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Assume that a finite-dimensional pure state $|\psi\rangle\in \mathcal{H}\simeq \mathbb{C}^m$, $m<\infty$, is the (unique) frustration-free ground state of a local parent Hamiltonian and suppose that the locality notion is given in terms of a connected set of neighbourhoods $\{\mathcal{N}_k\}$. My question is the following one: Is it true that any unitary $U$ satisfying $$U|\psi\rangle\langle \psi|U^\dagger=|\psi\rangle\langle \psi|$$ can be decomposed into a finite product of invariance-satisfying unitaries acting only on the neighbourhoods $\{\mathcal{N}_k\}$, that is $U$ can be written as $U=\prod_{i=1}^N U_{\mathcal{N}_{k_i}}$, where every $U_{\mathcal{N}_{k_i}}$ acts only on the neighbourhood $\mathcal{N}_{k_i}$ and it is such that $U_{\mathcal{N}_{k_i}}|\psi\rangle\langle \psi|U_{\mathcal{N}_{k_i}}^\dagger=|\psi\rangle\langle \psi|$ ?

Any (partial) answer/comment/reference is very welcome.

Thanks in advance.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Aug 8, 2016 at 12:50

3 Answers 3

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Isn't simple translation symmetry an example?

E.g. suppose you have a one-dimensional ring of $L$ spins described by $|\psi\rangle = \sum_{\{\sigma_i\}} \psi_{\sigma_1,\sigma_2,\cdots,\sigma_L} \; |\sigma_1,\sigma_2,\cdots,\sigma_N\rangle$. Then this wavefunction could be invariant under the unitary transformation $\psi_{\sigma_1,\sigma_2,\cdots,\sigma_{L-1},\sigma_L} \to \psi_{\sigma_2,\sigma_3,\cdots,\sigma_L,\sigma_1}$. However you clearly cannot do this locally, unless I misunderstand your characterization.

An even more visceral counter-example would be (spatial) inversion symmetry.

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Consider the toric code Hamiltonian situated on a spherical geometry. This has a unique ground state. Expand the sphere to an infinite radius. Consider a string excitation of the ground state, and loop the string around the sphere (an infinite number of local operations) such that it meets itself, returning our system to its ground state. We know by analogy to the topologically protected ground states of the toric code on an infinite toroidal geometry that such an evolution is not possible with a finite number of local operations. Thus we have described a unitary evolution of the system, with the ground state as its eigenstate, which can not be expressed as a finite decomposition of local operations.

That being said, I may just be cheating by how I am taking the thermodynamic limit, and you may have to clarify such considerations in your question.

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  • $\begingroup$ Cheating it is. $\endgroup$ Aug 3, 2016 at 15:14
  • $\begingroup$ This is correct. However, I forgot to write in the OP that I'm working on a finite-dimensional Hilbert space (my fault, sorry!). I edited the OP accordingly. $\endgroup$
    – Ludwig
    Aug 3, 2016 at 15:17
  • $\begingroup$ No worries, it was helpful for me to formulate this answer anyways. $\endgroup$ Aug 3, 2016 at 15:32
  • $\begingroup$ This is complete cheating. If you want to decompose an operation on an infinite system into a product on local operations, it is clear that you need to ask for an infinite number of them. This has nothing to do with the toric code, the same works for a product state. $\endgroup$ Aug 3, 2016 at 18:30
  • $\begingroup$ You're right Norbert. Somehow I thought I was on the right track, but I should have given this some more careful thought. $\endgroup$ Aug 4, 2016 at 7:47
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Here's one idea:

Say $U|\psi\rangle = |\psi\rangle$ for $UU^\dagger = U^\dagger U = I$.

Then if we consider the exponential form $U = \exp(iG)$ with $G = G^\dagger$ as usual, $|\psi\rangle$ must necessarily be in the kernel of $G$, $G|\psi\rangle = 0$.

On the other hand, having $U$ of the product form $U = \prod_k{U_{{\mathcal N}_k} }$ for mutually disjoint neighborhoods is equivalent to $G = \sum_k{G_{{\mathcal N}_k}}$ with $[G_{{\mathcal N}_j}, G_{{\mathcal N}_k}] = 0$ for any ${\mathcal N}_j$, ${\mathcal N}_k$ involved.

But obviously not all $G$ that have $|\psi\rangle$ in their kernel are of this decomposable form. So not all $U$ such that $U|\psi\rangle = |\psi\rangle$ can be of the product type $U = \prod_k{U_{{\mathcal N}_k} }$.

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  • $\begingroup$ @Jacquard I think your question is related with the problem of the geometrical structure of quantum states. For a general density matrix $\rho=U_{0}\sigma U_{0}^{+}$, it can be easily verified that the $U$ keeps $\rho$ invariant is given by $U_{0}gU_{0}^{+}$, with $g\sigma=\sigma g$. This is very similar to the Generalized Hopf Fibration of mixed state proposed by Montgomery in "Heisenberg and Isoholonomic Inequalities" and a similar work later in "DYNAMIC DISTANCE MEASURES ON SPACES OF ISOSPECTRAL MIXED QUANTUM STATES". $\endgroup$
    – XXDD
    Aug 4, 2016 at 6:50
  • $\begingroup$ So there is no gurantee that the $U$ is something 'local', it really depends on the density matrix of the system $\rho$. For your question concerning only pure states, $g$ can be decomposed into a product as $g=U(1)\otimes U(N-1)$, but considering $U=U_{0}gU_{0}^{+}$, it's still not 'local'. $\endgroup$
    – XXDD
    Aug 4, 2016 at 6:53
  • $\begingroup$ Sorry I forgot to mention that $g$ should be unitary. $\endgroup$
    – XXDD
    Aug 4, 2016 at 6:58
  • $\begingroup$ Thanks for the answer. Are you considering disjoint neighbourhood topologies? If not, why are you decomposing $U$ as the product of local unitaries acting on disjoint neighbourhoods? $\endgroup$
    – Ludwig
    Aug 4, 2016 at 7:00
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    $\begingroup$ @X.Dong: I'm looking for a counterexample concerning a connected neighbourhood structure. It's easy to find counterexamples if the neighbourhoods are not connected. $\endgroup$
    – Ludwig
    Aug 4, 2016 at 14:36

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