression for it?

Answer 1

Presumably, you mean for $H$ to be a vector space over $\Bbb C$. Note that $\operatorname{Herm}(H^k)$ is a vector space over $\Bbb R$ (as opposed to $\Bbb C$), and that $\dim \operatorname{Herm}(H^k) = k^2$. From there, we can simply apply your earlier reasoning to find that $$ \mathrm{dim}(\mathrm{Sym}(\mathrm{Herm}(H^k)^{\otimes N})) = \binom{N + k^2 - 1}{k^2 - 1} $$ Noting that this too is a vector space over $\Bbb R$.

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If $H$ is a vector space over $\Bbb R$, then $\operatorname{Herm}(H^k)$ is a vector space over $\Bbb R$ with dimension $\frac 12 k(k+1)$.