Basis Invariant Lifting Tensor

The main problem with the copy lifting tensor is the fact that it requires a choice of basis to define. In order to get a lifting tensor which does not depend on a choice of basis we add a third condition requiring the lifting tensor is symmetric in the sense of figure 5.5.

U U U^†

U^†

Figure 5.5: The definition of basis independence. In this figure there is a change of basis performed on all the bonds by the unitary U and its complex conjugate transpose. If this change of basis doesn’t effect the lifting tensor then we say the lifting tensor is basis independent. In this diagram each unitary has one bond connected to it at the top and one connected to it at the bottom. Making this clear is important so we can distinguish the unitary from its transpose. For the bonds corresponding to physical bulk sites this indicated by the slightly shifted bonds connected to the corresponding unitary and dual.

The reason that we are interested in this definition is that if we introduce the identity 1 = U U^† into any bond in the MERA then this is just another equivalent representation of the original MERA. But after we insert the copy lifting tensors there is no natural way to equate the two resulting bulk MERA without moving back into the MERA picture.

When considering basis independence as a condition of the lift the insertion of the identity can be viewed not as a wholesale change of the bulk MERA but rather as just a change of basis on the physical site.

This definition is motivated by the insertion of identity symmetry in the MERA. The insertion of identity symmetry corresponds to inserting I = U U^† along a bond where U is a unitary. These unitaries can then be absorbed into different tensors in the network changing its presentation but not the state that is represented. Since tensor contraction is required to compute any properties of the MERA this has no effect on physical mea-surements, however when we perform the lifting procedure with the copy tensor these different presentations give rise to different bulk MERA states and therefore different physical outcomes. By adding this symmetry requirement we introduce a natural

equiva-lence between the different bulk MERAs, the insertion of identity symmetry manifesting itself as a change of basis on the corresponding bulk sites of the lifted MERA. Physically this also means that we can transform between the resulting bulk MERAs without leaving the bulk MERA picture.

The maximal set of solutions to the equation in figure 5.5 is spanned by all values of α and β in the equation below:

L_a,b,c,d= αδ_a,bδ_c,d+ βδ_a,dδ_b,c (5.8) This equation is represented graphically form in figure 5.6. The reason that I consider only these two terms in equation (5.8) is due to an result arising from the study of certain non-abelian symmetric tensors - specifically the U (χ) symmetry group - that proves this statement. This proof is built off material in chapter 6 (and appendix C) and therefore relegated to appendix B. For now I will just note that it is clearly basis invariant as per the definition in figure 5.5, unlike the copy lifting tensor where a basis is explicitly chosen. Further we can now see the sense by which figure 5.5 makes the insertion of identity, I = U U^† before lifting equivalent to a change of basis on the bulk degrees of freedom. We also expect both α and β to be non zero as otherwise we find a stronger symmetry for the lifting tensor then we initially required 5.5.

α β

Figure 5.6: A spanning set of basis independent tensors which as discussed in Appendix B. For any values of α and β this tensor satisfies the conditions in figure 5.5, and any tensor which satisfies the conditions of figure 5.5 has a unique value of α and β. By enforcing these conditions on the lifting tensor axioms like in figure 5.5 (as done in figure 5.7) α and β can be fixed to compute single a basis independent lifting tensor.

The graphical form in figure 5.6 make it easy to see that both lifting axioms can be si-multaneously satisfied. For both terms in the sum connecting the two middle bonds will give us back something proportional to the identity, this is useful in proving the second

axiom. For the first axiom we can see that if we contract the lifting tensor with its dual then we get one term which is proportional to a projector and another term which is proportional to the identity (see figure 5.7a) and so the projector term is an eigenvector of the resulting matrix. To prove these assumptions we show the lifting axioms when the basis independent lifting tensors are expanded, giving rise to figure 5.7.

α β

b) a)

α β ^* α ^* β β

α

Figure 5.7: The result of expanding the two lifting tensor axioms when using the choice of basis independent lifting tensor in figure 5.6. a) The first lifting axiom from figure 5.3a, and b) is the second lifting axiom from figure 5.3b. Notice that all terms in the sum are proportional to the corresponding right hand sides in figure 5.3, closed loops corresponding to a multiplicative factor of the bond dimension χ.

The first axiom require that the trace over the matrix M = LL^† (contracted over the introduced bulk degrees of freedom) to have an eigenoperator that is the (vectorised) identity, shown in figure 5.7a. Because it is clear that the matrix M , arising from tracing over bulk physical degrees of freedom, is a sum of a projector like term plus an identity term then M acting on the state in the non-zero eigenspace of the projector will return something proportional to that state. Therefore we expect that M does have an eigenop-erator which is the (vectorised) identity. Solving the resulting equation (5.9) also places restrictions on the values α and β can take, dependent on the virtual bond dimension, χ, appearing whenever closed loops appeared in figure 5.7.

|α|²χ + (αβ^∗+ βα^∗) + |β|²χ = 1 (5.9) Secondly if we connect the two introduced physical sites, as shown in figure 5.7b, we find

that both terms in the sum are proportional to the identity. The requirements this places on α and β can then be simplified to the expression in equation 5.10.

α + βχ = 1 (5.10)

Rearranging equation (5.10) we can see that β = χ⁻¹(1 − α). Then restricting ourselves to only consider the solutions to equations (5.9) where α and β are positive real numbers we obtain equation (5.11).

α²χ + 2χ⁻¹(1 − α)α + χ⁻¹(1 − α)² = 1 ⇒ α² χ − χ⁻¹
= 1 − χ⁻¹ ⇒ α = 1

√1 + χ (5.11)

We find β is just β = χ⁻¹(1 − α) = χ⁻¹



1 −√1 + χ⁻¹

. The results of this calculation gives us a second lifting tensor option to consider that, unlike our previous suggestion, is basis independent. ¹

5.5 Relation Between Bulk and Boundary Operators