# Why is $\rho$ NPT if and only if $\rho^{\otimes N}$ is NPT?

In Horodecki et al. (1998), to prove that distillability implies having a negative partial transpose (being NPT). The authors use the fact that "a state $$\rho$$ is NPT if and only if $$\rho^{\otimes N}$$ is".

A state $$\rho$$ being "NPT" means here that that the operator $$\rho^{T_B}$$ with matrix elements $$(\rho^{T_B})_{ij,k\ell}\equiv\langle ij|\rho^{T_B}|k\ell\rangle=\langle i\ell|\rho|kj\rangle\equiv \rho_{i\ell,kj},$$ is not positive (and therefore not a state).

This is taken in the paper as an elementary fact, and not explicitly proven.

How can we prove that $$\rho^{T_B}$$ is not positive if and only if $$(\rho^{\otimes n})^{T_B}$$ isn't?

• You don't mean $\rho^{T_B}<0$, do you? (And similarly for the statements in the text above.) "NPT" means "not PPT". (Note that $\rho^{T_B}$ must have a positive eigenvalue, as $\mathrm{tr}\,\rho^{T_B}=1$.) Jan 18 '20 at 14:56
• @NorbertSchuch yes, of course you are right. Fixed.
– glS
Jan 20 '20 at 9:24

The short answer is that $$(\rho^{\otimes N})^{T_B}=(\rho^{T_B})^{\otimes N}$$.
More explicitly, if $$\rho=\sum_{ii'jj'}\rho_{ii',jj'}|i\rangle\!\langle i'|\otimes |j\rangle\!\langle j'|$$, then we can write $$\rho^{\otimes N}=\sum_{I I' JJ'}\rho_{II',JJ'}\bigotimes_{k=1}^N \Big(|i_k\rangle\!\langle i_k'|\otimes |j_k\rangle\!\langle j_k'|\Big),$$ where $$I\equiv(i_1,...,i_N)$$ and same for $$I',J,J'$$. Alternatively, we can write this state highlighting the bipartite structure still present in $$\rho^{\otimes N}$$ as $$\rho^{\otimes N}= \sum_{I I' JJ'}\rho_{II',JJ'} \left(\bigotimes_{k=1}^N |i_k\rangle\!\langle i_k'|\right) \otimes \left(\bigotimes_{k=1}^N |j_k\rangle\!\langle j_k'|\right).$$ The partial transpose operator then acts on it as \begin{align} (\rho^{\otimes N})^{T_B} = \sum_{I I' JJ'}\rho_{II',JJ'} \left(\bigotimes_{k=1}^N |i_k\rangle\!\langle i_k'|\right) \otimes \left(\bigotimes_{k=1}^N |j_k'\rangle\!\langle j_k|\right).\end{align} In short, this means that the partial transpose effectively operates by changing $$\rho_{II',JJ'}\to \rho_{II',J'J}$$. But the coefficients $$\rho_{II',JJ'}$$ also have a known structure, given by the tensor product operation: $$\rho_{II',JJ'}=\prod_{k=1}^N \rho_{i_k i_k',j_k j_k'},$$ and the coefficients on the RHS of this expression are exactly what you would get computing explicitly the matrix elements of $$(\rho^{T_B})^{\otimes N}$$, thus the result follows:
\begin{align} (\rho^{T_B})^{\otimes N} &= \left(\sum_{ii',jj'}\rho_{ii',jj'}\big(|i\rangle\!\langle i'|\otimes|j'\rangle\!\langle j|\big)\right)^{\otimes N}\\ &= \sum_{II'JJ'}\rho_{II',JJ'} \bigotimes_{k=1}^N \Big(|i_k\rangle\!\langle i_k'|\otimes |j_k'\rangle\!\langle j_k|\Big) \\ &\simeq \sum_{II'JJ'}\rho_{II',JJ'} \left(\bigotimes_{k=1}^N |i_k\rangle\!\langle i_k'|\right) \otimes \left(\bigotimes_{k=1}^N |j_k'\rangle\!\langle j_k|\right) =(\rho^{\otimes N})^{T_B}. \end{align}