Consider a simple periodic 1D chain with four sites with periodic boundary condition. The Hamiltonian reads

$$ H = t c_1^\dagger c_2 + c_2^\dagger c_3 + t c_3^\dagger c_4 + c_4^\dagger c_1 + h.c. $$

where $t$ is the hopping strength. In terms of matrix, it is simply

$$ H = \left[\begin{array}{cccc} 0 & t & 0 & 1 \\ t & 0 & 1 & 0 \\ 0 & 1 & 0 & t \\ 1 & 0 & t & 0 \end{array}\right] $$

with eigenvalues $[-1-t,-1+t,1-t,1+t]$. For $|t|<1$, the ground state will be a two-body state, which fills the lowest two eigenlevels such that the ground state energy is $E=-2$.

Now let's assume the operators $c_i$ are hardcore-bosonic, such that we can make the following Jordan-Wigner type of transformation

$$ c^\dagger_j = (X_j + i Y_j)/2\\ c_j = (X_j - i Y_j)/2 $$

where $X_j,Y_j$ are the Pauli operators at site $j$. As one can check, the operators satisfy $[c_i^\dagger,c_j]=0$ if $i\neq j$ and $\left\{c_i^\dagger,c_j\right\}=1$ if $i=j$. In this representation, we shall make the following identification

$$ c_1^\dagger c_2 = \frac{X_1+iY_1}{2}\otimes\frac{X_2-iY_2}{2}\otimes I_3\otimes I_4 $$

similarly for other terms, such that the Hamiltonian reads

$$ \begin{align*} H =\ &t \frac{X_1+iY_1}{2}\otimes\frac{X_2-iY_2}{2}\otimes I_3\otimes I_4\\ &+ I_1\otimes\frac{X_2+iY_2}{2}\otimes \frac{X_3-iY_3}{2}\otimes I_4 \quad \\ &+ t I_1\otimes I_2\frac{X_3+iY_3}{2}\otimes\frac{X_4-iY_4}{2} \\ &+ \frac{X_1-iY_1}{2}\otimes I_2\otimes I_3\otimes\frac{X_4+iY_4}{2} \\ &+ h.c. \end{align*} $$

which is a $16\times 16$ matrix that contains all the possible many-body state. One could also diagonalize it to obtain the ground state energy, and for $|t|<1$, I found $E=-2\sqrt{1+t^2}$, which is different from the previous result.

One could repeat the calculation for chains of length $2L$, and I found that the two methods wil give the same ground state energy if $L$ is odd, while there seems to be discrepancy if $L$ is even. This is quite strange to me, and I could not find any mistake. Any help is greatly appreciated!

  • $\begingroup$ Hardcore bosons have a different spectrum than soft bosons $\endgroup$ Jan 7, 2021 at 14:23
  • $\begingroup$ @JahanClaes Sure, and I explicitly use the commutator for the hard boson here. But this does not explain why the spectrum changes when I change from hard boson representation to spin representation, when L is even. $\endgroup$
    – fagd
    Jan 7, 2021 at 18:25
  • $\begingroup$ Fermions? Bosons? Hardcore bosons? (And which of those where?) The mapping from spins/hardcore bosons to fermions can have antiperiodic boundary conditions depending on the parity. $\endgroup$ Jan 7, 2021 at 22:19
  • $\begingroup$ @NorbertSchuch The system is hardcore boson, and this is the basis for the first Hamiltonian. Then I try to transform it to spins, which is the basis for the second Hamiltonian. There is no fermion involved here. This is supposed to be trivial, and I thought the boundary condition does not play a role here. Am I making a mistake here? $\endgroup$
    – fagd
    Jan 8, 2021 at 0:10
  • $\begingroup$ You say (correctly) that $H$ is a $16\times16$ matrix, but the matrix you wrote explicitly is $4\times4$. Clearly they are not the same matrix. $\endgroup$
    – fqq
    Jan 8, 2021 at 1:48

1 Answer 1


In your second approach, you correctly solve the hardcore boson problem (in essence, hardcore bosons are exactly two-level spin systems).

In your first approach, on the other hand, you diagonalize the single-particle problem. As an approach to solve the many-particle problem, where you subsequently fill all the negative energy modes, this only works for non-interacting particles, either bosons or fermions. It will, however, not work for hardcore bosons (which are bosons with an infinite on-site repulsion).

However, you are also not solving free bosons in your first approach: In that case, you would have to put an infinite number of bosons in each mode with negative energy, leading to an energy $-\infty$. (Differently speaking, for free bosons, the single-mode energies of the Hamiltonian must all be positive).

So what you are doing is solving the free fermion problem, where you fill each negative energy mode with one fermion.

So how does this compare to the hard-core boson problem (and why do you get the same result for odd $L$)?

When you do the Jordan-Wigner transformation from fermions to spins ( = hard-core bosons), you will get pretty much the same Hamiltonian (probably with a overall minus sign, but your free fermion energies are symmetric around zero, so this is no big business). However, the term across the boundary will be different: It will be either periodic or antiperiodic, depending on the total fermion parity of the ground state. If I am not mistaken, it should be periodic for an odd ground state parity. This is what is going on for odd $L$: Half of the modes are filled in the ground state, so there is an odd number of fermions, and the fermionic problem does indeed map to the hardcore bosons, and thus has the same energy.

  • $\begingroup$ Thanks for the reply, it at least gives some explanation for the problem. So essentially there is no single-body method to solve the many-body hardcore boson problem, because it is intrinsically interacting problem? I thought after we take into account the onsite repulsion, the hardcore problem can be treated as noninteracting. $\endgroup$
    – fagd
    Jan 8, 2021 at 18:32
  • $\begingroup$ @fagd Nope, it cannot. That's why these problems are hard. (Not "super-hard", because they don't have a sign problem and you can do Monte Carlo, as opposed to interacting fermions, but still hard.) And what do you mean by "it at least gives some explanation for the problem" -- what is missing? $\endgroup$ Jan 8, 2021 at 20:06
  • $\begingroup$ Yes, I do have another question. Based on the logic above, if I have even L and periodic boundary condition, then the spin Hamiltonian can be mapped to neither hardcore boson nor fermion. How about if I have even L and anti-periodic boundary condition? Can it be mapped to hard-core boson problem? $\endgroup$
    – fagd
    Jan 8, 2021 at 20:09
  • $\begingroup$ You have explained it very well, and really appreciate that. But it is counter-intuitive, as I thought the boundary condition is irrelevant because the transformation for the boson case is local. Now I realize it is actually not the case. And in fact, I have tested that if the boundary condition is open, then regardless of whether L is even or odd, the spectrum is the same. $\endgroup$
    – fagd
    Jan 8, 2021 at 20:13
  • 1
    $\begingroup$ @fagd Pretty much. Well, almost: In one dimension, there is no difference between fermions and hard-core bosons, except for the boundary conditions. So if you get your boundary conditions right, you can solve your hard-core bosons by mapping them to free fermions. But this approach only works in one dimension. $\endgroup$ Jan 8, 2021 at 21:17

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