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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?

Xiao-Gang Wen
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Greg Graviton
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2 Answers2

8

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.

David Bar Moshe
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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.

Xiao-Gang Wen
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