For example, let's consider a 1D spin-1/2 ferromagnetic (FM) Heisenberg chain $H=-J\sum_{i=1}^{N}\mathbf{S}_i\cdot\mathbf{S}_{i+1}$ with periodic boundary conditions. Now we want to study its low energy excitations via the following two approaches:

(1)Jordan-Wigner (JW) transformation, we get $H\approx H_f=\sum \omega_kf_k^\dagger f_k-\frac{NJ}{4}$, where $f_k$ are JW fermionic operators;

(2)Holstein and Primakoff (HP) transformation, we get $H\approx H_b=\sum \omega_kb_k^\dagger b_k-\frac{NJ}{4}$, where $b_k$ are HP bosonic operators.

Where in both of the above two expressions, $\omega_k=J(1-\cos k)$ (setting lattice constant unity). Then the ground states of $H_f$ and $H_b$ are the vacuum states and both exactly equal to the exact FM ground state of the original Hamiltonian $H$. And the ground state energy of $H_f$ and $H_b$ (the constant terms $-\frac{NJ}{4}$) is exactly equal to the exact ground state energy of the original Hamiltonian $H$. Note that the approximation $\approx$ indicates that we have assumed the spin fluctuation (i.e., $\left \langle \hat{n}_i\right \rangle\ll 1$) to be small and hence dropped the higher order terms (i.e., interactions between JW fermions or HP bosons), which is justified since the ground states are the vacuum states containing no JW fermions or HP bosons ($\left \langle \hat{n}_i\right \rangle=0$), implying that this approximation is at least self consistant. Furthermore, $\omega_k\approx Jk^2/2$ as $k\rightarrow 0$ corresponds to the Goldstone mode.

If we believe that the above two pictures are both correct, then I get some questions: (1) Can the elementary excitations of a system be either fermions or bosons which may depends on the theory we adopt? Is there some duality relation or deep connection between the fermionic and bosonic approaches? (2) From $H_f$, can we infer that the gapless Goldstone mode is a fermion? As we always say a Goldstone boson instead of a Goldstone fermion.

Thank you very much.

  • 2
    $\begingroup$ Isn't it an artifact of the one-dimensionality of the lattice ? For example, JW transformation isn't that useful in higher dimension due to the strings. And in 1D, bosons and fermions are essentially the same. $\endgroup$
    – Adam
    Commented May 27, 2014 at 18:17
  • $\begingroup$ @Adam Thanks for your comment. "And in 1D, bosons and fermions are essentially the same." I'm sorry that I do not know about this fact, can you explain it more specifically or could you provide me some related references? Thank you very much. $\endgroup$
    – Kai Li
    Commented May 27, 2014 at 19:10
  • 3
    $\begingroup$ This loose statement is related to the fact that in 1D the statistic is not well defined since you can't exchange particles without making them go through each other. Fermions can be described by bosons through bosonisation for example. Any good textbook on 1D quantum fluid discuss that. $\endgroup$
    – Adam
    Commented May 27, 2014 at 19:18

3 Answers 3


I should say that you have 3 related questions, namely 1) To what extent can we trust the approximations based on HP and Jw transformations, 2) The nature of the low excitation spectrum and 3) The relation with Goldstone modes.

We shall look first at the Holstein-Primakoff method. The spin ladder operators for at a site $j$ are given by

$S^-_j = \sqrt{2S}b_j^\dagger\sqrt{1-\frac{n_j}{2S}}$

and its adjoint, where $S$ is the spin of you're model, in this case we have $S=1/2$. You're making the approximation $S_j^-=\sqrt{2S}b_j^\dagger$, or in other words expanding the square root and discarding non-linear terms, which should be good as long as $\langle n_j \rangle << S=1/2$. Spin one-half is not really the best case to use HP because is the one with greatest error in the linear approximation. Nevertheless let's continue. To study the low energy spectrum we introduce excitation (called magnons) with thermal distribution according to BE statistics $\langle n_k\rangle =(e^{\beta\omega_k}-1)^{-1}$ and see the correction to the magnetization $\Delta S(T)=S-\langle S_j\rangle$ at each site. By translational invariance we have $\langle n_j\rangle =\frac{1}{N}\sum_j \langle n_j\rangle$. Passing to momentum representation as usual we get

$\Delta S(T)=\int \frac{dk}{2\pi}\frac{1}{e^{\beta\omega_k}-1}$

Is easy to see that the integral diverges at low momenta as $\Delta S \propto \int_\epsilon \frac{dk}{k^2}\propto\frac{1}{k}$. This is just an instance of Mermin-Wagner theorem that says that in 1 and 2 dimensions there is no spontaneous symmetry breaking because the corresponding massless Goldstone bosons have infrared divergencies. You can check tha in 3D the correction goes as $\Delta S\propto T^{3/2}$. I see you're interested in the zero temperature limit. For fermion theories the Luttinger-Ward theorem gives the conditions for which the finite temperature results hold in the zero temperature limit. For bosons is somewhat harder because you have to deal with bose condensation. For the simple case of the Heisenberg model in 1D the classic result from Coleman can be extended without much trouble, as he himself notes, namely a proibition of spontaneous symmetry breaking in 1D an consequently absence of Goldstone modes.

So this answers question 3) regarding the Goldstone modes (they do not exist) and shows that although Holstein-Primakoff seens reasonable it gives results which are difficult to interpret as soon as one talks about excitations.

What about the JW transformation? It works greatly in 1D. In fact I think it is instructive to work all the terms. There is a convention of signals in the transforms, but I get for the full Hamiltonian (in lattice space, and disregarding terms that depend only on $n_j$ and ignoring boundary because I'm concerned with the $N\rightarrow \infty$ limit.)

$H_f=\sum_j -J\frac{1}{2}(f_j^\dagger f_{j+1}+f_{j+1}^\dagger f_j) -Jn_{j+1}n_j$

with $J>0$. In momentum space the first term is the kinectic one you wrote. The second one is easy to see corresponds to an attractive interaction. Therefore as soon as you put excitations you need to worry about the fermions forming bound states.

In fact, the one-dimension Heisenberg model is exactly solvable by Bethe Ansatz, and one can show that the low energy spectrum is made of gapped bosons, which from the JW point of view are bound states. If you want to understand the finite $N$ model the Bethe Ansatz is even better, since you can construct the exact energies and corresponding eigenstates.

In resume, HP is not really trustworthy in this case, it is better to look at JW, but in low dimensions basically every interaction is strong no matter how weak the coupling, so it pays to look beyond the first terms in perturbation theory. And there is no Goldstone mode, boson or fermion, because of the infrared divergence.

Nevertheless it is well known that in one dimensional systems we do not have spin-statistics theorem, viz. because there is no consistent definition of spin. Therefore there is a mapping from bosons to fermions. This article discuss the equivalence between fermions and bosons. In case you want further discussions I would recommend the great book by Giamarchi "Quantum Physics in one dimension". You''ll find a lot about Luttinger liquids, bosonization and there is a short introduction to Bethe Ansatz, complete with low energy excitations.

For even further discussions of Heisenberg model in 1D I really like "The theory of magnetism made simple", by Daniel Mattis. Not really made that simple though.

For a relation between the bosons and fermions in the context of Heinsenberg model, check this paper from Luscher where he discusses the antiferromagnet as a lattice regularization of the Thirring Model which Coleman had shown being equivalent to the Sine-Gordon model. It may be possible that the ferromagnet case you're interested also possess a similar relation.


The answer to the apparent contradiction of these two transformations (the excitations seem to be either bosonic or fermionic) comes the fact that the spins are not equivalent to the fermions, because they have string attached to them, to respect the commutative nature of spins on different sites, see JW transformation on wiki.

Therefore, even though the Hamiltonian is expressed in terms of fermions (without strings, which is a specificity of one-dimensional systems), the spin correlation functions are not fermionic (the strings insure that).

The spins excitation are bosonic in essence, as can also be seen from the exact mapping from spins to hardcore bosons (no more than one boson per site).

In fact, via fermionization, you can express one-dimensional bosonic systems with fermionic operator, even though the correlation functions will respect bosonic commutation relations, thanks again the the strings attached to the fermions.


There's an important subtlety hidden in the higher-order boson interaction terms in the HP transformation. If you include interaction terms to all orders, you find that in any dimension, a spin-1/2 system is exactly dual to a system of hard-core bosons via the HP transformation. In one dimension (only), it's also dual to a system of fermions via the JW transformation. By concatenating these two transformations as you suggest, we find that in one dimension, fermions are exactly equivalent to hard-core bosons (and both are equivalent to spin-1/2 chains). Heuristically, this is because there's no way to braid particles of either type around each other in 1D, so we don't need to worry about the minus sign from fermionic exchange. Bosonization makes this connection clear: in 1D, bosonic and fermionic theories can be mapped between each other relatively easily.

It's therefore just a philosophical question whether the excitations of a spin-1/2 chain should be thought of as bosonic or fermionic - both pictures are useful in different contexts. Roughly speaking, the bosonic picture is usually more useful for numerics because the Hilbert space has a simpler tensor product structure. The fermionic picture is sometimes more useful for analytic calculations because the fermions are more weakly interacting (sometimes even free, as in the transverse Ising and $XY$ chains). On the other hand, we often use bosonization to map fermionic systems to bosonic ones which can be easier to work with analytically. And in the Bethe ansatz, the bosonic excitations are much more conceptually natural than the fermionic ones.


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