8
$\begingroup$

When bosonizing an interacting spinless Luttinger liquid, the action can be written as \begin{equation} S=\frac{K}{2\pi}\int dx d\tau\ (\partial_\mu\phi)^2 = \frac{1}{2\pi K}\int dx d\tau\ (\partial_\mu\theta)^2, \end{equation} where $$K=\sqrt{\frac{v_F+g_4/\pi+g_2/\pi}{v_F+g_4/\pi-g_2/\pi}}$$ is the Luttinger parameter, which is one for free fermions. The convention for the $\phi$ field is such that its compactification radius is $R=1$. Alternatively $K$ can be absorbed into the definition of the fields to change $R$ to $R=\sqrt{K}$.

The first equation has an apparent symmetry under $K\to 1/K$ and $\theta\to\phi$, and the free fermion case lies right on the self-dual point $K=1$ ($R=1$).

This duality looks very similar to the $T$-duality for a compact free boson CFT. In fact on page 157 of Fradkin's book (pdf available online), it was explicitly pointed out that "in string theory this transformation is known as T-duality and the Luttinger parameter is known as the compactification radius (see e.g. Polchinski (1998) and Di Francesco et al. (1997))."

However, in CFT it is well known that the $T$-duality takes $R\to 1/(2R)$ (or $K\to 1/(4K)$ using the convention above) and the self-dual point is $R^*=1/\sqrt{2}$, rather than $R=1$. Moreover, assuring this is not just a naive convention issue, there is an emergent $SU(2)\times SU(2)$ symmetry at this self-dual radius $R^*$, which is not the case for a free fermion theory.

So I am really confused whether the $K=1$ case is self-dual under $\theta\to\phi$, and if it is, whether it has anything to do with the $T$-duality. Is the statement in Fradkin's book quoted above wrong? What is the relation between this "$\theta$-$\phi$ duality" and $T$-duality?

$\endgroup$
  • $\begingroup$ Hi @pathintegral, please stop making trivial edits to bump the question into the front page. Doing so only introduces noise for reviewers. Thank you. $\endgroup$ – AccidentalFourierTransform Aug 20 '18 at 15:57
  • $\begingroup$ @AccidentalFourierTransform I like keeping my questions and answers well-versed by editing them whenever I find confusing phrases. It has nothing to do bumping my question to the front page. I'd appreciate it if you could let me know if I am violating a policy unintentionally. $\endgroup$ – pathintegral Aug 20 '18 at 17:51
5
$\begingroup$

If you looks at this paper, figure 1: https://www.sciencedirect.com/science/article/pii/0550321388902490 you can clearly see that the self-dual point of the compact boson branch of the $c=1$ moduli space (which is $SU(2)_1$) at $R = 1/\sqrt{2}$ is not the free fermion CFT.

The free fermion CFT does not actually have the $\phi, \theta$ exchange symmetry because if treated properly (and using your normalization) these fields must have different periodicity. Indeed, one of them will have periodicity $4\pi$, say $\theta$, so that the fermion operator $e^{i \theta/2}$ is local. The duality exchanges which field gives rise to the fermion.

I also recommend you read I. Affleck's lectures from the 1988 Les Houches school on fields, strings, and critical phenomena (and all of the lectures contained inside). They are much more careful than the modern references.

$\endgroup$
  • $\begingroup$ Thanks for the answer and for the suggested reading. I will read carefully and add additional comments. In the meantime could you elaborate a bit on why $\theta$ has period $4\pi$? A heuristic answer will be great. $\endgroup$ – pathintegral Aug 12 '18 at 20:37
  • $\begingroup$ Basically it comes down to the explanation of the spin of the excitation created by vertex operators (in the big yellow for instance). $e^{i \theta}$ is a boson, but $e^{i\theta/2}$ is a fermion. We want the fermions to be local particles, so the vertex operator which creates it needs to be a local operator, and this imposes the periodicity constraint. I'm not being careful with normalization, and the devil's in the details of course. $\endgroup$ – Ryan Thorngren Aug 12 '18 at 20:44
  • $\begingroup$ BTW, $R=\sqrt{2}$ (KT point) in figure 1 of the paper you mentioned above is the self-dual point of the orbifold branch. The self dual point of the compact boson branch has $R=\sqrt{2}/2$. $\endgroup$ – pathintegral Aug 12 '18 at 20:57
  • $\begingroup$ Thanks, you're right. A confusing thing about this is that there is a "KT point" for classical KT which is $\hat SU(4)$ and one for "quantum KT", ie. the XXZ multicritical point, which is $SU(2)_1$. Also note that Majorana^2 != Dirac as most would erroneously claim. It's important to be clear which branch you're on... $\endgroup$ – Ryan Thorngren Aug 12 '18 at 21:04
  • 1
    $\begingroup$ Thanks again for the answer. Indeed without specifying the radii for the fields any $K$ can just be mapped to any $K’$ by rescaling the fields accordingly. I was able to explicitly show the radius for $\theta$ and $\phi$ are different for free fermion case. I will try to spell out the details later in a separate answer. Glad I solved this puzzle with your help. $\endgroup$ – pathintegral Aug 15 '18 at 4:36
2
$\begingroup$

Inspired by the answer by Ryan, I was able to work out the details identifying the $\theta$-$\phi$ duality as the $T$-duality.

Let us begin with \begin{align} S=\frac{1}{2\pi}\int dx d\tau\ (\partial_\mu \phi)^2, \end{align} where the compactification radius \begin{align}R_{\phi}=\sqrt{K}\end{align} and put it on a cylinder with circumference $L=1$.

We know that the spectrum for such a compactified boson is contributed by collective motions of the string and a sector of harmonic oscillations of the string. The equation of motion has (see Big Yellow Book) the following solution $\phi = \phi_0 + vt + 2\pi \sqrt{K}mx + \mathrm{oscillations}$, where $m\in\mathbb{Z}$ is a winding number around the cylinder. The speed $v$ is related to the canonical momentum $\Pi_{\phi}$, which is quantized as $\Pi_{\phi}=n/\sqrt{K},~n\in \mathbb{Z}$. Thus such a state is characterized by $(n,m)$, namely \begin{align} \phi = \phi_0 + \frac{n\pi t}{\sqrt{K}} + 2\pi \sqrt{K}mx + \mathrm{oscillations}. \end{align}

We can ask what primary fields creates this state. We know that the $(n,0)$ states, which are eigenstates of $\Pi_{\phi}$, is created by the vertex operator \begin{align} \mathcal{V}_{n}=e^{i{n}\phi{\sqrt{K}}}, \end{align} but it is less obvious to find a field that creates a winding number $m$ in $\phi$. It turns out that it involves the dual field $\theta$ introduced in bosonization, with $S=\frac{1}{2\pi}\int dx d\tau \ (\partial_\mu\theta)^2$. We know from bosonization that $\theta$ and $\phi$ satisfy \begin{align} \partial_{t}\theta= -\partial_{x}\phi(=-2\pi \sqrt{K}m), \end{align} so $m$ is the actually the quantum number of the canonical momentum $\Pi_{\theta}= -2\sqrt{K} m$. The operator corresponding to $(n,m)$ state is thus \begin{align} \mathcal{V}_{(n,m)}=e^{i{n}\phi{\sqrt{K}}+i2\sqrt{K}m\theta}. \end{align} We see that the compactification radius for the $\theta$ field is \begin{align} R_{\theta} = \frac{1}{2\sqrt{K}}. \end{align} The $n\to m$, $\theta\to\phi$, $K\to 1/(4K)$ symmetry of $\mathcal{V}_{(n,m)}$ is well known as the $T$-duality of the theory. The self-dual point is at $K=1/2$. At this point $R_{\phi}=R_{\theta}=1/\sqrt{2}$.

From this it is clear that the free Dirac fermion case with $K=1$ does not have the $\theta$-$\phi$ symmetry since their radii is different.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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