Jordan-Wigner transformation is a powerful tool, mapping between models with spin-1/2 degrees of freedom and spinless fermions. The key idea is that there is a simple mapping between the Hilbert space of a system with a spin-1/2 degree of freedom per site and that of spinless fermions hopping between sites with single orbitals. One can associate the spin-up state with an empty orbital on the site and a spin-down state with an occupied orbital.

Bosonization/fermionization is also a powerful tool, mapping between 1+1d bosonic field theory to 1+1d fermionic field theory. There is a nontrivial correspondence between operators of two sides in 1+1d.


  1. Are we aware the exact relations between the twos in 1+1d: Jordon-Wigner transformation v.s. Bosonization?

  2. Can one use one to prove the other?

  3. Do both have subtle restrictions for 1+1d open chain or in 1+1d closed ring?

  4. Higher dimensional analogy in $d+1$d in general?


2 Answers 2


I take a different stance than Qmechanic: bosonization is `simply' the continuum version of the Jordan-Wigner transformation. Of course Qmechanic is right in that field theories are much more subtle than lattices theories. Nevertheless, the fact that JW is so simple does not mean it is not relevant when thinking of bosonization, in fact the opposite holds: it makes bosonization much easier to follow, as for example discussed by Fisher and Glazman.

To make my statement more concrete, I would say the following diagram commutes:

$$ \begin{array}{ccc} \textrm{fermionic chain} & \xrightarrow{\textrm{Jordan-Wigner}} & \textrm{spin chain} \\ \downarrow \small \textrm{continuum} & \circlearrowleft & \downarrow \small \textrm{continuum} \\ \textrm{fermionic field theory} & \xrightarrow{\textrm{bosonization}} & \textrm{bosonic field theory} \end{array}$$

(where in the spin-case the continuum limit would be taken using spin coherent state path integrals)

Of course one might end up with different field theory descriptions, but they would describe the same field theory. More exactly, there is a local mapping relating one to the other.

As an example, let us instead start from a spin chain. In particular, take the gapless spin-$\frac{1}{2}$ Heisenberg Hamiltonian $H = \sum \mathbf S_n \mathbf{\cdot S}_{n+1}$. Then:

$$ \begin{array}{ccc} \textrm{interacting fermions} & \xleftarrow{\textrm{Jordan-Wigner}} & H = \sum \mathbf S_n \mathbf{\cdot S}_{n+1} \\ \downarrow \small \textrm{continuum} & \circlearrowleft & \downarrow \small \textrm{continuum} \\ \textrm{interacting fermionic field theory} & \xrightarrow{\textrm{bosonization}} & \begin{array}{c} \textrm{Wess-Zumino-Witten }SU(2)_1 \\ || \\ \textrm{Luttinger liquid $K=\frac{1}{2}$ } \end{array} \end{array}$$ (where the LL description comes from the bosonization and the WZW description comes from the continuum limit of the spin model) and both resulting field theories are indeed equivalent after a local re-identification of operators. In particular they have the same scaling dimensions for local operators, e.g. the smallest scaling dimension for WZW $SU(N)_1$ is $\frac{N-1}{N} = \frac{1}{2}$ and for the LL is $\frac{1}{4K} = \frac{1}{2}$.


I.1) Jordan-Wigner (JW) transformation. Let there be given a bosonic Heisenberg algebra of canonical commutation relations (CCR)

$$[a_i,a_j^{\dagger}] ~=~\delta_{ij} {\bf 1}, \qquad [a_i,a_j] ~=~0, \qquad [a_i^{\dagger},a_j^{\dagger}] ~=~0, \qquad i,j~\in~\{1,\ldots, N\},\tag{1}$$

$$ n_i~\equiv~a^{\dagger}_i a_i \qquad\qquad\text{(no sum over $i$)}. \tag{2} $$


$$ [n_i, a_j]~=~-\delta_{ij}a_j, \qquad [n_i, a^{\dagger}_j]~=~\delta_{ij}a^{\dagger}_j,\tag{3} $$ $$ \{(-1)^{n_i}, a_i\}_+ ~=~ 0, \qquad\{(-1)^{n_i}, a^{\dagger}_i\}_+ ~=~ 0, \qquad\qquad\text{(no sum over $i$)};\tag{4} $$

$$ [(-1)^{n_i}, a_j] ~=~ 0, \qquad [(-1)^{n_i}, a^{\dagger}_j] ~=~ 0, \qquad\text{if}\qquad i~\neq ~j .\tag{5} $$

I.2) The JW transformation is defined as

$$ c_k ~\equiv~ (-1)^{\sum_{i=1}^{k-1}n_i} a_k, \qquad c_k^{\dagger} ~\equiv~ (-1)^{\sum_{i=1}^{k-1}n_i} a_k^{\dagger}, \qquad n_k~=~c^{\dagger}_k c_k. \tag{6}$$

Then we have a fermionic Heisenberg algebra of canonical anticommutation relations (CAR)

$$\{c_k,c_{\ell}^{\dagger}\}_+ ~=~\delta_{k\ell} {\bf 1}, \qquad \{c_k,c_{\ell}\}_+ ~=~0, \qquad \{c_k^{\dagger},c_{\ell}^{\dagger}\}_+ ~=~0, \qquad k,\ell~\in~\{1,\ldots, N\}.\tag{7}$$

II.1) Fermionization. Here we will just discuss the simplest prototype. Let there be give a chiral/holomorphic boson $\varphi(z)$ in Euclidean 2D CFT with OPE

$$ {\cal R} \varphi(z) \varphi(w)~\sim~-{\bf 1}~{\rm Ln}(z-w),\qquad z,w~\in~\mathbb{C}; \tag{8}$$

with primary momentum current

$$ j ~\equiv~i\partial \varphi ;\tag{9}$$

and with chiral stress-energy-momentum (SEM) tensor

$$T~\equiv~ \frac{1}{2} :j^2:~ .\tag{10}$$ The bosonic equal-radius-commutator-relations read

$$ [\varphi(z),\varphi(w)]~=~-i\pi {\bf 1}~{\rm sgn}(\arg z- \arg w)\qquad\text{for}\qquad |z|~=~|w| ,\tag{11}$$

$$ [j(z),\varphi(w)]~=~2\pi {\bf 1}~\delta(\arg z- \arg w)\qquad\text{for}\qquad |z|~=~|w| .\tag{12}$$

II.2) The chiral/holomorphic fermions are defined via the vertex operator

$$ \psi_{\pm} ~\equiv~ :e^{\pm\varphi}:~;\tag{13} $$ with number current

$$ j~\equiv~\pm :\psi_{\pm}\psi_{\mp}:~;\tag{14} $$

and with chiral SEM tensor

$$ T~\equiv~\frac{1}{2}:\psi_{\pm} \stackrel{\leftrightarrow}{\partial} \psi_{\mp}:~.\tag{15}$$

The OPEs become

$$ {\cal R} \psi_{\pm}(z)\psi_{\mp}(-z)~=~ \frac{\bf 1}{2z} ~\pm~ j(0) ~+~2z ~T(0) ~+~{\cal O}(z^2), \tag{16}$$

$$ {\cal R} \psi_{\pm}(z)\psi_{\pm}(-z)~=~2z ~{\bf 1} ~+~{\cal O}(z^2) .\tag{17}$$ The fermionic equal-radius-anticommutator-relations read

$$ \{ \psi_{\pm}(z),\psi_{\mp}(w)\}_+~=~ 2\pi i{\bf 1} ~\delta(\arg z- \arg w)\qquad\text{for}\qquad |z|~=~|w|, \tag{18}$$ $$ \{ \psi_{\pm}(z),\psi_{\pm}(w)\}_+ ~=~0 \qquad\text{for}\qquad |z|~=~|w|.\tag{19}$$

For $\arg z\neq \arg w$, the eqs. (18/19) follow directly from eqs. (11), (13), and the truncated BCH formula:

$$ e^Ae^B~=~e^{C}e^Be^A, \qquad C~\equiv[A,B], \qquad \text{if}\qquad [A,C]~=~0~=~[B,C]. \tag{20}$$

The delta function in eq. (18) follows from the simple pole in eq. (16).

The bosonic equal-radius-commutator-relations read

$$ [\varphi(z),\psi_{\pm}(w)]~=~\pm \pi ~{\rm sgn}(\arg z- \arg w)~\psi_{\pm}(w)\qquad\text{for}\qquad |z|~=~|w| ,\tag{21}$$ $$ [j(z),\psi_{\pm}(w)]~=~\pm 2\pi i ~\delta(\arg z- \arg w)~\psi_{\pm}(w)\qquad\text{for}\qquad |z|~=~|w| .\tag{22}$$

III) We interpret OP's main question as the following.

Can the fermionization (18/19) with $(j, \psi_+, \psi_-)$ be proven via the JW transform (7) with $(n_k, c_k, c^{\dagger}_k)$?

Answer: There is clearly an analogy between the discrete and the continuous model. However the JW transform (7) is a triviality, while the fermionization (18/19) is a non-trivial result in operator-valued distribution theory. It is not worth the effort to chase a triviality in an otherwise sophisticated proof.

IV) Note that a single chiral/holomorphic boson is just a prototype input for fermionization (18/19). It can be Wick-rotated to the 1+1D Minkowski plane. There is also antichiral/antiholomorphic version. Also there are different versions depending on topology/boundary conditions. In case of several chiral bosons, one needs co-cycle prefactors often based on the JW/Klein transformation.

V) There is no higher-dimensional analogue of fermionization per se, although e.g. superstring theory famously decompose the $10=5\times 2$ dimensional Euclidean target space into a product of 5 2D planes, and applies the fermionization in each 2D plane.


  1. S. Mandelstam, Soliton operators for the quantized sine-Gordon equation, Phys. Rev. D 11 (1975) 3026.

  2. J. Polchinski, String Theory, Vol. 2, 1998; p. 11-12.

  • 1
    $\begingroup$ Reviews: J. von Delft & H. Schoeller, Bosonization for Beginners --- Refermionization for Experts, arxiv.org/abs/cond-mat/9805275 $\endgroup$
    – Qmechanic
    Sep 13, 2017 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.