How to split a 2-local unitary operator through singular value decomposition?

Question

I’m studying the paper Expressive power of tensor-network factorizations for probabilistic modeling by Glasser et al. In equation S6 (page 2 of supplementary material, excerpt of paper figure below) it is claimed that a 2-local unitary $\rm U$ may be split into a contraction of two 1-local gate tensors through a singular value decomposition (SVD). How this is done, however, is not clear to me.

In tensor notation, the entries of a unitary $\rm U$ acting on two qudits may be written $U^{ij}_{kl}$, with the first upper and lower index referring to the first qudit, and the second upper and lower index referring to the second qudit. Splitting $\rm U$ into a contraction of two 1-local gate tensors $\rm A$ and $\rm B$ would mean that $$U^{ij}_{kl}=\sum_\alpha A^{i\alpha}_kB^j_{l\alpha}\tag{1}$$ or, equivalently $${\rm U}=\sum_\alpha{\rm A}^{\kern-0.085em\alpha}\otimes{\rm B}_\alpha\tag{2}$$ with $\rm A$ acting on the first qudit and $\rm B$ acting on the second. Now, when $\rm U$ is decomposed into $\rm C\Sigma\rm D^\dagger$, we have in tensor notation that $$U^{ij}_{kl}=\sum_{\gamma\delta}\sigma_{\gamma\delta}C^{ij}_{\gamma\delta}(D^\dagger)^{\gamma\delta}_{kl}.\tag{3}$$ This is much different from the first decomposition, as $\rm C$ and $\rm D$ still act on both qudits.

Can somebody explain how the decomposition should be carried out so as to arrive at eq. 1 or 2?

Answer 1

If I understand what you are asking for (I don't know much about tensor networks), both equations are singular value decompositions of $U$, just with respect to different indices (and in the latter case highlighting the singular values, which in the first equation is "hidden" in $A,B$).

There are two things to notice here. First of all, given an arbitrary matrix $A_{ij}$ you can write its SVD as (assuming summation over repeated indices, and without paying attention to whether indices are above or below the tensors): $$A_{ij}= s_k B^k_i \bar C^k_j,\tag1$$ where $s_k\ge0$ and $B^k_i \bar B^\ell_i=\delta_{k\ell}$, $C^k_j \bar C^\ell_j=\delta_{k\ell}$. But another way to write the SVD is to "hide" the singular values in the matrices in the decomposition, thus writing $$A_{ij}=D^k_i \bar E^k_j,\tag2$$ where now $D_i^k \bar D_i^\ell=s_k \delta_{k\ell}$ and $E_i^k \bar E_i^\ell=s_k \delta_{k\ell}$. You can always do this, being the singular values ensured to be non-negative. The connection between the objects in (2) and (1) is that $D_i^k\equiv\sqrt{s_k} B_i^k$ and $E_j^k\equiv \sqrt{s_k} C^k_j$.

Another point is that, if you have a tensor with more than two indices, there are multiple ways to apply the SVD. You can choose any way to separate the set of indices of the tensor in two groups, and then apply the SVD between these two groups.

For example, you can think of $U^{ij}_{kl}$ as a matrix "sending the indices $(k,l)$ into the indices $(i,j)$" (hopefully you know what this means formally), which is what you are doing in your second equation (also note that there is need to use two intermediate indices as you are doing, one is enough)/

Alternatively, you can think of $U^{ij}_{kl}$ as "sending $(j,l)$ to $(i,k)$", which gives you the SVD as in your first equation. Note that uou can also do other things here: e.g. separate $(i,l)$ and $(j,k)$.