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The partial transpose of an operator $M$ with respect to subsystem $A$ is given by $$ M^{T_A} := \left(\sum_{abcd} M^{ab}_{cd} \underbrace{|a\rangle \langle b| }_{A}\otimes \underbrace{|c \rangle \langle d|}_B\right)= \left(\sum_{abcd} M^{ab}_{cd} |b\rangle \langle a| \otimes |c \rangle \langle d|\right). $$ Is there an identity for the partial transpose of the product of operators,i.e. $$ (AB)^{T_A}= \quad ? $$ I suspect that in general $(AB)^{T_A} \neq B^{T_A} A^{T_A} $ or even $\neq A^{T_A}B^{T_A}$.

What if $A=B$?

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Your suspicion is correct, even when $A=B$. Consider the Hilbert space of two qubits and let $^{T_A}$ denote the partial transpose with respect to one of them. Suppose that

$$ A=B=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. $$

Then

$$ A^{T_A}=B^{T_A}=\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix}. $$

and we see that $(AB)^{T_A} = I^{T_A} = I$, but $A^{T_A}B^{T_A} = B^{T_A}A^{T_A} = 2 A^{T_A}$ is a rank one matrix.


I am not aware of a simple identity connecting $(AB)^{T_A}$ and $A^{T_A}B^{T_A}$ or $B^{T_A}A^{T_A}$.

A useful computational aid is offered by tensor notation and in some cases tensor networks. Note that operators such as $A$ and $B$ that act on the Hilbert space of a bipartite system can be thought of as tensors with four indices $A_{ijkl}$ and $B_{ijkl}$. Their usual matrix representation is obtained by grouping pairs of indices for example $A_{(ij)(kl)}$, $B_{(ij)(kl)}$ so that $(ij)$ indexes the rows and $(kl)$ indexes the columns. In this view partial transpose swaps the first indices from each pair $A_{(ij)(kl)}^{T_A} = A_{(kj)(il)}$. Thus

$$ (AB)_{(ij)(mn)} = \sum_{kl} A_{(ij)(kl)}B_{(kl)(mn)}\\ (AB)_{(ij)(mn)}^{T_A} = \left(\sum_{kl} A_{(ij)(kl)}B_{(kl)(mn)}\right)^{T_A} = \sum_{kl} A_{(mj)(kl)}B_{(kl)(in)}\\ (A^{T_A}B^{T_A})_{(ij)(mn)} = \sum_{kl} A^{T_A}_{(ij)(kl)}B^{T_A}_{(kl)(mn)} = \sum_{kl} A_{(kj)(il)}B_{(ml)(kn)} $$

which shows that changing the order of matrix multiplication and partial transpose affects which elements are multiplied together.

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    $\begingroup$ Thank you for your answer, I’ll investigate if I can progress using tensor networks, fingers crossed. Entanglement negativity seems like a not so easy to compute entanglement measure after all... I will give some time for others to contribute before assigning you the accepted answer. $\endgroup$ – FriendlyLagrangian Mar 8 at 16:23

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