A bit of a more mathematical/physical question. Given a Gauged MPS (let’s suppose left orthogonalized for simplicity) and an MPO, does a gauge for the MPO exist such that it preservers the gauge of the MPS? Does such gauge exists for any MPO? If not, what minimal conditions must the MPO satisfy to have such gauge? Or is it simply not possible?
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I think a unitary MPO preserves the gauge (imagine taking a unitary circuit and then compressing it into an MPO).
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I’ve realized that actually it is mostly an algorithm depended question, like the densitymatrix yields always a right orthogonalized MPS. While if we think about simply contracting each tensor of the MPS and of the MPO, I think that the MPS will always have a not defined gauge, at least what I found analytically is that if a tensor \psi of an MPS is left unitary, i.e.
\psi_{\alpha \beta}^i \psi_{\alpha \beta'}^i = \delta_{\beta,\,\beta'}
Where I’m using the convention that greek letters are link indices while latin ones are site indices, than a Tensor O to preserve this property should be written as:
O_{\alpha \beta}^{i i'} = U_{\alpha \gamma}^i V_{\gamma \beta}^{i'}
Where both U and V are left unitary, in the sense that
U_{\alpha \gamma}^i U_{\alpha \gamma'}^i = \delta_{\gamma, \gamma'}
Or, changing notation, and calling j = (\alpha,i) as the rows of U as a matrix
U^\dagger U = Id
and also
V^\dagger V = Id
where V has \gamma as row index. The problem is that this is possible only if dim(\gamma) is greater than dim(i',\beta), and since dim(\gamma)=\min(dim(i,\alpha),dim(i',\beta)), we arrive to the conclusion that for that to be true dim(\alpha)\ge dim(\beta). But if we take the leftmost tensor of an MPO we have that \dim(\alpha)=1, so we see that such condition can’t be true.