# Ground state degeneracy: Spin vs Fermionic language

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.

Edit

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.

• The degeneracy is the same in the two models, corresponding to a particle-hole symmetry in the fermionic language. – Adam Oct 15 '17 at 13:50
• 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., $h > h_c = 1$. – Titas Chanda Oct 15 '17 at 15:10
• @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. – Norbert Schuch Oct 15 '17 at 16:56
• 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. – Ruben Verresen Oct 15 '17 at 19:17
• 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. – Titas Chanda Oct 16 '17 at 8:54

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.
• For periodic case, following the fermionic Hamiltonian that I mentioned, one can easily calculate the spectrum. And in that case, I am unable to understand from where this degeneracy is coming. 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. Please correct me if I am wrong. – Titas Chanda Oct 16 '17 at 8:45