Schmidt decomposition in larger Hilbert spaces

Question

Consider a bipartite quantum system described by the density operator, $\hat{\rho}$, an operator acting on the Hilbert space $\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$. This matrix will be a vector (that in quantum physics is denoted with a double bra or ket $|\rho\rangle\rangle$) in the Liouville space, $\mathcal{L}=\mathcal{H}\otimes\mathcal{H}^{\dagger}$. Since the Liouville space is also a Hilbert space (when equipped with the Hilbert-Schmidt inner product) I can use the Schmidt decomposition on $|\rho\rangle\rangle$: $$ |\rho\rangle\rangle=\sum_{k}\alpha_{k}|\sigma_{k}\rangle\rangle\otimes|\Gamma_{k}\rangle\rangle.$$ My question is where does the Schmidt basis belong to? Does this make sense: $|\sigma_{k}\rangle\rangle\in\mathcal{H}_{A}\otimes\mathcal{H}_{A}^{\dagger}$ and $|\Gamma_{k}\rangle\rangle\in\mathcal{H}_{B}\otimes\mathcal{H}_{B}^{\dagger}$?

Answer 1

One problem here is that $|\rho\rangle\rangle$ is not a state anymore in the sense that $\langle\langle \rho|\rho\rangle\rangle=\langle\rho|\rho\rangle_{\rm HS}={\rm tr}(\rho^2)$ which is $<1$ unless $\rho$ is already a pure state.

That aside, you can of course normalize things ($|\rho\rangle\rangle\to \frac{|\rho\rangle\rangle}{{\rm tr}(\rho^2)}$) and apply the usual Schmidt decomposition; identifying $\mathcal H\otimes\mathcal H^\dagger\simeq (\mathcal H_A\otimes\mathcal H_A^\dagger)\otimes(\mathcal H_B\otimes\mathcal H_B^\dagger)$ this yields orthonormal systems $\{|\sigma_k\rangle\rangle\}_k\subset\mathcal H_A\otimes\mathcal H_A^\dagger$, $\{|\Gamma_k\rangle\rangle\}_k\subset\mathcal H_B\otimes\mathcal H_B^\dagger$ with the desired property. However, again, these $|\sigma_k\rangle\rangle,|\Gamma_k\rangle\rangle$ need not correspond to density matrices on $\mathcal H_A$, $\mathcal H_B$, respectively.