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

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?

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 $\Bbb R$.
If $H$ is a vector space over $\Bbb R$, then $\operatorname{Herm}(H^k)$ is a vector space over $\Bbb R$ with dimension $\frac 12 k(k+1)$.
• You'll have to clarify what you mean by density operators. If you mean "density operator" in the QM sense of "PSD operator with trace $1$", then you should note that these do not form a subspace. – Omnomnomnom Jul 23 '15 at 15:25
• Thank you for the clear answer. I do mean density operators in the QM sense. If they do not form a (sub)space, then does the question of dimensionality even make sense? The density operators are a subset of the Hermitian operators, so is it correct to say that the dimension of the density operators is equal to (or perhaps smaller than) the Hermitian operators? It seems you have a small typo in your answer by the way, it should be $\binom{N+k^2-1}{k^2-1}$, right? – Kenneth Goodenough Jul 23 '15 at 16:04
• Alternatively, you could try to talk about the dimension of the space of density operators as a manifold. This gives you the answer $$\binom{N + k^2 - 1}{k^2 - 1} -1$$ – Omnomnomnom Jul 23 '15 at 16:11