U(1) symmetry breaking within the superluid phase

  • Thread starter mavipranav
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  • #1
mavipranav
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Hi

With the Bose-Hubbard Hamiltonian (BHH) being invariant under a U(1)[tex]\equiv[/tex]O(2) symmetry transformation, it is said that the hopping-term in the BHH tends to break the U(1) symmetry as the system leaves the insulating phase. This is not clear to me.

However within the mean-field description of the BHH, the above statement makes sense because the MF-BHH is no longer invariant under the U(1) transformation due to the superfluid order-parameter term (which comes from the hopping-term) not being gauge invariant. But what is not clear is how this same conclusion is reached even for the original hamiltonian.

Thanks in advance,
Mavi
 

Answers and Replies

  • #2
DrDu
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Where "is it said"? Are you referring to a specific publication?
 
  • #3
mavipranav
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Where "is it said"? Are you referring to a specific publication?

Well, as I mentioned before and to be more explicit, I can argue phenomenologically that the U(1) is broken in the SF phase as follows: since U(1) symmetry is nothing but charge conservation (see for e.g. Fradkin's book "Field theories on Condensed matter") and since there is spontaneous creation and destruction of charges in the SF phase, the symmetry is broken here. And since it is the hopping-term which takes the system to the SF phase, these hopping terms are responsible for the breaking of the U(1) symmetry of the state.

Now I can argue in the above way because I know the physics of the model and its different phases. But I was wondering if there is a way to show, by simple group-theoretic arguments, just from the Hamiltonian and knowing nothing else, that the hopping-terms prefer a state with a broken symmetry. Seemingly it can be seen immediately (e.g. as mentioned in Sachdev's book "Quantum Phase transitions"), but I could find no sufficient explanation.

Thanks,
Mavi
 
  • #4
DrDu
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Well, it is characteristic of a system with a spontaneously broken symmetry, that the hamiltonian is fully symmetric under the group but that there are different ground states which are not symmetric under that group. So the hopping terms do not really break the symmetry of the hamiltonian.
 
  • #5
mavipranav
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Well, it is characteristic of a system with a spontaneously broken symmetry, that the hamiltonian is fully symmetric under the group but that there are different ground states which are not symmetric under that group. So the hopping terms do not really break the symmetry of the hamiltonian.

Yes, this is clear. My question was how does one see, mathematically or heuristically just with the BHH expression being given, that the hopping-terms prefer a state with a broken symmetry? For example, the ferromagnetic Heisenberg Hamiltonian has -S.S term which favours parallel alignment of the spins to minimize the energy, meaning a state with broken O(3) symmetry.
 
  • #6
DrDu
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The hopping term alone (limit of no on-site repulsion) can be diagonalized introducing anihilation operators for momentum eigenstates, something like [tex] a_k= \sum_j b_j \exp(ikj)[/tex]. The state with k=0 is lowest in energy, so all boson would condense in that state. The relevant term in the hamiltonian is then something like [tex]E_0 a^+_0 a_0 [/tex] which you should compare to the -S.S term in case of ferromagnetism.
 

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