# Is there an identity for the partial transpose of a product of operators?

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 $$A=B$$?

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.

• 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. Mar 8, 2021 at 16:23