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?

  • $\begingroup$ 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 $\mathrm{tr}\,\rho^{T_B}=1$.) $\endgroup$ Commented Jan 18, 2020 at 14:56
  • $\begingroup$ @NorbertSchuch yes, of course you are right. Fixed. $\endgroup$
    – glS
    Commented Jan 20, 2020 at 9:24

1 Answer 1


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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.