Let us first consider the Hamiltonian of well-known 1D periodic Ising model as $$ H = -\frac{1}{2} \sum_{j=1}^N \sigma^x_j \sigma^x_{j+1} + \frac{h}{2}\sum_{j=1}^N \sigma^z_j. $$ Now, in the thermodynamic limit, $H$ has two-fold degenerate ground state in the ferromagnetic phase, i.e., for $h < h_c = 1$. Now, if we use Jordan-Wigner transformation, we get $$ H = -\frac{1}{2} \sum_{j=1}^{N}(c^{\dagger}_jc_{j+1} + c^{\dagger}_jc_{j+1}^{\dagger} + \mbox{h.c.}) + h \sum_{j = 1}^Nc^{\dagger}_jc_j. $$ This can be easily diagonalized by successive Fourier and Bogoliubov transformation (edit-starts) and can be written as $$ H = \sum_{p} \omega_p \eta^{\dagger}_p\eta_p + \mbox{some constant}. $$ Clearly, the ground state will be the vacuum state $|\Omega\rangle$ of the Bogoliubov operators $\{\eta_p\}$, which is, upto my understanding, unique (?) (edit ends). So my questions are following:

(1) Why these two Hamiltonian gives two different ground state degeneracy?

(2) Is there any possible way to find out the ground state degeneracy of the Spin model from fermionic language?

(3) Fermionic Hamiltonian has degeneracy in ground state energy, only if there is a gap closing, i.e., for $h = h_c = 1$. Is this a generic statement about fermionic Hamiltonians, (edit starts) those can be written as the above diagonal form in terms of Bogoluibov operators? (edit ends) Or could there be a degeneracy in the fermionic Hamiltonian without any quantum criticality?

As per I understand, since the fermionic ground state is the vacuum state of the corresponding Bogoliubov operators (i.e., state with zero bogolon quasi-particle), it has to be unique and degeneracy can appear only if there is a spectral gap closing.


After Prof. Norbert Schuch's answer, I want to add that I am neglecting "a-cyclic" terms after JW-transformation, making it "c-cyclic" for large N, same as initial calculations of the LSM paper (we get the degeneracy for $N \rightarrow \infty$, if $0 < h < 1$). In this paper, authors have used above two Hamiltonians for their results. They have also mentioned that

A subtle difference occurs when thinking in terms of the spin model (7) since in the ferromagnetic phase the ground state of the spin model is twofold degenerate, while the fermionic model always has a unique ground state.

But this should not be the case, as these two Hamiltonians are connected by unitary, and I got confused. In my own calculations, I am also unable to find the origin this degeneracy for fermionic case, with periodic boundary conditions.

In case of open boundary condition, this degeneracy is somewhat easy to perceive, as one can have zero-energy Majorana edge modes ($f$) for $h < h_c = 1$, such that $|\Omega\rangle$ and $f^{\dagger}|\Omega\rangle$, both have the same energy.

Edit 2

After putting little bit thought into this matter, finally I can understand Prof. Schuch’s point. Since, the a-cyclic term was neglected after JW-transformation, we are stuck to odd parity case. So, we are getting unique ground state. To fully characterize the orginal spin model, we must consider fermionic Hamiltonian with both parities. In that case, we can explain degenerate ground state for periodic boudary condition. This surely clears all of my confusion regarding this matter.

  • $\begingroup$ The degeneracy is the same in the two models, corresponding to a particle-hole symmetry in the fermionic language. $\endgroup$
    – Adam
    Commented Oct 15, 2017 at 13:50
  • $\begingroup$ Prof. @Adam ... Can you elaborate? Or can you give some references? It will be helpful. Since particle-hole symmetry will also be there for non-degenerate range, i.e., $h > h_c = 1$. $\endgroup$ Commented Oct 15, 2017 at 15:10
  • $\begingroup$ @TitasChanda This is not correct: The JW-trafo gives a fermionic model with periodic OR antiperiodic boundary condition, depending on the number of fermions (odd/even). In the sym-broken phase, those two sectors have degenerate energies. $\endgroup$ Commented Oct 15, 2017 at 16:56
  • $\begingroup$ With regards to your question (2), it might be helpful to realize that with open boundary conditions, both models have the same ground state degeneracy. This is always the case, because in the open boundary case the Jordan-Wigner transformation relates both models directly without requiring additional non-local boundary terms. $\endgroup$ Commented Oct 15, 2017 at 19:17
  • $\begingroup$ Dr. @RubenVerresen, in case of open boundary, it is somewhat easy to perceive, as one can have zero-energy Majorana edge modes ($f$) for $h < 1$, such that $|{\Omega}\rangle$ and $f^{\dagger}|{\Omega}\rangle$, both have the same energy. But with periodic condition, I am unable to perceive the origin of the degeneracy. For example, this paper mentions that fermionic ground state is unique, while for spin model it is two-fold degenerate. I just took their Hamiltonians for posting the question. $\endgroup$ Commented Oct 16, 2017 at 8:54

1 Answer 1


Your claim is not correct: After a Jordan-Wigner transformation, the Ising model is mapped to a free fermion chain which has either periodic or anti-periodic boundary conditions, depending whether the number of fermions in the system is even (antiperiodic) or odd (periodic).

In case $h<1$, both of these have (almost) the same ground state energy -- this gives two ground states, one with even and one with odd fermion number, corresponding to the $+$ and $-$ superposition of the symmetry broken states, respectively.

(I think this should address all three questions: There is no discrepancy between the two models.)

  • 1
    $\begingroup$ To confirm Schuch's (correct) comments, one has to go through the JW transformation carefully: note that funny things happen at the boundary. Alternatively, check out the seminal Lieb-Schultz-Mattis paper: math.ucdavis.edu/~bxn/lieb1961.pdf ( e.g. check Eq. 2.11 and 2.11' ) $\endgroup$ Commented Oct 15, 2017 at 19:20
  • $\begingroup$ @RubenVerresen Regarding the LSM paper: I've seen various references to it regarding the TFIM, but I haven't found where they would include a transverse field. (Of course, it doesn't change the math much since it is a local & diagonal term, but maybe I am just overlooking it?) $\endgroup$ Commented Oct 15, 2017 at 19:25
  • $\begingroup$ Yes, you are right. I looked into that a while ago, and it seems Pfeuty (1970) was the first to apply the method of LSM to the quantum Ising chain [ref: sciencedirect.com/science/article/pii/0003491670902708 ] He also acknowledges the work of McCoy (1968) who was the first to use the LSM method to try and tackle the highly non-trivial spin-spin correlation functions [ref: journals.aps.org/pr/abstract/10.1103/PhysRev.173.531 ]. McCoy did this for the XY chain, and Pfeuty then similarly did it for the quantum Ising chain. $\endgroup$ Commented Oct 15, 2017 at 21:08
  • $\begingroup$ Prof. Norbert Schuch, Dr. @RubenVerresen, Thank you for the answers. I was initially neglecting the "a-cyclic" terms in the boundary, making it "c-cyclic" similar to LSM paper for large N. In Barouch-Mccoy paper-1 (ref), XY model has been mapped to Kitaev p-wave chain. I was following their calculation. But then I see this paper, where the authors did the same mapping, and commented about the uniqueness of fermionic ground states. And I got confused. $\endgroup$ Commented Oct 16, 2017 at 8:28
  • 1
    $\begingroup$ @TitasChanda Right, so, to get the facts straight: in the spin language, the ground state is degenerate both with open and periodic boundary conditions. In the fermionic case, the degeneracy is there with open boundaries, but the ground state is unique with periodic boundaries. This difference is physical: in the spin language, there is spontaneous symmetry breaking, where the degeneracy doesn't care about edges, whereas in the fermionic language, we have a 'symmetry protected topological phase', which has degeneracies living on its edges. $\endgroup$ Commented Oct 16, 2017 at 12:03

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