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There are two procedures that I know of for finding the von Neumann entanglement entropy of a bipartite system split between $A$ and $B$ - a bad way and a better way. I have in mind computationally calculating the entanglement entropy for a spin chain of even length $L$.

The bad way is to take the outer product $\rho = |\psi \rangle \langle \psi |$, trace out the $B$ subsystem to get $\rho_A = Tr_B[\rho]$, and diagonalize $\rho_A$ to get the entanglement spectrum $\{r_i\}$. One can rapidly plug in the entanglement spectrum to get the von Neumann entanglement entropy. Though this procedure is how entanglement entropy is defined, there's a better operational way to compute it below.

The better way is to reshape $|\psi\rangle$ according to the following isomorphism: $|\psi\rangle = \sum_{i,j} c_{ij} |A_i\rangle |B_j \rangle \cong \sum_{i,j} c_{ij} |A_i\rangle \langle B_j |$. This allows a singular value decomposition to get the Schmidt coefficients $\{\lambda_i\}$. The entanglement spectrum obeys $r_i = |\lambda_i|^2$, and so we can get the von Neumann entanglement entropy as above.

The better way was better because it avoided large memory usage. For a spin-1/2 chain, the bad way required constructing the $2^L$ by $2^L$ matrix $\rho$ (a total of $4^L$ entries). On the other hand, vector-matrix isomorphism above only has a total of $2^L$ entries - much better. This also means the better way is much faster to calculate the entanglement spectrum, since it's for a smaller matrix.


The above example is to motivate my question. I recently learned about the concept of logarithmic negativity which is defined on density matrices. That is, given a density matrix $\rho_{AB}$ describing subsystems $A$ and $B$, the procedure is to take $\log( | \rho_{AB}^{T_B} |)$ where $T_B$ is the partial transpose on subsystem $B$.

In particular, for my problem, I'm considering a pure state $|\psi\rangle$ with corresponding density matrix $\rho$ that I write in a tripartite form involving partitions $A$, $B$, $C$. I use the vector-matrix isomorphism above to get $M = \sum_i c_i |A_i\rangle |B_i\rangle \langle C_i|$ and construct $\rho_{AB} = Tr_C[\rho]$ via an appropriate matrix multiplication of $M$ with $M^\dagger$. I then take the partial transpose to get $\rho_{AB}^{T_B}$. I then find the spectrum of $\rho_{AB}^{T_B}$ and compute the logarithmic negativity.

My trouble is that if the subsystem $C$ is quite small, $\rho_{AB}$ is a massive matrix.

For example, if there's only one qubit in $C$, then $\rho_{AB}$ is a $2^{L-1}$ by $2^{L-1}$ matrix with $4^{L-1}$ entries. This is hard to store and especially hard to find the spectrum.

Is there any way to compute the entanglement negativity without constructing and diagonalizing $\rho_{AB}^T$, similarly to how one could calculate the entanglement entropy of a pure state slightly obliquely in "the better way" above?

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  • $\begingroup$ Your "better way" is the right way to compute $\rho_A=CC^\dagger$, with $C$ the "reshaped $|\psi\rangle$". $\endgroup$ Commented Sep 1, 2022 at 20:59
  • $\begingroup$ Do I understand correctly that your question is "how can I compute the entanglement negativity of a low-rank state $\rho_{AB}=\sum \lvert\psi_i\rangle\langle\psi_i\vert$ more efficiently than building $\rho_{AB}$ and diagonalizing $\rho_{AB}^{T_B}$? $\endgroup$ Commented Sep 1, 2022 at 21:00
  • $\begingroup$ @NorbertSchuch Thanks, I think that's my question. I'd forgotten that if I traced over a small subsystem $C$ that $\rho_{AB}$ would be low-rank. Does the low-rank nature help? $\endgroup$
    – user196574
    Commented Sep 1, 2022 at 21:07
  • $\begingroup$ Abridged version of my question: I'm imagining 20 qubits on a line. To get half-chain entanglement entropy, I only need to manipulate a $2^{10}$ by $2^{10}$ matrix. For entanglement negativity, if $A$ and $B$ are halves of the system, there's a mapping to Renyi-$1/2$ which also only requires $2^{10}$ by $2^{10}$ matrices. However, if $A$ is 10 spins and $B$ is 9 spins, then I need to manipulate a $2^{19}$ by $2^{19}$ (but rank 2, I suppose) matrix. Is there a nicer way to store and diagonalize (or entirely avoid storing and diagonalizing) the partial transpose of this matrix? $\endgroup$
    – user196574
    Commented Sep 1, 2022 at 21:07
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    $\begingroup$ @NorbertSchuch It's discussed on page 7, section 5 of arxiv.org/pdf/quant-ph/0102117.pdf. It's that for a state with Schmidt coefficients $\lambda_i$ chosen $\geq 0$, the negativity is a function of $\sum_i \lambda_i$. I said Renyi-$1/2$ just because $\lambda_i$ is the square root of the eigenvalues of the reduced density matrix, and so this same quantity is used to calculate Renyi-$1/2$. $\endgroup$
    – user196574
    Commented Sep 1, 2022 at 21:19

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