We need to get some feel for how these dualities manifest in our theories. To get a sense for the role of the extraU(1)factors, and as a first check on our conventions, we will look at the purely Abelian casep=k−p= 1first. Then we will see the basic mechanism by which the non-Abelian duality works.
3Note that in particular, in our conventions we would always have|k|> p, which is the inequality we saw previously ensured that states never violate the unitarity bound fork >0.
The Abelian Case
Consider these theories, respectively coupled to the currents of a single boson and a single fermion:
LB = n+ 1
4π a∧da+aµJBµ, LF = − 1
4πb∧db+ 1
2πb∧dc+ n
4πc∧dc+bµJFµ.
Now suppose one naively integrates outcin the theory LF; its equation of motion is nc=−band hence we get
L0F = bµJFµ− 1 1−n+11
! 1
4πb∧db.
Lettingκ = n+ 1, we learn that the bosonic theory with the inverse level 1/κshould be dual to the fermionic theory with1 −1/κ, up to a parity transformation. This is indeed what we have found, right down to the parity issue – recall that on one side of the duality, we added∂z derivatives, whilst on the other we added∂¯zderivatives. This is our first confirmation of these dualities!
Furthermore, as we mentioned above, the spins ofΦandΨ, as measured by J0 (see Section5.3), are respectively
JΦ0 =− 1
2κ and JΨ0 = 1 2 −1
2
1− 1 κ
= + 1 2κ, so in particularJΨ0 =−JΦ0.
Two Particle Abelian Wavefunctions and Spectral Flow
To get a better sense of what is going on, it is very helpful to look at the case of two Abelian anyons in a harmonic trap, which is exactly solvable [129]. We could do this by following the approach of Section5.5.1, but it is helpful to instead consider the problem from first principles.
Consider the configuration space of two identical particles in two dimensions. The space is given byR2×R2/Z2. Let us focus on the relative degrees of freedom,R2/Z2, where theZ2 quotient identifies the pointsx ∼ −x. We can create a Hilbert space for these particles by fibering a one-dimensional complex line over this space, so let us start by doing this. However, it is clear that there is a singular point in this space, namely at the origin, and as a result we have a choice as to precisely what gluing of these lines we make: there can be monodromy around the origin.
Concretely, if we consider a wavefunction which is section of this bundle,χ(r, θ), then it may behave likeχ(r, θ +π) = U χ(r, θ)for some unitary operator U. We may freely choose the operatorU which sits here; each choice clearly defines a superselection rule in the theory. Here, since the Hilbert space is one-dimensional,U = exp(iπ/k)is simply a phase. It is, of course, thestatistical phaseof the particles. For bosons,1/k= 0and for fermions1/k= 1.
Suppose the particles are free, except that we put the theory in a harmonic trap. In this case, the Hamiltonian would be
H =− ~2 2m
∂2
∂r2 +1 r
∂
∂r + 1 r2
∂
∂θ 2!
+1
2mω2r2.
However, it is convenient to change to work with a single-valued function χ(r, θ) =ˆ exp(−iθ/k)χ(r, θ), and then the Hamiltonian becomes
Hˆ =−~2 2m
∂2
∂r2 + 1 r
∂
∂r + 1 r2
∂
∂θ + i k
2! + 1
2mω2r2.
This takes exactly the form of the standard two-dimensional simple harmonic oscil-lator, with solutionsχ(r, θ) = exp(ilθ)Rˆ n(r), except that the appearances of the angular momentalare shifted by1/k.
This means that almost all of the spectrum of the relative degrees of freedom is im-mediately obvious from the standard approach to the harmonic oscillator. Assuming that|1/k| ≤1, the spectrum for angular momental 6= 0is
E
ω = 1 + 2q+
l+ 1 k
where q = 0,1,2, . . . and l =±2,±4, . . . . (6.20) However, thel = 0sector is slightly subtle, since in fact there is a continuum of energy eigenvalues associated to square-integrable wavefunctions unless one imposes more precise boundary conditions on the behaviour at the origin. As we discussed in some detail back in Section5.5.1, our choice for the bosonic theory is equivalent to making
ˆ
χ ∼ r1/k at the origin for these wavefunctions. With this convention, the remaining states have energy
E
ω = 1 + 2q+ 1
k where q = 0,1,2, . . . for l = 0, . . . . (6.21) Some nice features of this spectrum are now visible. Firstly, suppose we smoothly increase1/kfrom0to1. We see that the resulting spectrum smoothly interpolates be-tween the spectrum of a boson and of a fermion, where the fermion state which matches
a boson state always has itsleigenvalue one larger. Thus the picture of flux attachment works very neatly in this case.
Suppose we instead decreased1/kfrom0to−1. Now something a little odd happens:
thel= 0states described by (6.21) go the wrong way. In particular, although almost all of them end up lining up with a free fermion state as they should, theq = 0, l = 0state heads down toE/ω= 0. At this point, of course, the state then becomes logarithmically non-normalizable, and a Jackiw-Pi vortex appears. We are not going to pursue this line of reasoning again, however.
We are really here to see what form bosonization takes from this perspective. It is easy enough to capture. Suppose we consider fermions with a phase shift of1/k−1. We can read off their spectrum from (6.20) simply by looking at oddl, as noted above.
But then the−1within the phase shift conspires with the summation over odd integers to produce a summation over even integers at the phase shift1/k. In other words, we obtain perfect agreement with a bosonic spectrum at phase shift1/k.
We also see that the bosonic state of angular momentum l arises from a fermionic one of angular momentuml+ 1. This ties in with what we have seen in terms of chiral operators; the fermionic dual ofΦ†Φ†wasΨ†∂Ψ†. The above shows that this extends in an elegant way to the rest of the spectrum.
The Basic Non-Abelian Case
One can perform the same manipulations as we did forU(1)for the more general case of
Nf bosons+U(p)k,k−p+np ←→Nf fermions+U(k−p)−k,−p×U(1)n (6.22) provided one is not too bothered about having a fractionalU(1)level. This duality then becomes the following:
Nf bosons+U(p)k,k−p+np ←→Nf fermions+U(k−p)−k,−p−(k−p)/n
Conveniently, even though theU(1)level here is not necessarily an integer, substituting it naively into our formulae gives the same answer as working in the theory with the extra gauge fields and extending our previous analysis to cover this situation.
As we saw for the U(1) case, the fact that we have investigated chiral states on one side and anti-chiral states on the other implies that in fact we have been looking at the left-hand theory together with a parity inversion of the right-hand theory. This parity inversion leaves the dimensions invariant, so we need not worry about it further
– we will quote results from our previous work but apply them to the true pair of dual theories. With all this in mind, let us see if we can verify the non-Abelian duality (6.22).
Let us begin on the left-hand side of (6.22). Consider a bosonic operator transform-ing in the representationR ofU(p). This has an associated Young diagram λ, and is composed of|λ|fundamental scalars withmderivatives. Then by what we have seen, this has the dimension
∆B=|λ|+m+ P 1
2λi(λi −1)−(i−1)λi
k +1
2|λ|(|λ| −1)×1 p
1
k−p+np − 1 k
.
Assume for the moment thatR is such thatλ has at most ˆk columns. Then we can consider a fermionic operator in the representationR˜associated to the Young diagram λT, living on the right-hand side of (6.22). Firstly, notice that because of the change in statistics and the change in representation, one can leave the derivatives in exactly the same place without causing any problems with vanishing symmetrizations or the like.
Further, this means that the operator transforms under theSU(Nf)global symmetry in the same representation as the bosonic one did. But now we can compute the dimension of this operator. We find that
∆F = |λT|+m+ P 1
2λTi (λTi −1)−(i−1)λTi
−k + 1
2|λT|(|λT| −1)× 1 k−p
1
−p−(k−p)/n − 1
−k
= |λ|+m+
P (i−1)λi− 12λi(λi−1)
−k +1
2|λ|(|λ| −1)× 1−n k(k−p+np)
= ∆B
confirming that the dimensions of these BPS operators agree exactly!
One may easily verify that the angular momenta J0 also agree between these two operators. This is a consequence of the fact that the fundamental representation of U(p)has the quadratic Casimirpand the identity
p
2(ˆk+p) = 1
2− kˆ 2(ˆk+p)
which shows that the true spin of an isolated boson in aU(p)theory is identical to that of a fermion in aU(ˆk)theory whose bare spin is1/2.