# Edge theory of FQHE - Unable to produce Green's function from anticommutation relations and equation of motion?

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I'm studying the edge theory of the fractional quantum Hall effect (FQHE) and I've stumbled on a peculiar contradiction concerning the bosonization procedure which I am unable to resolve. Help!

In particular, consider the first few pages of X.G. Wen's paper "Theory of the edge states in fractional quantum Hall effects". Here, Wen defines a fermionic field $\Psi(x,t)$ in (1+1) dimensions in terms of a bosonic field $\phi(x,t)$ as

$$\Psi(x,t) \propto e^{i\frac{1}{\nu}\phi(x,t)} .$$

The number $\nu$ is the filling fraction of the FQHE, which we shall set to $\nu=1/3$ for simplicity. The bosonic field fulfills the somewhat strange commutation relations

$$[\phi(x,y),\phi(y,t)] = i\pi\nu\,\text{sgn}(x-y)$$

which are necessary to make $\Psi(x,t)$ anticommute like a proper fermion

$$\lbrace \Psi(x,t), \Psi^\dagger(y,t) \rbrace = \delta(x-y) .$$

Moreover, the bosonic field satisfies the equation of motion

$$(\partial_t-v\partial_x) \phi(x,t) = 0 .$$

Some operator algebra shows that $\Psi(x,t)$ is also a solution to this equation of motion. However, it appears to me that these two requirements, anticommutation and the equation of motion, already fix the Green's function of the fermion!

However, Wen goes on to note that these fermions have the Green's function (equation (2.12) in the paper)

$$G(x,t) = \langle T(\Psi^\dagger(x,t) \Psi(0)) \rangle = \exp[\langle\frac1{\nu^2}\phi(x,t)\phi(0)\rangle] \propto \frac{1}{(x-vt)^{1/\nu}} .$$

I do not understand how this can be. After all, from the anticommutation relations and the equation of motion, we can calculate the Green's function to be

$$G(x,t) \propto \frac{1}{x-vt} .$$

To do this, define Fourier modes $\Psi_k, \Psi^\dagger_k$, obtain the usual anticommutation relations for these, solve the equation of motion, and transform back into real space. The result will be as noted, and the exponent $1/\nu$ will be missing.

Where did the exponent $1/\nu$ go? What's wrong with calculating the Green's function from the anticommutation relations and the equation of motion?

Maybe there's something going on inside the $\delta(x-y)$ part of the anticommutation relations? If so, what exactly? Or maybe something about the ground state? Or something about the bosonization procedure as a whole?

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Greg Graviton
asked Oct 23, 2011
Are you properly taking into account the normal ordered product in the definition of $\Psi$?

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user José Figueroa-O'Farrill
@JoséFigueroa-O'Farrill: No, but that shouldn't affect the anticommutation relations and the equation of motion? The ground state may be different, but I have a hard time seeing how a different ground state might give rise to a different exponent in the Greens function.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Greg Graviton
Actually they do. This sort of calculation is typical in conformal field theory, a subject I am much more comfortable with than the FQHE. In CFT, the bosonic fields are not local, but their currents are. The commutation relations of the bosons can be read from their operator product expansion, which is logarithmic. The calculation of the two point function of normal-ordered exponentials can be found in many places; e.g., §6.3.2 in "Conformal field theory" by Di Francesco, Mathieu and Sénéchal.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user José Figueroa-O'Farrill

## 2 Answers

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I understand that you want to compute the fermion propagator in the operator formalism (in contrast to the path integral formalism where the same result can be obtained). Then following José's remark, the fermionization formula is correct, i.e., gives the canonical anti-commutation relations iff it is normal ordered:

$\psi(z) = :\exp(i \frac{1}{\nu}\phi(z)): = \exp(i \frac{1}{\nu}\phi_+(z)) \exp(i \frac{1}{\nu}\phi_-(z))$

where $\phi_+(z)$ ($\phi_-(z)$ ) contains only the dependence the creation (anihilation) field components.

The fermion propagator formula given in the question is a consequence of the product formula of two normally ordered exponentials:

$:\exp(ia\phi(z_1))::\exp(ib\phi(z_2)): =:\exp(ia\phi(z_1)+ib\phi(z_2)):exp(-ab\langle \phi(z_1) \phi(z_2)\rangle)$.

This formula can be easily verified independently for each mode using the Baker–Campbell–Hausdorff formula

Now, the computation is with respect to the boson vacuum and this is the reason that the fermion propagator has the power dependence.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user David Bar Moshe
answered Oct 24, 2011 by (4,195 points)
So, basically, you're saying that $|0\rangle$ is the vacuum state for the boson modes, $\phi_{-}(z)|0\rangle=0$, and that it is very different from the usual ground state (Fermi sea) for fermions, i.e. we have $\Psi_k|0\rangle\neq 0$? I would be interested in a back-of-the-envelope computation that shows how an "unusual" fermion ground state can give rise to a different power dependence of the Greens function.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Greg Graviton
(addendum: The idea is that I want to forget about all the bosons once I've got my fermion anticommutation relations and the equations of motion, and the ground state.)

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Greg Graviton
Expanding the last statement in the answer: The bosonic theory is a free field representation of an interacting fermionic theory. Please see Schulz, Cuniberti, Pieri arxiv.org/abs/cond-mat/9807366. The solution by means of bosonization is actually a Bogolyubov transformation (that mixes anihilation and creation operators thus transforms the "free fermion" vacuum state into the exact vacuum state.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user David Bar Moshe
cont. Now, If one wants to work exclusively with fermion operators, it is possible to express the Bose operators as bilinears in the Fermi operators, please see for example Rao and Sen: arxiv.org/abs/cond-mat/0005492 section 2.1., but this is just reversing the Bogolyubov transformation.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user David Bar Moshe
Thanks a lot, in particular for the last reference, which I think is one of the clearest introductions to bosonization with operators. It shows how the Fermion ground state differs from the usual one. While not directly answering my question, you've helped me a lot and I'm going to accept your answer.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Greg Graviton
+ 5 like - 0 dislike

Here, I would like to make some additional remarks.

1) In eq (2.11) of the referred paper, the correlation of the boson field is given $\langle\phi(x,t)\phi(0)\rangle =-\nu \ln(x-vt)$. This allows us to calculate $G(x,t) = \langle T(\Psi^\dagger(x,t) \Psi(0)) \rangle = \exp[\langle\frac1{\nu^2}\phi(x,t)\phi(0)\rangle] \propto \frac{1}{(x-vt)^{1/\nu}}$.

2) It is not correct to write $\lbrace \Psi(x,t), \Psi^\dagger(y,t) \rbrace = \delta(x-y)$, since here $\Psi(x,t)$ is not the bare electron operator. $\Psi(x,t) = exp(i\phi(x,t)/\nu)$ is only the projection of the bare electron operator into the low energy subspace. So we have $\Psi(x,t) \Psi^\dagger(y,t) = (-)^{1/\nu}\Psi^\dagger(y,t) \Psi(x,t) =-\Psi^\dagger(y,t) \Psi(x,t)$ when $1/\nu =$ odd integer. But $\lbrace \Psi(x,t), \Psi^\dagger(y,t) \rbrace = \delta(x-y)$ is not correct.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Xiao-Gang Wen
answered May 30, 2012 by (3,475 points)
So, basically you are saying that for $x≠y$ we do have $\lbrace \Psi(x,t), \Psi^\dagger(y,t) \rbrace = 0$, but for $x=y$ the delta function is different from the usual one? Is the form of this delta function known?

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Greg Graviton
@Greg: You are right. For x=y, the anti commutator may not be a $\delta$-function.

This post imported from StackExchange Physics at 2014-04-04 16:15 (UCT), posted by SE-user Xiao-Gang Wen

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