# Hybrid argument for quantum states

The usual hybrid argument tells us that if two efficiently sampled ensembles are computationally indistinguishable based on a single sample, then, computational indistinguishability holds even for polynomially-many independent samples.

My question is based on the fact of a quantum state induces a probability distribution, and is as follows:

Can the hybrid argument be extended to the fact that if two quantum states $$\rho_0,\rho_1$$ are computationally indistinguishable to all QPT adversaries $$\mathcal{A}$$, then by giving any adversary $$\mathcal{A}$$ polynomially-many copies of the quantum states $$\rho_0^{\otimes p(\lambda)}, \rho_1^{\otimes p(\lambda)}$$, then the distributions of the outputs of $$\mathcal{A}$$ are still indistinguishable? (I.e., $$\rho_0^{\otimes p(\lambda)}, \rho_1^{\otimes p(\lambda)}$$ are computational indistinguishable?)

Assuming $$\mathcal{A}$$ can by itself generate copies of $$\rho_0$$ and $$\rho_1$$, then yes this is the case.
\begin{align*} &\| \mathcal{A}(\rho_0^{\otimes p}) - \mathcal{A}(\rho_0^{\otimes p})\| \\ &\leq \sum_i \| \mathcal{A}(\rho_0^{\otimes (p-i)}\otimes \rho_1^{\otimes i}) - \mathcal{A}(\rho_0^{\otimes (p-i-1)}\otimes \rho_1^{\otimes i+1}) \|\\ &\leq \sum_i \| \mathcal{A}_i(\rho_1) - \mathcal{A}_i(\rho_0)\|\\ \end{align*} where $$\mathcal{A}_i$$ prepares $$p-i-1$$ copies of $$\rho_0$$ and $$i$$ copies of $$\rho_1$$ and on input $$\rho$$ outputs $$\mathcal{A}(\rho_0^{\otimes (p-i-1)}\otimes \rho \otimes \rho_1^{\otimes i})$$.
If $$\mathcal{A}$$ cannot construct copies of $$\rho_0$$ or $$\rho_1$$ by itself, then you need to assume nonuniform indistinguishability of $$\rho_0$$ and $$\rho_1$$, so that the state $$\rho_0^{\otimes (p-i-1)}\otimes\rho_1^{\otimes i}$$ can be given as quantum advice to $$\mathcal{A}_i$$.